Resonance Study Material Class 10

ELECTRICITY 

Pre-requisite : Before going through this chapter, you should be thorough with the basic concepts of the chapter explained in X NCERT.

 = Permittivity of the medium and  = 0r r = relative Permittivity or dielectric constant of the medium. Coulomb’s law is based on physical observation and it is not logically derived from any other concept.

CONSERVATION OF ELECTRIC CHARGE Whenever two bodies are charged by rubbing, one gets positively charged and the other gets negatively charged. The net charge on the two bodies, however, remains zero–the same as that before rubbing. In other words, charge is conserved. It can neither be created nor be destroyed. The only thing that happens on rubbing is that charged particles (electrons) get transferred from one body to the other. In some phenomena, charged particles are created. But even then the conservation of charge holds. For example, a free neutron converts itself into an electron and the proton taken together is also zero. So, there is no change in the conversion of a neutron to an electron and a proton.

COULOMB’S LAW Charles Augustine de Coulomb studied the interaction forces of charged particles in detail in 1784. He used a torsion balance. On the basis of his experiments he established Coulomb’s law. According to this law the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges. In mathematical terms, the force that each of the two point charges q1 and q2 at a distance r apart exerts on the other can be expressed as– F= k

r2

This force is repulsive for like charges and attractive for unlike charges.

1 4  0 ,

here 0 is absolute permittivity of free space. The force is directed along the line joining the centres of the two charged particles. For any other medium except air, free space or vacuum coulomb’s law reduces to F

1 q1q2 = 4 r 2

1.

Find out the electrostatics force between two point charges placed in air (each of +1 C) if they are separated by 1m .

Sol. Fe =

kq1q2 r

=

2

9  10 9  1 1

= 9×10 9 N

12

Note : From the above result we can say that 1 C charge is too large to realize. In nature, charge is usually of the order of C 2.

A particle of mass m carrying charge q 1 is revolving around a fixed charge –q 2 in a circular path of radius r. Calculate the period of revolution and its speed also.

4 2mr 1 q1q2 2 Sol. = mr = ' 4 0 r 2 T2

T2 =

or

( 4 0 )r 2 ( 4 2mr ) q1q2  0mr q1q2

T = 4r

and also we can say that

q1q 2 4 0r 2

=

mv 2 r

V=



q1q2 4 0mr

PROPERTIES OF ELECTRIC FIELD INTENSITY

q1q2

Where k is a constant of proportionality. k =

ILLUSTRATIONS

(i) It is a vector quantity. Its direction is the same as the force experienced by positive charge. (ii) Electric field due to positive charge is always away from it while due to negative charge always towards it. (iii) Its S.. unit is Newton/Coulomb. (iv) Electric force on a charge q placed in a region of electric field at a point where the electric field







intensity is E is given by F  qE . Electric force on point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge. PAGE # 1 1

(v) It obeys the superposition principle, that is, the field intensity at a point due to a system of charges is vector sum of the field intensities due to individual point charges.

4.



Sol. As force on a charge q in an electric field E is

    E  E1  E 2  E 3 + .....

  F q = qE

A

Five point charges, each of value q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value – q coulomb placed at the centre of the hexagon?

  | Fq |  | W |

i.e., E =

L E

q

|q|E = mg

W mg = 10 N/C., in downward direction. |q|

The position where the resultant force on a charged particle becomes zero is called equilibrium position.

O

q

F

i.e.,

E

q

ELECTROSTATIC EQUILIBRIUM

D q

Fe

So according to given problem

(vi) It is produced by source charges. The electric field will be a fixed value at a point unless we change the distribution of source charges. 3.

Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge –10 c and mass 10 mg.

q

(a) Stable Equilibrium :

C

-q

A charge is initially in equilibrium position and is displaced by a small distance. If the charge tries to return back to the same equilibrium position then this equilibrium is called position of stable equilibrium.

q A

B

(b) Unstable Equilibrium : If charge is displaced by a small distance from its equilibrium position and the charge has no tendency to return to the same equilibrium position. Instead it goes away from the equilibrium position.

Sol. If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on – q at O would be zero (as the forces due to individual charges will balance each other), i.e.,

FR  0



(c) Neutral Equilibrium : If charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium.



Now if f is the force due to sixth charge and F due to remaining five charges.

  F + f =0 or, |F| = |f| =

i.e.

  F =–f

1 q2 1 q q = 4 0 L2 4 0 L2

1 q2   = = FNet FCO 4  2 along CO  L

5.

Two equal positive point charges 'Q' are fixed at points B(a, 0) and A(–a, 0). Another test charge q0 is also placed at O(0, 0). Show that the equilibrium at 'O' is (i) stable for displacement along X-axis. (ii) unstable for displacement along Y-axis.

Sol. (i)









Initially FAO + FBO = 0  | FAO | = | FBO | =

KQq0 a2

When charge is slightly shifted towards + x axis by a small distance x, then.

  | FAO | < | FBO | PAGE # 2 2

Therefore the particle will move towards origin (its original position) hence the equilibrium is stable. (ii) When charge is shifted along y axis

(ii) The electric intensity at a point is the number of lines of force streaming through per unit area normal to the direction of the intensity at that point. The intensity will be more where the density of lines is more.

After resolving components net force will be along y axis so the particle will not return to its original position so it is unstable equilibrium. Finally the charge will move to infinity.

ELECTRIC LINES OF FORCE (ELOF) The line of force in an electric field is an imaginary line, the tangent to which at any point on it represents the direction of electric field at the given point.

(iii) Number of lines originating (terminating) from (on) is directly proportional to the magnitude of the charge.

(a) Properties : (i) Line of force originates out from a positive charge and terminates on a negative charge. If there is only one positive charge then lines start from positive charge and terminate at . If there is only one negative charge then lines start from  and terminates at negative charge.

(iv) ELOF of resultant electric field can never intersect with each other. (v) Electric lines of force produced by static charges do not form close loop. (vi ) E lec tr ic lin es of fo rc e en d o r s ta rt perpendicularly on the surface of a conductor. (vii) Electric lines of force never enter into conductors. 6.

If number of electric lines of force from charge q are 10 then find out number of electric lines of force from 2q charge.

Sol. No. of ELOF  charge 10  q  20  2q So number of ELOF will be 20. 7.

A charge + Q is fixed at a distance of d in front of an infinite metal plate. Draw the lines of force indicating the directions clearly. PAGE # 3 3

Sol. There will be induced charge on two surfaces of conducting plate, so ELOF will start from +Q charge and terminate at conductor and then will again start from other surface of conductor.

ELECTRIC POTENTIAL ENERGY Consider a charge Q placed at a point P as shown in figure. If another charge q of the same sign is now brought from a very far away distance (infinity) to point O near P, then charge q will experience a force of repulsion due to charge Q. If charge q is still pushed towards P, work is done. This work done is the potential energy of the system of these two charges.

ELECTRIC FLUX

Q 

Consider some surface in an electric field E . Let us select a small area element dS on this surface. The electric flux of the field over the  

area element is given by dE = E . ds Direction of dS is normal to the surface. It is along

nˆ or or or

dS

E

dE = EdS cos  dE = (E cos ) dS dE = En dS

r P

O

direction of dS .

q From infinity

Thus, the electric potential energy of a system of charges is defined as the amount of work done in bringing the various charges from infinite separation to their present positions to form the required system. It is denoted by U. For the system of two charges separated by distance r as shown in figure, the electric potential energy is given by : U=

where En is the component of electric field in the

q

kQq r

Electric potential energy is the from of energy, therefore it is measured in joule (J).

SUPER CONDUCTOR AND ITS APPLICATIONS

If the electric field is uniform over that area then

 

E = E  S (a) Physical Meaning : The electric flux through a surface inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal the surface. It is a property of electric field (b) Unit : (i) The SI unit of electric flux is Nm2 C–1 (gauss) or J m2 C–1. (ii) Electric flux is a scalar quantity. (It can be positive, negative or zero) 8.

Prof. K. Onnes in 1911 discovered that certain metals and alloys at very low temperature lose their resistance considerably. This phenomenon is known as super-conductivity. As the temperature decreases, the resistance of the material also decreases, but when the temperature reaches a certain critical value (called critical temperature or transition temperature), the resistance of the material completely disappears i.e. it becomes zero. Then the material behaves as if it is a super-conductor and there will be flow of electrons without any resistance whatsoever. The critical temperature is different for different material. It has been found that mercury at critical temperature 4.2 K, lead at 7.25 K and niobium at critical temperature 9.2 K become super-conductor.

The electric field in a region is given by, Applications of super conductors :

 3  4  3 E  E 0 i  E 0 j with E0 = 2.0 × 10 N/C. Find the 5 5 flux of this field through a rectangular surface of area 0.2m2 parallel to the Y–Z plane. Sol.

 

E = E  S 2  3  4  E 0 i  E 0 j  . 0.2ˆi = 240 N  m 5 5  C

=

 

(i) Super conductors are used for making very strong electromagnets. (ii) Super conductivity is playing an important role in material science research and high energy particle physics. (iii) Super conductivity is used to produce very high speed computers. (iv) Super conductors are used for the transmission of electric power.

PAGE # 4 4

CELL It converts chemical energy into electrical energy. Electrochemical cells are of three types : (a) Primary cell (b) Secondary cell (c) Fuel cell (a) Primary Cell :

also so that current flows through resistor R. If r is the internal resistance of the cell and V is the reading shown by voltmeter, then I=

E Rr

 E = IR + Ir Here, IR = V the potential difference So, E = V + r

It is an electrochemical cell, which cannot be recharged, but the chemicals have to be replaced after a long use. The reactions taking place in the cell are irreversible. Eg. : Daniel cell, Lechlanche cell, Dry cell etc. (b) Secondary Cell :

 E = I (R + r)

r=

 V = IR or I =

E–V I

.........(i)

V R

So for equation (i)

Electrical energy can be converted into chemical energy and chemical energy can be converted into electrical energy in these cells, i.e. secondary cells can be recharged after use. Chemical reaction taking place in these cells are reversible.

r=

(E – V)R V

........(ii)

GROUPING OF CELLS Example : Lead accumulator, Edison cell (alkali cell) and iron nickel cell. (a) Cells in Series : (c) Electro Motive Force of a Cell (E.M.F.) : It is the maximum potential difference between the two electrodes of the cell when no current is drawn from the cell or cell is in the open circuit. (d) Potential Difference of a Cell : It is the difference of potential between two terminals of the cell when current is drawn from it or the cell is in closed circuit. (e) Internal Resistance of a Cell : It is the resistance offered to the flow of current inside the cell i.e. internal resistance is the resistance offered to the flow of current by electrolyte. Internal resistance decreases with the increase of the area of plates and also with the decrease of the distance between plates.

Enrn

E1,r1 E2,r2 E3,r3

B

Eeq,req A

B

Equivalent EMF Eeq = E1 + E2 + ......... + En (write EMF’s with polarity) Equivalent internal resistance req = r1 + r2 + r3 + r4 + ........ rn If n cells each of emf E, are arranged in series and if r is internal resistance of each cell, then total emf = nE E,r E,r E,r

E,r

A

B Upto n

I R

Determination of internal resistance of a cell :

So current in the circuit,

I=

nE R  nr

There may by two cases : (i) If nr > R, then I = Connect a voltmeter to a cell through key K1. Also connect a resistor R to cell through K2. First put in key K1. The reading shown by voltmeter gives us the e.m.f. of the cell since negligible current flows through cell due to high resistance of the voltmeter. Insert key K2

E = current due to one cell. r

So, Series combination is not advantageous. Note : If polarity of m cells is reversed, then equivalent e.m.f. = (n–2m) E while the equivalent resistance is still nr + R, so current in R will be i=

(n  2m)E nr  R

PAGE # 5 5

(b) Cells in Parallel :

The combination of cells is equivalent to single cell of

If m cells each of emf E and internal resistance r be connected in parallel and if this combination is connected to an external resistance then the emf of the circuit is E.

emf = mE and internal resistance =

Internal resistance of the circuit =

Current I =

r . m

mr n

mE mr R n

For maximum current, nR = mr

mr = internal resistance of the equivalent n

or R = battery. Imax =

E and I =

r R m



mE mR  r

using mn = N in above equation we get number of rows n =

There may by two cases : (i) If mR > r, then I =

nE mE  . 2r 2R

E = current due to one cell. R

So, parallel combination is not advantageous.

9.

9 cells, each having the same emf and 3 ohm internal resistance, are used to draw maximum current through an external resistance of 3 ohm. find the combination of cells. Sol. For the condition of maximum current number of rows n=

Nr R

If emf and internal resistances of each cell are different, then, Eeq =

E1 / r1  E2 / r2  .....En / rn 1/ r1  1/ r2  .....1/ rn

for two cells E =

E1r2  E2r1 (Use emf with polarity) r1  r2

E1 E2 E3

r1

En

rn

r2 r3

Nr R

so n =

93 =3 3

so combination will be like 3 rows and 3 cells in each row.

BATTERY Battery is an arrangement that creates a constant potential difference between its terminals. It is a combination of a number of cells in series. The impact of battery :

(c) Cells in Multiple Arc : n = number of rows m = number of cells in each row.

mn = N (total number of identical cells) :

With the discovery of voltaic cell, it was soon realised that if one constructs a number of cells and joins the negative terminal of one with the positive terminal of the other and so on, then the emf (which is the potential difference between the electrodes in an open circuit) of the combination of cells will be the sum of the emf’s of the individual cells. This observation led to a burst of scientific activity in 1802. Humphrey Davy, an English chemist, made a battery of 60 pairs of zinc and copper plates. The large emf thus produced, was used to get high current, which could melt iron and platinum wires. By 1807, he had a battery of almost 300 plates with which he was able to decompose chemical salts. This led to the discovery of new elements. PAGE # 6 6

By 1808, Davy had assembled 2,000 pairs of plates. W ith this battery, he created electric arcs and succeeded in extracting the elements like barium, calcium and magnesium from their compounds. Thus, electricity took a front seat in exploring the nature of matter.

or

R  r 4

or

R r  4 R r

 4  0.15 % = 0.60% EFFECT OF STRETCHING OF A WIRE ON RESISTANCE

ELECTRICAL RESISTANCE The property of a substance by virtue of which it opposes the flow of electric current through it is termed as electrical resistance. Electrical resistance depends on the size, geometry, temperature and internal structure of the conductor.

In stretching, the density of wire usually does not change. Therefore Volume before stretching = Volume after stretching

 1A 1   2 A 2

We known that, vd = eE 

m

=

and

eV m

If information of lengths before and after stretching

I = Anevd

R2   2    R1   1 

V m  I Ane 2 

2 A  1 1 A 2

 =

R 2  A1    R1  A 2 

m Ane 2 

 A

2

If information of radius r1 and r2 is given then use

V m  R= I Ane 2 

 R =

A1  2  A 2 1

is given, then use

eV = Ane m

Ane2 V  I= m

R=

R 2  2 A1   R1  1 A 2

  =

 is called resistivity (it is also called specific

m

4

(a) Reciprocal of resistivity of a conductor is called its conductivity. It is generally represented by  .

ne 2 

ne 2 

r    1   r2 

CONDUCTIVITY :

RA 

m

resis tance), and  =

2

=

1 ,  is called 

conductivity. Therefore current in conductors is proportional to potential difference applied across

(b)  

1 

(c) Unit : ohm 1. metre 1

EFFECT OF TEMPERATURE ON RESISTANCE AND RESISTIVITY

its ends. This is Ohm's Law. Units:    1m 1 also called siemens m –1. 10. If a copper wire is stretched to make its radius decrease by 0.15%, Find the percentage increase in resistance (approximately). Sol. Due to stretching resistance changes are in the ratio

R 2  r1    R1  r2 

4

The resistance of a conductor depends upon the temperature. As the temperature increases, the random motion of free electrons also increases. If the number density of charge carrier electrons remains constant as in the case of a conductor, then the increase of random motion increases the resistivity. The variation of resistance with temperature is given by the following relation

R t  R 0 1  t   t 2



 PAGE # 7 7

where Rt and R0 are the resistance at t0C and 00C respectively and  and  are constants.

The

constant  is very small so its may be assumed negligible.



R t  R 0 1  t 

or



R t  R0 R0  t

This constant  is called as temperature coefficient of resistance of the substance. If R0 = 1 ohm, t = 10C, then

  R t  R 0  Thus, the temperature coefficient of resistance is equal to the increase in resistance of a conductor having a resistance of one ohm on raising its temperature by 10C. The temperature coefficient of resistance may be positive or negative. From calculations it is found that for most of the metals the value of  is nearly

1 0 / C . Hence 273

substituting  in the above equation

t   R t  R 0 1   273  

T  273  t   R0    R0 273 273  

The resistivity of alloys increases with the rise of temperature but less than that of metals. On applying pressure on pure metals, its resistivity decreases but on applying tension, the resistivity increases. The resistance of alloys such as eureka, manganin etc., increases in smaller amount with the rise in temperature. Their temperature coefficient of resistance is negligible. On account of their high resistivity and negligible temperature coefficient of resistance these alloys are used to make wires for resistance boxes, potentiometer, meter bridge etc., The resistance of semiconductors, insulators, electrolytes etc., decreases with the rise in temperature. Their temperature coefficients of resistance are negative. On increasing the temperature of semi conductors a large number of electrons get free after breaking their bonds. These electrons reach the conduction band from valence band. Thus conductivity increases or resistivity decreases with the increase of free electron density. 11. A wire has a resistance of 2 ohm at 273 K and a resistance of 2.5 ohm at 373 K. What is the temperature coefficient of resistance of the material? Sol.  

R  R0 2.5  2  R 0 T  T0  2  373  273 

where T is the absolute temperature of the conductor.



Rt  T



Thus, the resistance of a pure metallic wire is directly proportional to its absolute temperature. The graph drawn between the resistance R t and temperature t is found to be a straight line

0 .5  2.5  10  3 / 0K 200

WHEATSTONE BRIDGE Wheatstone bridge is an arrangement of four resistors in the shape of a quadrilateral which can be used to measure unknown resistance in terms of the remaining three resistances.

Rt R0 tºC

The resistivity or specific resistance varies with temperature. This variation is due to change in resistance of a conductor with temperature. The dependence of the resistivity with temperature is represented by the following equation.

 t   0 1  t  With the rise of temperature the specific resistance or resistivity of pure metals increases and that of semi-conductors and insulators decrease.

The arrangement of Wheatstone bridge is shown in figure below. Out of four resistors, two resistances R1, R2 and R3, R4 are connected in series and are joined in parallel across two points a and c. A battery of emf E is connected across junctions a and c and a galvanometer (G) between junction b and d. The keys K1 and K2 are used for the flow of current in the various branches of bridge. Principle of Wheatstone Bridge : When key K1 is closed, current i from the battery is divided at junction a in two parts. A part i1 goes through R1 and the rest i2 goes through R3. When key K2 is closed, galvanometer shows a deflection.

PAGE # 8 8

(a) Ammeter : Ammeter is an electrical instrument which measures the strength of current in ‘ampere’ in a circuit. Ammeter is a pivoted coil galvanometer which is always connected in series in circuit so that total current (to be measured) may pass through it. For an ammeter of good quality, the resistance of its coil should be very low so that it may measure the strength of current accurately (without affecting the current passing through the circuit). The resistance of an ideal ammeter is zero (practically it should be minimum). So, to minimize the effective resistance of an ammeter, a low value resistance (shunt) as per requirement is connected in parallel to the galvanometer to convert it to ammeter of desired range. In electric circuit, the positive terminal of an ammeter is connected to positive plate and negative terminal is connected to negative plate of battery.

The direction of deflection depends on the value of potential difference between b and d. When the value of potential at b and d is same, then no current will flow through galvanometer. This condition is known as the condition of balanced bridge or null deflection condition. This situation can be obtained by choosing suitable values of the resistances. Thus, in null deflection state, we have : Va – Vb = Va – Vd or i1 R1 = i2 R3 ...(i) Similarly : Vb – Vc = Vd – Vc or i1 R2 = i2 R4 ...(ii) On dividing equation (i) by (ii), we get

i2 R 3 i1 R1 i1 R 2 = i2 R 4

or

Desired value of shunt depends on the range (measurable maximum current) of ammeter converted from galvanometer. If pivoted galvanometer of resistance G is to measure current i (as an ammeter) then from figure.

i

 R2    R1 

R4 = (R3) 

For better accuracy of the bridge one should choose resistances R1, R2, R3 and R4 of same order.

GALVANOMETER Galvanometer is a simple device, used to detect the current, to find direction of current and also to compare the currents. With the help of galvanometer we make two important devises known as Ammeter and voltmeter as discussed below.

G

is ig G = (i – ig) S

ig

S or

i

ig G S=

(i  ig )

Where ig is an amount of current required for full deflection in galvanometer. By using a low value of resistance S (shunt) in parallel to the galvanometer (resistance G), the effective resistance of

R1 R 3  R 2 R 4 ...(iii)

Equation (iii) states the condition of balanced bridge. Thus, in null deflection condition the ratio of resistances of adjacent arms of the bridge are same. The resistor of unknown resistance is usually connected in one of the arm of the bridge. The resistance of one of the remaining three arms is adjusted such that the galvanometer shows zero deflection. If resistance of unknown resistor is R4. Then

ig

converted ammeter RA = 

GS becomes very low.. (G  S)

NOTE : Shunt : If anyhow, the flowing current through galvanometer becomes more than its capacity, the coil has possibility of burning due to heat produced by flowing current. Secondly, its pointer may break up due to impact with ‘stop pin’ as its proportional deflection as per amount of flowing current. In order to minimize these possibilities a low resistance wire (or strip) is connected in parallel with galvanometer, which is known as shunt. (b) Voltmeter : It is an electrical instrument which measures the potential difference in ‘volt’ between two points of electric circuit. It’s construction is similar as that of ammeter. The only difference between ammeter and voltmeter is that ammeter has its negligible (approximately zero) resistance so that it may measure current of circuit passing through it more accurately giving the deflection accordingly, while the voltmeter passes negligible current through itself so that potential difference developed due to maximum current passing PAGE # 9 9

through circuit may be measured. Therefore, an appropriate value of high resistance is required to be connected in series of galvanometer to convert it into a voltmeter of desired range. Voltmeter is connected in parallel to the electric circuit.

 V1  V2 + V3  V4 = 0. Boxes may contain resistor or battery or any other element (linear or nonlinear). It is also known as KVL 12. Figure shows, current in a part of electrical circuit, what will be the value of current (i) ?

If a galvanometer of resistance G is to be converted into a voltmeter of range V, then required value of high resistance RH will be

1A

2A

V = ig (RH + G)

V   or RH =  I  – G  g

P

2A Q

3 1.

R S

3A

A

i

V i ig RH

Connecting this value of high resistance in the series of galvanometer, it will be converted to a voltmeter of range V. After connecting high resistance RH in series of galvanometer of resistance G, the effective resistance of voltmeter becomes RV = (RH + G) very high (high in comparison to G). Ideal voltmeter has infinite resistance of its own. When ideal voltmeter is connected parallel to a part of an electric circuit, it passes zero amount of current through itself from the circuit so that measurement of potential difference across the points of connection may be perfectly accurate.

1A

2A

G Sol.

P

2A

i2

i1

R

Q

3A

3 1. i3

A

S i

From KCL, current at junction P, i1 = 2 + 3 = 5 A From KCL, current at junction Q, i2 = i1 + 1 = 5 + 1 = 6 A From KCL, current at junction R, i3 = i2 – 2 = 6 – 2 = 4 A From KCL, current at junction S, i = i3 – 1.3 = 4 – 1.3 = 2.7 A 13. In the circuit shown, calculate the value of R in ohm that will result in no current through the 30 V battery.

KIRCHHOFF'S LAWS (a) Kirchhoff’s Current Law (Junction law) : This law is based on law of conservation of charge. It states that "The algebraic sum of the currents meeting at a point of the circuit is zero" or total current entering a junction is equal total current leaving the junction.   in =  out. It is also known as KCL (Kirchhoff's current law).

Sol. Applying KVL in loop CEFDC 50 – iR – 20 i = 0 i =

(b) Kirchhof f ’s Voltage Law (Loop law) :

50 20  R

i

C

A

“The algebraic sum of all the potential differences along a closed loop is zero. IR + EMF =0”. The closed loop can be traversed in any direction. While traversing a loop if potential increases, put a positive sign in expression and if potential decreases put a negative sign.

E

50V i

R

20 B

10

D

i

F

Potential drop across R = Potential drop across AB iR = 30 

50 .R = 30 20  R

 R = 30 

PAGE # 1010

8.

EXERCISE Charge and coulomb’s Law : 1.

Conductivity of superconductor is : (A) infinite (B) very large (C) very small (D) zero

12 positive charges of magnitude q are placed on a circle of radius R in a manner that they are equally spaced. A charge +Q is placed at the centre. If one of the charges q is removed, then the force on Q is : (KVPY/2010) (A) zero qQ

(B) 2.

3.

4.

Two charges of +1 C & + 5 C are placed 4 cm apart, the ratio of the force exerted by both charges on each other will be (A) 1 : 1 (B) 1 : 5 (C) 5 : 1 (D) 25 : 1 A body has 80 microcoulomb additional electrons on it will (A) 8 x 10–5 (C) 5 x 1014

5.

W q

away from the position of the removed

qQ

(D)

40R2

towards the position of the removed

charge. In a neon discharge tube 2.8 × 1018 Ne+ ions move to the right per second while 1.2 ×1018 electrons move to the left per second. Therefore , the current in the discharge tube is : (IJSO/Stage-I/2011) (A) 0.64 A towards right (B) 0.256 A towards right (C) 0.64 A towards left (D) 0.256 A towards left

Electric filed and Potential :

F 2

(B) 2F

F (D) 4

(C) 4F

7.

40R2

charge.

9.

Two particles having charges q1 and q2 when kept at a certain distance, exert force F on each other. If distance is reduced to half, force between them becomes : (A)

6.

11qQ

(C)

Which of the following relation is wrong ? (A) Q = It

(D) V =

away from the position of the removed

charge.

of charge. Number of be : (B) 80 x 1015 (D) 1.28 x 10–17

1Coulomb (B) 1 ampere = 1Second (C) V = Wq

40R2

10. If Q = 2 coloumb and force on it is F = 100 newton, then the value of field intensity will be: (A) 100 N/C (B) 50 N/C (C) 200 N/C (D) 10 N/C

4 Coulomb of charge contains............................. 25 electrons : (A) 1015 (B) 1018 20 (C) 10 (D) none of these 5 charges each of magnitude 10–5 C and mass 1 kg are placed (fixed) symmetrically about a movable central charges of magnitude 5 × 10–5C and mass 0.5 kg as shown. The charges at P1 is removed. The acceleration of the central charge is : (KVPY/2009)

11. In the electric field of charge Q, another charge is carried from A to B. A to C, A to D and A to E, then work done will be -

A B Q + centre

P1 P2

P5

C D

O P3

P4

1 [Given OP1 = OP2 = OP3 = OP4 = OP5 1 m ; 4 0 = 9 × 109 in SI units] (A) 9 m s–2 upwards (C) 4.5 m s–2 upwards

(B) 9 m s–2 downwards (D) 4.5 m s–2 downwards

(A) (B) (C) (D)

E

minimum along path AB. minimum along path AD. minimum along path AE. zero along all the paths.

PAGE # 1111

12. A negatively charged particle initially at rest is placed in an electric field that varies from point to point. There are no other fields. Then : (KVPY/2008)

16. In the given circuit, the equivalent resistance between points A and B will be.

(A) the particle moves along the electric line of force passing through it. (B) the particle moves opposite to the electric line of force passing through it. (C) the direction of acceleration of the particle is tangential to the electric line of force at every instant. (D) the direction of acceleration of the particle is normal to the electric line of force at every instant. 13. Two charges +q and –q are placed at a distance b apart as shown in the figure below.

(KVPY/2009)

B P

A

C b/2 +q

–q b

The electric field at a point P on the perpendicular bisector as shown as : (A) along vector  A

(B) along vector  B

(C) along vector  C

(D) Zero

14. Two charges +Q and _2Q are located at points A and B on a horizontal line as shown below :

The electric field is zero at a point which is located at a finite distance : (KVPY/2011) (A) On the perpendicular bisector of AB (B) left of A on the line (C) between A and B on the line (D) right of B on the line

(A)

8 R 3

(C) 6R

(B) 4R (D) 10R

17. Resistance of a conductor of length 75 cm is 3.25 . What will be the length of a similar conductor whose resistance is 13.25 ? (A) 305.76 cm (B) 503.76 cm (C) 200 cm (D) 610 cm 18. A piece of wire of resistance 4 is bent through 1800 at its mid point and the two halves are twisted together, then resistance is : (A) 1  (B) 2  (C) 5  (D) 8  19. In how many parts (equal) a wire of 100  be cut so that a resistance of 1  is obtained by connecting them in parallel ? (A) 10 (B) 5 (C) 100 (D) 50 20. If a wire of resistance 1  is stretched to double its length, then the resistance will become : (A)

1  2

(B) 2 

(C)

1  4

(D) 4 

21. Two copper wires, one of length 1 m and the other of length 9 m, are found to have the same resistance. Their diameters are in the ratio : (A) 3 : 1 (B) 1 : 9 (C) 9 : 1 (D) 1 : 3 22. Reading of ammeter in ampere for the following circuit is :

Resistance : 15. There are three resistance 5, 6and 8connected in parallel to a battery of 15 V and of negligible resistance. The potential drop across 6resistance is : (A) 10 V (B) 15 V (C) 20 V

(D) 8 V (A) 0.8 (C) 0.4

(B) 1 (D) 2 PAGE # 1212

23. Two resistors are joined in series, then their equivalent resistance is 90  . When the same resistors are joined in parallel, the equivalent resistance is 20  . The resistances of the two resistors will be : (A) 70 , 20  (B) 80 , 10  (C) 60 , 30  (D) 50 , 40  24. In the ladder network shown, current through the resistor 3  is 0.25 A. The input voltage ‘V’ is equal to

29. The resistance of a wire of cross-section ‘a’ and length ‘  ’ is R ohm. The resistance of another wire of the same material and of the same length but cross-section ‘4a’ will be (A) 4R (C)

(B)

R 16

R 4

(D) 16 R

30. In the following circuit the value of total resistance between X and Y in ohm is : X

r

r r

Y

(A) 10 V

(B) 20 V

(C) 5 V

15 (D) V 2

25. The reading of voltmeter is

(A) 50V (C) 40V

10

(B) 60 V (D) 80 V

r

3 )R

(C) 

r to  r

(B) ( 3 – 1)R (D) 50 r

32. In case of the circuit arrangement shown below, the equivalent resistance between A and B is : (IAO/Jr./Stage-I/2009)

2

26. 1.4A

(A) 0.4 (C) 0.6

r

31. Wires A and B are made from the same material. Wire A has length 12m and weight 50 g, while wire B is 18 m long and weighs 40 g. Then the ratio (RA / RB) of their resistances will be : (IAO/Jr./Stage-I/2008) (A) 16 / 45 (B) 4 / 5 (C) 8 / 15 (D) 4 / 9

A 25

(A) (1 +

r

r

A

B

1.4A 5

(B) 1 (D) 1.2

27. Three identical bulbs are connected in parallel with a battery. The current drawn from the battery is 6 A. If one of the bulbs gets fused, what will be the total current drawn from the battery ? (A) 6A (B) 2A (C) 4A (D) 28. A uniform wire of resistance R is uniformly compressed along its length, until its radius becomes n times the original radius. Now, the resistance of the wire becomes : (A) R/n (B) R/n4 2 (C) R/n (D) n R

(A) 10 (C)

40 W 3

(B) 2.5 W (D) None of the above

33. The net resistance between points P and Q in the circuit shown in fig. is

(A) R/2

(B) 2R/5

(C) 3R/5

(D) R/3 PAGE # 1313

34. A wire of resistance 10.0 ohm is stretched so as to increase its length by 20%. Its resistance then would be : (IAO/Sr/Stage-I/2008) (A) 10.0 ohm (C) 14.4 ohm

38. In the circuit arrangement shown, if the point A and B are joined by a wire the current in this wire will be : (IJSO/Stage-I/2011)

(B) 12.0 ohm (D) 10.2 ohm

A

35. In the circuit shown below, all the resistances are equal, each equal to R. The equivalent resistance between points A and C is :

B

(IAO/Sr./Stage-I/2009)

B

24Volt

C R

R

R

R

(A) 1A. (C) 4A.

R R

R

39. In the following circuit, each resistor has a resistance of 15  and the battery has an e.m.f. of 12 V with

R

A

D

(A) R (C) R /2

(B) 2A. (D) zero.

(B) 4R (D) none of the above

negligible internal resistance. (IJSO/Stage-II/2011)

36. A battery or 10 V and negligible internal resistance is connected across the diagonally opposite corner of a cubical network consisting of 12 resistors each of resistance 1 . The total current 1 in the circuit external to the network is :

(KVPY/2007)

A 10V (A) 0.83 A

(B) 12 A

(C) 1 A

(D) 4 A

37. Figure (a) below shows a Wheatstone bridge in which P, Q, R, S are fixed resistances, G is a galvanometer and B is a battery. For this particular case the galvanometer shows zero deflection. Now, only the positions of B and G are interchanged,. as shown in figure (b). The new deflection of the galvanometer. (KVPY/2010)

(A) is to the left. (B) is to the right. (C) is zero. (D) depends on the values of P, Q, R, S

When a resistor of resistance R is connected between D & F, no current flows through the galvanometer (not shown in the figure) connected between C & F. Calculate the value of R. (A) 10  (B) 15  (C) 5  (D) 30  40. The circuit given below is for the operation of an industrial fan. The resistance of the fan is 3 ohms. The regulator provided with the fan is a fixed resistor and a variable resistor in parallel. (IJSO/Stage-II/2011)

Under what value of the variable resistance given below, Power transferred to the fans will be maximum? The power source of the fan is a dc source with internal resistnace of 6 ohm. (A) 3  (B) 0 (C)  (D) 6 

PAGE # 1414

41. When all the resistances in the circuit are 1 each, then the equivalent resistance across points A & B will be : (IJSO/Stage-II/2011)

 46. In the circuit shown below,

(IAO/Sr./Stage-I/2007)

10V X

Y

(A) current flowing in the circuit is 200 mA (B) power supplied by the battery is 2 watt (C) current from X to Y is zero (D) potential difference across 10 is equal to zero

(A) 5/6 

(B) 1/2 

(C) 2/3 

(D) 1/3 

47. We are given n resistors, each of resistance R. The ratio of the maximum to minimum resistance that can be obtained by combining them is : (KVPY/2008) (A) nn (B) n (C) n2 (D) logn Cell :

42. A cylindrical copper rod has length L and resistance R. If it is melted and formed into another rod of length 2L. the resistance will be : (KVPY/2011) (A) R (B) 2R (C) 4R (D) 8R

 43. There are four resistors of 12 ohm each. Which of the following values is/are possible by their combinations (series and / or parallel) ? (IJSO/Stage-I/2008) (A) 9 ohm (B) 16 ohm (C) 12 ohm (D) 30 ohm

 44. In case of the circuit shown below, which of the following statements is/are true ?

(IJSO/Stage-I/2009)

+ – 1 •

A R1

B R2

•

(A) R(E – V) (C)

(E  V ) E R

EV R (E  V ) R (D) V (B)

49. 24 cells, each having the same e.m.f. and 2 ohm internal resistance, are used to draw maximum current through an external resistance of 3 ohm. The cells should be connected (A) in series (B) in parallel (C) in 4 rows, each row having 6 cells (D) in 6 rows, each row having 4 cells

4 R3

2

3

(A) R1, R2 and R3 are in series. (B) R2 and R3 are in series. (C) R2 and R3 are in parallel. (D) The equivalent resistance of the circuit is R1+

48. A cell of emf E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell is given as :

R 2R3 R 2  R3

 45. A current i reaching at a point in a circuit gets branched and flows through two resistors R1 and R2. Then, the current through R1 varies as : (IAO/Jr./Stage-I/2007) (A) R1 (B) R2 (C) (R1+ R2) (D) 1l (R1 + R2)

50. The cells are joined in parallel to get the maximum current when (A) external resistance is very large as compared to the total internal resistance (B) internal resistance is very large as compared to the external resistance (C) internal resistance and external resistance are equal (D) emf of each cell is very large 51. In secondary cells : (A) Chemical changes can be reversed by heating electrodes (B) Chemical changes can be reversed by passing electric current (C) Current is produced by photo chemical reactions (D) None of these PAGE # 1515

52. Three types of electric cells which provide current are : (A) Button cell, solar cell & secondary cell (B) Solar cell, electrolytic cell, electro chemical cell (C) (A) and (B) both are correct (D) Neither (A) nor (B) is correct

Electric Energy and Power :

53. In which of the following cells, the potential difference between the terminals of a cell exceeds its emf.

58. Two identical heater wires are first connected in series and then in parallel with a source of electricity. The ratio of heat produced in the two cases is : (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

57. An electric iron of heating element of resistance 88  is used at 220 volt for 2 hours. The electric energy spent, in unit, will be : (A) 0.8 (B) 1.1 (C) 2.2 (D) 8.8

59. You are given three bulbs 25 W, 40 W and 60 W . Which of them has the lowest resistance? (A) 25 watt bulb (B) 40 watt bulb (C) 60 watt bulb (D) insufficient data

(A) a (C) c

(B) b (D) d

54. A cell, an ammeter and a voltmeter are all connected in series. The ammeter reads a current I and the voltmeter a potential difference V. If a torch bulb is connected across the voltmeter, then. (IJSO/Stage-I/2009) (A) both I and V will increase (B) both I and V will decrease (C) I will increase but V will decrease (D) I will decrease but V will increase 55. In the process of electrostatic induction. (IJSO/Stage-II/2011) (A) a conductor is rubbed with an insulator. (B) a charge is produced by friction. (C) negative and positive charges are separated. (D) electrons are ‘sprayed’ on the object. 56. Consider the circuit below. The bulb will light up if : (KVPY/2009) S1

S2

60. If R1 and R2 are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, then : (A) R1 = 2 R2 (B) R2 = 2 R1 (C) R2 = 4 R1 (D) R1 = 4 R2 61. If two bulbs, whose resistance are in the ratio of 1 : 2, are connected in series. The power dissipated in them has the ratio of : (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 4 62. When a voltage of 20 volt is applied between the two ends of a coil, 800 cal/s heat is produced. The value of resistance of the coil is :(1 calorie = 4.2 joule) : (A) 1.2  (B) 1.4  (C) 0.12  (D) 0.14  63. You are given two fuse wires A and B with current rating 2.5 A and 6 A respectively. Which of the two wires would you select for use with a 1100 W, 220 V room heater ? (A) A (B) B (C) A and B (D) none of these 64. An electric current of 2.0 A passes through a wire of resistance 25  . How much heat (in joule) will be developed in 1 minute ? (A) 6 (B) 6000 (C) 50 (D) 10 65 . Two bulbs, one of 200W and the other of 100W, are connected in series with a 100 V battery which has no internal resistance. Then, (KVPY/2009) 100V

~

S3

200W

(A) S1 S2 and S3 are all closed. (B) S1 is closed but S2 and S3 are open. (C) S1 and S3 are closed but S2 is open. (D) none of these

100W

(A) the current passing through the 200W bulb is more than that through the 100W bulb. (B) the power dissipation in the 200W bulb is more than that In the 100 W bulb. (C) the voltage drop across the 200W bulb is more than that across the 100W bulb. (D) the power dissipation In the 100W bulb is more than that in the 200W bulb. PAGE # 1616

66. An electric heater consists of a nichrome coil and runs under 220 V, consuming 1 kW power. Part of its coil burned out and it was reconnected after cutting off the burnt portion. The power it will consume now is : (KVPY/2010) (A) more than 1 kW. (B) less that 1 kW, but not zero. (C) 1 kW. (D) 0 kW.

