1. AITS-1 ijso

ALL INDIA IJSO(STAGE-I) TEST SERIES MOCK TEST PAPER # 1 (ONLINE MODE)

Time : 2 Hr.

Date : 06-09-2014

Max. Marks : 240

GENERAL INSTRUCTIONS 1.

In addition to this question paper, you are given a separate answer sheet.

2.

Fill up all the entries carefully in the space provided on the OMR sheet ONLY IN BLOCK CAPITALS. Incomplete/incorrect/carelessly filled information may disqualify your candidature.

3.

A student has to write his/her answers in the OMR sheet by darkening the appropriate bubble with the help of HB Pencil as the correct answer(s) of the question attempted.

4.

Paper carries 80 questions each of 3 marks.

5.

Any rough work should be done only on the blank space provided at the end of question paper.

6.

For each correct answer gets 3 marks, each wrong answer gets a penalty of 1 mark.

7.

Blank papers, clip boards, log tables, slide rule, calculators, mobiles or any other electronic gadgets in any form is "NOT PERMISSIBLE".

PCCP Head Office: Address : J-2, Jawahar Nagar, Main Road, Kota (Rajasthan)-324005 Contact. No. : +91-0744-2434727, 8824078330 Website : www.pccp.resonance.ac.in E-mail : [email protected]

1.

If pq – qr = (p + r)r–q ; p > r > q  prime number least than 11 then p + q is equal to : (A) r(r – q) (B) r(q – p) (C) r(p + q) (D) pq

2.

Find the sum of all the coefficients of the polynomial (x – 2002)3 (x – 2001)2 (x – 2000) (x – 3)3 (x – 2)2 (x –1) (A) 2000 (B) 1 (C) 0

3.

If

a 1 b c abc 1 d e 1 = , = 2, = , = 3 and = , then what is the value of ? b 3 c d 2 e def f 4

(A)

4.

3 4

(B)

If a + b + c = 0 then

a

2

27 4

(C)

 b2  c 2

(B) 4

27 8

(D)

3 8

2



a 2b 2  b 2 c 2  c 2 a 2

(A) 0 5.

(D) 2002

=? (C)

(D) None of these

Consider the expression

(a 2  a  1)(b 2  b  1)(c 2  c  1)(d 2  d  1)(e 2  e  1) abcde where a, b, c, d, e are positive numbers. The minimum value of expression is ? (A) 3 (B) 10 (C) 100 (D) 243 6.

Pandavas won a hen in the war of the mahabharat. They brought it on the Ist Jan 2002. This hen gave birth to the 7 new hens on the very first day. After it every new hen in respective of its age everyday gave birth (only once in a lifeteine) to 7 new hen. This process continued throughout the year, but no any hen had been died so far. On the 365th day all the pandows shared equally all the hens among all the five brother. The remaining (it these can not be shared equally) hens were donated to krishna. The no of hens which the krishna had recieved is (A) 3 (B) 2 (C) can’t be determined (D) none of these

7.

The sum of all the real roots of the equation |x – 2|2 + |x – 2| – 2 = 0 is : (A) 2 (B) 3 (C) 4

8.

The real numbers x1, x2 , x3 satisfying the equation x3 – x2 + bx + y = 0 are in A.P. find the intervals in which b and y lie, respectively : (A) (–  ,

1 1 ] [– , ) 3 27

 1 1  1 1  ,  (C)   ,     3 3   27 27 

9.

(D) None of these

(B) (–  , 3)

(D) None of these

If the polynomials f(x) = x2 + 6x + p and g(x) = x2 + 7x + q have a common factor, then which of the following is true ? (A) p2 + q2 + 2pq + 6p – 7q = 0 (B) p2 + q2 – 2pq + 7p – 6q = 0 2 2 (C) p + q – 2pq + 6p – 7q = 0 (D) p2 + q2 + 2pq + 7p – 6q = 0 Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 2

10.

When a fourth degree polynomial f(x) is divided by (x + 6), the quotient is Q(x) and the remainder is – 6. And when f(x) is divided by [Q(x) + 1], the quotient is (x + 6) and the remaindr is R(x). Find R(x) (A) 12 + x (B) –(x + 12) (C) 0 (D) 3

11.

Find the remainder when x33 is divided by x2 – 3x – 4 x

 4 33  1   4 33  4    (A)  5  x +  5     

 4 33  1   (B)    5 

 4 33  1    (C)  5  x +  

 4 33  4   x + (C)   5  

 4 33  1     5   

 4 33  4    +   5    4 33  1     5   

12.

If f(x) = x2 + 6x + a, g(x) = x2 + 4x + b, h(x) = x2 + 14x + c and the LCM of f(x) g(x) and h(x) is (x + 8) (x + 6) (x – 2), then find a + b + c (a , b ,c are constants) (A) 20 (B) 16 (C) 32 (D) 10

13.

If the ordered pair (sin, cos) satisfies the system of equations mx + ny + a + b = a – b and nx + my + 2b = 0, then find the value of Q where O  Q  90° (m  n) (A) 30° (B) 15° (C) 60° (D) cannot be determined

14.

A hybrid mango tree, whose life span is 10 year, starts giving fruits from the first year onwards. In the nth year it produces 11n raw mangoes. But during the first half of the tree’s life, every year a certain number, which is constant, fail to ripen into fruits, during the second half of the tree’s life every year the number of raw fruits that fail to ripen is half the corresponding number in the first half of the tree’s life. In the fourth year of the tree’s life, it produces 36 ripe mangoes. How many mangoes ripen during the 9th year of the tree’s life ? (A) 100 (B) 96 (C) 95 (D) 86

15.