71. The circuit shown has 3 identical light bulbs A, B, C and 2 identical batteries E1, E2 . When the switch is open, A and B glow with equal brightness. When the switch is closed: (KVPY/2007)

67. In the following circuit, the 1 resistor dissipates power P. If the resistor is replaced by 9. the power dissipated in it is : (KVPY/2011)

(A) A and B will maintain their brightness and C will be dimmer than A and B. (B) A and B will become dimmer and C will be brighter than A and B. (C) A and B will maintain their brightness and C will not glow. (D) A, B and C will be equally bright.

C E1

A

S E2

B

72. A student connects two lamps in the circuit shown. The emf of the two batteries is different. (IJSO/Stage-II/2011)

(A) P (C) 9P

(B) 3P (D) P/3

68. A neon lamp is connected to a voltage a.c. source. The voltage is gradually increased from zero volt. It is observed that the neon flashes at 50 V. The a.c. source is now replaced by a variable dc source and the experiment is repeated. The neon bulb will flash at : (IAO/Sr./Stage-I/2008) (A) 50V (B) 70V (C) 100V (D) 35V 69. A certain network consists of two ideal and indentical voltage sources in series and a large number of ideal resistor. The power consumed in one of the resistor is 4W when either of the two sources is active and other is replaced by a short circuit. The power consumed by same resistor when both sources are simultaneously active would be : (IJSO/Stage-II/2011) (A) 0 or 16W (B) 4W or 8W (C) 0 or 8W (D) 8W or 16W

Which of the following statements are correct? i. When keys 1, 2, 3 and 4 are closed, bulbs A and B will both glow ii. When key 2 and 4 are closed bulb A will glow iii.When 1 and 4 are closed, bulb A will glow iv.When 2, 3 and 4 are closed, both A and B will glow (A) only ii (B) only iv (C) i, ii and iv (D) ii and iii 73. Figure below shows a portion of an electric circuit with the currents in ampere and their directions. The magnitude and direction of the current in the portion PQ is : (KVPY/2011)

Circuit and Other : 70. A galvanometer can be converted into a voltmeter by connecting (A) A high resistance in series with the galvanometer (B) A high resistance in parallel with the galvanometer (C) a low resistance in series with the galvanometer (D) a low resistance in parallel with the galvanometer (A) 0A (C) 4A from Q to P

(B) 3A from P to Q (D) 6A from Q to P PAGE # 1717

MOLE CONCEPT ATOMS The symbol of Copper is Cu (from its Latin name Cuprum)

All the matter is made up of atoms. An atom is the smallest particle of an element that can take part in a chemical reaction. Atoms of most of the elements are very reactive and do not exist in the free state (as single atom).They exist in combination with the atoms of the same element or another element. Atoms are very, very small in size. The size of an atom is indicated by its radius which is called "atomic radius" (radius of an atom). Atomic radius is measured in "nanometres"(nm). 1 metre = 109 nanometre or 1nm = 10-9 m. Atoms are so small that we cannot see them under the most powerful optical microscope. 

It should be noted that in a "two letter" symbol, the first letter is the "capital letter" but the second letter is the small letter

Symbol Derived from English Names

Note : Hydrogen atom is the smallest atom of all , having an atomic radius 0.037nm. (a) Symbols of Elements : A symbol is a short hand notation of an element which can be represented by a sketch or letter etc. Dalton was the first to use symbols to represent elements in a short way but Dalton's symbols for element were difficult to draw and inconvenient to use, so Dalton's symbols are only of historical importance. They are not used at all.



English Name of the Element Hydrogen Helium Lithium Boron Carbon Nitrogen Oxygen Fluorine Neon Magnesium Aluminium Silicon Phosphorous Sulphur Chlorine Argon Calcium

Symbol H He Li B C N O F Ne Mg Al Si P S Cl Ar Ca

Symbols Derived from Latin Names

It was J.J. Berzelius who proposed the modern system of representing en element. The symbol of an element is the "first letter" or the "first letter and another letter" of the English name or the Latin name of the element. e.g. The symbol of Hydrogen is H. The symbol of Oxygen is O. There are some elements whose names begin with the same letter. For example, the names of elements Carbon, Calcium, Chlorine and Copper all begin with the letter C. In such cases, one of the elements is given a "one letter "symbol but all other elements are given a "first letter and another letter" symbol of the English or Latin name of the element. This is to be noted that "another letter" may or may not be the "second letter" of the name. Thus, The symbol of Carbon is C. The symbol of Calcium is Ca. The symbol of Chlorine is Cl.

English Name of the Element

Symbol

Latin Name of the Element

Sodium

Na

Natrium

Potassium

K

Kalium

(b) Significance of The Symbol of an Element : (i) Symbol represents name of the element. (ii) Symbol represents one atom of the element. (iii) Symbol also represents one mole of the element. That is, symbol also represent 6.023 × 1023 atoms of the element. (iv) Symbol represent a definite mass of the element i.e. atomic mass of the element. Example : (i) Symbol H represents hydrogen element. (ii) Symbol H also represents one atom of hydrogen element. (iii) Symbol H also represents one mole of hydrogen atom. (iv) Symbol H also represents one gram hydrogen atom. PAGE # 18

An anion bears that much units of negative charge as there are the number of electrons gained by the neutral atom to form that anion. e.g. A nitrogen atom gains 3 electrons to form nitride ion, so nitride ion bears 3 units of negative charge and it is represented as N3-. Note : Size of a cation is always smaller and anion is always greater than that of the corresponding neutral atom.

IONS An ion is a positively or negatively charged atom or group of atoms. Every atom contains equal number of electrons (negatively charged) and protons (positively charged). Both charges balance each other, hence atom is electrically neutral.



(a) Cation :

(c) Monoatomic ions and polyatomic ions : (i) Monoatomic ions : Those ions which are formed from single atoms are called monoatomic ions or simple ions. e.g. Na+, Mg2+ etc.

If an atom has less electrons than a neutral atom, then it gets positively charged and a positively charged ion is known as cation. e.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc. A cation bears that much units of positive charge as there are the number of electrons lost by the neutral atom to form that cation.

(ii) Polyatomic ions : Those ions which are formed from group of atoms joined together are called polyatomic ions or compound ions. e.g. Ammonium ion (NH4+) , hydroxide ion (OH–) etc. which are formed by the joining of two types of atoms, nitrogen and hydrogen in the first case and oxygen and hydrogen in the second.

e.g. An aluminium atom loses 3 electrons to form aluminium ion, so aluminium ion bears 3 units of positive charge and it is represented as Al3+.

(d) Valency of ions : The valency of an ion is same as the charge present on the ion. If an ion has 1 unit of positive charge, its valency is 1 and it is known as a monovalent cation. If an ion has 2 units of negative charge, its valency is 2 and it is known as a divalent anion.

(b) Anion : If an atom has more number of electrons than that of neutral atom, then it gets negatively charged and a negatively charged ion is known as anion. e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS Bivalent Electronegative

Monovalent Electronegative 1. Fluoride

F–

1. Sulphate 2. Sulphite

SO 3

3. Sulphide

S

4. Thiosulphate

S 2O3

5. Zincate

ZnO2

6. Oxide

O

7. Peroxide

O2

8. Dichromate

Cr 2O7

–

9. Carbonate

–

10. Silicate

2CO 3 2SiO 3

Br I

6. Hydroxide 7. Nitrite 8.Nitrate

H

– –

OH

–

NO2 – NO3

9. Bicarbonate or Hydrogen carbonate HCO3 10. Bisulphite or Hydrogen sulphite 11. Bisulphide or Hydrogen sulphide 12. Bisulphate or Hydrogen sulphate 13. Acetate



2-

–

Cl

3. Bromide 5. Hydride

2-

SO 4

–

2. Chloride 4. Iodide

Trivalent Electronegative

HSO3 – HS

22-

1. Nitride

N

3-

Tetravalent Electronegative 1. Carbide

C

4-

3-

2. Phosphide

P

3. Phosphite

PO 3

4. Phosphate

PO 4

33-

2-

222-

–

HSO4 – CH3COO

Note : Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present in them. PAGE # 19

Example :

LAWS OF CHEMICAL COMBINATION

Water is a compound of hydrogen and oxygen. It can be obtained from various sources (like river, sea, well etc.) or even synthesized in the laboratory. From whatever source we may get it, 9 parts by weight of water is always found to contain 1 part by weight of hydrogen and 8 parts by weight of oxygen. Thus, in water, this proportion of hydrogen and oxygen always remains constant.

The laws of chemical combination are the experimental laws which led to the idea of atoms being the smallest unit of matter. The laws of chemical combination played a significant role in the development of Dalton’s atomic theory of matter. There are two important laws of chemical combination. These are:  (i) Law of conservation of mass (ii) Law of constant proportions (a) Law of Conservation of Mass or Matter :

(c) Law of Multiple Proportions :

This law was given by Lavoisier in 1774 . According to the law of conservation of mass, matter can neither be created nor be destroyed in a chemical reaction. Or The law of conservation of mass means that in a chemical reaction, the total mass of products is equal to the total mass of the reactants. There is no change in mass during a chemical reaction. Suppose we carry out a chemical reaction between A and B and if the products formed are C and D then, A + B  C + D Suppose 'a' g of A and 'b' g of B react to produce 'c' g of C and 'd' g of D. Then, according to the law of conservation of mass, we have, a+b = c+d Example : When Calcium Carbonate (CaCO 3) is heated, a chemical reaction takes place to form Calcium Oxide (CaO) and Carbon dioxide (CO2). It has been found by experiments that if 100 grams of calcium carbonate is decomposed completely, then 56 grams of Calcium Oxide and 44 grams of Carbon dioxide are formed.

Note : The converse of Law of definite proportions that when same elements combine in the same proportion, the same compound will be formed, is not always true. According to it, when one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other, bears a simple ratio to one another. Simple ratio means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3 e.g. Carbon and oxygen when combine, can form two oxides that are CO (carbon monoxide), CO2 (carbon dioxide). In CO,12 g carbon combined with 16 g of oxygen. In CO2,12 g carbon combined with 32 g of oxygen. Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 g) bear simple ratio of 16 : 32 or 1 : 2



Note : The law of multiple proportion was given by Dalton in 1808.

Sample Problem : 1.

Since the total mass of products (100g ) is equal to the total mass of the reactants (100g), there is no change in the mass during this chemical reaction. The mass remains same or conserved. (b) Law of Constant Proportions / Law of Definite Proportions : Proust, in 1779, analysed the chemical composition (types of elements present and percentage of elements present ) of a large number of compounds and came to the conclusion that the proportion of each element in a compound is constant (or fixed). According to the law of constant proportions: A chemical compound always consists of the same elements combined together in the same proportion by mass. 

Note : The chemical composition of a pure substance is not dependent on the source from which it is obtained.

Carbon is found to form two oxides, which contain 42.8% and 27.27% of carbon respectively. Show that these figures illustrate the law of multiple proportions. Sol. % of carbon in first oxide = 42.8 % of oxygen in first oxide = 100 - 42.8 = 57.2 % of carbon in second oxide = 27.27  % of oxygen in second oxide = 100 - 27.27 = 72.73 For the first oxide Mass of oxygen in grams that combines with 42.8 g of carbon = 57.2  Mass of oxygen that combines with 1 g of carbon = 57.2  1.34 g 42.8 For the second oxide Mass of oxygen in grams that combines with 27.27 g of carbon = 72.73  Mass of oxygen that combines with 1 g of carbon = 72.73  2.68 g 27.27

Ratio between the masses of oxygen that combine with a fixed mass (1 g) of carbon in the two oxides = 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence, this illustrates the law of multiple proportion. PAGE # 20

(d) Law of Reciprocal Proportions :

(e) Gay Lussac’s Law of Gaseous Volumes : Gay Lussac found that there exists a definite relationship among the volumes of the gaseous reactants and their products. In 1808, he put forward a generalization known as the Gay Lussac’s Law of combining volumes. This may be stated as follows :

According to this law the ratio of the weights of two element A and B which combine separately with a fixed weight of the third element C is either the same or some simple multiple of the ratio of the weights in which A and B combine directly with each other. e.g.

When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the product, if these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure. e.g. Combination between hydrogen and chlorine to form hydrogen chloride gas. One volume of hydrogen and one volume of chlorine always combine to form two volumes of hydrochloric acid gas.

The elements C and O combine separately with the third element H to form CH4 and H2O and they combine directly with each other to form CO2. 4 H2 CH4

12 C 12

H2O

CO2

16 O 32

In CH4, 12 parts by weight of carbon combine with 4 parts by weight of hydrogen. In H2O, 2 parts by weight of hydrogen combine with 16 parts by weight of oxygen. Thus the weight of C and O which combine with fixed weight of hydrogen (say 4 parts by weight) are 12 and 32 i.e. they are in the ratio 12 : 32 or 3 : 8. Now in CO2, 12 parts by weight of carbon combine directly with 32 parts by weight of oxygen i.e. they combine directly in the ratio 12 : 32 or 3 : 8 which is the same as the first ratio. 

H2 (g) + Cl2 (g) 2HCl (g) 1vol. 1 vol. 2 vol. The ratio between the volume of the reactants and the product in this reaction is simple, i.e., 1 : 1 : 2. Hence it illustrates the Law of combining volumes. (f) Avogadro’s Hypothesis : This states that equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. This hypothesis has been found to explain elegantly all the gaseous reactions and is now widely recognized as a law or a principle known as Avogadro’s Law or Avogadro’s principle. The reaction between hydrogen and chlorine can be explained on the basis of Avogadro’s Law as follows :

Note : The law of reciprocal proportion was put forward by Ritcher in 1794.

2.

Sample Problem : Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that these data illustrate the law of reciprocal proportions.

Hydrogen + Chlorine

Sol

In NH3, 17.65 g of H combine with N = 82.35 g

 1g of H combine with N =

82.35 17.65

88.90 11.10

1 vol.

2 vol.

n molecules.

n molecules.

2n molecules.(By Avogadro's Law)

1 molecules. 2

1 molecules. 2

1 molecules. (By dividing throughout by 2n)

1 Atom

1 Atom

1 Molecule

(By experiment)

(Applying Avogadro's hypothesis)

g = 4.67 g

In H2O, 11.10 g of H combine with O = 88.90 g

 1 g H combine with O =

Hydrogen chloride gas

1 vol.

g = 8.01 g

 Ratio of the weights of N and O which combine

It implies that one molecule of hydrogen chloride gas is made up of 1 atom of hydrogen and 1 atom of chlorine. (i) Applications of Avogadro’s hypothesis :

with fixed weight (=1g) of H = 4.67 : 8.01 = 1 : 1.72

(A) In the calculation of atomicity of elementary gases.

In N2O3, ratio of weights of N and O which combine with each other = 36.85 : 63.15 = 1 : 1.71

e.g.

Thus the two ratio are the same. Hence it illustrates the law of reciprocal proportions.

2 volumes of hydrogen combine with 1 volume of oxygen to form two volumes of water vapours. Hydrogen + Oxygen  Water vapours 2 vol. 1 vol. 2 vol. PAGE # 21

Applying Avogadro’s hypothesis

GRAM-ATOMIC MASS

Hydrogen + Oxygen  Water vapours 2 n molecules n molecules 2 n molecules

1 or 1 molecule molecule 2

1 molecule

1 Thus1 molecule of water contains molecule of 2

The atomic mass of an element expressed in grams is called the Gram Atomic Mass of the element. The number of gram -atoms =

oxygen. But 1 molecule of water contains 1 atom of oxygen. Hence.

1 molecule of oxygen = 1 atom of 2

e.g. Calculate the gram atoms present in (i) 16g of oxygen

oxygen or 1 molecules of oxygen = 2 atoms of oxygen i.e. atomicity of oxygen = 2.

and (ii) 64g of sulphur.

(B) To find the relationship between molecular mass and vapour density of a gas.

(i) The atomic mass of oxygen = 16.

Density of gas Vapour density (V.D.) = Density of hydrogen =

Mass of a certain volume of the gas Mass of the same volume of hydrogen at the same temp. and pressure

 Gram-Atomic Mass of oxygen (O) = 16 g. No. of Gram-Atoms =

V.D. =

64 64 = Gram Atomic Mass of sulphur = =2 32

Atomic Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Mass of n molecules of the gas Mass of n molecules of hydrogen

=

Mass of 1 molecule of the gas Mass of 1 molecule of hydrogen

=

Molecular mass of the gas Molecular mass of hydrogen

Molecular mass 2 (since molecular mass of hydrogen is 2) Hence, Molecular mass = 2 × Vapour density =

ATOMIC MASS UNIT The atomic mass unit (amu) is equal to one-twelfth (1/12) of the mass of an atom of carbon-12.The mass of an atom of carbon-12 isotope was given the atomic mass of 12 units, i.e. 12 amu or 12 u. The atomic masses of all other elements are now expressed in atomic mass units.

RELATIVE ATOMIC MASS The atomic mass of an element is a relative quantity and it is the mass of one atom of the element relative to one -twelfth (1/12) of the mass of one carbon-12 atom. Thus, Relative atomic mass =

Mass of one atom of the element 1 12

 mass of one C  12 atom

[1/12 the mass of one C-12 atom = 1 amu, 1 amu = 1.66 × 10–24 g = 1.66 × 10–27 kg.] Note : One amu is also called one dalton (Da).

16 =1 16

(ii) The gram-atoms present in 64 grams of sulphur.

If n molecules are present in the given volume of a gas and hydrogen under similar conditions of temperature and pressure.



Mass of the element in grams Gram Atomic mass of the element

Element Symbol Hydrogen H Helium He Lithium Li Beryllium Be Boron B Carbon C Nitrogen N Oxygen O Fluorine F Neon Ne Sodium Na Magnesium Mg Aluminium Al Silicon Si Phosphorus P Sulphur S Chlorine Cl Argon Ar Potassium K Calcium Ca

Atomic mass 1 4 7 9 11 12 14 16 19 20 23 24 27 28 31 32 35.5 40 39 40

RELATIVE MOLECULAR MASS The relative molecular mass of a substance is the mass of a molecule of the substance as compared to one-twelfth of the mass of one carbon -12 atom i.e., Relative molecular mass =

Mass of one molecule of the substance 1 12

 mass of one C  12 atom

The molecular mass of a molecule, thus, represents the number of times it is heavier than 1/12 of the mass of an atom of carbon-12 isotope. PAGE # 22

GRAM MOLECULAR MASS

The molecular mass of a substance expressed in grams is called the Gram Molecular Mass of the substance . The number of gram molecules Mass of the subs tance in grams

=

Gram molecular mass of the subs t ance

e.g. (i) Molecular mass of hydrogen (H2) = 2u.

 Gram Molecular Mass of hydrogen (H2) = 2 g . (ii) Molecular mass of methane (CH4) = 16u

 Gram Molecular Mass of methane (CH4) = 16 g.

EQUIVALENT MASS (a) Definition : Equivalent mass of an element is the mass of the element which combine with or displaces 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine. (b) Formulae of Equivalent Masses of different substances : (i) Equivalent mass of an element = Atomic wt. of the element Valency of the element (ii) Eq. mass an acid =

Mol. wt. of the acid Basicity of the acid

e.g. the number of gram molecules present in 64 g of methane (CH4).

Basicity is the number of replaceable H+ ions from one molecule of the acid.

64 64 = Gram molecular mass of CH = = 4. 4 16

(iii) Eq. Mass of a base =

(a) Calculation of Molecular Mass : The molecular mass of a substance is the sum of the atomic masses of its constituent atoms present in a molecule. Ex.1 Calculate the molecular mass of water. (Atomic masses : H = 1u, O = 16u). Sol. The molecular formula of water is H2O.

Molecular mass of water = ( 2 × atomic mass of H) + (1 × atomic mass of O) = 2 × 1 + 1 × 16 = 18 i.e., molecular mass of water = 18 amu. Ex.2 Find out the molecular mass of sulphuric acid. (Atomic mass : H = 1u, O = 16u, S = 32u). Sol. The molecular formula of sulphuric acid is H2SO4.

 Molecular mass of H2SO4 = (2 × atomic mass of H) + ( 1 × atomic mass of S) + ( 4 × atomic mass of O) = (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98 i.e., Molecular mass of H2SO4= 98 amu.

FORMULA MASS The term ‘formula mass’ is used for ionic compounds and others where discrete molecules do not exist, e.g., sodium chloride, which is best represented as (Na+Cl–)n, but for reasons of simplicity as NaCl or Na+Cl–. Here, formula mass means the sum of the masses of all the species in the formula. Thus, the formula mass of sodium chloride = (atomic mass of sodium) + (atomic mass of chlorine) = 23 + 35.5 = 58.5 amu

Mol. wt. of the base Acidity of the base

Acidity is the number of replaceable OH– ions from one molecule of the base (iv) Eq. mass of a salt =

Mol. wt. of the salt Number of metal atoms  valency of metal

(v) Eq. mass of an ion =

Formula wt. of the ion Charge on the ion

(vi) Eq. mass of an oxidizing/reducing agent =

Mol wt. or At. wt No. of electrons lost or gained by one molecule of the substance

Equivalent weight of some compounds are given in the table :

S.No.

Compound

Equivalent weight

1

HCl

36.5

2

H2SO4

49

3

HNO3

63

4

45 COOH

5 COOH

.2H2O

63

6

NaOH

40

7

KOH

56

8

CaCO3

50

9

NaCl

58.5

10

Na2CO3

53

PAGE # 23

SOME IMPORTANT RELATIONS AND FORMULAE In Latin, mole means heap or collection or pile. A mole of atoms is a collection of atoms whose total mass is the number of grams equal to the atomic mass in magnitude. Since an equal number of moles of different elements contain an equal number of atoms, it becomes convenient to express the amounts of the elements in terms of moles. A mole represents a definite number of particles, viz, atoms, molecules, ions or electrons. This definite number is called the Avogadro Number (now called the Avogadro constant) which is equal to 6.023 × 1023.

(i) 1 mole of atoms = Gram Atomic mass = mass of 6.023 × 1023 atoms (ii) 1 mole of molecules = Gram Molecular Mass = 6.023 x 1023 molecules (iii) Number of moles of atoms

=

(iv) Number of moles of molecules

A mole is defined as the amount of a substance that contains as many atoms, molecules, ions, electrons or other elementary particles as there are atoms in exactly 12 g of carbon -12 (12C).

=

(a) Moles of Atoms :

=

(i) 1 mole atoms of any element occupy a mass which is equal to the Gram Atomic Mass of that element. e.g. 1 Mole of oxygen atoms weigh equal to Gram Atomic Mass of oxygen, i.e. 16 grams. (ii) The symbol of an element represents 6.023 x 1023 atoms (1 mole of atoms) of that element. e.g : Symbol N represents 1 mole of nitrogen atoms and 2N represents 2 moles of nitrogen atoms. (b) Moles of Molecules : (i) 1 mole molecules of any substance occupy a mass which is equal to the Gram Molecular Mass of that substance.

Mass of element in grams Gram Atomic Mass of element

Mass of substance in grams Gram Molecular Mass of substance

(v) Number of moles of molecules No. of molecules of element

N 

Avogadro number

NA

Ex.3 To calculate the number of moles in 16 grams of Sulphur (Atomic mass of Sulphur = 32 u). Sol. 1 mole of atoms = Gram Atomic Mass. So, 1 mole of Sulphur atoms = Gram Atomic Mass of Sulphur = 32 grams. Now, 32 grams of Sulphur = 1 mole of Sulphur So, 16 grams of Sulphur = (1/32) x 16 = 0.5 moles Thus, 16 grams of Sulphur constitute 0.5 mole of Sulphur. 22.4 litre

e.g. : 1 mole of water (H2O) molecules weigh equal to Gram Molecular Mass of water (H2O), i.e. 18 grams.

In term of volume 23

(ii) The symbol of a compound represents 6.023 x 10 23 molecules (1 mole of molecules) of that compound.

6.023 × 10 (NA) Atoms

23

1 Mole

e.g. : Symbol H 2O represents 1 mole of water molecules and 2 H2O represents 2 moles of water molecules. 

Note : The symbol H2O does not represent 1 mole of H2 molecules and 1 mole of O atoms. Instead, it represents 2 moles of hydrogen atoms and 1 mole of oxygen atoms.



Note : The SI unit of the amount of a substance is Mole. (c) Mole in Terms of Volume : Volume occupied by 1 Gram Molecular Mass or 1 mole of a gas under standard conditions of temperature and pressure (273 K and 1atm. pressure) is called Gram Molecular Volume. Its value is 22.4 litres for each gas. Volume of 1 mole = 22.4 litre (at STP)



Note : The term mole was introduced by Ostwald in 1896.

6.023 × 10 (NA) molecules

In terms of particles

In terms of mass

1 gram atom of element

1 gram molecule of substance

1 gram formula mass of substance

PROBLEMS BASED ON THE MOLE CONCEPT Ex.4 Calculate the number of moles in 5.75 g of sodium. (Atomic mass of sodium = 23 u) Sol. Number of moles Mass of the element in grams

=

Gram Atomic Mass of element

=

5.75 23

= 0.25 mole

or, 1 mole of sodium atoms = Gram Atomic mass of sodium = 23g. 23 g of sodium = 1 mole of sodium. 5.75 g of sodium =

5.75 23

mole of sodium = 0.25 mole

PAGE # 24

Ex.5 What is the mass in grams of a single atom of chlorine ? (Atomic mass of chlorine = 35.5u) Sol. Mass of 6.022 × 1023 atoms of Cl = Gram Atomic Mass of Cl = 35.5 g.

 Mass of 1 atom of Cl =

35.5 g 6.022  10 23

= 5.9 × 10–23 g.

Ex.6 The density of mercury is 13.6 g cm . How many moles of mercury are there in 1 litre of the metal ? (Atomic mass of Hg = 200 u). –3

Sol. Mass of mercury (Hg) in grams = Density (g cm–3)× Volume (cm3) = 13.6 g cm–3 × 1000 cm3 = 13600 g.  Number of moles of mercury Mass of mercury in grams 13600 = Gram Atomic Mass of mercury = = 68 200

Ex.7 The mass of a single atom of an element M is 3.15× 10–23 g . What is its atomic mass ? What could the element be ? Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms = mass of 1 atom × 6.022 × 1023 = (3.15 × 10–23g) × 6.022 × 1023 = 3.15 × 6.022 g = 18.97 g.  Atomic Mass of the element = 18.97u Thus, the element is most likely to be fluorine. Ex.8 An atom of neon has a mass of 3.35 × 10–23 g. How many atoms of neon are there in 20 g of the gas ? Sol. Number of atoms =

20 Total mass = = 5.97 × 1023 Mass of 1 atom 3.35  10 – 23

Ex.9 How many grams of sodium will have the same number of atoms as atoms present in 6 g of magnesium ? (Atomic masses : Na = 23u ; Mg =24u) Sol. Number of gram -atom of Mg Mass of Mg in grams

1

6

= Gram Atomic Mass = = 24 4

 Gram Atoms of sodium should be =

1 4

 1 Gram Atom of sodium = 23 g 1 4

gram atoms of sodium = 23 ×

1

= 5.75 g

4

Ex.10 How many moles of Cr are there in 85g of Cr2S3 ? (Atomic masses : Cr = 52 u , S =32 u) Sol. Molecular mass of Cr2S3 = 2 × 52 + 3 × 32 = 104 + 96 = 200 u. 200g of Cr2S3 contains = 104 g of Cr.

 85 g of Cr2S3 contains =

104  85 200

Thus, number of moles of Cr =

g of Cr = 44.2g

44.2 52

Ex.11 What mass in grams is represented by (a) 0.40 mol of CO2, (b) 3.00 mol of NH3, (c) 5.14 mol of H5IO6 (Atomic masses : C=12 u, O=16 u, N=14 u, H=1 u and I = 127 u) Sol. Weight in grams = number of moles × molecular mass. Hence, (a) mass of CO2 = 0.40 × 44 = 17.6 g (b) mass of NH3 = 3.00 × 17 = 51.0 g (c) mass of H5IO6 = 5.14 × 228 = 1171.92g Ex.12 Calculate the volume in litres of 20 g of hydrogen gas at STP. Sol. Number of moles of hydrogen Mass of hydrogen in grams

=

Gram Molecular Mass of hydrogen

=

20 = 10 2

 Volume of hydrogen = number of moles × Gram Molecular Volume. = 10 ×22.4 = 224 litres. Ex.13 The molecular mass of H 2 SO 4 is 98 amu. Calculate the number of moles of each element in 294 g of H2SO4. Sol. Number of moles of H2SO4 =

294 98

=3.

The formula H 2SO 4 indicates that 1 molecule of H2SO4 contains 2 atoms of H, 1 atom of S and 4 atoms of O. Thus, 1 mole of H2SO4 will contain 2 moles of H,1 mole of S and 4 moles of O atoms Therefore, in 3 moles of H2SO4 : Number of moles of H = 2 × 3 = 6 Number of moles of S = 1 × 3 = 3 Number of moles of O = 4 × 3 = 12 Ex.14 Find the mass of oxygen contained in 1 kg of potassium nitrate (KNO3). Sol. Since 1 molecule of KNO 3 contains 3 atoms of oxygen, 1 mol of KNO 3 contains 3 moles of oxygen atoms.  Moles of oxygen atoms = 3 × moles of KNO3 =3×

1000

= 29.7 101 (Gram Molecular Mass of KNO3 = 101 g)  Mass of oxygen = Number of moles × Atomic mass = 29.7 × 16 = 475.2 g Ex.15 You are asked by your teacher to buy 10 moles of distilled water from a shop where small bottles each containing 20 g of such water are available. How many bottles will you buy ? Sol. Gram Molecular Mass of water (H2O) = 18 g  10 mol of distilled water = 18 × 10 = 180 g. Because 20 g distilled water is contained in 1 bottle, 180 g

of distilled water is contained in =

= 0.85 . bottles = 9 bottles.

 Number of bottles to be bought = 9 PAGE # 25

180 20

Ex.16 6.022 × 1023 molecules of oxygen (O2) is equal to how many moles ? Sol. No. of moles = No. of molecules of oxygen Avogadro' s no. of molecules

N 

N

=

A

6.023  10 23 6.023  10 23

(iii) If the simplest ratio is fractional, then values of simplest ratio of each element is multiplied by smallest integer to get the simplest whole number for each of the element.

=1

(iv) To get the empirical formula, symbols of various elements present are written side by side with their respective whole number ratio as a subscript to the lower right hand corner of the symbol.

PERCENTAGE COMPOSITION The percentage composition of elements in a compound is calculated from the molecular formula of the compound. The molecular mass of the compound is calculated from the atomic masses of the various elements present in the compound. The percentage by mass of each element is then computed with the help of the following relations. Percentage mass of the element in the compound

(v) The molecular formula of a substance may be determined from the empirical formula if the molecular mass of the substance is known. The molecular formula is always a simple multiple of empirical formula and the value of simple multiple (n) is obtained by dividing molecular mass with empirical formula mass. n

=

Total mass of the element

=

× 100

Molecular mass

Ex.17 What is the percentage of calcium in calcium carbonate (CaCO3) ? Sol. Molecular mass of CaCO 3 = 40 + 12 + 3 × 16 = 100 amu. Mass of calcium in 1 mol of CaCO3 = 40g.

Percentage of calcium =

40  100

Ex-20 A compound of carbon, hydrogen and nitrogen contains these elements in the ratio of 9:1:3.5 respectively. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula ? Sol.

32  100 98

Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu. H2O has two atoms of hydrogen. So, total mass of hydrogen in H2O = 2 amu. 2  100 18

= 11.11 %

Similarly, percentage of oxygen =

16  100 18

Mass Atomic Relative Number Ratio Mass of Atoms

Carbon

9

12

Hydrogen

1

1

Nitrogen

3.5

14

= 88.88 %

The following steps are involved in determining the empirical formula of a compound : (i) The percentage composition of each element is divided by its atomic mass. It gives atomic ratio of the elements present in the compound. (ii) The atomic ratio of each element is divided by the minimum value of atomic ratio as to get the simplest ratio of the atoms of elements present in the compound.

9  0.75 12

Simplest Ratio 0.75×4 = 3

1  1 1

1×4=4

3.5  0.25 14

0.25 ×4 = 1

Empirical ratio = C3H4N Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54

= 32.65 %

Ex.19 W hat are the percentage compositions of hydrogen and oxygen in water (H2O) ? (Atomic masses : H = 1 u, O = 16 u)

Percentage of H =

Element

= 40 %

100 Ex.18 What is the percentage of sulphur in sulphuric acid (H2SO4) ? Sol. Molecular mass of H2SO4 = 1 × 2 + 32 + 16 × 4 = 98 amu.

 Percentage of sulphur =

Molecular Mass Empirical Formula Mass

n=

Molecular Mass 108 2 Empirical Formula Mass = 54

Thus, molecular formula of the compound = (Empirical formula)2 = (C3H4N)2 = C6H8N2 Ex.21 A compound on analysis, was found to have the following composition : (i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen = 69.50%, (iv) Hydrogen = 6.22%. Calculate the molecular formula of the compound assuming that whole hydrogen in the compound is present as water of crystallisation. Molecular mass of the compound is 322. Sol.

Element Percentage

Atomic mass

Relative Number of atoms

Simplest ratio

Sodium

14.31

23

14 .31  23

0.622

0.622 2 0.311

Sulphur

9.97

32

9 .97  32

0.311

0.311 1 0.311

Hydrogen

6.22

1

6.22

6.22  20 0.311

Oxygen

69.50

16

4.34

4.34  14 0.311

6 . 22 1 69. 50  16



PAGE # 26

The empirical formula = Na2SH20O14 Empirical formula mass = (2 × 23) + 32 + (20 × 1) + (14 × 16) = 322 Molecular mass = 322 Molecular formula = Na2SH20O14 Whole of the hydrogen is present in the form of water of crystallisation. Thus, 10 water molecules are present in the molecule. So, molecular formula = Na2SO4. 10H2O

Normality of a solution is defined as the number of gram equivalents of the solute dissolved per litre (dm3) of given solution. It is denoted by ‘N’. Mathematically, Number of gram equivalent s of solute

N =

N=

Weight of solute in gram / equivalent weight of solute Volume of the solution in litre

Strength in grams per litre

(a) Strength in g/L :

N=

The strength of a solution is defined as the amount of the solute in grams present in one litre (or dm3) of the solution, and hence is expressed in g/litre or g/dm3.

Equivalent mass of solute

=

S E

If ‘w’ gram of the solute is present in V cm3 of a given solution. 1000

w

Weight of solute in gram

N = Equivalent mass of the solute ×

Volume of solution in litre

(b) Molarity : Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm3) of solution. It is denoted by ‘M’. Mathematically,

w

N = Equivalent mass of the solute × N=

Volume of the solution in litre

V

e.g. A solution of sulphuric acid having 0.49 gram of it dissolved in 250 cm 3 of solution will have its normality,

Number of moles of solute

M=

Volume of the solution in litre

N can be calculated from the strength as given below :

CONCENTRATION OF SOLUTIONS

Strength in g/L =

(d) Normality :

0.49 49

1000

×

250

1000 V

= 0.04

(Eq. mass of sulphuric acid = 49). Mass of solute in gram/Gram Molecular Mass of solute Volume of solution in litre

M can be calculated from the strength as given below : M=

1000

w Molecular

×

mass

V e.g. a solution of sulphuric acid having 4.9 grams of it dissolved in 500 cm3 of solution will have its molarity, w

M=

Normality

Deci normal

1 2

1000

Molecular mass

4.9 98

1000

×

500

×

Centi normal

1 10

1 100

Some Important Formulae :

Molecular mass of solute

If ‘w’ gram of the solute is present in V cm of a given solution , then

M=

Semi normal

Strength in grams per litre 3

M=

Solution

V

= 0.1

(c) Formality : In case of ionic compounds like NaCl, Na2CO3 etc., formality is used in place of molarity. The formality of a solution is defined as the number of gram formula masses of the solute dissolved per litre of the solution. It is represented by the symbol ‘F’. The term formula mass is used in place of molecular mass because ionic compounds exist as ions and not as molecules. Formula mass is the sum of the atomic masses of the atoms in the formula of the compound. Mass of solute in gram/Formula Mass of solute Volume of solution in litre

(i) Milli equivalent of substance = N × V where , N  normality of solution V  Volume of solution in mL (ii) If weight of substance is given, milli equivalent (NV) =

w  1000 E

Where, W  Weight of substance in gram E  Equivalent weight of substance (iii) S = N × E S  Strength in g/L N  Normality of solution E  Equivalent weight (iv) Calculation of normality of mixture : N N HCl is mixed with 50 ml of H SO . 10 5 2 4 Find out the normality of the mixture. Sol. Milli equivalent of HCl + milli equivalent H 2SO 4 = milli equivalent of mixture { where, V3 =V1 + V2 ) N1 V1 + N2 V2 = N3 V3

Ex.22 100 ml of

 1  1   100     50   N3 × 150  10 5    

N3 =

20 150

=

2 15

= 0.133

PAGE # 27

Ex.23 100 ml of

N

HCl is mixed with 25 ml of

10 Find out the normality of the mixture.

N 5

NaOH.

Sol. Milli equivalent of HCl – milli equivalent of NaOH = milli equivalent of mixture N1 V1 – N2 V2 = N3 V3 { where, V3 =V1 + V2 )  1  1   100  –   25  = N3 × 125  10 5    

N3 = 

1 25

Note : 1 milli equivalent of an acid neutralizes 1 milli equivalent of a base. (e) Molality : Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams of the solvent. It is denoted by ‘m’. Mathematically,

Number of moles of the solute × 1000 ‘m’ can Weight of the solvent in gram be calculated from the strength as given below : m=

Strength per 1000 gram of solvent m= Molecular mass of solute If ‘w’ gram of the solute is dissolved in ‘W’ gram of the solvent then m=

w 1000 × Mol. mass of the solute W

e.g. A solution of anhydrous sodium carbonate (molecular mass = 106) having 1.325 grams of it, dissolved in 250 gram of water will have its molality m= 

1.325 1000  = 0.05 106 250

Ex.25 Find the molarity and molality of a 15% solution of H2SO4 (density of H2SO4 solution = 1.02 g/cm3) (Atomic mass : H = 1u, O = 16u , S = 32 u) Sol. 15% solution of H2SO4 means 15g of H2SO4 are present in 100g of the solution i.e. Wt. of H2SO4 dissolved = 15 g Weight of the solution = 100 g Density of the solution = 1.02 g/cm3 (Given) Calculation of molality : Weight of solution = 100 g Weight of H2SO4 = 15 g Wt. of water (solvent) = 100 – 15 = 85 g Molecular weight of H2SO4 = 98

 15 g H2SO4 =

15 98

= 0.153 moles

Thus ,85 g of the solvent contain 0.153 moles . 1000 g of the solvent contain=

0.153 85

× 1000 = 1.8 mole

Hence ,the molality of H2SO4 solution = 1.8 m Calculation of molarity : 15 g of H2SO4 = 0.153 moles Wt. of solution

Vol. of solution = =

100 1.02

Density of solution

= 98.04 cm3

This 98.04 cm3 of solution contain H2SO4 = 0.153 moles 1000 cm3 of solution contain H2SO4

0.153 =

98.04

× 1000 = 1.56 moles

Note :

Hence the molarity of H2SO4 solution = 1.56 M

Relationship Between Normality and Molarity of a Solution : Normality of an acid = Molarity × Basicity Normality of base = Molarity × Acidity

(f) Mole Fraction :

Ex.24 Calculate the molarity and normality of a solution containing 0.5 g of NaOH dissolved in 500 cm 3 of solvent. Sol. Weight of NaOH dissolved = 0.5 g Volume of the solution = 500 cm3 (i) Calculation of molarity : Molecular weight of NaOH = 23 + 16 + 1 = 40 Weight of solute/ molecular weight of solute Volume of solution in litre 0.5/40

Molarity = =

500/1000

= 0.025

(ii) Calculation of normality : Normality Weight of solute/ equivalent weight of solute = Volume of solution in litre 0.5/40 = = 0.025 500/1000

The ratio between the moles of solute or solvent to the total moles of solution is called mole fraction.

Moles of solute mole fraction of solute = =

Moles of solution



n nN

w/m w/m  W/M

Moles of solvent Mole fraction of solvent =

Moles of solution



N nN =

W/M w/m  W/M where,

n  number of moles of solute N  number of moles of solvent m  molecular weight of solute M  molecular weight of solvent w  weight of solute W  weight of solvent PAGE # 28

Ex.26 Find out the mole fraction of solute in 10% (by weight) urea solution. weight of solute (urea) = 10 g weight of solution = 100 g weight of solvent (water) = 100 – 10 = 90g

Moles of solution 10 / 60

w/m w/m  W/M 

=

10 / 60  90 / 18

2KClO3(s)  2KCl(s) + 3O2(g) 2 mole KClO3  3 mole O2

 3 mole O2 formed by 2 mole KClO3 2

Moles of solute mole fraction of solute =

Sol. Decomposition of KClO3 takes place as,

=

= 0.032

Note : Sum of mole fraction of solute and solvent is always equal to one.



 2.4 mole O 2 will be formed by  3  2.4    mole KClO3 = 1.6 mole KClO3 Mass of KClO3 = Number of moles × molar mass = 1.6 × 122.5 = 196 g (ii) Calculations based on mass-mass relationship: In making necessary calculation, following steps are followed (a) Write down the balanced chemical equation.

STOICHIOMETRY (a) Quantitative Relations in Chemical Reactions : Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the chemical equation and on the relationship between mass and moles. N2(g) + 3H2(g)  2NH3(g)

(b) Write down theoretical amount of reactants and products involved in the reaction. (c) The unknown amount of substance is calculated using unitary method. Ex.28 Calculate the mass of CaO that can be prepared by heating 200 kg of limestone CaCO3 which is 95% pure. Sol. Amount of pure CaCO 3 =

95  200 = 190 kg 100

A chemical equation can be interpreted as follows -

= 190000 g

1 molecule N2 + 3 molecules H2  2 molecules NH 3 (Molecular interpretation)

CaCO3(s)  CaO(s) + CO2(g)

1 mol N2 + 3 mol H2  2 mol NH3 (Molar interpretation)

100 g CaCO3  56 g CaO

28 g N2 + 6 g H2  34 g NH3 (Mass interpretation) 1 volume N2 + 3 volume H2  2 volume NH3 (Volume interpretation) Thus, calculations based on chemical equations are divided into four types (i) Calculations based on mole-mole relationship. (ii) Calculations based on mass-mass relationship. (iii) Calculations based on mass-volume relationship. (iv) Calculations based on volume -volume relationship. (i) Calculations based on mole-mole relationship : In such calculations, number of moles of reactants are given and those of products are required. Conversely, if number of moles of products are given, then number of moles of reactants are required. Ex.27 Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO 3). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2). How many moles and how many grams of KClO 3 are required to produce 2.4 mole O2.

1 mole CaCO3  1 mole CaO

100 g CaCO3 give 56 g CaO 56  190000 g CaCO3 will give= × 190000 g CaO 100 = 106400 g = 106.4 kg Ex.29 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO 2 ) with aqueous hydrochloric acid according to the reaction MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O How many grams of HCl will react with 5 g MnO2 ? Sol. 1 mole MnO2 reacts with 4 mole HCl or 87 g MnO2 react with 146 g HCl

 5 g MnO2 will react with =

146 × 5 g HCl = 8.39 g HCl 87

Ex.30 How many grams of oxygen are required to burn completely 570 g of octane ? Sol. Balanced equation

2C8H18 + 25O2 2 mole 2 × 114

16CO2 + 18H2O

25 mole 25 × 32

First method : For burning 2 × 114 g of the octane, oxygen required = 25 × 32 g For burning 1 g of octane, oxygen required =

25  32 g 2  114

Thus, for burning 570 g of octane, oxygen required =

25  32 × 570 g = 2000 g 2  114 PAGE # 29

Mole Method : Number of moles of octane in 570 grams

Ex-33 What quantity of copper (II) oxide will react with 2.80 litre of hydrogen at STP ?

570 = 5.0 114

Sol. CuO + 1 mol 79.5 g

For burning 2.0 moles of octane, oxygen required = 25 mol = 25 × 32 g For burning 5 moles of octane, oxygen required =

25  32 × 5.0 g = 2000 g 2 .0

Proportion Method : Let x g of oxygen be required for burning 570 g of octane. It is known that 2 × 114 g of the octane requires 25 × 32 g of oxygen; then, the proportion. x 25  32 g oxygen = 570 g oc tane 2  114 g oc tan e

2Fe2O3 + 8H2SO4

4 mole 4 × 120 g

8 mole 8 × 98 g

4 × 120 g of FeS2 yield H2SO4 = 8 × 98 g 1000 g of FeS2 will yield H2SO4 =

8  98 × 1000 4  120

= 1633.3 g (iii) Calculations involving mass-volume relationship : In such calculations masses of reactants are given and volume of the product is required and vice-versa. 1 mole of a gas occupies 22.4 litre volume at STP. Mass of a gas can be related to volume according to the following gas equation PV = nRT PV =

w RT m

NH4Cl(s)

=

79.5 × 2.80 g = 9.93 g 22.4

Ex-34 Calculate the volume of carbon dioxide at STP evolved by strong heating of 20 g calcium carbonate. Sol. The balanced equation is -

CaCO3

NH3(g) + HCl(g) 1 mol

 53.5 g NH4Cl give 1 mole NH3 1  26.75 g NH4Cl will give × 26.75 mole NH3 53.5 = 0.5 mole PV = nRT 1 ×V = 0.5 × 0.0821 × 300 V = 12.315 litre

CaO +

1 mol 100 g

CO2

1 mol = 22.4 litre at STP

22.4 × 20 = 4.48 litre 100

Ex.35 Calculate the volume of hydrogen liberated at 27ºC and 760 mm pressure by heating 1.2 g of magnesium with excess of hydrochloric acid. Sol. The balanced equation is

Mg + 2HCl

MgCl2

+

24 g

H2 1 mol

24 g of Mg liberate hydrogen = 1 mole 1.2 g Mg will liberate hydrogen = 0.05 mole PV = nRT 1 × V = 0.05 × 0.0821 × 300 V = 1.2315 litre (iv) Calculations based on volume volume relationship : These calculations are based on two laws : (i) Avogadro’s law (ii) Gay-Lussac’s Law e.g. N 2(g)

+

1 mol 1 × 22.4 L

3H 2(g) 3 mol 3 × 22.4 L

2NH 3(g) (Avogadro's law) 2 mol 2 × 22.4 L

(under similar conditions of temperature and pressure, equal moles of gases occupy equal volumes) N2(g)

+

1 vol

Ex-32. What volume of NH3 can be obtained from 26.75 g of NH4Cl at 27ºC and 1 atmosphere pressure. Sol. The balanced equation is 1 mol 53.5 g

22.4 litre of hydrogen at STP reduce CuO = 79.5 g 2.80 litre of hydrogen at STP will reduce CuO

=

Ex.31 How many kilograms of pure H 2SO 4 could be obtained from 1 kg of iron pyrites (FeS2) according to the following reactions ? 4FeS2 + 11O2  2Fe2O3 + 8SO2 2SO2 + O2  2SO3 SO3 + H2O  H2SO4 Sol. Final balanced equation, 4FeS2 + 15O2 + 8H2O

+ H2O

100 g of CaCO3 evolve carbon dioxide = 22.4 litre 20 g CaCO3 will evolve carbon dioxide

25  32  570 = 2000 g 2  114

x=

Cu

H2

1 mol 22.4 litre at NTP

3H2(g) 3 vol

2NH3(g) (Gay- Lussac's Law) 2 vol

(under similar conditions of temperature and pressure, ratio of coefficients by mole is equal to ratio of coefficient by volume). Ex-36 One litre mixture of CO and CO2 is taken. This is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. The volume are measured under the same conditions. Find the composition of mixture by volume. Sol. Let there be x mL CO in the mixture , hence, there will be (1000 – x) mL CO2. The reaction of CO2 with red hot charcoal may be given as -

CO2(g) + C(s) 1 vol. (1000 –x)

2CO(g) 2 vol. 2(1000 – x)

Total volume of the gas becomes = x + 2(1000 – x) x + 2000 – 2x = 1600 x = 400 mL

 volume of CO = 400 mL and volume of CO2 = 600 mL PAGE # 30

Ex-37 What volume of air containing 21% oxygen by volume is required to completely burn 1kg of carbon containing 100% combustible substance ? Sol. Combustion of carbon may be given as,

C(s) + O2(g) 1 mol 12 g

CO2(g)

1 mol 32 g

USEFUL FORMULAE FOR VOLUMETRIC CALCULATIONS (i) milliequivalents = normality × volume in millilitres. (ii) At the end point of titration, the two titrants, say 1 and 2, have the same number of milliequivalents, i.e., N1V1 = N2V2, volume being in mL.

 12 g carbon requires 1 mole O2 for complete combustion 1  1000 mole O2 for 12 combustion, i.e. , 83.33 mole O2

 1000 g carbon will require

m.e. . 1000 (iv) No. of equivalents for a gas =

(iii) No. of equivalents =

Volumeat STP equivalent volume( vol. of 1eq. at STP)

Volume of O2 at STP = 83.33 × 22.4 litre

(v) Strength in grams per litre = normality × equivalent

= 1866.66 litre

weight.

 21 litre O2 is present in 100 litre air 100 1866.66 litre O2 will be present in × 1866.66 litre air 21 = 8888.88 litre or 8.89 × 103 litre

Ex-38 An impure sample of calcium carbonate contains 80% pure calcium carbonate 25 g of the impure sample reacted with excess of hydrochloric acid. Calculate the volume of carbon dioxide at STP obtained from this sample. Sol. 100 g of impure calcium carbonate contains = 80 g pure calcium carbonate 25 g of impure calcium carbonate sample will contain =

80 × 25 = 20 g pure calcium carbonate 100

The desired equation is CaCO3 + 2HCl

(vi) (a) Normality = molarity × factor relating mol. wt. and eq. wt. (b) No. of equivalents = no. of moles × factor relat ing mol. wt. and eq. wt. Ex.39 Calculate the number of milli equivalent of H2SO4 present in 10 mL of N/2 H2SO4 solution. 1 × 10 = 5. 2 Ex.40 Calculate the number of m.e. and equivalents of

Sol. Number of m.e. = normality × volume in mL =

NaOH present in 1 litre of N/10 NaOH solution. Sol. Number of m.e. = normality × volume in mL =

1 × 1000 = 100 10

Number of equivalents =

CaCl2 + CO2 + H2O 22.4 litre at STP

1 mol 100 g

100 g pure CaCO3 liberate = 22.4 litre CO2. 20 g pure CaCO3 liberate =

22.4  20 100

= 4.48 litre CO2

VOLUMETRIC CALCULATIONS

no. of m.e. 100 = = 0.10 1000 1000

Ex.41 Calculate number of m.e. of the acids present in (i) 100 mL of 0.5 M oxalic acid solution. (ii) 50 mL of 0.1 M sulphuric acid solution. Sol. Normality = molarity × basicity of acid (i) Normality of oxalic acid = 0.5 × 2 = 1 N m.e. of oxalic acid = normality × vol. in mL = 1 × 100 = 100. (ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N m.e. of sulphuric acid

= 0.2 × 50 = 10

The quantitative analysis in chemistry is primarily carried out by two methods, viz, volumetric analysis and gravimetric analysis.In the first method the mass of a chemical species is measured by measurement of volume, whereas in the second method it is determined by taking the weight.

Ex.42 A 100 mL solution of KOH contains 10

The strength of a solution in volumetric analysis is generally expressed in terms of normality, i.e., number of equivalents per litre but since the volume in the volumetric analysis is generally taken in millilitres (mL), the normality is expressed by milliequivalents per millilitre.

Again, strength in grams/litre = normality × eq. wt.

milliequivalents of KOH. Calculate its strength in normality and grams/litre. no. of m.e. Sol. Normality = volume in mL

=

10  0 .1 100

 strength of the solution = N/10

=

1  56 = 5.6 gram/litre. 10

  molecular wt. 56  eq. wt. of KOH    56  acidity 1  

PAGE # 31

Ex.43 What is strength in gram/litre of a solution of H2SO4, N 12 cc of which neutralises 15 cc of NaOH 10 solution ?

Sol. m.e. of NaOH solution =

1 × 15 = 1.5 10

m.e. of 12 cc of H2SO4 = 1.5 1.5 12 Strength in grams/litre = normality × eq. wt.

 normality of H2SO4 =

1.5 × 49 grams/litre 12 = 6.125 grams/litre.

=

  molecular wt. 98  eq. wt. of H2 SO 4    49  basicity 2   Ex.44 What weight of KMnO4 will be required to prepare N solution if eq. wt. of KMnO4 is 31.6 ? 10 Sol. Equivalent weight of KMnO4 = 31.6

250 mL of its

1 10 Volume of solution (V) = 250 ml

Normality of solution (N) =

W

NEV ; W = 1  31 .6  250 31 .6  0.79 g 10 1000 1000 40

Ex.45 100 mL of 0.6 N H2SO4 and 200 mL of 0.3 N HCl were mixed together. What will be the normality of the resulting solution ? Sol. m.e. of H2SO4 solution = 0.6 × 100 = 60 m.e. of HCl solution = 0.3 × 200 = 60  m.e. of 300 mL (100 + 200) of acidic mixture = 60 + 60 = 120.

m.e. total vol. 120 2 = = N. 300 5

Normality of the resulting solution =

Ex.46 A sample of Na2CO3. H2O weighing 0.62 g is added to 100 mL of 0.1 N H2SO4. Will the resulting solution be acidic, basic or neutral ?

0.62 Sol. Equivalents of Na2CO3. H2O = = 0.01 62 124    62   eq. wt. of Na 2 CO 3 .H2 O  2   m.e. of Na2CO3. H2O = 0.01 × 1000 = 10 m.e. of H2SO4 = 0.1 × 100 = 10 Since the m.e. of Na2CO3. H2O is equal to that of H2SO4, the resulting solution will be neutral.

(a) Introduction : Volumetric analysis is a method of quantitative analysis. It involves the measurement of the volume of a known solution required to bring about the completion of the reaction with a measured volume of the unknown solution whose concentration or strength is to be determined. By knowing the volume of the known solution, the concentration of the solution under investigation can be calculated. Volumetric analysis is also termed as titrimetric analysis.

(b) Important terms used in volumetric analysis : (i) Titration : The process of addition of the known solution from the burette to the measured volume of solution of the substance to be estimated until the reaction between the two is just complete, is termed as titration. Thus, a titration involves two solutions: (a) Unknown solution and (b) Known solution or standard solution. (ii) Titrant : The reagent or substance whose solution is employed to estimate the concentration of unknown solution is termed titrant. There are two types of reagents or titrants: (A) Primary titrants : These reagents can be accurately weighed and their solutions are not to be standardised before use. Oxalic acid, potassium dichromate, silver nitrate, copper sulphate, ferrous ammonium sulphate, sodium thiosulphates etc., are the examples of primary titrants. (B) Secondary titrants : These reagents cannot accurately weighed and their solutions are to be standardised before use. Sodium hydroxide, potassium hydroxide, hydrochloric acid, sulphuric acid, iodine, potassium permanganate etc. are the examples of secondary titrants. (iii) Standard solution : The solution of exactly known concentration of the titrant is called the standard solution. (iv) Titrate : The solution consisting the substance to be estimated is termed unknown solution. The substance is termed titrate. (v) Equivalence point : The point at which the reagent (titrant) and the substance (titrate) under investigation are chemically equivalent is termed equivalence point or end point. (vi) Indicator : It is the auxiliary substance used for physical (visual) detection of the completion of titration or detection of end point is termed as indicator. Indicators show change in colour or turbidity at the stage of completion of titration. (c) Concentraion representation of solution (A) Strength of solution : Grams of solute dissolved per litre of solution is called strength of solution' (B) Parts Per Million (ppm) : Grams of solute dissolved per 10 6 grams of solvent is called concentration of solution in the unit of Parts Per Million (ppm). This unit is used to represent hardness of water and concentration of very dilute solutions. (C) Percentage by mass : Grams of solute dissolved per 100 grams of solution is called percentage by mass. (D) Percentage by volume : Millilitres of solute per 100 mL of solution is called percentage by volume. For example, if 25 mL ethyl alcohol is diluted with water to make 100 mL solution then the solution thus obtained is 25% ethyl alcohol by volume. (E) Mass by volume percentage :Grams of solute present per 100 mL of solution is called percentage mass by volume. For example, let 25 g glucose is dissolved in water to make 100 mL solution then the solution is 25% mass by volume glucose. PAGE # 32

(d) Classification of reactions involved in volumetric analysis (A) Neutralisation reactions The reaction in which acids and bases react to form salt called neutralisation. e.g., HCI + NaOH  NaCI + H2O H+(acid) + OH–(base)  H2O (feebly ionised) The titration based on neutralisation is called acidimetry or alkalimetry. (B) Oxidation-reduction reactions The reactions involving simultaneous loss and gain of electrons among the reacting species are called oxidation reduction or redox reactions, e.g., let us consider oxidation of ferrous sulphate (Fe2+ ion) by potassium permanganate (MnO 4– ion) in acidic medium. MnO4– + 8H+ + 5e–  Mn2+ + 4H2O (Gain of electrons or reduction) 5  [Fe2+  Fe3+ + e–] (Loss of electrons or oxidation) MnO4– + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O ________________________________________________________________ In the above reaction, MnO4– acts as oxidising agent and Fe2+ acts as reducing agent. The titrations involving redox reactions are called redox titrations. These titrations are also called according to the reagent used in the titration, e.g., iodometric, cerimetric, permanganometric and dichromometric titrations (C) Precipitation reactions : A chemical reaction in which cations and anions combine to form a compound of very low solubility (in the form of residue or precipitate), is called precipitation. BaCl2 + Na2SO4  BaSO4 + 2NaCl (white precipitate) The titrations involving precipitation reactions are called precipitation titrations. (D) Complex formation reactions : These are ion combination reactions in which a soluble slightly dissociated complex ion or compound is formed. Complex compounds retain their identity in the solution and have the properties of the constituent ions and molecules. e.g. CuSO4 + 4NH3  [Cu(NH3)4]SO4 (complex compound) AgNO3 + 2KCN  K[Ag(CN)2] + KNO3 (complex compound) 2CuSO4 + K4[Fe(CN)6]  Cu2[Fe(CN)6] + 2K2SO4 (complex compound) The titrations involving complex formation reactions are called complexometric titrations.

The determination of concentration of bases by titration with a standard acid is called acidimetry and the determination of concentration of acid by titration with a standard base is called alkalimetry. The substances which give different colours with acids and base are called acid base indicators. These indicators are used in the visual detection of the equivalence point in acid-base titrations. The acid-base indicators are also called pH indicators because their colour change according to the pH of the solution. In the selection of indicator for a titration, following two informations are taken into consideration : (i) pH range of indicator (ii) pH change near the equivalence point in the titration. The indicator whose pH range is included in the pH change of the solution near the equivalence point, is taken as suitable indicator for the titration. (i) Strong acid-strong base titration : In the titration of HCl with NaOH, the equivalence point lies in the pH change of 4–10. Thus, methyl orange, methyl red and phenolphthalein will be suitable indicators. (ii) Weak acid-strong base titration : In the titration of CH3COOH with NaOH the equivalence point lies between 7.5 and 10. Hence, phenolphthalein (8.3– 10) will be the suitable indicator. (iii) Weak base-strong acid titration : In the titration of NH4OH (weak base) against HCl (strong acid) the pH at equivalence point is about 6.5 and 4. Thus, methyl orange (3.1–4.4) or methyl red (4.2–6.3) will be suitable indicators. (iv) Weak acid-weak base titration : In the titration of a weak acid (CH3COOH) with weak base (NH4OH) the pH at the equivalence point is about 7, i.e., lies between 6.5 and 7.5 but no sharp change in pH is observed in these titrations. Thus, no simple indicator can be employed for the detection of the equivalence point. (v) Titration of a salt of a weak acid and a strong base with strong acid: H2CO3 + 2NaOH  Na2CO3 + 2H2O Weak acid Strong base Na 2CO3 when titrated with HCl, the following two stages are involved : Na2CO3 + HCl  NaHCO3 + NaCl (First stage) pH = 8.3, near equivalence point NaHCO3 + HCl  NaCl + H2CO3 (Second stage) pH = 4, near equivalence point For first stage, phenolphthalein and for second stage, methyl orange will be the suitable indicator. PAGE # 33

TITRATION OF MIXTURE OF NaOH, Na2CO3 AND NaHCO3 BY STRONG ACID LIKE HCl In this titration the following indicators are mainly used : (i) Phenolphthalein (weak organic acid) : It shows colour change in the pH range (8 – 10)

(ii) Methyl orange (weak organic base) : It shows colour change in the pH range (3.1 – 4.4). Due to lower pH range, it indicates complete neutralisation of whole of the base.

Let for complete neutralisation of Na2CO3, NaHCO3 and NaOH, x,y and z mL of standard HCl are required. The titration of the mixture may be carried by two methods as summarised below : Volume of HCl used with

Volume of HCl used

Mixture

Phenlphthalein from beginning

Methyl orange from beginning

Phenolphthalein from beginning

Methyl orange after first end point

1. NaOH + Na2CO3

z + (x/2)

(x + z)

z + x/2

x/2 (for remaining 50% Na2CO3)

2. NaOH + NaHCO3

z+0

(z + y)

z+0

y (for remaining 100% NaHCO3

3. Na2CO3 + NaHCO3

(x/2) + 0

(x + y)

(x/2) + 0

x/2 + y (for remaining 50% of Na2CO3 and 100% NaHCO3 are indicated)

An indicator is a substance which is used to determine the end point in a titration. In acid-base titrations organic substances (weak acids or weak bases) are generally used as indicators. They change their colour within a certain pH range. The colour change and the pH range of some common indicators are tabulated below: ________________________________________ Indicator pH range Colour change ________________________________________ Methyl orange 3.2 – 4.5 Orange to red Methyl red 4.4 – 6.5 Red to yellow Litmus 5.5 – 7.5 Red to blue Phenol red 6.8 – 8.4 Yellow to red Phenolphthalein 8.3 – 10.5 Colourless to pink ________________________________________

Theory of acid-base indicators : Two theories have been proposed to explain the change of colour of acid-base indicators with change in pH. 1. Ostwald's theory: According to this theory (a) The colour change is due to ionisation of the acidbase indicator. The unionised form has different colour than the ionised form. (b) The ionisation of the indicator is largely affected in acids and bases as it is either a weak acid or a weak base. In case, the indicator is a weak acid, its ionisation is very much low in acids due to common ions while it is fairly ionised in alkalies. Similarly if the indicator is a weak base, its ionisation is large in acids and low in alkalies due to common ions. Considering two important indicators phenolphthalein (a weak acid) and methyl orange (a weak base), Ostwald's theory can be illustrated as follows: PAGE # 34

Phenolphthalein: It can be represented as HPh. It ionises in solution to a small extent as:

This theory also explains the reason why phenolphthalein is not a suitable indicator for titrating a weak – base against strong acid. The OH ions furnished by a weak base are not sufficient to shift the equilibrium towards right hand side considerably, i.e., pH is not reached to 8.3. Thus, the solution does not attain pink colour. Similarly, it can be explained why methyl orange is not a suitable indicator for the titration of weak acid with strong base.

H+ + Ph–

HPh Colourless

Pink

Applying law of mass action, K=

[H ][Ph  ] [HPh]

The undissociated molecules of phenolphthalein are colourless while ph– ions are pink in colour. In presence of an acid, the ionisation of HPh is practically negligible as the equilibrium shifts to left hand side due to high concentration of H+ ions. Thus, the solution would remain colourless. On addition of alkali, hydrogen ions are removed by OH– ions in the form of water molecules and the equilibrium shifts to right hand side. Thus, the concentration of ph– ions increases in solution and they impart pink colour to the solution.

SOLUBILITY The solubility of a solute in a solution is always expressed with respect to the saturated solution. (a) Definition : The maximum amount of the solute which can be dissolved in 100g (0.1kg) of the solvent to form a saturated solution at a given temperature. Suppose w gram of a solute is dissolved in W gram of a solvent to make a saturated solution at a fixed temperature and pressure. The solubility of the solute will be given by -

Let us derive Hendetson's equation for an indicator HIn + H2O

H3+O + In–

'Acid form'

'Base form'

Mass of the solute

w Conjugate acid-base pair KIn =

W

[In  ][H3 O] [HIn]

[HIn] [In  ] +

pH = –log10 [H3 O] = –log10 [KIn] – log10

[In  ]

The solubility of a substance in liquids generally increases with rise in temperature but hardly changes with the change in pressure. The effect of temperature depends upon the heat energy changes which accompany the process.

[In ] (Henderson's equation for [HIn]

indicator) At equivalence point ; [In–] = [HIn] and pH = pKIn Methyl orange : It is a very weak base and can be represented as MeOH. It is ionised in solution to give Me+ and OH– ions. MeOH Orange

Me+ +OH– Red

Applying law of mass action, K=

× 100

(b) Effect of Temperature and Pressure on Solubility of a Solids :

[HIn]



pH = pKIn + log10

Mass of the solvent

For example, the solubility of potassium chloride in water at 20ºC and 1 atm. is 34.7 g per 100g of water. This means that under normal conditions 100 g of water at 20ºC and 1 atm. cannot dissolve more than 34.7g of KCl.

KIn = Ionization constant of indicator [H3+O] = KIn 

× 100 =

[Me  ][OH  ] [MeOH]

In presence of an acid, OH– ions are removed in the form of water molecules and the above equilibrium shifts to right hand side. Thus, sufficient Me+ ions are produced which impart red colour to the solution. On addition of alkali, the concentration of OH– ions increases in the solution and the equilibrium shifts to left hand side, i.e., the ionisation of MeOH is practically negligible. Thus, the solution acquires the colour of unionised methyl orange molecules, i.e. orange.



Note : If heat energy is needed or absorbed in the process, it is of endothermic nature. If heat energy is evolved or released in the process, it is of exothermic nature. (i) Effect of temperature on endothermic dissolution process : Most of the salts like sodium chloride, potassium chloride, sodium nitrate, ammonium chloride etc. dissolve in water with the absorption of heat. In all these salts the solubility increases with rise in temperature. This means that sodium chloride becomes more soluble in water upon heating. (ii) Effect of temperature on exothermic dissolution process : Few salts like lithium carbonate, sodium carbonate monohydrate, cerium sulphate etc. dissolve in water with the evolution of heat. This means that the process is of exothermic nature. In these salts the solubility in water decreases with rise in temperature.

PAGE # 35



Note : 1. While expressing the solubility, the solution must be saturated but for expressing concentration (mass percent or volume percent), the solution need not to be saturated in nature. 2. While expressing solubility, mass of solvent is considered but for expressing concentration the mass or volume of the solution may be taken into consideration.

EXERCISE 1.

minimum amount of water required to dissolve 4 g K2SO4 is (A) 10 g (C) 50 g 2.

(c) Ef fect of Temperature on the Solubility of a Gas

3. (ii) The solubility of gases in liquids increases on increasing the pressure and decreases on decreasing the pressure.

mass of solute  100 mass of solvent 12

×100 = 16 g

4.

Ex.48 4 g of a solute are dissolved in 40 g of water to form a saturated solution at 25ºC. Calculate the solubility of the solute. Mass of solvent

5.

6.

=

46 × (50g) = 23 g 100

(b) When the solution is cooled, the solubility of salt in water will decrease. This means, that upon cooling, it will start separating from the solution in crystalline form.

(B) 1.10% (D) 110%

Two elements A (at. wt. 75) and B (at. wt. 16) combine to yield a compound. The % by weight of A in the

(A) A2B (C) AB 7.

(B) A2B3 (D) AB2

No. of oxalic acid molecules in 100 mL of 0.02 N oxalic acid are (A) 6.023 × 1020 (C) 6.023 × 1022

8.

(b) What will happen if this solution is cooled ? Sol. (a) Mass of potassium chloride in 100 g of water in saturated solution = 46 g Mass of potassium chloride in 50 g of water in saturated solution.

The percentage of sodium in a breakfast cereal labelled as 110 mg of sodium per 100 g of cereal is -

compound was found to be 75.08. The empirical formula of the compound is -

Mass of solute = 4 g Mass of solvent = 40 g

Ex.49 (a) What mass of potassium chloride would be needed to form a saturated solution in 50 g of water at 298 K ? Given that solubility of the salt is 46g per 100g at this temperature.

In a compound AxBy -

(A) 11% (C) 0.110%

× 100

4 × 100 = 10 g 40

(D) 0.25 mole

(C) X × mole of A = y × mole of B = (x + y) × mole of AxBy (D) X × mole of A = y × mole of B

Mass of solute

Solubility =

8g of sulphur are burnt to form SO2, which is oxidised by Cl 2 water. The solution is treated with BaCl 2

(A) Mole of A = Mole of B = mole of AxBy (B) Eq. of A = Eq. of B = Eq. of AxBy

Thus, the solubility of potassium sulphate in water is 16 g at 60ºC.

Sol. Solubility =

(B) 200 (D) 18

(C) 0.75 mole

Ex.47 12 grams of potassium sulphate dissolves in 75 grams of water at 60ºC. What is the solubility of potassium sulphate in water at that temperature ?

75

Molarity of H2SO4 (density 1.8g/mL) is 18M. The molality of this solution is -

solution. The amount of BaSO4 precipitated is (A) 1.0 mole (B) 0.5 mole

SAMPLE PROBLEMS

=

(B) 25 g (D) 75 g

(A)36 (C) 500

(i) The solubility of a gas in a liquid decreases with the rise in temperature.

Sol. Solubility =

The solubility of K2SO4 in water is 16 g at 50ºC. The

(B) 6.023 × 1021 (D) 6.023 × 1023

Which of the following sample contains the maximum number of atoms (A) 1 mg of C4H10 (C) 1 mg of Na

9.

(B) 1 mg of N2 (D) 1 mL of water

The total number of protons, electrons and neutrons in 12 g of

12 6 C is

-

25

(A) 1.084 × 10 (C) 6.022 × 1022

(B) 6.022 × 1023 (D) 18

10. 4.4 g of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of molecules present in the container will be (A) 6.022 × 1023

(B) 1.2044 × 1023

(C) 2 mole

(D) 6.023 × 1024 PAGE # 36

11. Which is not a molecular formula ?

21. How many atoms are contained in a mole of Ca(OH)2 -

(A) C6H12O6

(B) Ca(NO3)2

(A) 30 × 6.02 × 1023 atoms/mol

(C) C2H4O2

(D) N2O

(B) 5 × 6.02 × 1023 atoms/mol

12. The hydrated salt, Na2SO4. nH2O undergoes 55.9% loss in weight on heating and becomes anhydrous.

(C) 6 × 6.02 × 1023 atoms/mol (D) None of these 22. Insulin contains 3.4% sulphur. The minimum

The value of n will be (A) 5

(B) 3

molecular weight of insulin is -

(C) 7

(D) 10

(A) 941.176 u

(B) 944 u

(C) 945.27 u

(D) None of these

13. W hich of the following mode of expressing concentration is independent of temperature -

23. Number of moles present in 1 m3 of a gas at NTP are -

(A) Molarity

(B) Molality

(A) 44.6

(B) 40.6

(C) Formality

(D) Normality

(C) 42.6

(D) 48.6

14. The haemoglobin of the red blood corpuscles of most

24. Weight of oxygen in Fe2O3 and FeO is in the simple

of the mammals contains approximately 0.33% of

ratio of -

iron by weight. The molecular weight of haemoglobin

(A) 3 : 2

(B) 1 : 2

is 67,200. The number of iron atoms in each molecule

(C) 2 : 1

(D) 3 : 1

of haemoglobin is (Atomic weight of iron = 56) (A) 2

(B) 3

(C) 4

(D) 5

15. An oxide of metal have 20% oxygen, the eq. wt. of metal oxide is -

25. 2.76 g of silver carbonate on being strongly heated yield a residue weighing (A) 2.16g

(B) 2.48 g

(C) 2.32 g

(D) 2.64 g

26. How many gram of KCl would have to be dissolved in

(A) 32

(B) 40

(C) 48

(D) 52

16. How much water is to be added to dilute 10 mL of 10 N HCl to make it decinormal ?

60 g of H2O to give 20% by weight of solution (A) 15 g

(B) 1.5 g

(C) 11.5 g

(D) 31.5 g

27. When the same amount of zinc is treated separately

(A) 990 mL

(B) 1010 mL

(C) 100 mL

(D) 1000 mL

17. The pair of compounds which cannot exist in solution is (A) NaHCO3 and NaOH (B) Na2SO3 and NaHCO3

with excess of H2SO4 and excess of NaOH, the ratio of volumes of H2 evolved is (A) 1 : 1

(B) 1 : 2

(C) 2 : 1

(D) 9 : 4

28. Amount of oxygen required for combustion of 1 kg of a

(C) Na2CO3 and NaOH

mixture of butane and isobutane is -

(D) NaHCO3 and NaCl

(A) 1.8 kg

(B) 2.7 kg

(C) 4.5 kg

(D) 3.58 kg

18. If 250 mL of a solution contains 24.5 g H2SO4 the molarity and normality respectively are (A) 1 M, 2 N

(B) 1M,0.5 N

(C) 0.5 M, 1N

(D) 2M, 1N

19. The mole fraction of NaCl, in a solution containing 1 mole of NaCl in 1000 g of water is (A) 0.0177

(B) 0.001

(C) 0.5

(D) 0.244

20. 3.0 molal NaOH solution has a density of 1.110 g/ mL. The molarity of the solution is (A) 2.9732

(B) 3.05

(C) 3.64

(D) 3.0504

29. Rakesh needs 1.71 g of sugar (C12H22O11) to sweeten his tea. What would be the number of carbon atoms present in his tea ? (A) 3.6 × 1022

(B) 7.2 × 1021

(C) 0.05 × 1023

(D) 6.6 × 1022

30. The total number of AlF3 molecule in a sample of AlF3 containing 3.01 × 1023 ions of F– is (A) 9.0 × 1024 (C) 7.5 × 10

(B) 3.0 × 1024

23

(D)1023

31. The volume occupied by one molecule of water (density 1 g/cm3) is (A) 18 cm3 (C) 6.023 × 10

(B) 22400 cm3 –23

(D) 3.0 × 10–23 cm3 PAGE # 37

32. 224 mL of a triatomic gas weigh 1 g at 273 K and 1 atm. The mass of one atom of this gas is (A) 8.30 × 10–23 g

(B) 2.08 × 10–23 g

(C) 5.53 × 10–23 g

(D) 6.24 × 10–23 g

33. The percentage of P2O5 in diammonium hydrogen phosphate is (A) 77.58

(B) 46.96

(C) 53.78

(D) 23.48

43. The number of molecules present in 11.2 litre CO2 at STP is (A) 6.023 × 1032 (C) 3.011 × 10

23

(B) 6.023 × 1023 (D) None of these

44. 250 ml of 0.1 N solution of AgNO3 are added to 250 ml of a 0.1 N solution of NaCl. The concentration of nitrate ion in the resulting solution will be (A) 0.1N

(B) 1.2 N

(C) 0.01 N

(D) 0.05 N

34. The mole fraction of water in 20% (wt. /wt.) aqueous 45. Amount of BaSO4 formed on mixing the aqueous

solution of H2O2 is (A) (C)

77 68 20 80

(B) (D)

68 77

solution of 2.08 g BaCl2 and excess of dilute H2SO4 is (A) 2.33 g

(B) 2.08 g

(C) 1.04 g

(D) 1.165 g

80 20

35. Which of the following has the maximum mass ? (A) 25 g of Hg (B) 2 moles of H2O (C) 2 moles of CO2

46. 2g of NaOH and 4.9 g of H2SO4 were mixed and volume is made 1 litre. The normality of the resulting solution will be (A) 1N

(B) 0.05 N

(C) 0.5 N

(D) 0.1N

47. 1g of a metal carbonate neutralises completely 200

(D) 4 g atom of oxygen 36. Total mass of neutrons in 7mg of 14C is (A) 3 × 1020 kg

(B) 4 × 10–6 kg

(C) 5 × 10–7 kg

(D) 4 × 10–7 kg

37. Vapour density of a metal chloride is 66. Its oxide contains 53% metal. The atomic weight of metal is (A) 21

(B) 54

(C) 26.74

(D) 2.086

38. The number of atoms in 4.25 g NH3 is approximately (A) 1 × 1023

(B) 1.5 × 1023

(C) 2 × 1023

(D) 6 × 1023

39. The modern atomic weight scale is based on (A) C12

(B) O16

(C) H1

(D) C13

40. Amount of oxygen in 32.2g of Na2SO4. 10H2O is (A) 20.8 g

(B) 22.4 g

(C) 2.24 g

(D) 2.08 g

mL of 0.1N HCl. The equivalent weight of metal carbonate is (A) 25

(B) 50

(C) 100

(D) 75

48. 100 mL of 0.5 N NaOH were added to 20 ml of 1N HCl and 10 mL of 3 N H2SO4. The solution is (A) acidic

(B) basic

(C) neutral

(D) none of these

49. 1M solution of H2SO4 is diluted from 1 litre to 5 litres , the normality of the resulting solution will be (A) 0.2 N

(B) 0.1 N

(C) 0.4 N

(D) 0.5 N

50. The volume of 7g of N2 at S.T.P. is (A) 11.2 L

(B) 22.4 L

(C) 5.6 L

(D) 6.5 L

51. One mole of calcium phosphide on reaction with excess of water gives -

41. Which of the followings does not change on dilution ? (A) Molarity of solution

(A) three moles of phosphine (B) one mole phosphoric acid

(B) Molality of solution

(C) two moles of phosphine

(C) Millimoles and milli equivalent of solute (D) Mole fraction of solute 42. Equal masses of O2, H2 and CH4 are taken in a container. The respective mole ratio of these gases in container is (A) 1 : 16 : 2

(B) 16 : 1 : 2

(C) 1 : 2 : 16

(D) 16 : 2 : 1

(D) one mole of P2O5 52. Mg (OH)2 in the form of milk of magnesia is used to neutralize excess stomach acid. How many moles of stomach acid can be neutralized by 1 g of Mg(OH)2 ? (Molar mass of Mg(OH)2 = 58.33) (A) 0.0171

(B) 0.0343

(C) 0.686

(D) 1.25

PAGE # 38

53. Calcium carbonate decomposes on heating

60. The volume of CO2 (in litres) liberated at STP when

according to the following equation -

10 g of 90% pure limestone is heated completely, is-

CaCO3(s)

(A) 22.4 L

(B) 2.24 L

(C) 20.16 L

(D) 2.016 L

How

CaO(s) + CO2(g)

many moles of CO 2 will be obtained by

decomposition of 50 g CaCO3 ? 3 (A) 2 1 (C) 2

61. A metal oxide has the formula Z2O3. It can be reduced

5 (B) 2

by hydrogen to give free metal and water. 0.1596 g of

(D) 1

reduction. The atomic mass of the metal is -

the metal requires 6 mg of hydrogen for complete

54. Sulphur trioxide is prepared by the following two

(A) 27.9

(B) 159.6

(C) 79.8

(D) 55.8

reactions -

Question number 62, 63, 64 and 65 are based on the

S8(s) + 8O2(g)  8SO2(g)

following information :

2SO2(g) + O2(g)  2SO3(g) How many grams of SO3 are produced from 1 mole

Q.

redox reaction. Following equations describe the

of S8 ? (A) 1280

(B) 640

(C) 960

(D) 320

Dissolved oxygen in water is determined by using a procedure -

I

2Mn2+(aq) + 4OH–(aq) + O2 (g)

I

MnO2(s)+2I–(aq)+4H+(aq)  Mn2+(aq)+I2(aq) + 2H2O(  )

phosphorous and hydrogen. The change in volume

III

2S 2O32 – (aq) + I2(aq)  2S 4 O 26 – (aq) + 2I–(aq)

when 100 mL of such gas decomposed is -

62. How many moles of S 2O 23 – are equivalent to each

55. PH 3 (g) decomposes on heating to produce

(A) + 50 mL

(B) + 500 mL

(C) – 50 mL

(D) – 500 mL

56. What amount of BaSO4 can be obtained on mixing 0.5 mole BaCl2 with 1 mole of H2SO4 ? (A) 0.5 mol

(B) 0.15 mol

(C) 0.1 mol

(D) 0.2 mol

57. In the reaction , CrO5 + H2SO4  Cr2(SO4)3 + H2O + O2 one mole of CrO5 will liberate how many moles of O2 ?

 2MnO2(s) + 2H2O(  )

mole of O2 ? (A) 0.5

B) 1

(C) 2

(D) 4

63. What amount of I2 will be liberated from 8 g dissolved oxygen ? (A) 127 g

(B) 254 g

(C) 504 g

(D) 1008 g

64. 3 × 10 moles O2 is dissolved per litre of water, then –3

(A) 5/2

(B) 5/4

what will be molarity of I– produced in the given

(C) 9/2

(D) None

reaction ? (A) 3 × 10–3 M

58. Calcium carbonate decomposes on heating

1  3  10 – 3 M 2 65. 8 mg dissolved oxygen will consume -

(C) 2 × 3 × 10–3 M

according to the equation CaCO3(s)  CaO(s) + CO2(g) At STP the volume of CO 2 obtained by thermal decomposition of 50 g of CaCO3 will be -

(B) 2.5 × 10–4 mol Mn2+

(B) 44 litre

(C) 10 mol Mn2+

(C) 11.2 litre

(D) 1 litre

(D) 2 mol Mn2+

oxygen, the following reaction takes place4FeCl3(s) + 3O2(g)  2Fe2O3(s) + 6Cl2(g) If 3 moles of FeCl3 are ignited in the presence of 2 moles of O2 gas, how much of which reagent is

(D)

(A) 5 × 10–4 mol Mn+2

(A) 22.4 litre

59. When FeCl 3 is ignited in an atmosphere of pure

(B) 4 × 3 × 10–3 M

66. 2 g of a base whose eq. wt. is 40 reacts with 3 g of an acid. The eq. wt. of the acid is(A) 40

(B) 60

(C) 10

(D) 80

67. Normality of 1% H2SO4 solution is nearly -

present in excess and therefore, remains unreacted ?

(A) 2.5

(B) 0.1

(A) 0.33 mole FeCl3 remains unreacted

(C) 0.2

(D) 1

(B) 0.67 mole FeCl3 remains unreacted

68. What volume of 0.1 N HNO3 solution can be prepared

(C) 0.25 mole O2 remains unreacted

from 6.3 g of HNO3 ?

(D) 0.50 mole O2 remains unreacted

(A) 1 litre

(B) 2 litres

(C) 0.5 litre

(D) 4 litres PAGE # 39

seminormal HCl solution to make it decinormal is (A) 200 mL (B) 400 mL (C) 600 mL

(D) 800 mL

70. 0.2 g of a sample of H2O2 required 10 mL of 1N KMnO4 in a titration in the presence of H2SO4. Purity of H2O2 is(A) 25% (B) 85% (C) 65%

(D) 95%

71. Which of the following has the highest normality ? (A) 1 M H2SO4 (C) 1 M H3PO4

(B) 1 M H3PO3 (D) 1 M HNO3

72. The molarity of 98% H2SO4(d = 1.8g/mL) by wt. is (A) 6 M (B) 18.74 M (C) 10 M

(D) 4 M

80. A 3 N solution of H2SO4 in water is prepared from Conc. H2SO4 (36 N) by diluting [KVPY-PartII-2007] (A) 20 ml of the conc. H2SO4 to 240 ml (B) 10 ml of the conc. H2SO4 to 240 ml (C) 1 ml of the conc. H2SO4 to 36 ml (D) 20 ml of the conc. H2SO4 to 36 ml 81. The solubility curve of KNO3 as a function of temperature is given below [KVPY-Part-II-2007] Solubility (g/100 ml water)

69. The volume of water to be added to 200 mL of

250 200 150 100 50 0

73. 0.7 g of Na2CO3 . xH2O is dissolved in 100 mL. 20 mL of which required to neutralize 19.8 mL of 0.1 N HCl. The value of x is (A) 4 (C) 2

(B) 3 (D) 1

74. 0.45 g of an acid of molecular weight 90 was neutralised by 20 mL of 0.5 N caustic potash. The basicity of the acid is (A) 1

(B) 2

(C) 3

(D) 4

75. 1 litre of 18 molar H2SO4 has been diluted to 100 litres. The normality of the resulting solution is (A) 0.09 N (B) 0.18 (C) 1800 N

(D) 0.36

N HCl is required to react completely with 10 1.0 g of a sample of limestone. The percentage purity

76. 150 ml of

of calcium carbonate is (A) 75% (C) 80%

(B) 50% (D) 90%

N N 77. 50 ml of HCl is treated with 70 ml NaOH. 10 10 Resultant solution is neutralized by 100 ml of

sulphuric acid. The normality of H2SO4 (A) N/50 (C) N/30

0

N HCl were added to 1 g calcium car10 bonate, what would remain after the reaction ?

(A) CaCO3

(B) HCl

(C) Neither of the two

(D) Part of both

40

60

80

100

Temperature (°C)

The amount of KNO3 that will crystallize when a saturated solution of KNO3 in 100 ml of water is cooled from 90°C to 30 °C, is (A) 16 g (B) 100 g (C) 56 g (D)160 g 82. The volume of 0.5 M aqueous NaOH solution required to neutralize 10 ml of 2 M aqueous HCl solution is : [KVPY-Part-I-2008] (A) 20ml (B) 40ml (C) 80ml (D) 120ml 83. 3.01×1023 molecules of elemental Sulphur will react with 0.5 mole of oxygen gas completely to produce [KVPY-Part-I-2008] (A) 6.02 × 1023 molecules of SO3 (B) 6.02 × 1023 molecules of SO2 (C) 3.01 × 1023 molecules of SO3 (D) 3.01 x 1023 molecules of SO2 84. The solubility of a gas in a solution is measured in three cases as shown in the figure given below where w is the weight of a solid slab placed on the top of the cylinder lid. The solubility will follow the order : [KVPY-Part-I-2008] w

(B) N/25 (D) N/10

78. 200 mL of

20

gas

solution

w

gas

solution

w

w

w

w

gas

solution

79. Equivalent mass of KMnO4, when it is converted to MnSO4 is (A) M/5

(B) M/3

(C) M/6

(D) M/2

(A) a > b > c (C) a = b = c

(B) a < b < c (D) a >b < c

PAGE # 40

85. The density of a salt solution is1.13 g cm–3 and it contains 18% of NaCI by weight. The volume of the solution containing 36.0 g of the salt will be : [KVPY-Part-II-2008] (A) 200 cm3 (B) 217 cm3 3 (C) 177 cm (D) 157cm3 86. One mole of nitrogen gas on reaction with 3.01 x 1023 molecules of hydrogen gas produces [KVPY-Part-I-2009] (A) one mole of ammonia (B) 2.0 x 1023 molecules of ammonia (C) 2 moles of ammonia (D) 3.01 × 1023 molecules of ammonia 87.

[KVPY-Part-I-2009]

Solubility g/I 250

KNO3

200 150

KCl

100 50 20

40

60 80 100 Temperature (ºC)

Given the solubility curves of KNO3 and KCl, which of the following statements is not true ? (A) At room temperature the solubility of KNO3 and KCI are not equal (B) The solubilities of both KNO3 and KCI increase with temperature (C) The solubility of KCI decreases with temperature (D) The solubility of KNO3 increases much more compared to that of KCl with increase in temperature 88. 10 ml of an aqueous solution containing 222 mg of calcium chloride (mol. wt. = 111) is diluted to 100 ml. The concentration of chloride ion in the resulting solution is [KVPY-Part-II-2009] (A) 0.02 mol/lit. (B) 0.01 mol/lit. (C) 0.04 mol/lit (D) 2.0 mol/lit. 89. Aluminium reduces manganese dioxide to manganese at high temperature. The amount of aluminium required to reduce one gram mole of manganese dioxide is [KVPY-Part-II-2009] (A) 1/2 gram mole (B) 1 gram mole (C) 3/4 gram mole (D) 4/3 gram mole 90. One mole of oxalic acid is equivalent to [IJSO-2009] (A) 0.5 mole of NaOH (B) 1 mole of NaOH (C) 1.5 mole of NaOH (D) 2 mole of NaOH



PAGE # 41

NUMBER

SYSTEM (vi) Real numbers : Numbers which can represent

(i) Natural numbers : Counting numbers are known as natural numbers. N = { 1, 2, 3, 4, ... }.

actual physical quantities in a meaningful way are known as real numbers. These can be represented on the number line. Number line is geometrical straight line with arbitrarily defined zero (origin).

(ii) Whole numbers :

(vii) Prime numbers : All natural numbers that have

All natural numbers together with 0 form the collection of all whole numbers. W = { 0, 1, 2, 3, 4, ... }.

one and itself only as their factors are called prime numbers i.e. prime numbers are exactly divisible by 1 and themselves. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23,...etc. If P is the set of prime number then P = {2, 3, 5, 7,...}.

(iii) Integers : All natural numbers, 0 and negative of natural numbers form the collection of all integers. I or Z = { ..., – 3, – 2, – 1, 0, 1, 2, 3, ... }. (iv) Rational numbers : These are real numbers which can be expressed in the

(viii) Composite numbers : All natural numbers, which are not prime are composite numbers. If C is the set of composite number then C = {4, 6, 8, 9, 10, 12,...}. 

(ix) Co-prime Numbers : If the H.C.F. of the given

p , where p and q are integers and q  0 . q e.g. 2/3, 37/15, -17/19. form of





All natural numbers, whole numbers and integers are rational. Rational numbers include all Integers (without any decimal part to it), terminating fractions ( fractions in which the decimal parts are terminating e.g. 0.75, – 0.02 etc.) and also non-terminating but recurring decimals e.g. 0.666....., – 2.333...., etc. Fractions : (a) Common fraction : Fractions whose denominator is not 10. (b) Decimal fraction : Fractions whose denominator is 10 or any power of 10. (c) Proper fraction : Numerator < Denominator i.e.

3 5

.

5 (d) Improper fraction : Numerator > Denominator i.e. . 3

(e) Mixed fraction : Consists of integral as well as 2 . 7 (f) Compound fraction : Fraction whose numerator and

fractional part i.e. 3



2/3 denominator themselves are fractions. i.e. . 5/7 Improper fraction can be written in the form of mixed fraction.

All real number which are not rational are irrational numbers. These are non-recurring as well as non-terminating type of decimal numbers. 2,

3

4 , 2 3 ,

numbers (not necessarily prime) is 1 then they are known as co-prime numbers. e.g. 4, 9 are co-prime as H.C.F. of (4, 9) = 1. 

Any two consecutive numbers will always be co-prime. (x) Even Numbers : All integers which are divisible by 2 are called even numbers. Even numbers are denoted by the expression 2n, where n is any integer. So, if E is a set of even numbers, then E = { ..., – 4, –2, 0, 2, 4,...}. (xi) Odd Numbers : All integers which are not divisible by 2 are c alled odd numbers. Odd numbers are denoted by the general expression 2n –1 where n is any integer. If O is a set of odd numbers, then O = {..., –5, –3, –1, 1, 3, 5,...}. (xii) Imaginary Numbers : All the numbers whose square is negative are called imaginary numbers. e.g. 3i, -4i, i, ... ; where i =

- 1.

(xiii) Complex Numbers : The combined form of real and imaginary numbers is known as complex numbers. It is denoted by Z = A + iB where A is real part and B is imaginary part of Z and A, B  R.  The set of complex number is the super set of all the sets of numbers.

Squares : When a number is multiplied by itself then the product is called the square of that number.

(v) Irrational Numbers :

For Ex. :

1 is neither prime nor composite number.

2 3 ,

47

3 etc.

Perfect Square : A natural number is called a perfect square if it is the square of any other natural number e.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.

PAGE # 4242

Ex.1 Find the smallest number by which 300 must be multiplied so that the product is a perfect square. Sol. Given number is 300, first we resolve it into prime factors.

Ex.3 Find the square root of 3 + Sol. Let 3+

2 300 2 150 25



2 pq =

5

5



4pq = 2

Clearly, 3 has no pair. Thus if we multiply it by 3 then product will be a perfect square.  Required smallest number is 3 . Ex.2 Find the smallest number by which 1575 must be divided so that the quotient becomes a perfect square. Sol. Given number is 1575, first we write it as the product of prime factors 3 1575 3 525

7

7 1





(p – q)2 = 7



p–q=

...(iv)

7

[By eqn (i)]

 p+q=3 

1 3 7 2 1 3 7 q= 2

p=

 

3 2 = 

 

[On adding (i) & (iv)] [On subtracting (i) & (iv)] 1

 3 7  3 7     2 

then the result is called the cube of that number. Perfect cube : A natural number is said to be a perfect cube if it is the cube of any other natural number. Ex.4 What is the smallest number by which 675 must be multiplied so that the product is a perfect cube. Sol. Resolving 675 into prime factor, we get

3 675

225 = 15 ...etc.

We can calculate the square root of positive numbers only. However the square root of a positive number may be a positive or a negative number. e.g.

[By squaring both sides ]

Cube : If any number is multiplied by itself three times

Square roots : The square root of a number x is that number which when multiplied by itself gives x as the product. As we say square of 3 is 9, then we can also say that square root of 9 is 3. The symbol use to indicate the square root of a number 81 = 9,

...(iii)

(p – q)2 = 9 – 2



 1575 = 3 × 3 × 5 × 5 × 7. Clearly, 7 has no pair, so if we divide it by 7 then quotient become a perfect square.

’ , i.e.

...(ii)

2





35

[By equating the parts]

(p – q)2 = (p + q)2 – 4 pq

300 = 2 × 2 × 3 × 5 × 5

175

[By squaring both sides]

...(i)

5

5

q

pq

p+q=3

75

5

p +

2 =p+q+2

3

1

is ‘

3 2 =

2.

25 = + 5 or – 5.

3 225 3 75 5

25

5

5 1

Properties of Square Roots : (i) If the unit digit of a number is 2, 3, 7 or 8, then it does not have a square root in N.

675 = (3 × 3 × 3) × 5 × 5 Grouping the factor in triplets of equal factors, we get

(ii) If a number ends in an odd number of zeros, then it does not have a square root in N.

675 = (3 × 3 × 3) × 5 × 5 We find that 3 occurs as a prime factor of 675 thrice but

(iii) The square root of an even number is even and

5 occurs as a prime factor only twice. Thus, if we multiply 675 by 5, 5 will also occur as a prime factor thrice and

square root of an odd number is odd.e.g. 81 = 9, = 16,

256

324 = 18 ...etc.

(iv) Negative numbers have no square root in set of real numbers.

the product will be 3 × 3 × 3 × 5 × 5 × 5, which is a perfect cube. Hence, we must multiply 675 by 5 so that the product becomes a perfect cube.

PAGE # 4343

Ex.5 What is the smallest number by which 18522 must be divided so that the quotient is a perfect cube ? Sol. Resolving 18522 into prime factors, we get

Laws of Surds :

2 18522 3 9261

n

n

(i)

 a

(ii)

n

(iii)

a  n b  n ab

n

3087

3

1029

(iv)

nm

7 7

343 49

(v)

n

7

7

18522 = 2 × 3 × 3 × 3 × 7 × 7 × 7 Grouping the factors in triplets of equal factors, we get 18522 = 2 × (3 × 3 × 3) × (7 × 7 × 7) Clearly, if we divide 18522 by 2, the quotient would be 3 × 3 × 3 × 7 × 7 × 7 = 33 × 73 which is a perfect cube. Therefore, we must divide 18522 by 2 so that the quotient ‘9261’ is a perfect cube.

a  nm a  m n a

a

np

a is given a special name Surd. Where ‘a’ is called radicand, rational. Also

Sol.

the symbol

Ex.8 Divide :

n

n

is called the radical sign and the index

6

3

8a 5 b  4a 2 b 2

3 8 3 a15 b 3  6 4 2 a 4b 4 = 4a b6 2ab .

n is called order of the surd. n

a is read as nth root of ‘a’ and can also be written 1 an

as

Sol.

3

6

24 3



200

3

24 by

6

(24) 3

(200)

200

6

2

216 . 625

Comparison of Surds :

.

It is clear that if x > y > 0 and n > 1 is a (+ve) integer

Identification of Surds : (i)

np

a p or, n a m  amp [Important for changing order of surds]

Ex.6 If x = 1 + 21/3 + 22/3, then find the value of x3 – 3x2 – 3x – 1. Sol. x = 1 + 21/3 + 22/3 x – 1 = (21/3 + 22/3) (x – 1)3 = ( 21/3 + 22/3)3 x3 – 3x2 + 3x – 1 = (21/3)3 + (22/3)3 + 3. 21/3. 22/3 (21/3 + 22/3) x3 – 3x2 + 3x – 1 = 2 + 22 + 3.21 (x – 1) x3 – 3x2 + 3x – 1 = 6 + 6 (x – 1) x3 – 3x2 + 3x – 1 = 6x x3 – 3x2 – 3x – 1 = 0. Ex.7 Simplify :

Any irrational number of the form

[Here order should be same]

a b

a n b  n

3

1

 n an = a

then

4 is a surd as radicand is a rational number..

Similar examples :

3

5 , 4 12 , 5 7 , 12 , ...

x >

n

n

y . e.g.

3

16 >

3

12 ,

5

36 >

5

25 and so

on. ,

Ex.9 Which is greater in each of the following : (ii) 2 + 3 is a surd (as surd + rational number will give a surd) Similar examples : 3 – 2 , 3  1, 3 3  1,... (iii)

3

(i) Sol. (i)



of 2 – 3 . Similar examples : 7  4 3 , 9 – 4 5 , 9  4 5 ,... 1

(iv)

3

3 is a surd as

Similar examples :

3 3

3

1  1 3 3  3 2   3 6  6 3    

5

8

1 and 2

(ii)

3

1 3

L.C.M. of 3 and 5 is 15.

7 – 4 3 is a surd as 7 – 4 3 is a perfect square



6 and

3

6  35 6 5  15 7776

5

8  35 8 3  15 512

15 7776  15 512



3

5 6  8 (ii) L.C.M. of 2 and 3 is 6.



3

6

5 , 4 5 6 , ...

 1   and 2

6

 1   3

2

(v) These are not a surds : (A)

3

8 , because

3

3

8  23

which is a rational

=

6

number. (B)

2  3 , because 2 + 3 is not a perfect square.

(C) 3 1 3 , because radicand is an irrational number..

1 and 8

as 8 < 9  So, 6

1 8

6

1 9

1 1  8 9

6

1 9

or,

1 > 2

3

1 . 3

PAGE # 4444

Ex.10 Arrange 2, 3 3 and Sol. L.C.M. of 2, 3, 4 is 12. 2



and

26

4

2 6  12 64 ,

3

3  34 3 4  12 81 ,

4

5

4 3

5 in ascending order..

5 3  12 125

As, 64 < 81 < 125. 

12

64  12 81  12 125

2 33 45 Conjugate Surds : 

R.F. of a  b and conjugate surds .

a  b type surds are called

Ex.11 (i) 2 – 3 is conjugate of 2  3 .



(ii) 5  1 is conjugate of 5 – 1. Sometimes conjugate surd and reciprocals are same.

Ex.12 (i) 2  3 , it’s conjugate is 2 – 3 , its reciprocal is 2 – 3 & vice versa. (ii) 5 – 2 6 , it’s conjugate is 5  2 6 , its reciprocal is 5  2 6 & vice versa.

Factors : ‘a’ is a factor of ‘b’ if there exists a relation such that a × n = b, where ‘n’ is any natural number. 

1 is a factor of all numbers as 1 × b = b.



Factor of a number cannot be greater than the number (infact the largest factor will be the number itself). Thus factors of any number will lie between 1 and the number itself (both inclusive) and they are limited. Multiples : ‘a’ is a multiple of ‘b’ if there exists a relation of the type b × n = a. Thus the multiples of 6 are 6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.



The smallest multiple will be the number itself and the number of multiples would be infinite.



NOTE : To understand what multiples are, let’s just take an example of multiples of 3. The multiples are 3, 6, 9, 12,.... so on. We find that every successive multiples appears as the third number after the previous. So if one wishes to find the number of multiples of 6 less than 255, we could arrive at the number through 255 = 42 (and the remainder 3). The remainder is of 6 no consequence to us. So in all there are 42 multiples. 255 =7 36 (and the remainder is 3). Hence, there are 7 multiples of 36.

If one wishes to find the multiples of 36, find

Factorisation : It is the process of splitting any number into form where it is expressed only in terms of the most basic prime factors. For example, 36 = 22 × 3 2. 36 is expressed in the factorised form in terms of its basic prime factors. Number of factors : For any composite number C, which can be expressed as C = ap × bq × cr ×....., where a, b, c ..... are all prime factors and p, q, r are positive integers, then the number of factors is equal to (p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So the factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9. Ex.13 If N = 123 × 34 ×52, find the total number of even factors of N. Sol. The factorised form of N is (22 × 31)3 × 34 × 52  26 × 37 × 52. Hence, the total number of factors of N is (6 + 1) (7 + 1) (2 + 1) = 7 × 8 × 3 = 168. Some of these are odd multiples and some are even. The odd multiples are formed only with the combination of 3s and 5s. So, the total number of odd factors is (7 + 1)(2 + 1) = 24. Therefore, the number of even factors = 168 – 24 = 144. Ex.14 A number N when factorised can be written N = a 4 × b 3 × c 7. Find the number of perfect squares which are factors of N (The three prime numbers a, b, c > 2). Sol. In order that the perfect square divides N, the powers of ‘a’ can be 0, 2 or 4, i.e. 3. Powers of ‘b’ can be 0, 2, i.e. 2. Power of ‘c’ can be 0, 2, 4 or 6, i.e. 4. Hence, a combination of these powers given 3 × 2 × 4 i.e. 24 numbers. So, there are 24 perfect squares that divides N. Ex.15 Directions : (i to iv) Answer the questions based on the given information. There are one thousand lockers and one thousand students in a school. The principal asks the first student to go to each locker and open it. Then he asks the second student go to every second locker and close it. The third student goes to every third locker, and if it is closed, he opens it, and it is open, he closes it. The fourth student does it to every fourth locker and so on. The process is completed with all the thousand students. (i) How many lockers are closed at the end of the process ? (ii) How many students can go to only one locker ? (iii) How many lockers are open after 970 students have done their job ? (iv) How many student go to locker no. 840 ?

PAGE # 4545

Sol. (i to iv) : Whether the locker is open or not depends on the number of times it is accessed. If it is accessed odd number of times, then it is open while if it is accessed even number of times then it is closed. How many times a locker will be accessed depends on the locker no. If it contains odd number of factors,, then it will be open and if it contains even number of factors. Then it will be closed. We know that a perfect square contains odd number of factors while a nonperfect square contains even number of factors. Thus the lockers with perfect square number will be open and the number of these perfect squares from 1 to 1000 determines the no. of open lockers. (i) No. of closed lockers = No. of non-perfect square numbers from 1 to 1000 = 1000 – 31 = 969. (ii) Upto 500 students they can go to two or more than two lockers, while the rest 500 can go to only one locker. (iii) The 31 perfect squares ( the last being 312 = 961) will be open while the lockers from 971 to 1000 is yet to be accessed last time so they all are open. The total being = 31 + 30 = 61 (iv) The no. of students that have gone to locker no. 840 is same as the no. of factors of 840. 840 = 23 × 3 × 5 × 7. So, the no. of factors = (3 + 1) (1 + 1) (1 + 1) (1 + 1) = 32.

LCM (least Common Multiple) : The LCM of given numbers, as the name suggests is the smallest positive number which is a multiple of each of the given numbers. HCF (Highest Common factor) : The HCF of given numbers, as the name suggests is the largest factor of the given set of numbers. Consider the numbers 12, 20 and 30. The factors and the multiples are : Factors 1, 2, 3, 4, 6, 12 1, 2, 4, 5, 10, 20 1, 2, 3, 5, 6, 10, 15, 30

Given numbers 12 20 30

Multiples 12, 24, 36, 48, 60, 72, 84, 96, 108, 120.... 20, 40, 60, 80, 100, 120..... 30, 60, 90, 120....

The common factors are 1 and 2 and the common multiples are 60, 120... Thus the highest common factor is 2 and the least common multiple is 60. Meaning of HCF is that the HCF is the largest number that divides all the given numbers. Also since a number divides its multiple, the meaning of LCM is that it is smallest number which can be divided by the given numbers.  HCF will be lesser than or equal to the least of the numbers and LCM will be greater than or equal to the greatest of the numbers.

Ex.16 Find a number greater then 3 which when divided by 4, 5, and 6 always leaves the same remainder 3. Sol. The smallest number which, when divided by 4, 5 and 6, leaves the remainder 3 in each case is LCM (4, 5 and 6) + 3 = 60 + 3 = 63. Ex.17 In a school 437 boys and 342 girls have been divided into classes, so that each class has the same number of students and no class has boys and girls mixed. What is the least number of classes needed? Sol. We should have the maximum number of students in a class. So we have to find HCF (437, 342) = 19. HCF is also the factor of difference of the number. 



437 342 + = 23 + 18 19 19 = 41 classes.

Number of classes =

For any two numbers x and y : x × y = HCF (x, y) × LCM (x, y). HCF and LCM of fractions : LCM of numerators LCM of fractions = HCF of deno min ators HCF of numerators HCF of fractions = LCM of deno min ators Make sure the fractions are in the most reducible form.

Ex.18 Find the least number which when divided by 6, 7, 8, 9 and 10 leaves remainder 1. Sol. As the remainder is same Required number = LCM of divisors + Remainder = LCM (6, 7, 8, 9, 10) +1 = 2520 + 1 = 2521. Ex.19 Six bells start tolling together and they toll at intervals of 2, 4, 6, 8, 10, 12 sec. respectively, find. (i) after how much time will all six of them toll together ? (ii) how many times will they toll together in 30 min ? Sol. The time after which all six bells will toll together must be multiple of 2, 4, 6, 8, 10, 12. Therefore, required time = LCM of time intervals. = LCM (2, 4, 6, 8, 10, 12) = 120 sec. Therefore after 120 s all six bells will toll together. After each 120 s, i.e. 2 min, all bell are tolling together.  30   1 Therefore in 30 min they will toll together   2  = 16 times 1 is added as all the bells are tolling together at the start also, i.e. 0th second.

Ex.20 LCM of two distinct natural numbers is 211. What is their HCF ? Sol. 211 is a prime number. So there is only one pair of distinct numbers possible whose LCM is 211, i.e. 1 and 211. HCF of 1 and 211 is 1. Ex.21 An orchard has 48 apple trees, 60 mango trees and 96 banana trees. These have to be arranged in rows such that each row has the same number of trees and all are of the same type. Find the minimum number of such rows that can be formed. Sol. Total number of trees are 204 and each of the trees are exactly divisible by 12. HCF of (48, 60, 96). 

204 = 17 such rows are possible. 12

PAGE # 4646

Division Algorithm : General representation of result is, Dividend Re mainder  Quotient  Divisor Divisor Dividend = (Divisor × Quotient ) + Remainder

NOTE : (i) (xn – an) is divisible by (x – a) for all the values of n. (ii) (xn – an) is divisible by (x + a) and (x – a) for all the even values of n. (iii) (xn + an) is divisible by (x + a) for all the odd values of n. Test of Divisibility : No.

Divisiblity Test

2

Unit digit should be 0 or even

3

The sum of digits of no. should be divisible by 3

4

The no formed by last 2 digits of given no. should be divisible by 4.

5

Unit digit should be 0 or 5.

6

No should be divisible by 2 & 3 both

8

The number formed by last 3 digits of given no. should be divisible by 8.

9

Sum of digits of given no. should be divisible by 9 The difference between sums of the digits at even & at odd places should be zero or multiple of 11. 25 Last 2 digits of the number should be 00, 25, 50 or 75.

11

Rule for 7 : Double the last digit of given number and subtract from remaining number the result should be zero or divisible by 7. Ex.22 Check whether 413 is divisible by 7 or not. Sol. Last digit = 3, remaining number = 41, 41 – (3 x 2) = 35 (divisible by 7). i.e. 413 is divisible by 7. This rule can also be used for number having more than 3 digits. Ex.23 Check whether 6545 is divisible by 7 or not. Sol. Last digit = 5, remaining number 654, 654 – (5 x 2) = 644; 64 – (4 x 2) = 56 divisible by 7. i.e. 6545 is divisible by 7. Rule for 13 : Four times the last digit and add to remaining number the result should be divisible by 13. Ex.24 Check whether 234 is divisible by 13 or not . Sol. 234, (4 x 4) + 23 = 39 (divisible by 13), i.e. 234 is divisible by 13. Rule for 17 : Five times the last digit of the number and subtract from previous number the result obtained should be either 0 or divisible by 17. Ex.25 Check whether 357 is divisible by 17 or not. Sol. 357, (7 x 5) – 35 = 0, i.e. 357 is divisible by 17.

Ex.28 Find the largest four digit number which when reduced by 54, is perfectly divisible by all even natural numbers less than 20. Sol. Even natural numbers less than 20 are 2, 4, 6, 8, 12, 14, 16, 18. Their LCM = 2 × LCM of first 9 natural numbers = 2 × 2520 = 5040 This happens to be the largest four-digit number divisible by all even natural numbers less than 20. 54 was subtracted from our required number to get this number. Hence, (required number – 54) = 5040  required number = 5094. Ex.29 Ajay multiplied 484 by a certain number to get the result 3823a. Find the value of ‘a’. Sol. 3823a is divisible by 484, and 484 is a factor of 3823a. 4 is a factor of 484 and 11 is also a factor of 484. Hence, 3823a is divisible by both 4 and 11. To be divisible by 4, the last two digits have to be divisible by 4. ‘a’ can take two values 2 and 6. 38232 is not divisible by 11, but 38236 is divisible by 11. Hence, 6 is the correct choice. Ex.30 Which digits should come in place of  and $ if the number 62684$ is divisible by both 8 and 5 ? Sol. Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 in never divisible by 8. So, 0 will replace $. Now, the number formed by the last three digits is 40, which becomes divisible by 8, if  is replaced by 4 or 8. Hence, digits in place of  and $ are 4 or 8 and 0 respectively. Ex.31 On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor. Dividend  Re mainder 15968  37  Sol. Divisor = = 179. Quotient 89 Ex.32 How many numbers between 200 and 600 are divisible by 4, 5, and 6 ? Sol. Every such number must be divisible by L.C.M. of 4, 5, 6, i.e.60. Such numbers are 240, 300, 360, 420, 480, 540 Clearly, there are 6 such numbers.

Rule for 19 : Double the last digit of given number and add to remaining number The result obtained should be divisible by 19. The method of finding the remainder without actually performing the process of division is termed as remainder theorem.

Ex.26 Check whether 589 is divisible by 19 or not. Sol. 589, (9 x 2) + 58 = 76 (divisible by 19), i.e. the number is divisible by 19. Ex27. Find the smallest number of six digits which is exactly divisible by 111. Sol. Smallest number of 6 digits is which is 100000. On dividing 100000 by 111, we get 100 as remainder.  Number to be added = (111 - 100) = 11. Hence, required number = 100011.



Remainder should always be positive. For example if we divide –22 by 7, generally we get –3 as quotient and –1 as remainder. But this is wrong because remainder is never be negative hence the quotient should be –4 and remainder is +6. We can also get remainder 6 by adding –1 to divisor 7 ( 7–1 = 6).

PAGE # 4747

Ex.33 Two numbers, x and y, are such that when divided by 6, they leave remainders 4 and 5 respectively. Find the remainder when (x2 + y2) is divided by 6. Sol. Suppose x = 6k1 + 4 and y = 6k2 + 5 x2 + y2 = (6k1 + 4)2 + (6k2 + 5)2 = 36k12 + 48k1 + 16 + 36k22 + 60k2 + 25 = 36k12 + 48k1 + 36k22 + 60k2 + 41 Obviously when this is divided by 6, the remainder will be 5. Ex.34 A number when divided by 259 leaves a remainder 139. What will be the remainder when the same number is divided by 37 ? Sol. Let the number be P. So, P – 139 is divisible by 259. P  139 Let Q be the quotient then, =Q 259  P = 259Q + 139 

Sol. 5 x 9 y 4 13 z 8 1 12 Now, 1169 when divided by 585 gives remainder = 584. To find the remainder of big number

 NOTE : (i) Binomial Expansion : (a + b)n = an +

(a – b)n = an –

P 259 Q  139 = 37 37

3 x 5 y 1 8 z 4 1 7  z = (8 × 1 + 7) = 15 ; y (5z + 4) = (5 × 15 + 4) = 79 ; x = (3y + 1) = (3 × 79 + 1) = 238. 8 238 5 29 6 Now, 3 5 4 1 2



Respective remainders are 6, 4, 2.

Ex.36 A number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. Then find out the number. Sol.

n 1!

n(n  1)

an–1b +

an–1b +

2! n(n  1) 2!

an – 2b2 + .... + bn, or

an– 2b2 – ... + (– 1)nbn.

Hence, first term is pure of a i.e an and last digit is pure of b, i.e. bn.

 259 is divisible by 37,  When 139 divided by 37, leaves a remainder of 28. Ex.35 A number being successively divided by 3, 5 and 8 leaves remainders 1, 4 and 7 respectively. Find the respective remainders if the order of divisors be reversed. Sol.

n 1!

(ii) Total number of terms in the expansion of (a + b)n is (n + 1). Ex.38 What is the remainder when 738 is divided by 48. 19

Sol.

19

19

49  48  1 7 38 72 = = = so by using 48 48 48 48 binomial expansion, we can say that 18 terms are completely divisible by 48 but the last term which is

 

 119 48

is not divisible. So, 119 = 1 is the remainder..

Ex.39 What is the remainder if 725 is divided by 4? Sol. 725 can be written (8–1)25. There are 26 terms in all and first 25 terms are divisible by 8, hence also by 4. The last term is (–1)25. Hence, (8 –1)25 can be written 8X – 1 or 4Y –1 ( where Y = 2X). So, 4Y – 1 divided by 4 leaves the remainder 3. Ex.40 What is the remainder if 345 is divided by 8 ? Sol. 345 can be written 922 × 3. 9 can be written as (8 + 1). Hence, any power of 9 can be written 8N + 1. In other words, any power of 9 is 1 more than a multiple of 8. Hence, (8N + 1) × 3 leaves remainder 3 when divided by 8. 16

4 x 5 y 2 6 z 3 1 4  z = (6 × 1 + 4) = 10  y = (5 × z + 3) = (5 × 10 + 53) = 53  x = (4 × y + 2) = (4 × 53 + 2) = 214 Hence, the required number is 214. Ex.37 In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors of 585) and got the remainders 4, 8 and 12. If he had divided number by 585, then find out the remainder.

Ex.41 What is the remainder when 1415

is divided by 5 ?

1516

Sol. 14 = (15 –1)odd = 15n + (–1)odd, i.e. a (multiple of 5) –1. Thus when divided by 5 the remainder will be (–1), i.e. 4. Ex.42 What is the remainder when 357 + 27 is divided by 28? Sol. 357 = (33)19  357 + 27 = (27)19 + 27 = (28 – 1)19 + 27 = 28M + (–1)19 + 27 [Expand by binomial theorem] = 28M – 1 + 27 = 28M + 26 When 28M + 26 divided by 28, the remainder is 26. Hence, the required remainder is 26.

PAGE # 4848

Ex.43 What is the remainder when 82361 + 83361 + 84361 + 85361 + 86361 is divided by 7? Sol. 82 361 + 83 361 + 84 361 + 85 361 + 86 361 = [(84 – 2) 361 + (84 – 1)361 + + 84361 + (84 + 1)361 + (84 + 2)361] Since, 84 is a multiple of 7, then the remainder will be when, (– 2)361 + (–1)361 + 1361 + 2361 is divided by 7. is (– 2)361 + (–1)361 + 1361 + 2361 = 0. So the remainder is zero.

We are having 10 digits in our number systems and some of them shows special characterstics like they, repeat their unit digit after a cycle, for example 1 repeat its unit digit after every consecutive power. So, its cyclicity is 1 on the other hand digit 2 repeat its unit digit after every four power, hence the cyclicity of 2 is four. The cyclicity of digits are as follows : Digit

Cyclicity

0, 1, 5 and 6

1

4 and 9

2

Ex.51 In (57)9 unit digit is 7 Ex.52 In (97)99 unit digit is 3 (vii) When there is 8 in unit’s place of any number. Since, in 81 unit digit is 8, in 82 unit digit is 4, in 83 unit digit is 2, in 84 unit digit is 6, after that unit’s digit repeats after a group of 4. (viii) When there is 9 in unit’s place of any number. Since, in 91 unit’s digit is 9, in 92 unit’s digit is 1, after that unit’s digit repeats after a group of 2. (ix) When there is zero in unit’s place of any number. There will always be zero in unit’s place. Ex.53 Find the last digit of (i) 357 (ii) 1359 57 gives the remainder 4 1. So, the last digit of 357 is same as the last digit of 31, i.e. 3.

Sol. (i) The cyclicity of 3 is 4. Hence,

(ii) The number of digits in the base will not make a difference to the last digit. It is last digit of the base which decides the last digit of the number itself. For 59 which gives a remainder 3. So the 4 59 last digit of 13 is same as the last digit of 33, i.e. 7.

1359, we find 2, 3, 7 and 8

4

So, if we want to find the last digit of 245, divide 45 by 4. The remainder is 1 so the last digit of 245 would be same as the last digit of 21 which is 2. To Find the Unit Digit in Exponential Expressions : (i) When there is 2 in unit’s place of any number. Since, in 21 unit digit is 2, in 22 unit digit is 4, in 23 unit digit is 8, in 24 unit digit is 6, after that the unit’s digit repeats. e.g. unit digit(12)12 is equal to the unit digit of, 24 i.e.6 Ex.44 In (32)33 unit digit is equal to the unit digit of 21 i.e. 2. (ii) When there is 3 in unit’s place of any number. Since, in 31 unit digit is 3, in 32 unit digit is 9, in 33 unit digit is 7, in 34 unit digit is 1, after that the unit’s digit repeats. Ex.45 In (23)13 unit digit be 3 Ex.46 In (43)46 unit digit be 9 (iii) When there is 4 in unit’s place of any number. Since, in 41 unit digit is 4, in 42 unit digit is 6, after that the unit’s digit repeats. Ex.47 In (34)14 unit digit is 6

Ex.54 Find the last digit of the product 723 x 813. Sol. Both 7 and 8 exhibit a cyclicity of 4. The last digits are 71 = 7 81 = 8 2 7 =9 82 = 4 73 = 3 83 = 2 4 7 =1 84 = 6 5 7 =7 85 = 8 The cycle would repeat itself for higher powers. 723 ends with the same last digit as 73, i.e. 3. 813 ends with the same last digit as 81, i.e. 8. Hence, the product of the two numbers would end with the same last digit as that of 3 × 8, i.e. 4. Ex.55Find unit’s digit in y = 717 + 734 Sol. 717 + 734 = 71 + 72 = 56, Hence the unit digit is 6 Ex.56What will be the last digit of (73 )75 6476

6476

76

Sol. Let (73 )75 = (73)x where x = 75 64 = (75)even power  Cyclicity of 3 is 4  To find the last digit we have to find the remainder when x is divided by 4. x = (75)even power = (76 – 1)even power , where n is divided by 4 so remainder will be 1. Therefore, the last digit of (73 )75

6476

will be 31 = 3.

33

Ex.48 In (34) unit digit is 4 (iv) When there is 5 in unit’s place of any number. Since, in 51 unit digit is 5, in 52 unit digit is 5 and so on. Ex.49 In (25)15 unit digit is 5 (v) When there is 6 in unit’s place of any number. Since, in 61 unit digit is 6, in 62 unit digit is 6 & so on. Ex.50 In (46)13 unit digit is 6.

,

(vi) When there is 7 in unit’s place of any number. Since, in 71 unit digit is 7, in 72 unit digit is 9, in 73 unit digit is 3, in 74 unit digit is 1, after that the unit’s digit repeats.

Ex.57What will be the unit digit of (87 )75 55

75 63

6355

.

55

Sol. Let (87 ) = (87)x where x = 75 63 = (75)odd  Cyclicity of 7 is 4.  To find the last digit we have to find the remainder when x is divided by 4. x = (75)odd power = (76 – 1)odd power where x is divided by 4 so remainder will be –1 or 3, but remainder should be positive always therefore, the last digit of (87 )75 Hence, the last digit is of (87 )75

6355

6355

will be 73 = 343. is 3. PAGE # 4949

To Find the Last Tw o Digits in Exponential Expressions :  We know that the binomial theorem : (a + b)n = an +

n

an–1b +

n(n  1)

an – 2b2 + .... + bn. 1! 2! (i) Last two digits of numbers ending in 1 : Let's start with some examples. Ex.58What are the last two digits of 31 786 ? Sol. 31 786 = (30 + 1) 786 = 30 786 +

786 (786  1)

+ 786 × 30 785 × 1

× 30 784 × 1 2 + .... + 1 786 .

2! Note that all the terms excluding last two terms will

end in two or more zeroes. The last two terms are 786 × 30 ×1 785 and 1 786. Now, the second last term will end with one zero and the tens digit of the second last term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second last term will be 80. The last digit of the last term is 1. So the last two digits of 31 786 are 81. Ex.59Find the last two digits of 41 2789. Sol. According to the previous example we can calculate the answer 61 (4 × 9 = 36. Therefore, 6 will be the tens digit and one will be the units digit). Ex.60 Find the last two digits of 71 56747. Sol. Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit) Ex.61 Find the last two digits of 51 456 × 61 567. Sol. The last two digits of 51 456 will be 01 and the last two digits of 61 567 will be 21. Therefore, the last two digits of 51 456 × 61 567 will be the last two digits of 01 × 21 = 21. (ii) Last two digits of numbers ending in 3, 7 or 9 : Ex.62 Find the last two digits of 19 266. Sol. 19 266 = (19 2)133. Now, 19 2 ends in 61 (19 2 = 361) therefore, we need to find the last two digits of (61)133. Once the number is ending in 1 we can straight away

(iii) Last two digits of numbers ending in 2, 4, 6 or 8 : There is only one even two-digit number which always ends in itself (last two digits) - 76 i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that 24 2 ends in 76 and 2 10 ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 24 34 will end in 76 and 24 53 will end in 24. Ex.65Find the last two digits of 2 543. Sol. 2 543 = (2 10)54 × 2 3 = (24) 54 (24 raised to an even power). 2 3 = 76 × 8 = 08. NOTE : Here if you need to multiply 76 with 2 n, then you can straightaway write the last two digits of 2 n because when 76 is multiplied with 2 n the last two digits remain the same as the last two digits of 2 n . Therefore, the last two digits of 76 ×2 7 will be the last two digits of 2 7 = 28. Ex.66Find the last two digits of 64 236. Sol. 64 236 = (2 6)236 = 2 1416 = (2 10)141 × 2 6 = 24 141 (24 raised to odd power) × 64 = 24 × 64 = 36. Now those numbers which are not in the form of 2 n can be broken down into the form 2 n × odd number. We can find the last two digits of both the parts separately. Ex.67Find the last two digits of 62 586. Sol. 62 586 = (2 × 31)586 = 2 586 × 31 586 = (2 10)58 × 2 6 × 31 586 = 76 × 64 × 81 = 84. Ex.68Find the last two digits of 54 380. Sol. 54 380 = (2 × 3 3)380 = 2 380 × 3 1140 = (2 10)38 × (3 4)285 = 76 × 81 285 = 76 × 01 = 76. Ex.69Find the last two digits of 56 283. Sol. 56 283 = (23 × 7)283 = 2849 × 7283 = (210)84 × 29 × (74)70 × 73 = 76 × 12 × (01)70 × 43 = 16. Ex.70 Find the last two digits of 78 379. Sol. 78 379 = (2 × 39)379 = 2 379 × 39 379 = (2 10)37 × 2 9 × (39 2)189 × 39 = 24 × 12 × 81 × 39 = 92.

get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tenth digit will be 8 and last digit will be 1). Factorial n : Product of n consecutive natural numbers Ex.63 Find the last two digits of 33 288.

is known as ‘factorial n’ it is denoted by ‘n!’.

Sol. 33 288 = (33 4)72. Now 33 4 ends in 21 (33 4 = 33 2 × 33 2 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21 72. By the previous method, the last two digits of 21 72 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)

472

factorial one. Hence 0! = 1 = 1! The approach to finding the highest power of x dividing

Ex.64Find the last two digits of 87 474

So, n! = n(n – 1)(n – 2).............321. e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.  The value of factorial zero is equal to the value of

2

4 118

474

.

2

Sol. 87 = 87 × 87 = (87 ) × 87 = (69 × 69) [The last two digits of 87 2 are 69] = 61 118 × 69 = 81 × 69 = 89.

118

× 69

y  y   y  y! is     2    3  ......., where [ ] represents just x x  x  the integral part of the answer and ignoring the fractional part.

PAGE # 5050

Ex.71 What is the highest power of 2 that divides 20! completely? Sol. 20! = 1 × 2 × 3 × 4 ×....× 18 × 19 × 20 = 1 × (21) × 3 × (22) × 5 × (21 × 31) × 7 × (23) × ..... so on. In order to find the highest power of 2 that divides the above product, we need to find the sum of the powers of all 2 in this expansion. All numbers that the divisible by 21 will contribute 1 to the exponent of 2 in the product 20

= 10. Hence, 10 numbers contribute 2 1 to the

21 product. Similarly, all numbers that are divisible by 22 will contribute an extra 1 to the exponent of 2 in the 20

product, i.e

= 5. Hence, 5 numbers contribute an 22 extra 1 to exponents. Similarly, there are 2 numbers that are divisible by 23 and 1 number that is divisible by 24. Hence, the total 1s contributed to the exponent of 2 in 20! is the sum of ( 10 + 5 +2 +1) = 18. Hence, group of all 2s in 20! gives 218 x (N), where N is not divisible by 2. If 20! is divided by 2x then maximum value of x is 18. Ex.72 What is the highest power of 5 that divides of x = 100! = 100 × 99 × 98 × ...... × 3 × 2 × 1. Sol. Calculating contributions of the different powers of 5, we have

100

= 20,

100

= 4. 5 52 Hence, the total contributions to the power of 5 is 24, or the number 100! is divisible by 524. 1

Ex.73 How many zeros at the end of first 100 multiples of 10. Sol. First 100 multiple of 10 are = 10 × 20 × 30 × ......× 1000 = 10100 (1 × 2 × 3 × .......× 100) = 10100 × 1024 × N = 10124 × N Where N is not divisible by 10 So, there are 124 zero at the end of first 100 multiple of 10. Ex.74 What is the highest power of 6 that divides 9!

9 9 = 1 and 2 = 0. Thus 6 6 answers we get is 1 which is wrong. True there is just one multiple of 6 from 1 to 9 but the product 2 × 3 = 6 and also 4 × 9 = 36, can further be divided by 6. Thus, when the divisor is a composite number find the highest power of its prime factors and then proceed. In this case, 9! can be divided by 27 and 34 and thus by 64 (In this case we need not have checked power of 2 as it would definitely be greater then that of 3).

Sol. By the normal method.

Ex.75What is the largest power of 12 that would divide 49! ? Sol. To check the highest power of 12 in 49!, we need to check the highest powers of 4 and 3 in it. Highest power of 3 in 49! = 22 Highest power of 2 in 49! = 46 46 = 23 2  Highest power of 12 will be 22. (Since the common power between 3 and 4 is 22).

 Highest power of 4 in 49! =

Ex.76 How many zeros will be there at the end of 36!36! ? Sol. Highest power of 5 in 36! is 8. So there will be 8 zeros at the end of 36! So at the end of 36!36! , there will be 8 × 36! zeros.

The number system that we work in is called the ‘decimal system’. This is because there are 10 digits in the system 0-9. There can be alternative system that can be used for arithmetic operations. Some of the most commonly used systems are : binary, octal and hexadecimal. These systems find applications in computing. Binary system has 2 digits : 0, 1. Octal system has 8 digits : 0, 1, 2..., 7. Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B, C, D, E, F. After 9, we use the letters to indicate digits. For instance, A has a value 10, B has a value 11, C has a value 12,... so on in all base systems. The counting sequences in each of the systems would be different though they follow the same principle. Conversion : conversion of numbers from (i) decimal system to other base system. (ii) other base system to decimal system. (i) Conversion from base 10 to any other base : Ex.77 Convert (122)10 to base 8 system.

8 122 8 Sol. 8

15 1 0

2 7 1

The number in decimal is consecutively divided by the number of the base to which we are converting the decimal number. Then list down all the remainders in the reverse sequence to get the number in that base. So, here (122) 10 = (172)8. Ex.78 Convert (169)10 in base 7

7 169 7 24 7 3 Sol. 0

1 3 3

Remainder

(169)10 =(331)7 Ex.79 Convert (0.3125)10 to binary equivalent. Sol. Integer 2  0.3125 = 0.625 0 2  0.625 = 1.25 1 2  0.25 = 0.50 0 2  0.50 = 1.00 1 Thus (0.3125)10 = (0.1010)2

PAGE # 5151

Ex.80 Convert (1987.725)10  (........)8 Sol. First convert non-decimal part into base 8. Ex.85 If a – b = 2, 8 1987 8 248 3 8 8

31 3 0

a a

0 7 3

and

b b _____ then find the value of a, b and c. cc 0

Sol. These problems involve basic number

(1987)10 = (3703)8 Now we have to convert (0.725)10 (........)8

(i) aa + bb = 11(a + b) (ii) aa, bb are two-digit numbers.

Multiply

Hence, their sum cannot exceed 198. So, c must be 1. (iii) Hence, cc0 = 110. This implies a + b = 10 or a = 6

0.725 × 8 = [5.8] 0.8 × 8 = [6.4] 0.4 × 8 = [3.2] 0.2 × 8 = [1.6] 0.6 × 8 = [4.8]

...5 ...6 ...3 ...1 ...4

and b = 4. Such problems are part of a category of problems called alphanumerices.

Keep on accomplishing integral parts after

a 3b

multiplication with decimal part till decimal part is zero.

a c Ex.86 If _____ a a 9



(0.725)10 = (0.56314...)8



(1987.725)10 = (3703.56314...)8

then find a, b and c if each of them is

distinctly different digit. (ii) Conversion from any other base to decimal system Ex.81 Convert (231)8 into decimal system. Sol. (231)8 , the value of the position of each of the numbers ( as in decimal system) is :

Sol. (i) since the first digit of (a 3 b) is written as it is after subtracting ac carry over from a to 3. (ii) there must be a carry over from 3 to b, because if no carry over is there, it means 3 – a = a. 

1 = 80 × 1

2a = 3

3 2 which is not possible because a is a digit. For a carry



3 = 81 × 3 2 = 82 × 2 Hence, (231)8 = (80 × 1 + 81 × 3 + 82 × 2)10 (231)8 = (1 + 24 + 128)10

a=

over 1, 2 – a = a a=1 (iii) it means b and c are consecutive digit (2, 3),

(231)8 = (153)10

(3, 4),.... (8, 9)

Ex.82 Convert (0.03125)10 to base 16. Sol. 16  0.3125 = 0.5

0

16  0.5 = 8.0

8

So (0.03125)10 = (0.08)16

1a4 x3b Ex.87

If

8c8 s 72 t 5d8

Ex.83 Convert (761.56)8  (......)16 Sol. In such conversion which are standard form conversions, it is easier to

then, find the value of a, b, c, d, s and t, where all of

(761.56)8  (.....)2  (.....)16

them are different digits.

Converting every digit in base 8 to base 2, (111110001.10110)2  (1F1.B8)16 Ex.84 Convert (3C8.08)16 to decimal Sol. (3C8.08)16 = 3  162 + C  161 + 8  16 + 0  16–1 + 8  16–2

Sol. Let us consider 1 a 4 × 3 = s72. a × 3 results in a number ending in 6. As 16 and 26 is ruled out, a is 2. Thus, s = 3, t = 4 Now 1 a 4 × b = 8c8 ; b = 2 or 7 Again 2 is ruled out because in that case, product would be much less than 800.

= 768 + 192 + 8 + 0 + 0.03125 = (968.03125)10

 b = 7.

So (3C8.08)16 = (968.03125)10

Hence, a = 2, b = 7, c = 6, d = 8, s = 3 and t = 4.

PAGE # 5252

1.

If the number 357y25x is divisible by both 3 and 5, then find the missing digit in the unit’s place and the thousand place respectively are : (A) 0, 6 (B) 5, 6 (C) 5, 4 (D) None of these

2.

There are four prime numbers written in ascending order. The product of the first three is 385 and that of the last three is 1001. The last number is : (A) 11 (B) 13 (C) 17 (D) 19

3.

Find the square root of 7 – 4 3 . (A) 2 – 3

(B) 5 – 3

(C) 2 – 5

(D) None of these

4.

The number of prime factors of (3 × 5)12 (2 × 7)10 (10)25 is : (A) 47 (B) 60 (C) 72 (D) 94

5.

How many three-digit numbers would you find, which when divided by 3, 4, 5, 6, 7 leave the remainders 1, 2, 3, 4, and 5 respectively ? (A) 4 (B) 3 (C) 2 (D) 1

6.

Six strings of violin start vibrating simultaneously and they vibrate at 3, 4, 5, 6,10 and 12 times in a minute, find : i. After how much time will all six of them vibrate together ? ii. How many times will they vibrate together in 30 min ? (A) 60 sec, 31 times (B) 60 min, 31 times (C) 120 sec, 15 times (D) None of these

7.

The HCF of 2 numbers is 11 and their LCM is 693. If their sum is 176, find the numbers. (A) 99,77 (B) 110, 66 (C) 88,77 (D) 121, 44

8.

If P is a prime number, then the LCM of P and (P + 1) is : (A) P(P +1) (B) (P + 2)P (C) (P + 1)(P – 1) (D) None of these

9.

Find out (A + B + C + D) such that AB x CB = DDD, where AB and CB are two-digit numbers and DDD is a threedigit number. (A) 21 (B) 19 (C) 17 (D) 18

10. Three pieces of cakes of weights 4

1 3 Ibs, 6 Ibs and 2 4

1 Ibs respectively are to be divided into parts of equal 5 weights. Further, each must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained ? (A) 54 (B) 72 (C) 20 (D) 41 7

11. How many natural numbers between 200 and 400 are there which are divisible by i. Both 4 and 5? ii. 4 or 5 or 8 or 10 ? (A) 9, 79 (B) 10, 80 (C) 10, 81 (D) None of these 12. 4 61  4 62  4 63  4 64 is divisible by : (A) 3 (C) 11

(B) 10 (D) 13

13. If x is a whole number, then x2 (x2 – 1) is always divisible by : (A) 12 (B) 24 (C) 12 – x (D) Multiple of 12 14. If 653 xy is exactly divisible by 80 then the find the value of (x + y) (A) 2 (B) 3 (C) 4 (D) 6 15. Find the unit digit of (795 – 358). (A) 6 (B) 4 (C) 3 (D) None of these 16. When a number P is divided by 4 it leaves remainder 3. If the twice of the number P is divided by the same divisor 4 than what will be the remainder ? (A) 0 (B) 1 (C) 2 (D) 6 17. If (232 +1) is divisible by a certain number then which of the following is also divisible by that number. (A) (216 – 1) (B) 216 + 1 96 (C) 2 + 1 (D) None of these 18. When 1! + 2! + 3! + ... + 25! is divided by 7, what will be the remainder ? (A) 0 (B) 5 (C) 1 (D) None of these 19. A number when divided by 342 gives a remainder 47. When the same number is divided by 19, what would be the remainder ? (A) 3 (B) 5 (C) 9 (D) None of these 20. What is the remainder when 9875347 × 7435789 × 5789743 is divided by 4 ? (A) 1 (B) 2 (C) 3 (D) None of these 21. P is a prime number greater than 5. What is the remainder when P is divided by 6? (A) 5 (B) 1 (C) 1 or 5 (D) None of these 22. What is the remainder when 587 is divided by 15? (A) 0 (B) 5 (C) 10 (D) None of these PAGE # 5353

23. What is the remainder when 763 is divided by 344? (A) 1 (C) 6

(B) 343 (D) 338

24. What is the remainder when 3040 is divided by 17? (A) 1 (B) 16 (C) 13

(D) 4

25. What is the remainder when 7413 – 4113 + 7513 – 4213 is divided by 66? (A) 2

(B) 64

(C) 1

(D) 0

26. A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. W hen it is successively divided by 5 and 4, then the respective remainders will be : (A) 1, 2

(B) 2, 3

(C) 3, 2

(D) 4, 1

27. How many zeros will be there at the end of the product

2!2!  4!4!  6!6!  8!8!  10!10! ? (A) 10! + 6! (C) 10! + 8! + 6!

(B) 2 (10!) (D) 6! + 8! + 2 (10!)

28. W hen Sholey screened on the TV there was a commercial break of 5 min after every 15 min of the movie. If from the start of the movie to the end of the movie there was in all 60 min of commercials that was screened what is the duration the movie ? (A) 180 min (B) 195 min (C) 169 min

(D) 165 min

29. Which of the following numbers is exactly divisible by all prime numbers between 1 and 17 ? (A) 345345 (B) 440440 (C) 510510

(D) 515513

Directions : (32 to 36) Read the following information carefully and answer the questions given below. In a big hostel, there are 1,000 rooms. In that hostel only even numbers are used for room numbers, i.e. the room numbers are 2, 4, 6, ...., 1998, 2000. All the rooms have one resident each. One fine morning, the warden calls all the residents and tells them to go back to their rooms as well as multiples of their room numbers. When a guy visits a room and finds the door open, he closes it, and if the door is closed, he opens it, All 1,000 guys do this operation. All the doors were open initially. 30. The last room that is closed is room number ? (A) 1936 (B) 2000 (C) 1922 (D) None of these 31. The 38th room that is open is room number : (A) 80 (B) 88 (C) 76 (D) None of these

32. If only 500 guys, i.e. residents of room number 2 to 1000 do the task, then the last room that is closed is room number (A) 2000 (B) 1936 (C) 1849 (D) None of these 33. In the case of previous question, how many rooms will be closed in all ? (A) 513 (B) 31 (C) 13 (D) 315 34. If you are a lazy person, you would like to stay in a room whose number is : (A) more than 500 (B) more than 1000 (C) 500 (D) 2000 35. A 4-digit number is formed by repeating a 2-digit number such as 2525, 3232 etc. Any number of this form is exactly divisible by : (A) 7 (B) 11 (C) 13 (D) Smallest 3-digit prime number 36. If we write all the whole numbers from 200 to 400, then how many of these contain the digit 7 only once ? (A) 32 (B) 34 (C) 35 (D) 36 37. If (12 + 22 + 32 + .....+ 102) = 385, then the value of (22 + 42 + 62 +...... + 202). (A) 770 (B) 1155 (C) 1540 (D) (385 × 385) 38. How many four-digit numbers are there such that the digits are repeated at least once, i.e. all the digits do not occur only once ? (A) 9000 (B) 4536 (C) 4464 (D) None of these 39. Find the total number of prime factors in the expression (4)11 × (7)5 × (11)2. (A) 37 (B) 33 (C) 26 (D) 29 40. The largest number which exactly divides the product of any four consecutive natural numbers is : (A) 6 (B) 12 (C) 24 (D) 120 41. The largest natural number by which the product of three consecutive even natural numbers is always divisible, is : (A) 6 (B) 24 (C) 48 (D) 96 42. The sum of three consecutive odd numbers is always divisible by : I. 2 II. 3 III. 5 IV. 6 (A) Only I (B) Only II (C) Only I and III (D) Only II and IV 43. A number is multiplied by 11 and 11 is added to the product. If the resulting number is divisible by 13, the smallest original number is : (A) 12 (B) 22 (C) 26 (D) 53

PAGE # 5454

44. A 3-digit number 4a3 is added to another 3-digit number 984 to give the four-digit number 13b7, which is divisible by 11. Then ,(a + b) is : (A) 10 (B) 11 (C) 12 (D) 15 45. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is : (A) 1220 (B) 1250 (C) 22030 (D) 220030 46. In a 4-digit number, the sum of the first two digits is equal to that of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number ? (A) 5 (B) 8 (C) 1 (D) 4 47. Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product ? (A) 1050 (B) 540 (C) 1440 (D) 1590 48. Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/ 3 of the mints, but returned four because she had a monetary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl? (A) 38 (B) 31 (C) 41 (D) 48 49. In a number system, the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes : (A) 406 (B) 1086 (C) 213 (D) 691 50. A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came and erased one number. The average of the remaining numbers is 35 (A) 7 (C) 9

7 . What was the number erased? 17

(B) 8 (D) None of these

51. Let D be a recurring decimal of the form D = 0. a1 a2 a1 a2 a1 a2 ....., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D? (A) 18 (B) 108 (C) 198 (D) 288

52. W hat is the value of the following expression

 1   1   1    1          (2 2  1)    ( 4 2  1)    (6 2  1)   .....   (20 2  1)  ?         (A)

9 19

(B)

10 19

(C)

10 21

(D)

11 21

53. Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is : (A) n (B) n + 1 (C) k × n, where k is a function of n 2 (D) n +   7

54. Let N = 553 + 173 – 723 . N is divisible by : (A) both 7 and 13 (B) both 3 and 13 (C) both 17 and 7 (D) both 3 and 17 55. Convert the number 1982 from base 10 to base 12. The results is : (A) 1182 (B) 1912 (C) 1192 (D) 1292 56. The LCM of two numbers is 864 and their HCF is 144. If one of the numbers is 288, then the other number is : (A) 576 (B) 1296 (C) 432 (D) 144 57. The LCM of two numbers is 567 and their HCF is 9. If the difference between the two numbers is 18, find the two numbers : (A) 36 and 18 (B) 78 and 60 (C) 63 and 81 (D) 52 and 34 58. If the product of n positive real numbers is unity, then their sum is necessarily : (A) a multiple of n (C) never less than n

1 n (D) a positive integer

(B) equal to n –

59. How many even integers n, where 100  n  200, are divisible neither by seven nor by nine ? (A) 40 (B) 37 (C) 39 (D) 38 60. In a certain examination paper, there are n question. For j = 1, 2....n, there are 2n-j students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is : (A) 12 (B) 11 (C) 10 (D) 9 61. The number of positive n in the range 12  n  40 such that the product (n –1) (n – 2).... 3.2.1 is not divisible by n is : (A) 5 (B) 7 (C) 13 (D) 14

PAGE # 5555

62. The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each 1 watchman, one at a time. To each he gave of the 2 diamonds he had then, and 2 more besides. He escaped with one diamond. How many did he steal originally ? (A) 40 (B) 36 (C) 25 (D) none of these 63. A rich merchant had collected many gold coins. He did not want any body to know about him. One day, his wife asked, “How many gold coins do we hire?” After pausing a moment he replied, “Well ! if divide the the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the square of the two numbers. “ The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold coins the merchant has ? (A) 96 (B) 53 (C) 43 (D) 48 64. A child was asked to add first few natural numbers (that is 1 + 2 + 3.....) so long his patience permitted. As he stopped, he gave the sum as 575. When the teacher declared the result wrong, the child discovered he had missed one number in the sequence during addition. The number he missed was : (A) less than 10 (B) 10 (C) 15 (D) more than 15 65. 76n – 66n, where n is an integer > 0, is divisible by : (A) 13 (B) 127 (C) 559 (D) All of these 66. The value of

10  25  108  154  225

(A) 4 (C) 8

is :

(B) 6 (D) 10

67 When (55) 10 is represented in base 25 then the expression is : (A) (25)25 (B) (35)25 (C) (55)25 (D) none of these 68. Arrange the following numbers in ascending order 3 4 7 1 , , , . 7 5 9 2

(A)

4 7 3 1 , , , 5 5 9 2

(B)

3 1 7 4 , , , 7 2 9 5

(C)

4 7 1 3 , , , 5 9 2 7

(D)

1 3 7 4 , , , 2 7 9 5

69. Which of the following surds is greatest in magnitude : 6

17 , 2,12 25 , 3 4

(A) 6 17

(B)

(C)

(D)

3

4

12

25 2

70. What is the decimal equivalent of the hexadecimal 25 digit number (100.....001)16 ? (A) 223 + 1 (B) 224 + 1 (C) 292 + 1 (D) 296 + 1 71. The square root of a perfect square containing n digits has ............ digits. n 1 (A) (B) n/2 2 (C) A or B (D) None 72. If the decimal number 2111 is written in the octal system, then what is its unit place digit ? (A) 0 (B) 1 (C) 2 (D) 3 73. The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f,....is : (A) u (B) v (C) x (D) w 74. If n = 67 then find the unit digit of [3n + 2n ]. (A) 1 (B) 10 (C) 5 (D) None 75. The number of 2-digit numbers n such that 3 divides n – 2 and 5 divides n – 3 is : [KVPY 2007] (A) 5 (B) 6 (C) 7 (D) 10 76. How many 4-digit numbers are there with the property that it is a square and the number obtained by increasing all its digits by 1 is also a square ? [KVPY 2007] (A) 0 (B) 1 (C) 2 (D) 4

77. The number of positive fractions m/n such that 1/3 < m/n < 1 and having the property that the fraction remains the same by adding some positive integer to the numerator and multiplying the denominator by the same positive integer is : [KVPY 2007] (A) 1 (B) 3 (C) 6 (D) infinite 78. If a and b are any two real numbers with opposite signs, which of the following is the greatest ? [KVPY 2008] (A) (a–b)2 (B) (|a| – |b|)2 (C) |a 2 – b 2| (D) a 2 + b 2 79. The number (1024)1024 is obtained by raising (16)16 to the power n. What is the value of n ? [KVPY 2008] (A) 64 (B) 64 2 (C) 64 64 (D) 160 80. The number of integers a such that 1  a  100 and a a is a perfect square is : [KVPY 2008] (A) 50 (B) 53 (C) 55 (D) 56 81. The sum of 7 consecutive positive integers is equal to the sum of the next five consecutive integers. What is the largest among the 12 numbers ? [KVPY 2008] (A) 24 (B) 23 (C) 22 (D 21

PAGE # 5656

82. Let 0 < a < b < c be 3 distinct digits. The sum of all 3-digit numbers formed by using all the 3 digits once each is 1554. The value of c is : [KVPY 2008] (A) 3 (B) 4 (C) 5 (D) 6 83. If the decimal 0.d25d25d25 —— is expressible in the form n/27, then d + n must be : [KVPY 2008] (A) 9 (B) 28 (C) 30 (D) 34 84. Let S n be the sum of all integers k such that 2n < k < 2n+1, for n  1. Then 9 divides Sn if and only if : [KVPY 2009] (A) n is odd (B) n is of the form 3k+1 (C) n is even (D) n is of the form 3k + 2 85. The number of ways in which 1440 can be expressed as a product of two positive integers is : [IAO 2007] (A) 17 (B) 18 (C) 35 (D) 36 86. Let N = 28 , the sum of all distinct factors of N is : [IAO 2008] (A) 27 (B) 28 (C) 55 (D) 56 2

3

2009

87. The units digit of (1 + 9 + 9 + 9 + --------- + 9 ) is : [IAO 2009] (A) 0 (B) 1 (C) 9 (D) 3 88. The biggest among the following is : [IAO 2009] (A) 21/2 (B) 31/3 (C) 61/6 (D) 81/8 89. W hen a positive integer x is divided by 47, the remainder is 11. Therefore, when x2 is divided by 47, the remainder will be : [IAO 2009] (A) 7 (B) 17 (C) 27 (D) 37

90. When the number 72010 is divided by 25, the remainder will be : [IAO 2009] (A) 1 (B) 7 (C) 18 (D) 24 91. P, Q and R are three natural numbers such that P and Q are primes and Q divides PR. Then out of the following the correct statement is : [IJSO 2008] (A) Q divides R (B) P divides R (C) P divides QR (D) P divides PQ 92. The expression (5a – 3b)3 + (3b – 7c)3 – (5a – 7c)3 is divisible by : [IJSO 2008] (A) (5a + 3b + 7c) (B) (5a – 3b – 7c) (C) (3b – 7c) (D) (7c – 5a) 93. If a2 + 2b = 7, b2 + 4c = – 7 and c2 + 6a = – 14, then the value of (a2 + b2 + c2) is : [IJSO 2009] (A) 14 (B) 25 (C) 36 (D) 47 94. In the familiar decimal number system the base is 10. In another number system using base 4, the counting proceeds as 1, 2, 3, 10, 11, 12, 13, 20, 21 .... The twentieth number in this system will be: [IJSO 2010] (A) 40 (B) 320 (C) 210 (D) 110 95. If the eight digit number 2575d568 is divisible by 54 and 87, the value of the digit ‘d’ is: [IJSO 2011] (A) 4. (B) 7. (C) 0. (D) 8. 96. The number of distinct prime divisors of the number 5123 – 2533 – 2593 is :

[KVPY 2011]

(A) 4

(B) 5

(C) 6

(D) 7



PAGE # 5757

LO G A R I T H M (ii) If 0 < a < 1, then The logarithm of the number N to the base ' a ' is the exponent indicating the power to which the base ' a ' must be raised to obtain the number N. This number is designated as log a N. Hence: logaN = x  a x = N , a > 0, a  1 & N > 0

(a) loga x < 0 for all x > 1

Systems of Logarithm : There are two systems of logarithm which are generally used.

function.

(b) loga x = 0 for x = 1 (c) logax > 0 for all x satisfying 0 < x < 1 (d) x > y



logax < loga y i.e. loga x is a decreasing

 graph of y = loga x, 0 < a < 1. y

(i) Common logarithm : In this system base is always taken as 10. If a = 10, then we write log b rather than log 10 b.

y = logax, 0 < a < 1. (1,0)

y'

a  1, b > 0, b  1 and  is any real number then ;

log 102 = 0.3010 ; log 103 = 0.4771 ; n 2 = 0.693 ;

n 10 = 2.303

REMARK : The existence and uniqueness of the number log a N can be determined with the help of set of conditions, a > 0 and a  1 and N > 0.

(i)

loga (mn) = loga m + logan

[Where m and n are +ve numbers] in general log a(x1 x2 ......xn) = log ax1 + log a x2 + ........+ log a xn m (ii) loga   = logam – logan n

(iii) loga(m)n = n logam

Some Useful Results : (i) If a > 1 then (a) loga x < 0 for all x satisfying 0 < x < 1 (b) loga x = 0 for x = 1

(iv) logam 

log m b

log a b (v) logam . logma = 1

(vi) If ‘a’ is a positive real number and ‘n’ is a positive rational number, then

(c) loga x > 0 for x > 1

 loga x > loga y i.e. logax is an increasing

a loga n  n

function.

(vii) If ‘a’ is a positive real number and ‘n’ is a positive

 graph of y = loga x, a > 1

rational number, then

y

logaq np  p loga n q

y = logax, a > 1 x'

x

Let m & n are arbitrary positive numbers, a > 0,

REMEMBER

(d) x > y

0

x'

(ii) Natural logarithm : In this system the base of the logarithm is taken as ‘e’. Where ‘e’ is an irrational number lying between 2 and 3. If a = e, we write nb rather than log e b. Here ' e ' is called as Napiers base & has numerical value equal to 2.7182.

0

(1,0)

x

(viii) p loga q  qloga p Ex.1 If log3a = 4, find value of a. Sol.  

y'



log3a = 4 a = 34 a = 81. PAGE # 5858

Ex.2 Find the value of log Sol. Given log

Ex.7 Find the value of the following :

9 27 3 – log  log 8 32 4

3  9 27  9 27 3   log – log  log  log  4 8 32 4  8 32 

  

(x – 2)2 = 0 

log4

Ex.4 Evaluate : 3

2 – log 3 5

x=2

= 9. 3 log3 5

[am + n = am.an]

(2  3 ) x =



x = –1.

(2  3 ) (2  3 )

1 = (2  3 )

2 3



10 –3x = 10 –4  x = 4/3.

(sin 2 60   cos 2 60  )

tan 45  cot 45 

sec 30 

(iii) log 20 1

alogaN = N, a > 0, a  1 & N > 0 REMARKS :

[ loga a  1 ]

(i)

log a1 = 0 [As a1 = a]

(ii) loga a = 1 log11 13

and B = log9125 + 13

log11 13

4

= log 3 3 5 + 7

4 log35 + 7 log11 13 3

log11 13

...(i)

3 log35 + 7 log11 13 2 By (i) and (ii) we have, 3 4 log35 = B – log35 2 3 4 3 log35 < log35 3 2 A < B.

,

(iii) loga 0 = not defined [As an = 0 is not possible, where n is any number] (iv) loga (–ve no.) = not defined. [As in loga N, N will always be (+ ve)] (v) log 1/a a =  1

(vi)

1 loga b

log ba =

ex na

Ex.9 Find the value of the following

B = log 3 2 5 3 + 7 log11 13 B=

log11 7

(vii) a x =

and, B = log9125 + 13 log11 7





Fun dame ntal Log arit hmic Ide ntit y :

then find the relation between A and B.



(2  3 )

Sol. (i) For (i) and (ii) log is not defined because base is equal to 1. (iii) log 201 = 0.

3 2 – 2 3

A–

3)

(2  3 ) x = 2  3

(ii) log

5 = . 6

or,

 x = 20/3.



(i) log

3 2 = log55 – log2 2 2 3

or,

1024

(2 2 ) = 1024 2 3x/2 = 2 10

9 = . 5

= log 5 2 5 3 – log 23 2 2

A=

2

x

Ex.8 Find the value of the following :

–1

Ex.5 Find the value of log25125 – log84. Sol. Given log25125 – log84

or,

(ii) Let x = log2

= 9 × 5–1

Sol. A = log27625 + 7

(2  3 )

3)

(v) Let x = log 0.001 0.0001  (0.001)x = 0.0001

.

Ex.6 If A = log27625 + 7

1024

 4x = 8  x = 3/2.

Sol. (i) Let x = log 48  2 2x = 2 3

(iv) Let x = log( 2

2 – log 3 5 Sol. Given, 3 2 – log3 5 = 3 .3

=

(iv) log( 2

2

(iii) Let x = log 1/49 343  7 –2x = 7 3  x = – 3/2.

2



(iii) log 1/49343

 

Ex.3 If 2log4x = 1 + log4(x – 1), find the value of x. Sol. Given 2log4x = 1 + log4(x – 1)  log4x2 – log4(x – 1) = 1

x =1 x –1 x2 41 = x –1 x2 = 4x – 4 x2 – 4x + 4 = 0

(ii) log 2

(v) log 0.001 0.0001

 9 32 3   log     8 27 4  = log1 = 0. [ loga1 = 0]



(i) log 48

(i) 81 ...(ii)

1 log 53

Sol. (i) Let x = 81 = 3

log3 5 

(ii) Let x = 8

(ii) 8 1 log 53

1   log 2 3 121   3 

= (3 4 )log3 5

= 5 4 = 625. 1   log 2 3 121   3 

= 2 3 log 2

3



121  1

= 2.2log2 121 = 2(121) = 242. PAGE # 5959

Ex.10 Find the value of (i) log 3 4. log 4 5. log 5 6. log 67 . log 78. log 89.

The equality log a x = log a y is possible if and only if x = y i.e. loga x = loga y x = y

(ii) log tan1° + log tan2° + log tan3° + .... + log tan89°. Sol. (i) log 3 4. log 4 5. log 5 6. log 67 . log 78. log 89 log 4 log 5 log 6 log 7 log 8 log 9 log 3 . log 4 . log 5 . log 6 . log 7 . log 8 log 9 = log 3 = 2. (ii) log tan1° + log tan2° +.......+ log tan89° = log(tan1°. tan2°. tan3° .... tan45° ..... tan89°) = log (tan1° tan2° tan3° ... tan45° ... cot3° cot2° cot1°) = log (1.1.1 ....... 1) = log 1 = 0. Ex.11 Find the value of : (i) 2log3 5 – 5 log3 2 (ii) log 2

75 32 5 – 2 log 2 + log 2 16 243 9

Always check that the solutions should satisfy x > 0, y > 0, a > 0, a  1. Ex.13 Solve the following equations : (i) log 3(x + 1) + log 3 (x + 3) = 1 (ii) log 2 log 4 log 5 x = 0 (iii) log 3 [5 + 4 log 3 (x – 1)] = 2 Sol. (i) log 3(x + 1) + log 3 (x + 3) = log 3 (x + 1) (x + 3) = 1  (x + 1) (x + 3) = 3  x2 + 4x = 0  x = 0, – 4. Also x + 1 > 0 or x>–1 And x + 3 > 0 or x>–3 Hence solution is x = 0. (ii)   

(iii) log3 0. 1 Sol. (i) 2log3 5 – 5 log 3 2

log 2 log 4 log 5x = 0 log 4 log 5x = 1 log 5x = 4 x = 5 4 = 625.

log 25

(iii)   

Here 2log3 5 = 2 log 2 3 = 2log2 5



log3 2



= 5log3

2

Hence 2log3 5 – 5 log3 2 = 0. 32 75 5 (ii) log 2 – 2 log 2 + log 2 243 16 9

 75 32 81   = log 22 = 1 = log 2  .  16 243 25 

(iii) log3 0. 1 = log 31/9 = log 33 –2 = – 2 Ex.12 If log 615 = a, log 1218 = b, then find log 2524 in terms of a and b. log3 15 1  log3 5 Sol. log 615 = log 6 = 1  log 2 = a ...(i) 3 3 log3 18 2  log3 2 log 1218 = log 12 = 1  2 log 2 = b ...(ii) 3 3 From (i) and (ii) 2b log 32 = 2b  1 ab  a  2b  1 And log 35 = 2b  1 log3 24 1  3 log3 2 log 2524 = log 25 = 2 log 5 3 3

 2b  1 3    2b  1  =  ab  a  2b  1  2  2b  1   5b = 2 (a  2b  ab  1) .

log 3 [5 + 4log 3 (x – 1)] = 2 5 + 4 log 3 (x – 1) = 9 log 3 (x – 1) = 1 x – 1 = 3  x = 4.

Ex.14 Find the total number of digits in the number 6 100 (Given log 102 = 0.3010 ; log 103 = 0.4771). Sol. Let x = 6 100  log 10x = 100 log106  log 10x = 100 [log 102 + log 103] = 100 [0.3010 + 0.4771] = 100 (0.7781) = 77.81  The number of digits = 78. Ex.15 Prove that log 35 is irrational. Sol. Let log 35 is rational.



log35 =

p [where p and q are co-prime numbers] q

 3 p/q = 5  3 p = 5 q. which is not possible, hence our assumption is wrong and log 35 is irrational.

1.

Given log2 = 0.3010, then log 16 is : (A) 2.4080 (C) 0.2408

2.

(B) 1.2040 (D) 1.9030

If logxy = 100 and log2x = 10, then the value of y is : (A) 21000 (B) 2100 (C) 22000

(D) 210000

PAGE # 6060

3.

4.

(A) 0

(B) 1

(C) 2

(D) 10

7.

9 N

9 (B) N 

9 M

3 (C) M 

3 N

9 (D) N 

3 M

(A) 243

(B) 27

(C) 343

(D) 64

Find all the real values of x, which satisfy the equation

1 (B) 2,

2

1 (C)

 2.

1

(A) 2,

,

1

2 2 2

,1

1 and

2

2 2

(D) 2

(C)

1  s 

of

(C)

3 (1  x  y )

summation

1 1 1   ...  log2 x log3 x log43 x i s e q u a l t o

1 x  y 3

1 5

(C) 6

(D)

1 6

14. If log 2 = a and log3 = b then [log(1) + log(1 + 3) + log (1 + 3 + 5) + .......+ .....+ log (1 + 3 + 5 + 7 + ..... + 19)] – 2[log 1 + log2 + log3 + ....... log7] = p + qa + rb where p,q,r are constants. What is the value of p + 2q + 3r if all logs are in base 10 ? (A) 12 (B) 26 (C) 18 (D) Cannot be determined

2n  1 n  1

(B)

(C) 1 – n

(D) 0 following

(D)

(B)

(A)

the

(B) 3 (1 - x - y)

(A) 5

(B) 1

1 1 1   q r s

Va l u e

1 3(1  x  y )

15. The sum to 2n terms of the log3 1 – log33 + log3 9 – log3 27 ...... is :

If ap = bq = cr = ds, then loga (bcd) is equal to : 1 1 (A) p    q r

(A)

13. If logyx = 10, then logx 3 (y6) is equal to :

The value of x, when log3(log2 x) + 2 log9(log7 8) = 2, is :

2 5 2  log2 x  log2 x  4  

9.

12. If log30 3 = x and log30 5 = y, then log8 30 is equal to :

1 If log 3 M  3 log 3 N = 1+ log0.0085, then : 3 9 (A) M 



(A) 1 (B) 2 (C) 3 (D) No real value of x exists

q 1 q q (D) p = 1 q

x3

8.



(B) p =

(C) p = q = 1

6.

log log x equation: 2 27 3 4 = (log4x)2 + 5(log4 x) + 2.

log10 p + log10 q = log10 (p–q), then : (A) p = q = 0

5.

11. Find the number of different values of x that satisfy the

The value of [log10 (5 log10 100)]2 is :

:

series,

n  1 2n  1

(D) – n

16. When the curves y = log10x and y = x – 1 are drawn in the x-y plane. How many times do they intersect for values x  1 ? (A) Never (B) Once (C) Twice (D) More than twice 17. Solve : log (x + 4) +log (x – 4) = log 3.

1 (A) log x e

1 (B) e

(A) x = – 19 (C) x = 

1 (C) log ( 43!) x

43! (D) log e x

10. Solve for ‘x’ if 2logx 7 + log7x7+3log49x7 = 0. (A) x 

4 3

(C) x  7 4 / 3

(B) x  7 1/ 2 (D) (B) or (C) only

(B) x = + 19

19

(D) Cannot be determined 18. Find x from the equation ax  c– 2x = b3x + 1. 2 log b (A) log a  2 log c  3 log b log b (B) log a  2 log c  3 log b log b (C) log a  2 log c  3 log b (D) Cannot be determined

PAGE # 6161

19. If log 2 = 0.3010 and log 3 = 0.4771, then find the number of digits in 617 . (A) 12 (B) 14 (C) 18 (D) 17 20. Express 3 log 105 – 4 log10 x + log10y2 as the log of a single number. (A) log10

 25 y 2     x4   

(B) log10

 125 y 2    (C) log10  3   x 

 125 y 3   x4 

   

 125 y 2    (D) log10  4   x 

 a3   in terms of log a, log b and  c 5b 2   

21. Express the log  log c.

31. [(log x – log y)(logx2 + logy2)]  [(logx2 – logy2)(logx + log y)] is equal to : (A) 0 (B) 1 (C) log x/y (D) log xy 32. If log10 [log10(log10x)] = 0. (A) x = 103 (C) x = 155

(B) x = 1010 (D) None

(B) 2 (D) none of these

34. Find the number of integral pairs for (x, y) from the following equations :

22. Find x if log10 1250 + log1080 = x. (A) 5 (B) 4 (C) 8 (D) 7 23. If logx2 = a, logx3 = b, logx5 = c, then find the value of the following in terms of a, b and c. (i) logx16 (ii) logx75 (iii) logx60 (A) 4a, 2a + b, 2a + b + c (B) 4a, 2c + b, 2a + b + c (C) 4a, 2c + b, 2c + b + a (D) 4a, 2c + b, 2a + 3b + c 24. Find the value of log (1 + 2 + 3) = (A) log1 + log2 – log3 (B) log1 + log2 + log3 (C) log1 – log2 + log3 (D) – log1 + log2 + log3 (B) a rational number (D) whole number

xy 1 = (log x + log y), then : 2 2

(B) x = – y (D) 2x = yZ

27. Find the value of (yz)logy – logz × (zx)logz – logx × (xy)logx – logy = ? (A) 3 (B) 2 (C) 0 (D) 1 28. The number of solutions for the equation in logex + x – 1 = 0 is : (A) 1 (B) 2 (C) 4 (D) None of the above 29. If log 10N  2.5 then, find out total number of digits in N. (A) 3 (C) 5 (D) Cannot be determined

(B) 3 (D) 100

(A) 1 (C) 4

3 (D) log a – 5 log c – log b 2

(A) x = y (C) x = 2y

(A) 1 (C) 10

equation 2log 2 log 2 x + log 1/2 log 2 (2 2 x ) = 1 :

2 3 (C) log a – 5 log c – log b 3 2

26. If log

1    . When simplified has the value equal to log10 1   1999 

33. The number of solutions for real x, which satisfy the

3 log a – 5 log c – 2 log b 2 (B) 3 log a – 5 log c – 2 log b

(A)

25. log418 is : (A) an irrational number (C) natural number

1 1 1    30. log 10  1   + log 10  1   + log 10  1   + ... + 2 3 4   

log 100|x + y| = (A) 0 (C) 2

1 & log 10 y – log 10|x| = log 100 4 2 (B) 1 (D) none of these

35. How many positive real numbers x are there such that xx

x

(A) 1 (C) 4

x

 ?

 x x

[KVPY 2007] (B) 2 (D) infinite

36. Let logab = 4, logcd = 2 where a, b, c, d are natural numbers. Given that b – d = 7, the value of c – a is : [KVPY 2009] (A) 1 (B) – 1 (C) 2 (D) – 2 37. The value of log10(3/2) + log10 (4/3) + ------ up to 99 terms. [IAO 2008] (A) 0 (B) 2 (C) 2.5 (D) none of the above 38. If a, b  1, ab > 0, a  b and logba = logab, then ab = ? [IAO 2008] (A) 1/2 (B) 1 (C) 2 (D) 10 39. Find x, if 100.3010 = 2, 100.4771 = 3 and 10x = 45. [IAO 2009] (A) 0.6532 (B) 1.6532 (C) 1.6570 (D) 1.7781 40. If x < 0 and log7 (x2 – 5x – 65) = 0, then x is : [IJSO-2011] (A) –13 (B) –11 (C) – 6 (D) – 5

(B) 4

PAGE # 6262

NUTRITION It helps in mechanical churning and chemical digestion of food. The stomach of ruminant (cud chewing) animals have compound stomach and consists of 4 chambers rumen, reticulum; omasum and abomasum. Rumen and reticulum harbour numerous bacteria and Protozoa which help in digestion of cellulose called symbiotic digestion. Abomasum is true stomach as it secretes gastric juice containing HCl and pepsin.

DIGESTIVE SYSTEM It includes alimentary canal and digestive glands. Alimentary canal starts from mouth and ends into anus. •

•

Mouth or Buccal Cavity : An adult has 16 teeth on each jaw. In each half of jaws starting from middle to backward these are, incisors - 2, canine - 1, premolar - 2, molars - 3 (2 + 1 + 2 + 3). Dental formula is 2123/2123. Last molars are called wisdom teeth. Milk teeth start erupting after 6 months of birth and appear between 6 to 24 months. Dental formula of milk teeth is 2102/2102. Stomach : Inner mucosa of stomach is raised into larger number of longitudinal folds called gastric rugae.

•

Small Intestine : Villi and microvilli increase the surface area of digestion and absorption of food. Pancreatic duct release few enzymes which act on carbohydrates, fats and proteins.

•

Large Intestine : In some herbivores (like horse and ass), caecum is large and is a site of microbial digestion of cellulose. In man caecum very small vestigial organ and is called appendix.

Table : Vitamins Necessary for Normal Cell Functioning Deficiency Disease

Deficient Nutrient

Deficient Nutrient

1. Xerophthalmia

Vitam in A / Retinol

10. Megaloblastic anaemia

Folic acid and Vitam in B 12

2. Night-blindness

Vitam in A

11. Pernicious anaemia

Vitam in B 12 (Cyanocobalam ine)

3. Rickets (in children)

Vitam in D/ Sun-Shine Vitam in 12. Scurvy

4. Osteomalacia (adults) Vitam in D

•

Deficiency Disease

Vitam in C/ As corbic acid

13. Tetany

Calcium

5. Sterility

Vitam in E (Tocopherol)

14. Anaemia

Iron

6. Bleeding disease

Vitam in K (Phylloquinone)

15. Goitre

Iodine

7. Beri beri

Vitam in B 1 (Thiam ine)

16. Fluorosis

Exces s of fluorine

8. Cheilosis

Vitam in B 2 (Riboflavin)

17. Kw ashiorkor

Proteins

9. Pellagra

Vitam in B 3 (Niacin)

18. Marasmus

Proteins and food calories

Knowledge Boosters :

(a) Cardiac glands : secrete an alkaline mucus.

(1) Salivary Glands : It produces saliva. In man only three pairs of salivary glands are present.

(b) Pyloric glands : secrete an alkaline mucus.

(a) Parotid glands : largest glands present just below the external ear. In this glands, virus causes mumps disease.

1. Peptic / Zymogen cells - secrete pepsinogen, prorennin

(b) Submaxillary glands : these lie beneath the jawangles. (c) Sublingual glands : smallest glands which lie beneath the tongue and open at the floor of buccal cavity. (2) Gastric Glands: Present in the mucosa of the stomach. These are of 3 types :

(c) Fundic glands : each gland has 4 types of cells.

2. Oxyntic cells - secrete HCI 3. Goblet cells - secrete mucus 4. Argentaffin cells - produces serotonin, somatostatin and histamine (3) Liver : It is the largest gland and consists of a large right lobe, a small left lobe and two small lobes. On the right lobe lies gall bladder, which, stores bile juice secreted by the liver. PAGE # 63

• •

Bile juice contains no enzyme but possesses bile salts and bile pigments (bilirubin-yellow and biliverdin-green). Formation of glucose from excess organic acids. Storage of vitamins : A, D, E, K Synthesis of vitamin A from carotene.

• • • •

Secretions of blood anticoagulant named heparin. Synthesis of blood or plasma proteins, (fibrinogen and prothrombin) Secretion of bile, detoxification of harmful chemicals. Elimination of pathogens and foreign particles through phagocytic cells called Kupffer’s cells.

Table : Various Enzymes Involved in Digestion

5.

Stomach is protected from HCl and gastric juice because : (A) The two are dilute (B) Epithelial lining is resistant to them (C) Wall has neutralizing agents (D) Stomach lining is covered by mucus

6.

In a villus, some of the glycerol and fatty acids are combined to form fats, coated with proteins, and then transported to the : (A) Lacteal (B) Capillaries (C) Lumen of the small intestine (D) Lumen of the large intestine

7.

Among mammals, an herbivore has : (A) More teeth than a carnivore (B) Fewer teeth than a carnivore (C) Flatter teeth than a carnivore (D) Teeth that are more pointed than a carnivore

8.

Gastric juice has a pH of about : (A) 1 (B) 2 (C) 6 (D) 10

9.

Softness and deformities of bones like bowlegs and pigeon-chest are symptoms of the disease : (A) Rickets (B) Osteomalacia (C) Goitre (D) Beri-Beri

EXERCISE 1.

In human digestion is : (A) Intercellular (C) Extracellular

(B) Intracellular (D) Both A & B

2.

The enzyme rennin converts (A) Proteins to proteoses (B) Fats to fatty acids (C) Casein to paracasein (D) Proteins to peptones

3.

Digestion in Amoeba is : (A) Intercellular (B) Intracellular (C) Both of these (D) None of these

4.

Statements true or false is: (i) No absorption of food takes place in mouth and oesophagus (ii) Absorption of H2O, alcohol, simple salts, glucose and chloride takes place in the stomach to slight extent. (iii) Whole protein particles can be absorbed by pinocytosis. (A) (i) and (ii) true (iii) rarely true (B) All true (C) (i) and (iii) true B false (D) (i) and (ii) true (iii) false

PAGE # 64

10. In rabbit when a dilute solution of glucose reaches small intestine is : (A) Absorbed rapidly by the active transport with sodium ions (B) Absorbed by facilitated diffusion (C) Lost to outside with undigested food (D) Absorbed rapidly by active transport independent of sodium ions

18. B.M.R. for a normal human adult is : (A) 1600 Kcal/day (B) 2000 Kcal/day (C) 2500 Kcal/day (D) 3000 Kcal/day

11. Chymotrypsin acts on : (A) Carbohydrates (C) Fats

20. If all bacteria of intestine die, then what will be the effect on body: (A) Man will feel tired all the body (B) It will cause blindness (C) There will be no synthesis of vitamin -B Complex (D) Excretion will be effected

(B) Proteins (D) Starch

12. The vermiform appendix is made up of : (A) Striated muscles (B) Excretory tissue (C) Lymphatic tissue (D) Stem cells

21. The gall bladder is involved in

13. Digestion is completed in : (A) Ileum (B) Duodenum (C) Stomach (D) Rectum

[KVPY 2011]

14. One of the following is not an enzyme of digestive system : (A) Trypsin (B) Amylase (C) Enterogastrone (D) Enterokinase 15. Fat soluble vitamins are : (A) A, D and E (B) B, C and D (C) B,D and E (D) A, B and C 16. Sunshine vitamin is : (A) Vitamin A (C) Vitamin K

19. Cholesterol rich diet is undesirable because it : (A) Makes person obese (B) Causes heart ailments (C) Is difficult to remove it from body (D) All of the above

(B) Vitamin D (D) Vitamin E

17. Which of the vitamins is essential for normal vision : (A) Folic acid (B) Biotin (C) Riboflavin (D) Retinol

(A) (B) (C) (D)

synthesizing bile storing and secreting bile degrading bile producing insulin

22. In the 16th century, sailors who travelled long distances had diseases related to malnutrition, because they were not able to eat fresh vegetables and fruits for months at a time scurvy is a result of deficiency of [KVPY 2011] (A) carbohydrates (B) proteins (C) vitamin C (D) vitamin D 23. Several mineral such as iron, iodine, calcium and phosphorous are important nutrients. Iodine is found in [KVPY 2011] (A) thyroxine (B) adrenaline (C) insulin (D) testosterone



PAGE # 65

SERIES COMPLETION Series completion problems deals with numbers, Ex 5. alphabets and both together. While attempting to solve the question, you have to check the pattern of the series. Series moves with certain mathematical Sol. operations. You have to check the pattern. Type of questions asked in the examination : Ex 6. (i) Find the missing term(s). (ii) Find the wrong term(s). Sol.

NUMBER SERIES

8, 12, 21, 46, 95, ? (A) 188 (B) 214 (C) 148 (D) 216 (D) The pattern is + 22, + 32, + 52, + 72, ....... missing number = 95 + 112 = 216 3, 9, 36, 180, ? (A) 1080 (B) 900 (C) 720 (D) None of these (A) Each term is multiplied by 3, 4, 5 and so on respectively. Therefore, the next term would be 180 × 6 = 1080.

(a) Some Important Patterns : (b) Multiple Series : (i) a, a ± d, a ± 2d, a ± 3d.......(Arithmetic Progression)

A multiple series is a mixture of more than one series :

(ii) a, ak, ak2, ak3, ................(Geometric Progression)

a a a (iii) a, , 2 , 3 , .............(Geometric Progression) k k k

Ex 7.

Sol. (iv) Series of prime numbers – i.e. 2, 3, 5, 7, 11, ...... (v) Series of composite numbers – i.e. 4, 6, 8, 9, 10, 12, ................. Directions : (1 to 10) Find the missing numbers : Ex 1.

Sol.

16, 19, 22, 25, ? (A) 27 (B) 28 (C) 29 (D) 25 (B) As per series a, a + d, a + 2d,......... a = 16 d=3 a + 4d = 16 + 4 × 3

Ex 8.

Sol.

Ex 9. Ex 2.

Sol.

Ex 3.

Sol. Ex 4.

Sol.

9, 18, 36, ?, 144 (A) 70 (B) 56 (C) 54 (D) 72 (D) As per series, a, ak, ak2, ak3, ........ a = 9, k = 2 ak3 = 9 × 23 = 72 2, 6, 14, 26, ? (A) 92 (B) 54 (C) 44 (D) 42 (D) The pattern is +4, +8, +12, +16, ....... 240, ? , 120, 40, 10, 2 (A) 120 (C) 40

(B) 240 (D) 10

1 1 1 1 (B) The pattern is ×1, × , × , × , × 3 5 2 4  missing term = 240 × 1 = 240

Sol.

Ex 10.

Sol.

4, 7, 3, 6, 2, 5, ? (A) 0 (B) 1 (C) 2 (D) 3 (B) The sequence is a combination of two series I 4, 3, 2, ? II 7, 6, 5 The pattern followed in I is – 1, – 1, – 1  missing number = 2 – 1 = 1 14, 15, 12, 16, 9, 18, 4, 21, ? (A) 2 (B) 3 (C) – 3 (D) – 5 (C) The sequence is a combination of two series.  14, 12, 9, 4, (....) and  15. 16, 18, 21 The pattern followed in  is – 2, – 3, – 5, .......  missing number = 4 – 7 = – 3 1, 1, 4, 8, 9, ? ,16, 64 (A) 21 (B) 27 (C) 25 (D) 28 (B) (i) 1, 4, 9, 16 [12, 13, 22, 23, 32, 33.............] (ii) 1, 8, __, 64 mixed combination 3, 6, 24, 30, 63, 72, ?, 132 (A) 58 (B) 42 (C) 90 (D) 120 (D) The difference between the terms is given below as : 3

6

24 30

Difference 3

18

6

Difference

3

15

63

33 3

72 9 15

?

48

132 ?

?

Therefore, alternate difference between the difference is 3 and 15 respectively. Hence, the next term would be 72 + 48 = 120. PAGE # 66

Directions : (11 to 12) Find the wrong term(s) — Ex 11.

Sol.

Ex 13.

5, 8, 10, 12, 15, 18, 20, 23 (A) 8 (B) 12 (C) 15 (D) 18 (B) Sol.

12 28 64 140 37 (P) (Q) (R) Which number will come (A) 1412 (C) 696 (A)

Therefore, number 12 is wrong and should be Similarly replaced by 13. Ex 12.

Sol.

37

1, 3, 8, 19, 42, 88, 184 (A) 3 (B) 8 (C) 19 (D) 88 (D) 3

1 2

19

8 11

5 3

6

12

(Q)

(R)

(S)

(T)

78

164

340

696

1412

×2+8

×2+12

×2+16

×2+20

Therefore, the number 1412 will come in place of (T). Ex 14. 184

89 47

24

(P)

×2+4

42 23

(S) (T) in place of (T) ? (B) 164 (D) 78

95 48

Hence, number 88 is wrong and should be replaced by 89. or 1 × 2 + 1, 3 × 2 + 2, 8 × 2 +3, 19 × 2 + 4, 42 × 2 + 5, Sol. 89 × 2 + 6

2 9 57 3 (P) (Q) Which number will come (A) 113 (C) 3912

(T)

(A) Similarly, (P)

Directions : (13 to 14) In each of the following questions, a number series is given. After the series, below it in the next line, a number is given followed by (P), (Q), (R), (S) and (T). You have to complete the series starting with the number given following the sequence of the given series. Then answer the question given below it.

337 (R) (S) in place of (Q) ? (B) 17 (D) 8065

(Q)

×8–7

×7–6

(S)

673

113

17

3

(R)

3361 ×5–4

×6–5

Therefore, the number 113 will come in place of (Q).

ALPHABET SERIES (a) Pattern of Alphabets Show Variation Based on : (i) Position of the letters

(ii) Difference between the alphabets

(i) Position of alphabets : Alphabets in order :

Alphabets in reverse order :

Directions : (15 to 24) Find the missing term(s) : Ex 15.

Sol.

Ex 16.

Sol.

Ex 17.

B, E, H, ? (A) K (B) L Sol. (C) J (D) M (A) In the given series, every letter is moved three steps forward to obtain the corresponding letters of the next term. So, the missing term is K. Q, N, K, ?, E (A) H (B) I (C) J (D) G (A) In the given series, every letter is moved three Ex 18. steps backward to obtain the corresponding letters of the next term. So, the missing term is H.

A, Y, D, W, G, U, J, ? (A) R (B) T (C) S (D) P (C) The given sequence consists of two series : . A, D, G, J in which each letter is moved three steps forward to obtain the next term . Y, W, U, ? in which each letter is moved two steps backward to obtain the next term. So, the missing term would be S. AG, LR, WC, HN, ? (A) SX (C) SY

(B) RY (D) TX PAGE # 67

Sol.

(C) The first letter of each group and the second Ex 24. letter of each group differs by 11 letters between them. L A W H Sol. 12

1

11

23

11

11 R 18

G 7

Similarly,

Alphabetical positions

8

11

Difference in Alphabetical positions C N 3 14 11

11

 A  B  C   6  D  E  F   15  G  H  I   24    , 4 5 6 ,   1 2 3  7 8 9 Alphabetical positions Difference in Alphabetical positions

Therefore, the next group of letter would be SY. H

N

S

   10 11 12 

So, the missing term would be  J  K  L   33 Directions : (25 to 27) Find the wrong term (s) : Ex 25.

Y

And 11

Ex 19.

Sol.

Ex 20.

Sol.

Ex 21.

Sol.

Ex 22.

Sol.

Ex 23.

Sol.

(ABC) – 6, (DEF) – 15, (GHI) – 24, ? (A) (IJK) – 33 (B) (JKM) – 33 (C) (IJK) – 32 (D) (JKL) – 33 (D) In a given series Let A = 1, B = 2, C = 3, D = 4, E = 5, F = 6, and so on

11

Sol.

AD, EI, JO, PV, ? (A) VD (B) WC (C) WD (D) VE (C) The first letter of subsequent groups have a difference of 4, 5 and 6 places respectively, whereas the second letter of the subsequent groups has a difference of 5, 6, and 7 places respectively, Therefore, on following the same pattern, we get Ex 26. ‘WD’ as the next term which would replace the question mark. Sol. AB, BA, ABD, DBA, PQRS, ? (A) SRQP (B) SRPQ (C) SQRP (D) RSQP (A) The first term is reversed to get second term. The third term is reversed to get the fourth term. Similarly, to get the sixth term, we reverse the fifth Ex 27. term. So, the missing term would be SRQP. HEJ, JGL, LIN, NKP, ? (A) MOR (B) PNS Sol. (C) PMR (D) NPT (C) First letter of each group differs by 2 letters. Second letter of each group differs by 2 letters. Third letter of each group differs by 2 letters. All the letters differ in the forward direction. Hence, the next choice would be PMR. XYQ, ZAR, BCS, DET, ? (A) GFU (B) FUG (C) FZU (D) FGU (D) Here, first two terms of every group of letters are in continuation, like XY, ZA, BC, DE, and the third letter of each group is again in forward continuation, i.e. Q, R, S, T. Hence, the term replacing the question mark would be FGU.

ABD, DGK, HMS, NTB, SBL, ZKW (A) NTB (B) DGK (C) SBL (D) ZKW (A) First letter of first, second, third,.........terms is moved three, four, five, ........steps forward respectively. Similarly, second letter is moved five, six, seven,......steps forward respectively and third letter is moved seven, eight, nine,........steps forward respectively. Hence, NTB is the wrong term and should be replaced by MTB. EPV, FQW, GRX, HTY, ITZ (A) FQW (B) GRX (C) HTY (D) ITZ (C) In every term, first second and third letter is in alphabetical order to its next term respectively. Fourth term is not following the same rule. Hence, HTY is the wrong term and should be replaced by HSY. D4V, G10T, J20R, M43P, P90N (A) P90N (B) G10T (C) J20R (D) D4V (B) First letter of every term is moved three steps forward in each next term. Second number of every term of the pattern  × 2 + 1, × 2 + 2, × 2 + 3,............and third letter of every term is moved two steps backward. Hence, G10T is the wrong term and should be replaced by G9T.

LETTER REPEATING SERIES Pattern of such questions is that some letters in sequence are missing. (i) The letters may be in cyclic order (clockwise or anti-clockwise). (ii) To solve a problem, we have to select one of the alternative from the given alternatives. The alternative which gives a sequence form of letters is the choice.

17Z5, 15X4, 13V3, ?, 9R1 (A) 11S2 (B) 11T2 (C) 11U2 (D) 11T3 Directions : (28 to 32) Find the missing term(s) : (B) The first number & second letter of every term is moved two steps backward & the third number Ex 28. a a _ b a a _ b b b _ a of every term is moved one step backward. So, the (A) baa (B) abb missing term would be 11T2. (C) bab (D) aab

PAGE # 68

Sol.

(A) we proceed step by step to solve the above Ex 35. series: Steps :

1.

The first blank space should be filled in by 'b' so that we have two a's followed by two b's.

2.

Second blank space should be filled in by 'a' so that we have three a's followed by three b' s.

3. Ex 29.

The last blank space must be filled in by 'a' to keep the series in sequence. _ bca _ ca _ c _ b _ Ex 36. (A) aabbc (B) abbbc (C) aabcc (D) abbac

Sol.

(D)

Sol.

Sol.

Series is abc/ abc/ abc/ abc. So, pattern abc is repeated. Ex. 30

a_h_ _c_ ne_ h_ eac_ _ _ _ _ 21_4 3_5 __2 54 ____ _ _ _ _ The last five terms in the series are (A) 32524 (B) 43215 (C) 25314 (D) 32541 (B) By taking a = 2, c = 1, n = 4, h = 5 and e = 3, the numbers series runs as 21543 15432 54321 43215. If first digit of a group of five digits is placed as the last digit, we obtain the second group of five digits and so on. _ m y e _ _ y l x _ y l m _ _ l _ _ _ _ 4 6 _ 5 8 6 _ __ 5 7_ 6 5 8 __ _ _ _ The last five terms of the number series are (A) 46758 (B) 74658 (C) 76485 (D) 46785 (D) By taking e = 5, l = 4, m = 6, y = 7 and x = 8 the number series runs as 46758 67485 74658 46785. By taking the digits in the groups of five, we find that first digit of the first group (i.e. 4) is the third digit of the second group and the last two digits have interchanged their positions. The same rule applies in others groups also.

a _ abb _ aa _ ba _ a _ b (A) ababa (B) aabba (C) aabab (D) aaabb Sol(C) Series is aaabb/ aaabb/ aaabb. So, pattern Direction : (37) In the following question, three sequences aaabb is repeated. of letter/numbers are given which correspond to each other in some way. In the given question, you Ex 31. a _ c _ abb _ ca _ a have to find out the letter/numerals that come in (A) baca (B) bbca the vacant places marked by (?). These are given (C) bacc (D) bacb as one of the four alternatives under the question. Sol(A) Series is abc/ aabbcc/ aaa Mark your answer as instructed. Ex 32. bc _ b _ c _ b _ ccb Ex 37. C B _ _ D _ B A B C C B (A) cbcb (B) bbcb (C) cbbc (D) bcbc _ _ 2 3 5 4 _ _ ? ? ? ? Sol(A) Series is bccb / bccb / bccb. So, pattern bccb is p _ p q _ r _ q _ _ _ _ repeated (A) 4 5 5 4 (B) 4 3 3 4 (C) 4 2 2 4 (D) 2 5 5 2 Directions : (33 to 34) The question given below is based (C) Comparing the positions of the capital letters, on the letter series, In series, some letters are Sol. numbers and small letters, we find p corresponds missing. Select the correct alternative. If more than to C and 2 corresponds to p. So, p and 2 correspond five letters are missing, select the last five letters of the series. to C. q corresponds to A and 3 corresponds to q. So, q and 3 corresponds to A. Also, 5 corresponds Ex 33. xyzu _ yz _ v _ _ uv _ _ _ _ _ _ _ to D. So, the remaining number i.e., 4 corresponds (A) uvxyz (B) vuzyx to B. So, BCCB corresponds to 4, 2, 2, 4. (C) uvzyx (D) vuxyz Sol. (A) The series is x y z u v / y z u v x/ z u v x y/u v x y z MISSING TERMS IN FIGURES Thus the letters are written in a cyclic order. EX 34.

Sol.

abcd _ bc _ e _ _ de _ _ _ _ _ _ _ (A) deabc (B) edcba (C) decba (D) edabc (A) The series is a b c d e / b c d e a / c d e a b / b e a b c Thus the letters are written in a cyclic order.

Directions : (38 to 47) Find the missing number(s) :

Ex 38.

Direction : (35 to 36) There is a letter series in the first row and a number series in the second row. Each number in the number series stands for a letter in the letter series. Since in each of that series some term are missing you have to find out as to what Sol. those terms are, and answer the questions based on these as given below in the series.

6

9

15

8

12

20

4

6

?

(A) 5 (B) 10 (C) 15 (D) 21 (B) In the first row, 6 + 9 = 15 In the second row, 8 + 12 = 20  In the third row, missing number = 4 + 6 = 10. PAGE # 69

3 Ex 39.

10 2

6

Ex 43.

9

4 (A) 11 (C) 3 Sol.

(B) 6 (D) 2

(C) Clearly, in the  column,

6 4 8 3

18  3  27 We take x in place of ? 2 95 15  x Similarly in the  column, 3  9 ,x  15 5

Sol.

In the  column,

Ex 40.

Sol.

3C

27D

9E

7I

21K

3M

4D

?

7J

Ex 41.

Sol.

Ex 42.

2

Sol.

(A) 32 (B) 22 (C) 18 (D) 27 (B) In first figure] 5 × 4 + 6 = 26 In second figure] 8 × 3 + 5 = 29 missing number in third figure] 6 × 3 + 4 = 22

81 18

Sol.

88 ? 11

9 (B) 21 (D) 81

In second figure, 9 

6 ? 3 4

3

8 29 3 5

7

6

174

5

3

2 5 ?

336

3 2 9 2 7 9 6 4 5 (A) 140 (B) 150 (C) 200 (D) 180 (B) In first figure]8 × 5 × 3 + 3 × 2 × 9 = 120 + 54 = 174 In second figure]6 × 7 × 5 + 2 × 7 × 9= 210 + 126 = 336  missing number in third figure] 3 × 2 × 5 + 6 × 4 × 5 = 30 + 120 = 150

9

15

7

21

25

14

2 18 2

Ex 47.

= 84. = 81.

= 88 or x =

88  2 11

= 16.

Sol.

176

?

(A) 184 (B) 210 (C) 241 (D) 425 (A) The number at the bottom is the difference of squares of two numbers given at top In first figure] 112 – 92 = 121 – 81 = 40 In second figure] 152 – 72 = 225 – 49 = 176  In third figure] 252 – 212 = 625 – 441 = 184

3

Let the missing number In third figure be x.

x 2

5

8

40

(A) In first figure, 12

Then, 11

2

5

Ex 46.

(A) 16 (C) 61 Sol.

5

(A) 15 (B) 20 (C) 25 (D) 40 (B) Clearly In first figure] 6 × 3 – 4 × 2 = 18 – 8 = 10 In second figure] 9 × 5 – 5 × 3 = 45 – 15 = 30  In third figure] 6 × 5 – 2 × 5 = 30 – 10 = 20

11

(A) 16 (B) 9 (C) 85 (D) 112 (C) Hint ; 42 + 52 = 16 + 25 = 41 12 + 22 = 1 + 4 = 5 62 + 72 = 36 + 49 = 85

84 14 12

?

5 26 4 6

1

7

6

5

41 ? 5

6

30 3

Ex 44.

(A) 11E (B) 28G  (C) 35I (D) 48F (B) The letters in the first row form a series C, D, E (a series consecutive letters). The letters in the second row form a series I, K, M (a series of Ex 45. alternate letters). Similarly, the letters in the third row will form the series D, G, J (a series in which each letter is three steps ahead of the previous one). So, the missing letter is G. Also, the number in the second column is equal to the product of the numbers in the first and third Sol. columns. So, missing number is (4 × 7) i.e. 28. Thus, the answer is 28G.

4

5

5

5

33 6 3

4

7 48

5

5

3 ?

4

5

(A) 47 (B) 45 (C) 37 (D) 35 (D) In first figure, 6 × 3 + 3 × 5 = 33 In second figure, 5 × 4 + 4 × 7 = 48  In third figure, 5 × 4 + 3 × 5 = 35 PAGE # 70

4

17.

EXERCISE-1 Directions : (1 to 25) Find the missing numbers : 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

2, 8, 18, 32, ? (A) 62 (C) 50 16, 54, 195, ? (A) 780 (C) 816 14, 316, 536, 764, ? (A) 981 (C) 8110

(B) 60 (D) 46

3, 15, 35, ?, 99, 143 (A) 63 (C) 69

(B) 77 (D) 81

9, 16, 30, 58, ? (A) 104 (C) 116

(B) 114 (D) 118

3, 12, 27, 48, 75, 108, ? (A) 192 (C) 162

(B) 183 (D) 147

1, 4, 12, 30, ? (A) 60 (C) 64

(B) 62 (D) 68

94, 166, 258, ?, 4912 (A) 3610 (C) 1026

(B) 1644 (D) 516

22. (B) 180 (D) 452

7,19, 55, 163, ? (A) 387 (C) 527

(B) 329 (D) 487

1, 2, 9, 4, 25, 6, ? (A) 51 (C) 50

(B) 49 (D) 47

24.

16, 33, 67, 135, ? (A) 371 (C) 271

(B) 175 (D) 287

25.

101, 100, ?, 87, 71, 46 (A) 92 (C) 89

(B) 88 (D) 96

13.

6, 24, 60, 120, 210, 336, ?, 720 (A) 496 (B) 502 (C) 504 (D) 498

16.

(B) 581 (D) 481

2, 3, 6, 18, ?, 1944 (A) 154 (C) 108

100, 50, 52, 26, 28, ? 16, 8 (A) 30 (B) 36 (C) 14 (D) 32

15.

137, 248, 359, 470, ? (A) 582 (C) 571

21.

(B) 210 (D) 258

3, 1, 4, 5, 9, 14, 23, ? (A) 32 (C) 41

(B) 37 (D) 28

3, 6, 18, 72, 360, ? (A) 720 (C) 1600

(B) 1080 (D) 2160

78, 79, 81, ?, 92, 103, 119 (A) 88 (B) 85 (C) 84 (D) 83

(B) 195 (D) 103

20.

8, 11, 15, 22, 33, 51, ?, 127, 203 (A) 80 (B) 53 (C) 58 (D) 69

2, 12, 36, 80, 150, ? (A) 194 (C) 252

2, 9, 28, 65, ? (A) 121 (C) 126

1, 11, ?, 11, 11, 11, 16, 11 (A) 1 (B) 11 (C) 6 (D) 192

(B) 1048 (D) 9100

8, 24, 16, ?, 7, 14, 6, 18, 12, 5, 5, 10 (A) 14 (B) 10 (C) 7 (D) 5

(B) 112 (D) 108

19. (B) 802 (D) 824

12.

14.

18.

0, 6, 20, 42, 72, ? (A) 106 (C) 110

23.

Directions : (26 to 28) In each of the following questions, a number series is given. After the series, below it in the next line, a number is given followed by (P), (Q), (R), (S) and (T). You have to complete the series starting with the number given following the sequence of the given series. Then answer the question given below it. 26.

2 3 8 27 5 (P) (Q) (R) (S) (T) Which of the following numbers will come in place of (T) ? (A) 184 (B) 6 (C) 925 (D) 45

27.

5 18 48 7 (P) (Q) Which number will come (A) 172 (C) 270

28.

112 (R) (S) in place of (S) ? (B) 276 (D) 376

(T)

15 159 259 323 7 (P) (Q) (R) (S) (T) Which of the following numbers will come in place of (R) ? (A) 251 (B) 315 (C) 176 (D) 151 PAGE # 71

Directions : (29 to 35) Find the wrong term(s) — 29.

30.

31.

32.

33.

9, 11, 15, 23, 39, 70, 135 (A) 23 (B) 39 (C) 70 (D) 135 3, 9, 36, 72, 216, 864, 1728, 3468 (A) 3468 (B) 1728 (C) 864 (D) 216 2, 5, 11, 20, 30, 47, 65 (A) 5 (C) 30 121, 143, 165, 186, 209 (A) 143 (C) 186

(B) 20 (D) 47

(B) 165 (D) 209

9, 15, 24, 34, 51, 69, 90 (A) 15 (B) 24 (C*) 34 (A) 15 (C) 34

8.

9.

10.

11.

35.

(C) 69

(D) 132

105, 85, 60, 30, 0, – 45, – 90 (A) 85 (B) – 45 (C) 105 (D) 0

4.

5.

6.

7.

ZGL, XHN, VIQ, TJU, ? (A) RKX (C) RLZ

(B) RKY (D) RKZ

RML, VIJ, ZFH, DDF, ? (A) HDC (C) HCD

(B) CHI (D) DIC

LRX, DJP, VBH, NTZ, ? (A) ELS (C) GKS

(B) FMR (D) FLR

MAD, OBE, SCH, YDM, ? (A) HET (C) GET

(B) HES (D) UAE

2B, 4C, 8E, 14H, ? (A) 22L (C) 22K

(B) 24L (D) 2M

15.

16.

Directions : (1 to 24) Find the missing term(s) :

3.

(B) OET (D) PEV

(B) ZVX (D) VZX

17.

2.

CYD, FTH, IOL, LJP, ? (A) PET (C) OEY

ZSD, YTC, XUB, WVA, ? (A) VZZ (C) VWZ

EXERCISE-2

1.

(B) ELS (D) DLS

13.

9, 13, 21, 37, 69, 132, 261 (B) 37

JXG, HTJ, FPN, ?, BHY (A) EKS (C) DLR

MTH, QRK, UPN, YNQ, ? (A) CKT (B) ELT (C) CLT (D) EKT

(D) 51 (B) 24 (D) 51

(A) 21

(B) KBO (D) LBN

12.

14. 34.

DFK, FEL, HDM, JCN, ? (A) KBN (C) LBO

X, U, S, P, N, K, I, ? (A) J (C) M

(B) K (D) F

18.

1 BR, 2 EO, 6 HL, 15 KI, ? (A) 22 NF (B) 31 NF (C) 31 NE (D) 28 NF

Z, X, U, Q, L, ? (A) F (C) G

(B) K (D) E

19.

P3C, R5F, T8I, V12L, ? (A) Y17O (C) X17O

A, H, N, S, W, ? (A) A (C) B

(B) Y (D) Z

20.

Z 15 A, W 13 C, ?, Q 9 G, N 7 I (A) T 12 E (B) R 11F (C) T 11E (D) R 13 D

Q, T, V, Y, A, ? (A) B (C) D

(B) C (D) F

21.

B3M, E7J, H15G, K31D, ? (A) N65A (B) O63A (C) N63A (D) N63Z

X, A, D, G, J, ? (A) N (C) M

(B) O (D) P

22.

5X9, 8U12, 11R15, 14O18, ? (A) 17L21 (B) 17K21 (C) 17M21 (D) 17L23

Z, L, X, J, V, H, T, F, ?, ? (A) R, D (C) S, E

(B) R, E (D) Q, D

23.

6C7, 8F10, 11J14, 15O19, ? (A) 19U24 (B) 20U25 (C) 19U25 (D) 20U24

AZ, YB, CX, WD, ? (A) VE (C) EU

(B) UE (D) EV

24.

B2E, D5H, F12K, H27N, ? (A) J58Q (B) J56Q (C) J57Q (D) J56P

(B) X17M (D) X16O

PAGE # 72

Directions : (25 to 30) Find the wrong term(s) : 25.

26.

27.

28.

29.

30.

ECA, JHF, OMK, TQP, YWU (A) ECA (B) JHF (C) TQP (D) YWU

9.

10.

a _ cab _ a _ c _ b c (A) bbac (C) abba

(B) abab (D) bcba

ba _ cb _ b _ bab _ (A) acbb (C) cabb

(B) bcaa (D) bacc

DKY, FJW, HIT, JHS, LGQ (A) FJW (B) LGQ (C) JHJ (D) HIT

11.

DVG, FSI, HPK, JNM, LJO (A) DVG (B) JNM (C) HPK (D) LJO

a _ bc _ a _ bcda _ ccd _ bcd _ (A) abddbd (B) acbdbb (C) adbbad (D) bbbddd

12.

CDF, DEG, EFH, FHI (A) CDF (C) FHI

cc _ ccdd _ d _ cc _ ccdd _ dd (A) dcdcc (B) dcddc (C) dccdd (D) None of these

13.

a_baa_baa _ba (A) a a b (C) b b a

(B) b a b (D) b b b

babbb_b_b_bb (A) b b a (C) a b a

(B) b a a (D) a a a

m _ l _ ml _ m _ llm (A) lmmm (C) lmml

(B) lmlm (D) mllm

(B) DEG (D) EFH

ZLA, BMY, CNW, FOU, HPS (A) ZLA (B) BMY (C) FOU (D) CNW

14.

G4T, J10R, M20P, P43N, S90L (A) G4T

(B) J10R

(C) M20P

(D) P43N

15.

EXERCISE-3

Directions : (16 to 19) The questions given below are based on the letter series, In each of these series, some Directions : (1 to 15) Which sequence of letters when placed letters are missing. Select the correct alternative. If at the blanks one after the other will complete the more than five letters are missing, select the last given letter series ? five letters of the series. 1.

2.

3.

4.

a_baa_aa__ab (A) a a a a (C) b b a a

(B) b a a a (D) a b b a

_aabb_a_ab_b (A) b b a a (C) b a a b

(B) b a b a (D) a b a b

aab_aaa_bba_ (A) b a a (C) b a b

(B) a b b (D) a a b

a__b_a_ab_aa (A) a b a a b (C) b b a b b

(B) b b a b a (D) b a a b a

5.

abc _ d _ bc _ d _ b _ cda (A) bacdc (B) cdabc (C) dacab (D) dccbd

6.

a _ bbc _ aab _ cca _ bbcc (A) bacb (B) acba (C) abba (D) caba

7.

_bc__bb_aabc (A) acac (B) babc (C) abab (D) aacc _ b c c _ ac _ a a b b _ a b _ c c (A) aabca (B) abaca (C) bacab (D) bcaca

8.

16.

_ _ r _ ttp _ _ s _ tp _ _ _ s _ _ _ (A) rstqp (B) tsrqp (C) rstpq (D) None

17.

_ _ x _ zbxazyxabyz _ _ _ _ _ (A) abxzy (B) abzxy (C) abxyz (D) bxayz

18.

x _ xxy _ x _ xy _ yxx _ _ yy _ y (A) xyyyy (B) xxyyx (C) yxxyx (D) xyxyx

19.

_ _ r _ tqrptsrpqst _ _ _ _ _ (A) pqrts (B) pqtrs (C) pqrst (D) qrpst

Directions : (20 to 23) There is a letter series in the first row and a number series in the second row. Each number in the number series stands for a letter in the letter series. Since in each of that series some term are missing you have to find out as to what those terms are, and answer the questions based on these as given below in the series. 20.

ab_cd_a_abd_dba_ 1_3_32_1___4____ The last four terms in the series are (A) 1234 (B) 3112 (C) 3211 (D) 4312 PAGE # 73

21.

22.

23.

_bnt _ _nam _nab_ _a__ _ _ 13_25 3__ 5 24_32 5__ _ __ The last five terms in the series are (A) 13425 (B) 41325 (C) 34125 (D) 13452 n _ g f _ t _ f h t n _ _ t _ b _ f 1 3_ 2 4 5 0 _ 4 _ _ 3 _ _ _ _ _ _ The last five terms of the number series are (A) 50123 (B) 40321 (C) 40231 (D) 51302

3.

_ m i a x _ i r x a _ _ma _ _ _ _ _ _ 4 _ 5 _7 3 _ _ _6 _ _ _ _ _ _ _ _ _ _ The last five term of the letter series are (A) r m x i a (B) x m r a i (C) x r m a i (D) r m i x a

25.

_ A C _ B D _ CD C D 2 _4 1 _ 1 4 _ _ _ _ r s _ q r _ p ? ? ? ? (A) p q p q (B) p r p r (C) r q r q (D) r s r s

6.

(A) 235 (C) 144

(B) 141 (D) 188

12

18

30

16

6

(A) 18 (C) 9

5

12

_ A D A C B _ _ B D C C 2 4 _ _2 3 5 3 _ _ _ _ p _ _ q _ _ r s ? ? ? ? (A) p r s s (B) p s r r (C) r p s s (D) s r p p

36

18 27

?

(B) 12 (D) 6

6

6

21

7

4

5

(A) 14 (C) 32

A _ B A C _ D _ B C D C _ 4 _ 3 _ 2 _ 5 ? ? ? ? d c _ _ b a c b _ _ _ _ (A) 2 4 5 4 (B) 2 5 4 5 (C) 3 4 5 4 (D) 4 5 2 5

32 40

8

4

7. 26.

(B) 92 (D) 102

4.

Directions : (24 to 26) In each of the following questions, three sequences of letter/numbers are given which 5. correspond to each other in some way. In each question, you have to find out the letter/numerals that come in the vacant places marked by (?). These are given as one of the four alternatives under the question. Mark your answer as instructed. 24.

(A) 112 (C) 82

?

8

10 (B) 22 (D) 320

5

9

8

5

15

?

3

5

6

(A) 12 (C) 16

(B) 11 (D) 26

(A) 72 (C) 9

(B) 18 (D) 19

(A) 1 (C) 90

(B) 18 (D) 225

(A) 20 (C) 24

(B) 22 (D) 12

EXERCISE-4 8. Directions : (1 to 39) Find the missing term in the given figures

1. 9. (A) 36 (C) 25

(B) 9 (D) 64

2.

10. (A) 14 (C) 11

(B) 18 (D) 13

PAGE # 74

11.

7 12 15

11 49 8 54 4 ?

(A) 36 (C) 25

12.

18 24 32 12 14 16 3 ? 4 72 112 128

(A) 28 (C) 81

(B) 3 (D) 5

2 4

10

3

5

27 80

39 45

29

33

42

20.

43

43

(A) 69 (C) 50

30 70

59

31 44

35

43

48

15

56

(A) 1 (C) 3

(B) 142 (D) 198

7 3 4 74

12

3

6

4

1

140 4 10 7

3 (B) 14 (D) 22

4 9

3 24

9 ?

(A) 117 (C) 32

(B) 36 (D) 26

(A) 26 (C) 27

(B) 25 (D) 30

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?

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(A) 48 (C) 44

23.

17.

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8 11

20

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5 20

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1 3 5 35

80

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38

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40

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101

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2

3

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4 54 2

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5

(A) 78 (C) 94

51

6 3

?

5

4 (B) 82 (D) 86 PAGE # 75

48

2 25.

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3

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34.

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(A) 14 (C) 11

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6

HCA - 138 FED - 456 E?H - 87?

BIG - 792

36 49 26 64 25

9 81 21 16

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184

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10 20 39.

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4

9

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(A) 60 (C) 21 40.

(B) 50 (D) 25

Find the value of X in the following figure :

15

32.

4 33

(A) 12 (C) 14 33

27

(B) 9 (D) 10

(B) JNS (D) KRS

2 36

8 32

Find the missing letters from left to right.

(A) JSN (C) JRS

2

(A) 3 (C) 8

X

18

9

22

11

12

3

(B) 4 (D) 12 PAGE # 76

PUZZLE TEST Directions : (1 to 5) Read the following information carefully and answer the questions given below it. (i). Five professors (Dr. Joshi, Dr. Davar, Dr. Natrajan, Dr. Choudhary and Dr. Zia) teach five different subjects (zoology, physics, botany, geology and history) in four universities ( Delhi, Gujarat, Mumbai, and Osmania). Do not assume any specific order. (ii). Dr. Choudhary teaches zoology in Mumbai University . (iii). Dr. Natrajan is neither in Osmania University nor in Delhi University and he teaches neither geology nor history. (iv). Dr. Zia teaches physics but neither in Mumbai University nor in Osmania University. (v). Dr. Joshi teaches history in Delhi University. (vi). Two professors are from Gujarat University. (vii). One professor teaches only one subject and in one University only. Ex 1.

Ex 2.

Ex 3.

Ex 4.

Ex 5.

Sol. :

Who teaches geology ? (A) Dr Natrajan (C) Dr. Davar

(B) Dr. Zia (D) Dr. Joshi

Which university is Dr. Zia from ? (A) Gujarat (B) Mumbai (C) Delhi (D) Osmania Who teaches botany ? (A) Dr. Zia (C) Dr. Joshi

(B) Dr. Davar (D) Dr. Natrajan

Names Dr. Joshi Dr. Davar Dr. Natrajan Dr. Choudhary Dr. Zia

University Delhi Osmania Gujarat Mumbai

Subject History Geology Botany Zoology

Gujarat

Physics

On the basis of the above table, rest of the questions can be solved very easily. 1.

(C) Dr. Davar teaches geology.

2.

(A) Dr. Zia is from Gujarat university.

3.

(D) Dr. Natrajan teaches botany.

4.

(B) Dr. Davar is from Osmania University.

5.

(D) Dr. Natranjan - Gujarat University is the correct combination.

Ex 6.

Ramesh is taller than Vinay who is not as tall as Karan. Sanjay is taller than Anupam but shorter than Vinay. Who among them is the tallest ? (A) Ramesh (B) Karan (C) Vinay (D) Cannot be determined (D) In this question ranking of Karan is not defined. Consequently, either Ram or Karan occupies the top position with regard to height. Hence, option (d) is the correct choice.

Sol.

Directions : (7 to 11) Read the following information carefully and answer the questions given below it : There are five men A, B, C, D and E and six women P, Q, R, S, T and U. A, B and R are advocates; C, D, Which of the following combinations is correct ? P, Q and S are doctors and the rest are teachers. (A) Delhi University - Dr. Zia Some teams are to be selected from amongst (B) Dr. Choudhary - geology these eleven persons subject to the following (C) Dr. Davar - Mumbai University conditions : (D) Dr. Natranjan - Gujarat University A, P and U have to be together. (1 to 5) B cannot go with D or R. From the given information in the question : E and Q have to be together. From II, we get Dr. Choudhary teaches zoology in C and T have to be together. Mumbai University. D and P cannot go together. From III, We get Dr. Natrajan is neither in Osmania C cannot go with Q. nor in Delhi University. Therefore, he will be either If the team is to consist of two male advocates, two at Mumbai or Gujarat University. Similarly, as he Ex 7. lady doctors and one teacher, the members of the teaches neither geology nor history, therefore, he team are must be teaching physics or botany. ..........(1) (A) A B P Q U (B) A B P U S From IV, Dr. Zia  Physics but as he is not teaching (C) A P R S U (D) B E Q R S in either Mumbai or Osmania University, he must (B) The male advocates are A and B, lady doctors be teaching either in Delhi or Gujarat University...(2) Sol. are P, Q and S ; teachers are E, T and U. Form V, we get Dr Joshi teaches history in Delhi Now, A and B will be selected. University Form (1) and (2), we conclude that Dr Natarajan teaches botany. And from (1), (2) and VI, A, P and U have to be together. Now, we have to we get both Natarajan and Zia teach in Gujarat select one lady doctor more. It can be Q or S. But Q University. Finally, On summarisation we can and E have to be together. Since E is not selected, prepare the following table. so S will be selected. Thus, the team is A B P U S. Who is from Osmania University ? (A) Dr. Natrajan (B) Dr. Davar (C) Dr. Joshi (D) Dr. Zia

PAGE # 77

Ex 8.

Sol.

Ex 9.

If the team is to consist of one advocate, two Directions : (12 to 15) Read the following paragraph carefully : doctors, three teachers and C may not go with T, Four women A, B, C and D and three men E, F and the members of the team are : G play bridge, a game for four players. (A) A E P Q S U (B) A E P Q T U (i) The group consists of three married couples and a widow. (C) B E Q S T U (D) E Q R S T U (ii) Spouses are never partners in a game. (B) The advocates are A, B and R ; doctors are (iii) No more than one married couple ever plays in C, D, P, Q, S ; teachers are E, T and U. The team the same game. (iv) One day they played four games as follows. consists of 3 teachers i.e. E, T, U. Now, A, P and U A and E versus B and F. have to be together. E and Q have to be together. A and G versus D and F. Thus, the team is A E P Q T U. B and C versus F and G. C and E versus D and G. If the team is to consist of one male advocate, one male doctor, one lady doctor and two teachers, the Ex 12. members of the team are :

Sol.

(A) A C P T U

(B) A D E P T

(C) A D E P U

(D) B C E Q U

Ex 13.

(A) The male advocates are A and B ; male doctors are C and D ; lady doctors are P, Q and S ; teachers are E, T and U. If A is selected, P and U will be Ex 14. selected. D and P cannot go together. So, a male doctor C will be selected. C and T have to be together. Thus, the team is A C P T U. If B is Ex 15. selected, D will not be selected. So, male doctor C

Whom is E married to ? (A) A (C) C

(B) B (D) D

Whom is F married to ? (A) A (C) C

(B) B (D) D

Whom is G married to ? (A) A (C) C

(B) B (D) D

Which of the following is a widow ? (A) A (B) B (C) C (D) D

will be chosen. C and T have to be together. Now, the second teacher to be selected is E or U. But, U cannot go without A. So, E will be selected. E and Q have to be together. Thus, the team can also be B C E Q T. Ex 10.

If the team is to consist of one advocate, three doctors and one male teacher, the members of

Sol. : (12 to 15) From (iv), is married either to A or to C. If F is married to A, then G is married to B or to C. If G is married to B, then E is married to D ; if G is married to C, then E is married to B or to D. If F is married to C, then G is married to B ; then E is married to D. Hence, the married couples are : FA, GB, ED or FA, GC, EB or FA, GC, ED or FC, GB, ED. Of these, only FA, GB, ED does not contradict any of the statements.

the team are:

Sol.

(A) A D P S U

(B) C D R S T

(C) D E Q R S

(D) D E Q R T

12.

(D) E is married to D.

13.

(A) F is married to A.

(C) The advocates are A, B and R ; the doctors are 14. C, D, P, Q and S ; male teacher is E. Clearly, E will 15. be selected. E and Q have to be together. C and Q cannot be together. So, C will not be selected. P Ex 16. also cannot be selected because U is not selected. So, two other doctors D and S will be selected. P is not selected, so A will not be selected. D is selected, so B cannot be selected. Thus, the team is D E Q R S.

Ex 11.

If the team is to consist of two advocates, two Sol. doctors, two teachers and not more than three ladies, the members of the team are :

Sol.

(A) A B C P T U

(B) A C P R T U

(C) A E P Q R T

(D) B C E Q R T

(B) G is married to B. (C) C is a widow. A vagabond runs out of cigarettes. He searches for the stubs, having learnt that 7 stubs can make a new cigarette, good enough to be smoked, he gathers 49 stubs, If he smokes 1 cigarette every three - quarters of an hour, how long will his supply last ? (A) 5.25 hr (B) 6 hr (C) 4.5 hr (D) 3 hr (B) He has got =

49  7 cigarettes. 7

 The duration of time he will take to smoke these 3 7 cigarettes = 7  hr = 5.25 hr (i.e. 5 hr and 15 4

each of these combinations consists of four ladies.

min). Now note that after he has smoked these 7 cigarettes, he will collect 7 more stubs (one form each), form which he will be able to make another

B C E Q R T is incorrect because B and R cannot

cigarette. This will take him another

(A) A C P R T U and A E P Q R T are wrong because

go together.

3 hr (45 min) 4

to smoke. Therefore, total time taken = 6hr. PAGE # 78

Directions : (17 to 18) Read the following information and 1. answer the questions that follow. There are 70 clerks working with M/s. Jha Lal Khanna & Co. chartered accountants, of which 30 are female. 2. (i) 30 clerks are married. (ii) 24 clerks are above 25 years of age (iii) 19 Married clerks are above 25 years of age; among them 7 are males. 3. (iv) 12 males are above 25 years of age (v) 15 males are married. Ex 17.

Ex 18.

Sol.

17. 18.

How many unmarried girls are there ? (A) 12 (B) 15 (C) 18 (D) 10

4.

Who stays in locality Q ? (A) A (C) C

(B) B (D) E

What is E’s occupation ? (A) Business (B) Engineer (C) Lawyer (D) Doctor Agewise who among the following lies between A and C ? (A) Lawyer (B) Doctor (C) Cloth merchant (D) Engineer What is B’s occupation ? (A) Business (B) Engineer (C) Lawyer (D) Doctor

How many of these unmarried girls are above 25 ? (A) 12 (B) 15 (C) 4 (D) 0 5. What is C’s occupation ? (17 to 18) : From the given data, we can make the (A) Doctor (B) Lawyer following table with the help of which rest of the (C) Engineer (D) Business questions can be solved very easily. Directions : (6 to 10) Study the given information carefully Male (40) Female (30) and answer the questions that follow. Above 25 There are four people sitting in a row : one each from India, Japan, USA and Germany, but not in Married 7 12 that order, Unmarried 5 0 . They are wearing caps of different colours - green, Below 25 yellow, red and white, not necessarily in that order. married 8 3 II. One is wearing a kurta and one a T-shirt. III. The Indian is wearing a green cap and a jacket. unmarried 20 15 IV. The American is not seated at either end. 40 30 Total V. The persons with kurta and T-shirt are sitting next to each other. There are 15 unmarried girls. VI. The persons with kurta wears a red cap and sits next to the Japanese. In these 15 unmarried girls no one is above 25. VII. The Japanese wears a shirt and is not seated at either end. VIII. The man with white cap wears T-shirt and is EXERCISE seated at one end.

Directions : (1 to 5) Study the following information carefully and answer the questions given below it : There are five friends A, B, C, D and E. Two of them are businessmen while the other three belong to different occupations viz. medical, engineer and legal. One businessman and the lawyer stay in the same locality S, while the other three stay in three different localities P, Q and R. Two of these five persons are Hindus while the remaining three come from three different communities viz. Muslim, Christian and Shikh. The lawyer is the oldest in age while one of the businessmen who runs a factory is the youngest. The other businessman is a cloth merchant and agewise lies between the doctor and the lawyer. D is a cloth merchant and stays in locality S while E is a Muslim and stays in locality R. The doctor is a Christian and stays in locality P, B is a Shikh while A is a Hindu and runs a factory.

6.

Who wears the T-shirt ? (A) Indian (C) American

(B) Japanese (D) German

7.

Who is wearing a kurta ? (A) Indian (B) Japanese (C) American (D) German

8.

What is the colour of the cap worn by the Japanese? (A) Red (B) Green (C) Yellow (D) White

9.

Who precedes the man wearing T-shirt ? (A) Indian (B) Japanese (C) American (D) German

10.

Who precedes the man wearing jacket ? (A) Indian (B) German (C) Japanese (D) Cannot say

PAGE # 79

Directions : (11 to 15) Read the following information Directions : (19 to 23) Read the information given below carefully and answer the questions that follow. and answer the questions. I. There are six students ( A, B, C, D, E and F) in a The age and height of six children in a class are as group. Each student can opt for only three choices follows : out of the six which are music, reading, painting, (i) A is taller and older than B but shorter and badminton, cricket and tennis. younger than C. II. A, C and F like reading. (ii) D is taller than E who is not as tall as B. III. D does not like badminton, but likes music. (iii) The oldest is the shortest. IV. Both B and E like painting and music. (iv) The youngest would be fourth if the children V. A and D do not like painting, but they like cricket. stood in a line according to their height and one VI. All student except one like badminton. started counting from the tallest. VII. Two students like tennis. (v) D is younger than F but older than E who is VIII. F does not like cricket, music and tennis. older than C. 11.

12.

Which pair of students has the same combination 19. of choices ? (A) A and C (B) C and D (C) B and E (D) D and F 20. Who among the following students likes both tennis and cricket ? (A) A and B (B) C 21. (C) B and D (D) D

13.

How many students like painting and badminton ? (A) 1 (B) 2 (C) 3 (D) 4

14.

Who among the following do not like music ? (A) A , C and D (B) A, B and C (C) A, C and F (D) B, D and F

15.

Which of the following is the most popular choice? (A) Tennis (B) Badminton 23. (C) Reading (D) Painting

22.

16.

Who among them is the tallest ? (A) B (B) E (C) C (D) Data inadequate Who is older than B but younger than C ? (A) F (B) D (C) A (D) Data inadequate Which of the following statements is definitely true? (A) D is the most old person (B) B has the max. height (C) A is older than D (D) F is the shortest Which of the following is the correct order of height in descending order? (A) A, C, D, B, E, F (B) F, D, E, C, A, B (C) D, C, A, B, E, F (D) C, D, A, B, E, F W hose Rank in height cannot be positioned definitely ? (A) B (B) D (C) C (D) E

R earns more than H but not as much as T, M earns more than R. Who earns least among them? Directions : (24 to 28) Study the information given below (A) R (B) T and answer the questions that follow. (C) H (D) M (i) Six Plays P, Q, R, S, T and U are to be organised from Monday to Saturday i.e. 10 to 15 one play each 17. Harish is taller than Manish but shorter than day. Suresh. Manish is shorter than Anil but taller than (ii) There are two plays between R and S and one Raghu. Who among them is the shortest having play between P and R. regard to height ? (iii) There is one play between U and T and T is to (A) Anil (B) Manish be organised before U. (C) Raghu (D) Cannot be determined (iv) Q is to be organised before P, not necessarily Direction : (18) Examine the following statements : immediately. I. Either A and B are of the same age or A is older (v) The organisation does not start with Q. than B. The organisation would start from which play ? II. Either C and D are of the same age or D is older 24. (A) P (B) S than C. (C) T (D) None III. B is older than C. 18.

Which one of the following conclusions can be 25. drawn from the above statements ? (A) A is older than B (B) B and D are of the same age 26. (C) D is older than C (D) A is older than C

On which date is play T to be organised ? (A) 10th (B) 11th th (C) 12 (D) None The organisation would end with which play ? (A) P (B) Q (C) S (D) None

PAGE # 80

27.

Which day is play Q organised ? (A) Tuesday (B) Wednesday (C) Thursday (D) None

28.

Which of the following is the correct sequence of organising plays ? 32. (A) PTRUQS (B) QSTURP (C) SUTRQP (D) None

31.

If it is sure that Henna will go to the fair, then who among the following will definitely go ? (A) Rama (B) Shamma (C) Reena (D) Rama and Reena If Tina does not go to the fair, which of the following statements must be true ? (i) Henna cannot go (ii) Shamma cannot go (iii) Reena cannot go (iv) Rama cannot go (A) (i) and (ii) (B) (iii) and (iv) (C) (i), (iii) and (iv) (D) (i) and (iv)

Directions : (29 to 30) Read the following information carefully and answer the questions given below it. I. Seven books are placed one above the other in a particular way . II. The history book is placed directly above the civics book. Directions : (33 to 37) Read the following paragraph III. The geography book is fourth from the bottom carefully and choose the correct alternative. and the English book is fifth from the top. The office staff of XYZ corporation presently IV. There are two books in between the civics and consists of three females A, B, C and five males D, economics books. E, F, G and H. The management is planning to open a new office in another city using three males 29. To find the number of books between the civics and two females of the present staff. To do so they and the science books, which other extra piece of plan to separate certain individuals who do not information is required, from the following ? function well together. The following guidelines (A) There are two books between the geography were established and the science books. I. Females A and C are not to be together (B) There are two books between the mathematics II. C and E should be separated and the geography books . III. D and G should be separated (C) There is one book between the English and IV. D and F should not be part of a team. the science books. (D) The civics book is placed before two books 33. above the economics book. 30.

To know which three books are kept above the English book, which of the following additional pieces of information, if any, is required? (A) The economics book is between the English 34. and the science books. (B) There are two books between the English and the history books. (C) The geography book is above the English book. 35. (D) No other information is required.

Directions : (31 to 32) A five-member team that includes Rama, Shamma, Henna, Reena, and Tina, is planning to go to a science fair but each of them 36. put up certain conditions for going .They are as follows. I. If Rama goes, then at least one amongst Shamma and Henna must go. II. If Shamma goes, then Reena will not go. 37. III. If Henna will go, then Tina must go. IV. If Reena goes, then - Henna must go. V. If Tina goes, then Rama must go but Shamma cannot go. VI. If Reena plans not to go the fair, then Rama will also not go.

If A is chosen to be moved, which of the following cannot be a team ? (A) ABDEH (B) ABDGH (C) ABEFH (D) ABEGH If C and F are to be moved to the new office, how many combinations are possible ? (A) 1 (B) 2 (C) 3 (D) 4 If C is chosen to the new office, which number of the staff cannot be chosen to go with C ? (A) B (B) D (C) F (D) G Under the guidelines, which of the following must be chosen to go to the new office ? (A) B (B) D (C) E (D) G If D goes to the new office, which of the following is/are true ? I. C cannot be chosen II. A cannot be chosen III. H must be chosen. (B) II only (A) I only (C) I and II only (D) I and III only

PAGE # 81

Directions : (38 to 42) Study the following information Directions : (45 to 49) Read the following information carefully and answer the questions given below. carefully and answer the questions that follow : (i) There is a family of six persons- L, M, N, O, P A team of five is to be selected from amongst five and Q. They are professor, businessman, boys A, B, C, D and E and four girls P, Q, R and S. chartered account, bank manager, engineer and Some criteria for selection are : medical representative, not necessarily in that A and S have to be together order. (ii) There are two married couples in the family. P cannot be put with R. (iii) O, the bank manager is married to the lady D and Q cannot go together. professor. C and E have to be together. (iv) Q, the medical representative, is the son of M R cannot be put with B. and brother of P. Unless otherwise stated, these criteria are (v) N, the chartered accountant, is the daughter - in applicable to all the questions below : law of L. (vi) The businessman is married to the chartered 38. If two of the members have to be boys, the team acconuntant. will consist of : (vii) P is an unmarried engineer. (viii) L is the grandmother of Q (A) A B S P Q (B) A D S Q R (C) B D S R Q 39.

40.

(D) C E S P Q

45.

How is P related to Q. (A) Brother (B) Sister (C) Cousin (D) Either brother or sister

46.

Which of the following is the profession of M ? (A) Professor (B) Chartered accountant (C) Businessman (D) Medical representative

If R be one of the members, the other members of the team are : (A) P S A D

(B) Q S A D

(C) Q S C E

(D) S A C E

If two of the members are girls and D is one of the members, the members of the team other than D are : 47. (A) P Q B C (B) P Q C E (C) P S A B (D) P S C E

41.

If A and C are members, the other members of the 48. team cannot be : (A) B E S (C) E S P

42.

(B) D E S (D) P Q E

If including P at least three members are girls, the members of the team other than P are : (A) Q S A B (C) Q S C E

(B) Q S B D (D) R S A D

Directions : (43 to 44) Read the given information carefully and answer the questions that follow : Ratan, Anil, Pinku and Gaurav are brothers of Rakhi, Sangeeta, Pooja and Saroj, not necessarily in that order. Each boy has one sister and the names of bothers and sisters do not begin with the same

49.

44.

Pooja’s brother is (A) Ratan

(B) Anil

(C) Pinku

(D) Gaurav

Which of the following are brother and sister ? (A) Ratan and Pooja (C) Pinku and Sangeeta

(B) Anil and Saroj (D) Gaurav and Rakhi

Which of the following is one of the couples ? (A) QO (B) OM (C) PL (D) None of these How is O related to Q? (A) Father (C) Uncle

(B) Grandfather (D) Brother

Directions : (50 to 54) I. There is a group of six persons P,Q, R, S, T and U from a family. They are Psychologist, Manager, Lawyer, Jeweller, Doctor and Engineer. II. The Doctor is grandfather of U, who is a Psychologist. III. The Manager S is married to P. IV. R, the Jeweller is married to the Lawyer. V. Q is the mother of U and T. VI. There are two married couples in the family.

letter. Pinku and Gaurav are not Saroj’s or 50. Sangeeta’s brothers. Saroj is not Ratan’s sister. 43.

Which of the following is the profession of L ? (A) Professor (B) Charted accountant (C) Businessman (D) Engineer

51.

52.

What is the profession of T ? (A) Doctor (B) Jeweller (C) Manager (D) None of these How is P related to T ? (A) Brother (C) Father

(B) Uncle (D) Grandfather

How many male members are their in the family ? (A) One (B) Three (C) Four (D) Data inadequate

PAGE # 82

53.

What is the profession of P ? (A) Doctor (B) Lawyer (C) Jeweller (D) Manager

54.

Which of the following is one of the pairs of couples in the family ? (A) PQ (B) PR (C) PS (D) Cannot be determined

Direction : (55) The ages of Mandar, Shivku, Pawan and 56. Chandra are 32, 21, 35 and 29 years, not in order, Whenever asked they lie of their own age but tell the truth abut others. (i) Pawan says, “My age is 32 and Mandar’s age is not 35” (ii) Shivku says, “My age is not 2 9 and Pawan’s 57. age in not 21” (iii) Mandar says, “My age is 32.” 55.

What is Chandra’s age ? (A) 32 years (B) 35 years (C) 29 years (D) 21 years

Directions : (56 to 57) Answer the questions on the basis of the information given below. 5 friends Nitin, Reema, Jai, Deepti and Ashutosh are playing a 58. game of crossing the roads. In the beginning, Nitin, Reema and Ashutosh are on the one side of the road and Deepti and Jai are on the other side. At the end of the game, it was found that Reema and Deepti are on the one side and Nitin, Jai and Ashutosh are on the other side of the road. Rules of the game are as follows : I. One “movement” means only one person crosses the road from any side to the other side. II. No two persons can cross the road simultaneously from any side to the other side.

III. Two persons from the same side of the roads cannot move in consecutive “movements”. IV. If one person crosses the road in a particular movement, he or she cannot immediately move back to the other side. V. Jai and Reema did not take part in first 3 movements. W hat is the minimum possible number of movements that took place in the entire game ? (A) 3 (B) 4 (C) 5 (D) 6 If number of movements are minimised in the game, then which of the following combination of friends can never be together on one particular side of the road during the course of the game ? (A) Nitin, Reema amd Deepti (B) Nitin, Jai and Deepti (C) Deepti, Jai and Ashutosh (D) Ashutosh, Nitin and Deepti You have 12 similar looking coins. 11 of them weigh the same. One of them has a different weight, but you don’t know whether it is heavier or lighter. You also have a scale. You can put coins on both sides of the scale and it’ll tell you which side is heavier or will stay in the middle if both sides weigh the same. What is the minimum number of weighing required to find out the odd coin. (A) 3 (B) 4 (C) 5 (D) 6



PAGE # 83

CALENDAR AND CLOCK TEST Similarly, 200 years = 10 odd days = 03 odd days



300 years =

= 1 odd day.. 7 20  1 400 years = = 0 odd day (1 is added as 400 7 is a leap year) Similarly, 800, 1200, 1600, 2000, 2400 years contain 0 odd days.

We are to find the day of the week on a mentioned date. Certain concepts are defined as under. 

An ordinary year has 365 days.



In an ordinary year, first and last day of the year are same.



15

After counting the odd days, we find the day according to the number of odd days.

 A leap year has 366 days. Every year which is divisible by 4 is called a leap year. For example  1200, 1600, 1992, 2004, etc. are all leap years.

Sunday for 0 odd day, Monday for 1 odd day and so on as shown in the following table.



For a leap year, if first day is Monday than last day will be Tuesday for the same year.



In a leap year, February is of 29 days but in an ordinary year, it has only 28 days.



Year ending in 00's but not divisiable by 400 is not considered a leap year. e.g., 900, 1000, 1100, 1300, 1400, 1500, 1700, 1800, 1900, 2100 are not leap years.

Ordinary Year

Days

The day on which calendar started (or the very first day ) i.e., 1 Jan, 0001 was Monday.

January

31

3

February

28

0

March

31

3

April

30

2

May

31

3

Table : 1 (Odd days for week days) Days

Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Odd Days





Calendar year is from 1 Jan to 31 Dec. Financial year is from 1 April to 31 March.

ODD DAYS The no. of days exceeding the complete no. of weeks in a duration is the no. of odd days during that duration.

COUNTING OF ODD DAYS 

Every ordinary year has 365 days = 52 weeks +1 day.  Ordinary year has 1 odd day.



Every leap year 366 days = 52 weeks + 2 days.  Leap year has 2 odd days.

0

1

2

3

4

5

6

Table : 2 (Odd days for months in a year) Odd Leap year Days

Days

Odd Days

January

31

3

February

29

1

March

31

3

April

30

2

May

31

3

June

30

2

June

30

2

Total

181 days

6

Total

182 days

0

July

31

3

July

31

3

August

31

3

August

31

3

September

30

2

September

30

2

October

31

3

October

31

3

November

30

2

November

30

2

December

31

3

December

31

3

Total

184 days

1

Total

184 days

2

Table : 3 (Odd days for every quarter)





Odd days of 100 years = 5, Odd days of 200 years = 3, Odd days of 300 years = 1, Odd days of 400 years = 0.

M onths of years

I st three m onths 1 Jan to 31 M arch

Total days

90 / 91 Ord. / Leap

Odd days

6 /0 0 Ord. / Leap O dd day

Explanation : 100 years = 76 ordinary years + 24 leap years Ex 1. ( The year 100 is not a leap year) = 76 odd days + 2 × 24 odd days = 124 odd days. Sol. Odd days =

124 7

= 5 odd days.

IInd three IIIrd three m onths m onths 1 Apr to 1 July to 30 June 30 Sep. 91

92 1 Odd day

Iv th three Total year m onths 1 Jan to 1 Oct. to 31 Dec. 31 Dec. 92

365 / 366 Ord. / Leap

1 1 /2 Odd day Ord. / Leap

If it was Saturday on 17th December 1982 what will be the day on 22nd December 1984 ? Total number of odd days between 17 Dec.1982 to 17 Dec.1984 the number of odd days = 1+2 = 3. From 17 to 22 Dec. number of odd days = 5  3 + 5 = 8 odd days = 1 odd day.  Saturday + 1 odd day = Sunday. PAGE # 84

Ex 2. Sol.

Ans. Ex 3. Sol.

Ans. Ex 4. Sol.

Ans.

Ex 5. Sol.

Find the day of the week on 16 January, 1969. Ex 6. 1600 years have ‘0’ odd day. .....................(A) 300 years have ‘1’ odd day. ......................(B) Sol. 68 years have 17 leap years and 51 ordinary years. Thus = (17 × 2 + 51 × 1 ) = 85 odd days  ' 01' odd day ...(C) 16 January has = ' 02' odd days..(D) Adding (A) + (B) +(C) +(D), We get, 0 + 01 +01 +02 = 04 odd days Thursday

Ex 7.

Find the day of the week on 18 July, 1776 (leap Sol. year). Here 1600 years have ‘0’ odd day.....................(A) 100 years have ‘5’ odd days..............................(B) 75 years = (18 leap years + 57 ordinary years) = (18 × 2 + 57 × 1) = 93 odd days = (7 × 13 + 2) = ‘2’ odd days.............................(C) Now, the no. of days from 1st January to 18 July, 1776 = 182 + 18 = 200 days = (28 × 7 + 4) days = ‘4’ odd days.....................(D) Adding (A) + (B) +(C) +(D), We get, 0 + 5 + 2 + 4 = 04 odd days Thursday On what dates of October, 1975 did Tuesday fall ? For determining the dates, we find the day on 1st Oct, 1975. 1600 years have ‘0’ odd days.....................(A). 300 years have ‘01’ odd days.....................(B). 74 years have (18 leap years + 56 ordinary years) 2 × 18 + 1 × 56 = 92 odd days = ‘01’ odd days.............(C) Days from 1st January to 1st Oct., 1975 1st Jan – 30 June + 1st July to 1st Oct. 181 + 31 + 31 + 30 + 1 = 274 days = ‘01’ odd days......(D) (274/7= 01 days) Adding (A) + (B) +(C) +(D) = 0 + 01 +01 +01 = '03' odd days Wednesday( 1st Oct), hence 7,14,21,28 Oct. will Tuesday fall.

The year next to 1996 having the same Calendar will be 1996 1997 1998 1999 2000 2001 2002 2003 2 1 1 1 2 Total = 2 + 1 + 1 + 1 + 2 = 7= 0 odd days Hence, year 2001 will have the same calendar as year 1996. Prove that last day of a century cannot be Tuesday, Thursday or Saturday. 100 years have = 5 odd days  Last day of st century is Friday 200 years have = 10 odd days  Last day of IInd century is Wednesday = 3 odd days 300 years have = 15 odd days  Last day of rd century is Monday = 01 odd day 400 years have = (5 × 4 + 1) Last day of 4th century is Sunday = 21 odd days = 0 odd days Since the order keeps on cycling, we see that the last day of the century cannot be Tuesday, Thursday or Saturday.

Important Notes : 

Minute hand and hour hand coincides once in every hour. They coincide 11 times in 12 hours and 22 times in 24 hours.



They coincide only one time between 11 to 1 O’ clock. at 12 O’ clock.



Minute hand and hour hand are opposite once in

 Calendar for 1995 will serve for 2006, prove ? The Calendar for 1995 and 2006 will be the same ,if day on 1st January of both the years is the same.  This is possible only if the total odd days between 31st Dec. 1994 and 31st Dec.2005 is 0. [one day before both the years as we want to know the day on 1st January of both the years i.e. same]  During this period, we have 3 leap years (1996, 2000, 2004) and 08 ordinary years  (1995,1997,1998,1999, 2001, 2002, 2003,2005) Total odd days = (2 × 3 + 1 × 8) = 14 = 0 odd days (Thus Proved)

every hour. They do it 11 times in 12 hours and 22 times in 24 hours. They opposite only one time between 5 to 7 O’ clock. at 6 O’ clock. Both hands (minute and hour) are perpendicular twice in every hour. 22 times in 12 hours and 44 times in 24 hours. In one minute, hour hand moves 1/2º and minute hand moves 6º. In one hour, hour hand moves 30º and minute hand moves 360º. In an hour, minute hand moves 55 minutes ahead of hour hand.

PAGE # 85

Case-II When the time taken (20 + 15) = 35 min.

HANDS COINCIDE Ex.8 Sol.

 Minute hand is 55 min. ahead of hour hand in

At what time between 3 O’Clock and 4 O’Clock will the two hands coincide ? At 3 O’clock the distance between the two hands is 15 minutes when they coincide with each other the distance between the two hands will be 0 min. So, the time taken (15 + 0 ) = 15 minutes.  Minute hand is 55 min. ahead of hour hand in 60 min.  Minute hand is 1 min. ahead of hour hand in 60 min. 55

60 min. Minute hand is 35 min. ahead of hour hand in 60  35 420 2 = = 38 min. 55 11 11 2 Hence, the right time is 38 min. past 4. 11

MIRROR IMAGE OF CLOCK



to find the mirror image, time is subtracted from 11 : 60.

 Minute hand is 15 min. ahead of hour hand in 60  15 180 4 = = 16 min. 55 11 11



4 Hence the right time is 16 minute past 3. 11

HANDS ARE OPPOSITE Ex.9 Sol.

60  40 480 7 = = 43 min. 55 11 11

Hence, the right time is 43

23 : 60. Ex.11

The time in the clock is 4 : 46, what is the mirror image ?

Sol.

(11 : 60) – (4 : 46) = 7 : 14.

7 min. past 2. 11



Case-I

At what time between 4 O’clock and 5 O’clock will the hands are perpendicular ? At 4 O’clock the distance between the two hands is 20 min. When they are at 15 minutes distance, they are perpendicular to each other. When the time taken (20 – 15) = 5 min.  Minute hand is 55 min. ahead of hour hand in 60 min. Minute hand is 5 min. ahead of hour hand in

Both types of angles are 360º in total. If one angle is known, other can be obtained by subtracting from 360º.

Ex.13

At 4 : 30, what is the angle formed between hour

Sol.

hand and minute hand ? At 4 O’ clock angle between hour and min. hand is

HANDS ARE PERPENDICULAR

Sol.

If the time is between 11 O’clock to 1 O’clock, then to find the mirror image, time is subtracted from

At what time between 2 O’clock and 3 O’clock will Ex.12 The time in the clock is 12 : 35, then find its mirror the two hands be opposite ? image. At 2 O’clock the distance between the two hands is Sol. (23 : 60) – (12 : 35) = 11 : 25. 10 minutes. When they are at 30 minutes distance, they are opposite to each other. The time taken TO FIND THE ANGLE BETWEEN TWO HANDS (30 + 10 ) = 40 min.  Minute hand is 55 min. ahead of hour hand in  Angle are of two types : 60 min. Positive angle : It is obtained by moving from hour  Minute hand is 1 min. ahead of hour hand in hand to minute hand moving in clockwise direction. 60 min. Negative angle : It is obtained by moving from 55 minute hand to hour hand.  Minute hand is 40 minutes ahead of hour hand in

Ex.10

If the time is between 1 O’clock to 11 O’clock, then

of 120º. In 30 min. minute hand make an angle of 180º. So, the resultant angle is 180º – 120º = 60º. But in 30 min. hour hand will also cover an angle of 15º. Hence, the final angle between both hands is 60º – 15º = 45º. Short trick

60  5 60 5 = = 5 min. 55 11 11 5 Hence, the right time is 5 min. past 4. 11

PAGE # 86

9.

Ex.14

Sol.

A bus for Delhi leaves every thirty minutes from a bus stand. An enquiry clerk told a passenger that the bus had already left ten minutes ago and the next bus will leave at 9.35 A.M. At what time did the enquiry clerk 10. give this information to the passenger ? Bus leaves after every 30 minutes. The next bus will leave at 9 : 35 A.M. The last bus left at 9 : 35 – 0 : 30 = 9 : 05 A.M. but clerk said that bus had left 10 minutes earlier. 11. 9 : 05 + 0 : 10 = 9 : 15 A.M.

EXERCISE 1.

Find the day of the week on 26 January, 1950. (A) Tuesday (B) Friday (C) Wednesday (D) Thursday

2.

W hich two months in a year have the same calendar ? (A) June, October (B) April, November 13. (C) April, July (D) October, December

3.

4.

5.

12.

Are the years 900 and 1000 leap years ? (A) Yes (B) No (C) Can't say (D) None of these If it was Saturday on 17th November, 1962 what will be the day on 22nd November, 1964 ? (A) Monday (B) Tuesday 14. (C) Wednesday (D) Sunday Sangeeta remembers that her father's birthday was certainly after eighth but before thirteenth of December. Her sister Natasha remembers that their father's birthday was definitely after ninth but before fourteenth of December. On which date of December was their father's birthday ? (A) 10th (B) 11th (C) 12th (D) Data inadequate 15.

6.

Find the day of the week on 15 August, 1947. (A) Tuesday (B) Friday (C) Wednesday (D) Thursday

7.

Karan was born on Saturday 22nd March 1982. On what day of the week was he 14 years 7 months and 8 days of age ? (A) Sunday (B) Tuesday (C) Wednesday (D) Monday

8.

If on 14th day after 5th March be Wednesday, what day of the week will fall on 10th Dec. of the same year ? (A) Friday (B) Wednesday (C) Thursday (D) Tuesday

16.

If the day before yesterday was Saturday, what day will fall on the day after tomorrow ? (A) Friday (B) Thursday (C) Wednesday (D) Tuesday If February 1, 1996 is Wednesday, what day is March 10, 1996 ? (A) Monday (B) Sunday (C) Saturday (D) Friday If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month ? (A) Sunday (B) Monday (C) Wednesday (D) Friday Mohini went to the movies nine days ago. She goes to the movies only on Thursday. What day of the week is today ? (A) Thursday (B) Saturday (C) Sunday (D) Tuesday At what time are the hands of a clock together between 5 and 6 ? (A) 33

3 min. past 5 11

(B) 28

3 min. past 5 11

(C) 27

3 min. past 5 11

(D) 26

3 min. past 5 11

At what time between 9 and 10 will the hands of a clock be in the straight line, but not together ? (A) 16 minutes past 9 4 (B) 16 minutes past 9 11 6 (C) 16 minutes past 9 11 9 (D) 16 minutes past 9 11 At what time between 5 & 5 : 30 will the hands of a clock be at right angle ? 10 (A) 10 minutes past 5 11 5 (B) 11 minutes past 5 11 10 (C) 9 minutes past 5 11 9 (D) 10 minutes past 5 11 Ajay left home for the bus stop 15 minutes earlier than usual. It takes 10 minutes to reach the stop. He reached the stop at 8.40 a.m. What time does he usually leave home for the bus stop ? (A) 8.30 a.m. (B) 8.45 a.m. (C) 8.55 a.m. (D) Data inadequate

PAGE # 87

17.

The priest told the devotee, "The temple bell is 26. rung at regular intervals of 45 minutes. The last bell was rung five minutes ago. The next bell is due to be rung at 7.45 a.m." At what time did the priest give this information to the devotee ? (A) 7.40 a.m. (B) 7.05 a.m. (C) 6.55 a.m. (D) None of these

18.

There are twenty people working in an office. The first group of five works between 8.00 A.M. and 2.00 P.M. The second group of ten works between 10.00 27. A.M. and 4.00 P.M. And the third group of five works between 12 noon and 6.00 P.M. There are three computers in the office which all the employees frequently use. During which of the following hours the computers are likely to be used most ? (A) 10.00 A.M. –– 12 noon (B) 12 noon –– 2.00 P.M. (C) 1.00 P.M. –– 3.00 P.M. (D) 2.00 P.M. –– 4.00 P.M. 28.

19.

A tired worker slept at 7.45 p.m.. If he rose at 12 noon, for how many hours did he sleep ? (A) 5 hours 15 min. (B) 16 hours 15 min. (C) 12 hours (D) 6 hours 45 min.

20.

How many times are the hands of a clocks perpendicular in a day ? (A) 42 (B) 48 (C) 44 (D) 46

21.

If a clock shows 04: 28 then its mirror image will be ? (A) 07: 42 (B) 07: 32 (C) 08: 32 (D) 08: 42

22.

A watch, which gains uniformly, is 3 minutes slow at noon on Monday and is 3 minutes 48 seconds fast at 2 p.m. on the following Monday. What time it was correct ? (A) 2 p.m. On Tuesday 30. (B) 2 p.m. On Wednesday (C) 3 p.m. On Thursday (D) 1 p.m. On Friday.

29.

23.

How many times are the hands of a clocks coincide in a day ? (A) 10 (B) 11 (C) 12 (D) 22

24.

At what time between 2 and 3 O’ clock the hands of a clock will make an angle of 160º ? (A) 20 minutes past 2 (B) 30 minutes past 2 (C) 40 minutes past 2 (D) 50 minutes past 2

25.

Ashish leaves his house at 20 minutes to seven in the morning, reaches Kunal’s house in 25 minutes, they finish their breakfast in another 15 minutes and leave for their office which takes another 35 32. minutes. At what time do they leave Kunal’s house to reach their office ? (A) 7.40 am (B) 7.20 am (C) 7.45 am (D) 8.15 am

31.

The train for Lucknow leaves every two and a half hours from New Delhi Railway Station. An announcement was made at the station that the train for Lucknow had left 40 minutes ago and the next train will leave at 18. 00 hrs. At what time was the announcement made ? (A) 15.30 hrs (B) 17.10 hrs (C) 16.00 hrs (D) None of these A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour. If he begins his ascent at 8.00 a.m., at what time will he first touch a flag at 120 feet from the ground ? (A) 4 p.m. (B) 5 p.m. (C) 6 p.m. (D) None of these If the two incorrect watches are set at 12 : 00 noon at correct time, when will both the watches show the correct time for the first time given that the first watch gains 1 min in 1 hour and second watch loses 4 min in 2 hours : (A) 6 pm, 25 days later (B) 12 : 00 noon, 30 days later (C) 12 noon, 15 days later (D) 6 am 45 days later Rajeev and Sanjeev are too close friends Rajeev's watch gains 1 minute in an hour and Sanjeev's watch loses 2 minutes in an hour. Once they set both the watches at 12 : 00 noon, with my correct watch. When will the two incorrect watches of Rajeev and Sanjeev show the same time together? (A) 8 days later (B) 10 days later (C) 6 days later (D) can't be determined At a railway station a 24 hour watch loses 3 minutes in 4 hours. If it is set correctly on Sunday noon when will the watch show the correct time ? (A) 6 pm after 40 days (B) 12 noon after 75 days (C) 12 pm after 100 days (D) 12 noon after 80 days A swiss watch is being shown in a museum which has a very peculiar property. It gains as much in the day as it loses during night between 8 pm to 8 am. In a week how many times will the clock show the correct time ? (A) 6 times (B) 14 times (C) 7 times (D) 8 times A wrist watch which is running 12 minutes late on a Sunday noon is 16 minutes ahead of the correct time at 12 noon on the next Sunday. When is the clock 8 minutes ahead of time ? (A) Thursday 10 am (B) Friday noon (C) Friday 8 pm (D) Tuesday noon PAGE # 88

33.

34.

35.

A clock loses 2 minutes in a hour and another clock 38. gains 2 minutes in every 2 hours. Both these clocks are set correctly at a certain time on Sunday and both the clocks stop simultaneously on the next day with the time shown being 9 am and 10 : 06 AM. What is the correct time at which they stopped? (A) 9 : 54 am (B) 9 : 44 pm (C) 9 : 46 am (D) 9 : 44 am David sets his watch at 6 : 10 am on Sunday, which gains 12 minutes in a day. On Wednesday if this watch is showing 2 : 50 pm. What is the correct time ? 39. (A) 1 : 50 pm (B) 2 : 10 pm (C) 2 : 30 pm (D) 3 : 30 pm Ramu purchased a second hand Swiss watch which is very costly. In this watch the minute-hand and hour hand coincide after every 65

36

37.

3 minutes. 11

Kumbhakarna starts sleeping between 1 pm and 2 pm and he wakes up when his watch shows such a time that the two hands (i.e., hour-hand and minute-hand) interchange the respective places. He wakes up between 2 pm and 3 PM on the same night. How long does he sleep ? (A) 55

5 min 13

(B) 110

10 min 13

(C) 54

6 min 13

(D) None of these

A clock loses 3% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time that the clock will show exactly 14 days from the time it was set right ? (A) 1 : 36 : 48 (B) 1 : 40 : 48 (C) 1 : 41 : 24 (D) 10 : 19 : 12

How much time does the watch lose or gain per day ? Direction : (40 to 41) A 12 dial clock has its minute hand (A) 4 min (B) 5 min defective. W henever it touches dial 12, it (C) 4 min, 20 sec (D) none of these immediately falls down to 6 instead of running My watch was 8 minutes behind at 8 pm on Sunday smoothly (the hour hand remains unaffected during but within a week at 8 pm on Wednesday it was 7 that fall). It was set right at 12 ‘O’ clock in the noon. minutes ahead of time. During this period at which time this watch has shown the correct time : 40. What was the actual time when the minute hand of (A) Tuesday 10 : 24 am the clock touched dial 9 for the 5th time? (B) Wednesday 9 : 16 pm (A) 2 : 15 (B) 3 : 00 (C) It cannot show the correct time during this period (C) 5 : 15 (D) 6 : 45 (D) None of the above Out of the following four choices which does not 41. show the coinciding of the hour hand and minutehand : (A) 3 : 16 : 2 (B) 6 : 32 : 43 (C) 9 : 59 : 05 (D) 5 : 27 : 16

If the actual time is 10 : 10, what is the position of the hour hand in that defective clock ? (A) Between 2 and 3 (B) Between 4 and 5 (C) Between 10 and 11 (D) Between 3 and 4



PAGE # 89

CUBE AND DICE-TEST CUBES A cube is three dimensional figure, having 8 corners, 6 surfaces and 12 edges. If a cube is painted on all of its surfaces with any colour and further divided into various smaller cubes, we get following results. Smaller cubes with three surfaces painted will be present on the corners of the big cube. Here 3

2

2 2

1 1

3

2

3 2 2 3

2 1 1 2

2 1

1 2

3

2 2

3

2 3 1 2 1 2 2 3

3

2 2 3

n=

2 1 1

side of big cube 4  4 side of small cube 1

3

2 1 1 2

2

Ex 1.

How many smaller cubes have three surfaces

2

painted ?

3

(A) 4

(B) 8

(C) 16

(D) 24

2

Sol.

(B) Number of smaller cubes with three surfaces

Smaller cubes with two surface painted will be present on the edges of the big cube. Smaller cubes with one surface painted will be present on Ex 2. the surfaces of the big cube. Smaller cubes with no surface painted will be present inside the big cube.

painted = 8

If a cube is painted on all of its surfaces with a Sol. colour and then divided into smaller cubes of equal size then after separation, number of smaller cubes so obtained will be calculated as under : Ex 3. Number of smaller cubes with three surfaces painted = 8 Number of smaller cubes with two surfaces painted = (n – 2) × 12 Number of smaller cubes with one surfaces Sol. painted = (n – 2)2 × 6 Number of smaller cubes with no surfaces painted = (n – 2)3 Ex 4. Where n = No of divisions on the surfaces of the bigger cube

(D) Number of smaller cubes with two surfaces

length of edge of big cube = length of edge of one smaller cube

(D) Number of smaller cubes with no surface

TYPE I

Sol.

How many smaller cubes have two surfaces painted ? (A) 4

(B) 8

(C) 16

(D) 24

painted = (n – 2) × 12 = (4 – 2) × 12 = 24 How many smaller cubes have only one surface painted ? (A) 8

(B) 16

(C) 24

(D) 32

(C) Number of smaller cubes with one surface painted = (n – 2)2 × 6 = (4 – 2)2 × 6 = 4 × 6 = 24 How many smaller cubes will have no side painted ? (A) 18

(B) 16

(C) 22

(D) 8

painted = (n – 2)3 = (4 – 2)3 = (2)3 = 8

TYPE II If a cube is painted on all of its surfaces with single colour and then divided into various smaller cubes of equal size.

If a cube is painted on all of its surfaces with different colours and then divided into various

smaller cubes of equal size. Directions : ( 1 to 4) A cube of side 4 cm is painted black on all of its surfaces and then divided into various smaller cubes of side 1 cm each. The smaller Directions : ( 5 to 7 ) A cube of side 4 cm is painted black on the pair of one opposite surfaces, blue on the pair cubes so obtained are separated.

444  64 Total cubes of obtained = 1 1 1

of another opposite surfaces and red on remaining pair of opposite surfaces. The cube is now divided into smaller cubes of equal side of 1 cm each.

PAGE # 90

Ex 9.

Sol.

Ex 5.

Sol.

Ex 6.

Sol.

How many smaller cubes have two surfaces painted ? (A) 4 (B) 8 (C) 16 (D) 24 (C) Number of smaller cubes with two surfaces painted = Number of cubes present at the corners + Numbers of cubes present at 4 edges = 8 + (n – 2) × 4 = 8 + 8 = 16

How many smaller cubes have three surfaces painted ? (A) 4 (B) 8 (C) 16 (D) 24 (B) Number of smaller cubes with three surfaces painted = 8 (These smaller cubes will have all three surfaces painted with different colour blue, black and red.) How many smaller cubes have two surfaces painted ? Ex 10. (A) 4 (B) 8 (C) 16 (D) 24 (D) Number of smaller cubes with two surfaces painted = 24. And out of this (a) Number of cubes with two surfaces painted Sol. with black and blue colour = 8. (b) Number of cubes with two surfaces painted with blue and red colour = 8. (c) Number of cubes with two surfaces painted with black and red color = 8. Ex 11.

Ex 7.

How many smaller cubes have only one surface painted ? (A) 8 (B) 16 (C) 24 (D) 32 (D) Number of smaller cubes with one surface painted = Number of cubes present at the 8 edges + number of cubes present at the four surfaces= (n – 2) × 8 + (n – 2)2 × 4 = 2 × 8 + 4 × 4 = 16 + 16 = 32 How many smaller cubes will have no side painted ? (A) 18 (B) 16 (C) 22 (D) 8 (B) Number of smaller cubes with no side painted = Number of cubes on the two unpainted surfaces + number of cubes present inside the cube. = (n – 2)2 × 2 + (n – 2)3 = 4 × 2 + (2)3 = 8 + 8 = 16.

How many smaller cubes have only one surface painted ? (A) 8 (B) 16 Sol. (C) 24 (D) 32 Sol. (C) Number of smaller cubes with one surface painted = 24. And out of this (a) Number of cubes with one surface painted with black colour =8. TYPE IV (b) Number of cubes with one surface painted with blue colour = 8. If a cube is painted on its surfaces in such a way (c) Number of cubes with one surface painted that one pair of adjacent surfaces is left unpainted. with red colour = 8. Directions : (12 to 15 )A cube of side 4 cm is painted red on TYPE III the pair of one adjacent surfaces, green on the pair of other adjacent surfaces and two adjacent If a cube is painted on its surfaces in such a way surfaces are left unpainted. Now the cube is divided that one pair of opposite surfaces is left unpainted. into 64 smaller cubes of side 1 cm each. Directions : ( 8 to 11 ) A cube of side 4 cm is painted red on the pair of one opposite surfaces, green on the pair of another opposite surfaces and one pair of opposite surfaces is left unpainted. Now the cube is divided into 64 smaller cubes of side 1 cm each. Ex 8.

Sol.

How many smaller cubes have three surfaces painted ? (A) 0 (B) 8 (C) 16 (D) 20 (A) Number of smaller cubes with three surfaces Ex 12. painted = 0 (Because each smaller cube at the corner is attached to a surface which is unpainted.)

How many smaller cubes have three surfaces painted ? (A) 2 (B) 4 (C) 8 (D) 6 PAGE # 91

Sol.

Ex 13.

Sol.

Ex 14.

Sol.

Ex 15.

(A) Number of smaller cubes with three surfaces Type-II painted = Number of smaller cubes at two corners =2 Ex 17. The figures given below show the two different positions of a dice. Which number will appear How many smaller cubes have two surfaces opposite to number 2 ?. painted ? (A) 4 (B) 8 (C) 16 (D) 14 (D) Number of smaller cubes with two surfaces painted = Number of smaller cubes at four corners (A) 3 (B) 4 + Number of smaller cubes at 5 edges. (C) 5 (D) 6 = 4 + (n – 2) × 5 = 4 + 2 × 5 Sol. (C) The above question, = 4 + 10 = 14 where only two positions of a dice are given, can easily How many smaller cubes have only one surface be solved with the painted ? following method. (A) 8 (B) 16 Step I. The dice, when unfolded, will appear as shown in (C) 24 (D) 30 the figure given on the right side. (D) Number of smaller cubes with one surface painted = Number of smaller cubes at four Step II. Write the common number to both the dice in the middle block. Since common number is 4, hence surfaces + Number of smaller cubes at 6 edges + number 4 will appear in the central block. Number of smaller cubes at two corners. = (n – 2)2 × 4 + (n – 2) × 6 + 2 Step III. Consider the figure (i) and write the first number in = 4 × 4 + 2 × 6 + 2 = 16 + 12 = 28 + 2 = 30 the anti-clockwise direction of number 4, (common number) in block I and second number How many smaller cubes will have no side painted in block II. Therefore, numbers 3 and 2 being the ? first and second number to 4 in anticlockwise directions respectively, will appear in block I & II (A) 18 (B) 16 respectively. (C) 22 (D) 8

Sol.

(A) Number of smaller cubes with no surfaces Step IV. Consider figure (ii) and wire first and second number in the anticlock-wise direction to number painted = Number of smaller cubes from inside 4, (common number) in block (III) & (IV). Hence the big cube + Number of cubes at two surfaces + numbers 6 and 5 will appear in the blocks III and IV Number of cubes at one edge. respectively. 3 2 = (n – 2) + (n – 2) × 2 + (n – 2) 3 2 Step V. Write remaining number in the remaining block. = (2) + (2) × + 2 Therefore, number 1 will come in the remaining = 8 + 8 + 2 = 18 block. Now, from the unfolded figures we find that DICES number opposite to 6 is 3, number opposite to 2 is 5 and number opposite to 4 is 1. Therefore, option (C) is our answer. Type-I ( Short Trick : From the given dice, we will take the General Dice : In a general dice the sum of numbers common number as the base and then in its on the any two adjacent faces is ‘7’. respect move clockwise direction and write as follows : 4 – 2 – 3 Standard Dice : In a standard dice the sum of 4 – 5 – 6. numbers on the opposite faces is '7'. Here,we find that number opposite to 6 is 3, number Ex 16. Which number is opposite 4 in a standard dice opposite to 2 is 5 and number opposite to 4 is given below ? remaining number 1. Therefore, option (C) is our answer. )

1 5 4 (A) 1 (C) 5 Sol.

Ex 18. (B) 3 (D) Can’t be determined

Clearly , from the standard dice the sum of numbers on the opposite faces is '7', so number opposite to 4 is 3.

On the basis of two figures of dice, you have to tell what number will be on the opposite face of number 5?

(A) 1 (C) 4

(B) 2 (D) 6

PAGE # 92

Sol.

(D) The above question where only two positions Type-V of a dice are given, can easily be solved with the following method : If in the given dice, there are two numbers common, Ex 21. Which of the following dices is identical to the unfolded figure as shown here ? then uncommon numbers will always be opposite of each other. Therefore, option (D) is our answer.

Type-III Ex 19.

From the following figures of dice, find which number will come in place of ‘?’ (X)

Sol.

(A) 4 (B) 5 (C) 2 (D) 3 (D) If the above dice is unfolded, it will look like as the figure (i) given below. Sol.

Figure (i)

Now the number in place of ‘?’ can be obtained by making a slight change in the figure as given here. Now comparing figure (ii) with third dice as above, we get that number in place of ? is 3.

Figure (ii)

Sol.

(B)

(C)

(D)

(A) From the unfolded figure of dice, we find that number opposite to 2 is 4, for 5 it is 3 and for 1 it is 6. From this result we can definitely say that figure (B), (C) and (D) can not be the answer figure as numbers lying on the opposite pair of surfaces are present on the adjacent surfaces.

EXERCISE Directions : (1 to 5) A cube is coloured orange on one face, pink on the opposite face, brown on one face and silver on a face adjacent to the brown face. The other two faces are left uncoloured. It is then cut into 125 smaller cubes of equal size. Now answer the following questions based on the above statements. 1.

Type-IV Ex 20.

(A)

A dice has been thrown four times and produces following results. 2.

How many cubes have at least one face coloured pink ? (A) 1 (B) 9 (C) 16 (D) 25 How many cubes have all the faces uncoloured ? (A) 24 (B) 36 (C) 48 (D) 64

3.

How many cubes have at least two faces coloured ? (A) 19 (B) 20 (C) 21 (D) 23

4.

How many cubes are coloured orange on one face and have the remaining faces uncoloured ? (A) 8 (B) 12 (C) 14 (D) 16

Which number will appear opposite to the number 3? 5. (A) 4 (B) 5 (C) 6 (D) 1 (A) From the figures (i), (ii) and (iv) we find that numbers 6, 1, 5 and 2 appear on the adjacent surfaces to the number 3. Therefore, number 4 will be opposite to number 3.

How many cubes one coloured silver on one face, orange or pink on another face and have four uncoloured faces ? (A) 8 (B) 10 (C) 12 (D) 16

PAGE # 93

Directions : (6 to 11) A cube is painted red on two adjacent Directions : (17 to 21) The outer border of width 1 cm of a surfaces and black on the surfaces opposite to cube with side 5 cm is painted yellow on each side red surfaces and green on the remaining faces. and the remaining space enclosed by this 1 cm Now the cube is cut into sixty four smaller cubes of path is painted pink. This cube is now cut into 125 equal size. smaller cubes of each side 1 cm. The smaller 6. How many smaller cubes have only one surface cubes so obtained are now seperated. painted ? (A) 8 (B) 16 17. How many smaller cubes have all the surfaces (C) 24 (D) 32 uncoloured ? 7. How many smaller cubes will have no surface (A) 0 (B) 9 painted ? (C) 18 (D) 27 (A) 0 (B) 4 (C) 8 (D) 16 18. How many smaller cubes have three surfaces 8. How many smaller cubes have less than three coloured ? surfaces painted ? (A) 2 (B) 4 (A) 8 (B) 24 (C) 28 (D) 48 (C) 8 (D) 10 9.

10.

11.

How many smaller cubes have three surfaces 19. painted ? (A) 4 (B) 8 (C) 16 (D) 24

How many cubes have at least two surfaces coloured yellow ? (A) 24

(B) 44

How many smaller cubes with two surfaces painted have one face green and one of the 20. adjacent faces black or red ? (A) 8 (B) 16 (C) 24 (D) 28

(C) 48

(D) 96

(A) 0

(B) 1

How many smaller cubes have at least one surface painted with green colour ? 21. (A) 8 (B) 24 (C) 32 (D) 56

(C) 2

(D) 4

How many cubes have one face coloured pink and an adjacent face yellow ?

How many cubes have at least one face coloured ? (A) 27

(B) 98

(C) 48

(D) 121 Directions : (12 to 16) A cube of 4 cm has been painted on its surfaces in such a way that two opposite Directions : (22 to 31) A solid cube has been painted yellow, surfaces have been painted blue and two adjacent blue and black on pairs of opposite faces. The surfaces have been painted red. Two remaining cube is then cut into 36 smaller cubes such that surfaces have been left unpainted. Now the cube 32 cubes are of the same size while 4 others are is cut into smaller cubes of side 1 cm each. of bigger sizes. Also no faces of any of the bigger cubes is painted blue. 12. How many cubes will have no side painted ? (A) 18 (B) 16 22. How many cubes have at least one face painted (C) 22 (D) 8 blue ? 13. How many cubes will have at least red colour on (A) 0 (B) 8 its surfaces ? (C) 16 (D) 32 (A) 20 (B) 22 (C) 28 (D) 32 23. How many cubes have only one faces painted ? (A) 24 (B) 20 14. How many cubes will have at least blue colour on (C) 8 (D) 12 its surfaces ? (A) 20 (C) 24 15.

16.

(B) 8 (D) 32

24.

How many cubes will have only two surfaces painted with red and blue colour respectively ? 25. (A) 8 (B) 12 (C) 24 (D) 30 How many cubes will have three surfaces coloured ? (A) 3 (B) 4 26. (C) 2 (D) 16

How many cubes have only two faces painted ? (A) 24 (B) 20 (C) 16 (D) 8 How many cubes have atleast two faces painted ? (A) 36 (B) 34 (C) 28 (D) 24 How many cubes have only three faces painted ? (A) 8 (B) 4 (C) 2 (D) 0 PAGE # 94

27.

28.

29.

How many cubes do not have any of their faces 37. painted yellow ? (A) 0 (B) 4 (C) 8 (D) 16 38. How many cubes have at least one of their faces painted black ? (A) 0 (B) 8 (C) 16 (D) 20 39. How many cubes have at least one of their faces painted yellow or blue ? (A) 36 (B) 32 (C) 16 (D) 0 40.

30.

How many cubes have no face painted ? (A) 8 (B) 4 (C) 1 (D) 0

31.

How many cubes have two faces painted yellow and black respectively ? (A) 0 (B) 8 41. (C) 12 (D) 16

Directions : (32 to 35) Some equal cubes are arranged in the form of a solid block as shown in the adjacent figure. All the visible sufaces of the block (except the bottom) are then painted. 32.

How many cubes have one face painted ? (A) 9 (B) 24 (C) 22 (D) 20

34.

How many cubes have only two faces painted ? (A) 0 (B) 16 (C) 20 (D) 24

35.

How many cubes have only three faces painted ? (A) 4 (B) 12 (C) 6 (D) 20 Directions : (36 to 40) A cuboid of dimensions (6 cm  4 cm  1 cm) is painted black on both the surfaces of dimensions (4 cm 1 cm), green on the surfaces of dimensions (6 cm 4 cm). and red on the surfaces of dimensions (6 cm 1 cm). Now the block is divided into various smaller cubes of side 1 cm. each. The smaller cubes so obtained are separated. 36.

How many cubes will have all three colours black, green and red each at least on one side? (A) 16 (B) 12 (C) 10 (D) 8

If cubes having only black as well as green colour are removed then how many cubes will be left? (A) 4 (B) 8 (C) 16 (D) 30 How many cubes will have 4 coloured sides and 2 sides without colour? (A) 8 (B) 4 (C) 16 (D) 10 How many cubes will have two sides with green colour and remaining sides without any colour? (A) 12 (B) 10 (C) 8 (D) 4

Which alphabet is opposite D ?

(A) E (C) F 42.

How many cubes do not have any of the faces painted ? (A) 27 (B) 8 (C) 10 (D) 12

33.

How many cubes will be formed? (A) 6 (B) 12 (C) 16 (D) 24

(B) C (D) A

What should be the number opposite 4 ?

(i)

(ii)

(iii)

(A) 5 (C) 3

(B) 1 (D) 2

(i)

(ii)

43.

(iii) (iv) Which letter will be opposite to letter D ? (A) A (B) B (C) E (D) F Directions : (44 to 45) The figure (X) given below is the unfolded position of a cubical dice. In each of the following questions this unfolded figure is followed by four different figures of dice. You have to select the figure which is identical to the figure (X). PAGE # 95

50.

44.

(X)

(A)

(B) 51.

(B)

45.

Which symbol will appear on the opposite surface to the symbol x?

(D)

(A) 

(B) =

(C) 

(D) O

Three positions of the same dice are given below. Observe the figures carefully and tell which number will come in place of ‘?’

1 6 3

3 5 4

4 2 ?

(i)

(ii)

(iii)

(A) 1 (C) 3

(X) 52.

(A)

(C)

(B)

(B) 6 (D) 5

On the basis of the following figures you have to tell which number will come in place of ‘?’

3 6 1

4 2 6

? 1 5

(i)

(ii)

(iii)

(A) 2 (C) 6

(D)

(B) 3 (D) 4

Directions : (53 to 55) Choose from the alternatives, the boxes that will be formed when figure (X) is folded: Directions : (46 to 48) In each of the following questions, select the correct option for the question asked.

53.

(i) 46.

47.

48.

(X)

(ii)

Which number will come opposite to number 2? (A) 5 (B) 1 (C) 6 (D) 3 Which number will come opposite to number 6? (A) 1 (B) 5 (C) 4 (D) 3

(A)

(B)

(C)

(D)

Which number will come opposite to number 4? (A) 3 (B) 5 (C) 1 (D) 2 +

49.

On the basis of two figures of dice, you have to tell what 54. number will be on the opposite face of number 5?

(i)

(ii)

(A) 1 (C) 4

(B) 2 (D) 6

(X)

(A)

(C)

(B)

+

+

(D)

PAGE # 96

55.

59.

(X)

(i)

(A)

(C)

(ii)

(B) (iii) (iv) Which number is opposite to number 5? (A) 6 (B) 5 (C) 1 (D) 3

(D)

Directions : (60 to 64) Choose the cube from the options that will unfold to give the figure on the left Direction : (56) The six faces of a cube have been marked with numbers 1, 2, 3, 4, 5 and 6 respectively. This cube is rolled down three times. The three 60. positions are given. Choose the figure that will be formed when the cube is unfolded.

X

M

56. M

X

(A)

(A)

M

X

M (B)

(C)

(D)

(E)

(B) 1

4

8

3

61.

7 9

(C)

(D) 9

7 8

1

57.

Which number is opposite 3 in a standard dice given below ?

(A) 1 (C) 5 58.

(B) 4 (D) Can’t be determined

(A)

4

1

7

(B)

(C)

7 (D)

8

8 7 4 (E)

62. 8 D

Which number is opposite 4 ?

(A) 5 (C) 2

(B) 3 (D) 1

Directions : (59) In the following question four positions of the same dice have been shown. You have to see these figures and select the number opposite to the number as asked in each question.

8

8

(A)

(B)

D (C)

(D)

(E)

(D)

(E)

63. B B

(A)

(B)

(C)

PAGE # 97

66.

Which number/letter is opposite 2 ?

J

3

64.

J (A)

(B)

(A) A (C) 1

J (C)

(D)

Which letter is opposite Q ?

Q O P L N M (A) L (C) N

(B) C (D) 3

(E)

Directions : (65 to 68) In each of the following questions, a 67. diagram has been given which can be folded into a cube. The entries given in the squares indicate the entries on the face of the cube. In each question a number or a letter has been given . Of the four alternatives given below it, you have to find the one that would appear on the face opposite to it in the cube. 65.

I C A B 2

68.

Which number/letter is opposite O?

L N M 2 I O (A) L (C) N

(B) M (D) 2

Which letter is opposite R?

Q R S P U T (B) M (D) P

(A) P (C) T

(B) S (D) U



PAGE # 98

ANSWER KEY ELECTRICITY(PHYSICS) Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

A

A

C

C

C

B

A

D

A

B

D

C

A

B

B

Que.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

B

A

A

A

D

D

B

C

B

C

B

C

B

B

A

Que.

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

Ans.

A

D

B

C

D

B

C

D

A

B

D

C

ABCD CD

BD

Que.

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

C

D

C

B

B

C

B

C

C

B

B

D

C

B

62

63

64

65

66

67

68

69

70

71

72

73

C

B

B

D

A

A

B

A

A

C

D

D

14 C 29 A 44 D 59 A 74 B 89 D

15 B 30 D 45 A 60 D 75 D 90 D

Ans. ABD Que. 61 Ans. B

MOLE CONCEPT(CHEMISTRY) Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans.

1 B 16 A 31 D 46 B 61 C 76 A

2 C 17 A 32 C 47 B 62 D 77 A

3 D 18 A 33 C 48 C 63 A 78 C

4 B 19 A 34 B 49 C 64 B 79 A

5 C 20 A 35 C 50 C 65 A 80 A

6 B 21 B 36 B 51 C 66 B 81 D

7 A 22 A 37 C 52 B 67 C 82 B

8 D 23 A 38 D 53 C 68 A 83 D

9 A 24 D 39 A 54 B 69 D 84 B

10 B 25 A 40 B 55 A 70 B 85 C

11 B 26 A 41 C 56 A 71 C 86 B

12 D 27 A 42 A 57 D 72 B 87 C

13 B 28 D 43 C 58 C 73 C 88 C

NUMBER SYSTEM(MATHEMATICS) Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans. Q. Ans.

1 B 11 A 21 C 31 B 41 C 51 C 61 B 71 C 81 B 91 A&D

2 B 12 B 22 B 32 A 42 B 52 C 62 B 72 A 82 B 92 C&D

3 A 13 A 23 B 33 A 43 A 53 B 63 D 73 C 83 D 93 A

4 D 14 A 24 A 34 B 44 A 54 D 64 D 74 C 84 C 94 D

5 C 15 B 25 D 35 D 45 D 55 C 65 D 75 B 85 D 95 B

6 B 16 C 26 B 36 D 46 A 56 C 66 A 76 B 86 D 96 C

7 A 17 C 27 D 37 C 47 D 57 C 67 A 77 B 87 A

8 A 18 B 28 B 38 C 48 D 58 C 68 B 78 A 88 B

9 A 19 C 29 C 39 D 49 A 59 C 69 A 79 D 89 C

10 D 20 A 30 C 40 C 50 A 60 A 70 D 80 C 90 D

PAGE # 9999

LOGARITHM (MATHEMATICS)

1

Q.

2

3

4

5

6

7

8

9

10

Ans.

B

A

B

D

B

C

B

A

C

D

Q.

11

12

13

14

15

16

17

18

19

20

Ans.

A

A

B

B

D

D

B

C

B

D

Q.

21

22

23

24

25

26

27

28

29

30

Ans.

A

A

B

B

A

A

D

A

A

B

Q.

31

32

33

34

35

36

37

38

39

40

Ans.

B

B

A

B

B

A

D

B

B

C

NUTRITION(BIOLOGY)

Que s.

1

2

3

4

5

6

7

8

9

10

Ans.

C

C

B

D

D

A

C

B

A

A

Que s.

11

12

13

14

15

16

17

18

19

20

Ans. Que s. Ans.

B 21 B

C 22 C

A 23 A

C

A

B

D

A

D

C

SERIES COMPLETION(MENTAL ABILITY) EXERCISE-1 (Number Series) Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

C

D

D

A

C

D

B

C

C

C

D

C

C

B

D

Que.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

B

C

C

C

B

A

B

D

D

A

C

B

B

C

A

Que.

31

32

33

34

35

Ans.

C

C

C

D

D

EXERCISE- 2 (Alphabet Series) Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D

A

D

C

C

A

D

C

D

B

D

C

C

C

D

Que.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

C

A

B

C

C

C

A

B

A

C

D

B

C

D

B

EXERCISE- 3 (Letter Repeating Series) Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

D

C

B

D

C

B

25

26

Ans.

D

D

A

A

C

B

A

C

D

Que.

16

17

18

19

20

21

22

23

24

Ans.

C

A

A

A

C

D

D

D

A

B

D

PAGE # 100 100

EXERCISE- 4 (Missing Term In Figure) Que .

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans .

B

D

B

D

C

C

C

D

A

D

D

B

C

A

B

Que .

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

A

C

D

B

A

Ans .

B

C

A

B

B

A

C

A

D

D

Que .

31

32

33

34

35

36

37

38

39

40

Ans .

B

B

A

C

A

C

B

C

D

B

PUZZLE-TEST(MENTAL ABILITY) Que. Ans. Que. Ans. Que. Ans. Que. Ans.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

A

B

D

C

A

D

C

C

C

C

C

D

C

C

B

16

17

18

19

20

21

22

23

24

25

26

C

C

D

D

C

D

D

B

B

C

A

31

32

33

34

35

36

37

38

39

40

41

27 A 42

28 D 43

29 C 44

30 D 45

D

C

B

A

B

A

D

A

D

C

D

A

C

B

D

46

47

48

49

50

51

52

53

54

55

56

C

A

D

B

D

D

D

A

C

A

A

57 D

58 B

CALENDAR AND CLOCK-TEST(MENTAL ABILITY) Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D 16 B 31 D

C 17 B 32 B

B 18 B 33 D

D 19 B 34 B

D 20 C 35 A

B 21 B 36 A

C 22 C 37 C

B 23 D 38 A

C 24 C 39 D

C 25 B 40 A

A 26 D 41 C

B 27 C

C 28 B

B 29 B

A 30 D

Que. Ans. Que. Ans.

CUBE AND DICE TEST(MENTAL ABILITY) Que.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Ans.

D

C

C

D

A

C

C

D

B

B

C

A

C

D

B

Que.

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

Ans.

C

D

C

B

A

B

D

D

A

D

C

A

D

C

B

Que. Ans. Que. Ans. Que. Ans.

31

32

33

34

35

36

37

38

39

40

41 B 56 C

42 B 57 B

43 A 58 A

44 D 59 C

45 B 60 C

C

D

C

D

C

A

D

C

B

C

46 D 61 A

47 A 62 D

48 B 63 E

49 C 64 D

50 D 65 C

51 A 66 A

52 B 67 B

53 D 68 B

54 B

55 D

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