If 3 | x | + 5 | y | = 8 7 | x | – 3 | y | = 48 then find the values of x + y (A) 5 (B) – 4

16.

(C) 4

(D) the values does not ex ist

In the adjoining figure, find the area of shaded portion. Given O is the centre of the circle, seg AB = 16 cm, seg OE  seg BC, radius OE = 7 cm. Find ratio of area of the shaded portion to the unshaded portion. A E

B

(A) 27 : 2

(B) 16 : 11

O

C

(C) 2 : 11

(D) 8 : 11

Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 3

17.

In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they touch each other and also the sides of the triangle. Find the area of the triangle. (A) 4 3 + 6

18.

y 1 y 1

(B) y =

In the figure AD = DB, BE =

1 x 1 x

(D) 5 3 + 6

(C) xy + x – y + 1 = 0

(D) xy + x + y + 1 = 0

1 1 EC and CF = AF. If the area of ABC = 120 cm2, the area (in cm2) of 3 2

DEF is :

A

D B

(A) 21 20.

(C) 3 3 + 6

If sec– tan= x and y = cosec– cotthen (A) x =

19.

(B) 2 3 + 6

(B) 35

F C E

(C) 40

(D) 45

If a,b,c are positive real number and a+b+c=5, then find the maximum value of (1+a)(1+b)(1+c)

8 (A)   3

3

5 (B)   3

3

 1 (C)   3

3

 1 (D)   5

3

21.

Let the electrostatic force between two electrons (Fe) be x times the gravitational force (Fg) between them. Then, x is of the order of (A) 1040 (B) 1042 (C) 1038 (D) 1037

22.

A boy is standing on a truck which is moving with constant speed along a straight road. On a day when wind is negligible, the boy throws a ball vertically up with some velocity. The ball comes back and falls (A) into boy’s hand (B) Behind the boy (C) In Front of the boy (D) Behind or in front of the boy depending on the speed of the truck and ball.

23.

The pressure at the bottom of the four vessels filled with water to the same level is P1, P2, respectively. Then which of the following conclusion is correct.

(A) P1  P2  P3  P4

(B) P1  P2  P3  P4

(C) P1  P4  P2  P3

(D) P1  P2  P3  P4

P3 and P4

Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 4

A ball is thrown up vertically in still air with a velocity of 20ms-1. It comes back to ground. The velocity – time graph is (g=10ms-2).

24.

(A)

(B)

(C)

(D)

25.

Sound waves traveling in air enter water at an angle i with the normal. It gets refracted at angle r with (A) i > r (B) r > i (C) i = r (D) Sound waves do not get refracted

26.

A particle of mass 0.5kg travelling with a velocity of 2ms-1 experiences acceleration of 2ms-2 for 9s. The work done by the force on the particle during this period is (A) 99J (B) 101J (C) 190J (D)396J

27.

What is the reading of the spring balance shown in the figure below?

(A) 0 28.

(B) 2N

(C) 4N

(D) 6N

The real image of an extended object placed in front of a concave mirror is formed at a distance of 40 cm from the object. If the image is 3 times bigger than the object, the magnitude of focal length of the mirror is. (A) 15cm (B) 10cm (C) 20cmd (D) 5cm Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 5

29.

30.

A person suffering from short sight is advised to wear spectacles having concave lens of power-1.25D. What is the farthest distance he can see clearly without spectacles? (A) 60cm (B) 100cm (C) 120cm (D) 80cm Two wires made of same material have length l and 2l. If the masses of the wires are same, the ratio of the resistance of shorter wire to that of longer wire is (A) 1/2 (B) 2 (C) 1/4 (D) 4

31.

Current passing through a wire increases by 20%. Due to Joule heating the resistance increases by 20%. The percentage increase in the power is (A) 72.8% (B) 44% (C) 33% (D) 40%

32.

A bar magnet is placed on a table. There is n number of field lines connecting North pole to South pole of the magnet. Another identical magnet is placed on the first magnet with North Pole on North Pole and South Pole on South Pole. The number of field lines are now (A) n (B) n2 (C) n/2 (D)2n

33.

A conducting wire shown in the figure carries current I. Segments AB, BC and CD are of same length. The direction of the magnetic field at point P is given by B

A

P

C

(A) into the plane of the paper (C) towards right 34.

D

(B) out of the plane of the paper (D) towards left

In a nuclear reactor the fission process of each 235 U  atom gives out energy of 200MeV. According to Einstein’s equation the amount of mass getting converted to energy in this process is (A) 3.55  10 30 kg

(B) 3.55  10 38 kg

(C) 3.55  10 28 kg

(D) 3.55  10 27 kg

35.

A ball is projected at an angle of 450 with horizontal. In the absence of air resistance, the ball follows (A) Elliptical orbit (B) sinusoidal path (C) parabolic path (D) linear path

36.

60g of ice at 0ºC is added to 20g of water at 40º. The final temperature attained by the mixture is (given latent heat of melting of ice = 80 cal/g and specific heat of water is1cal/gºc) (A) 0ºC (B) 20ºC (C) 10ºC (D) 5ºC

37.

Wavelength is (i) The distance traveled by the wave in one period of oscillation of particles in the medium. (ii) The distance between two particles, which are in the same phase? (iii) Half of the distance between two particles, which are in the same phase? The correct definitions are (A) (i) and (iii) (B) (i) and (ii) (C) (i), (ii) and (iii) (D) (ii) and (iii) Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 6

38.

The distance between two spots A & B on the same bank of the river is 75km. Speed of the boat in still water is twice as much as that of the speed of the water current of the river. The boat travels in the river from A to B and returns back to the spot in 16 hour. What is the speed of the boat in still water? (A) 12.5kmph (B) 15kmph (C) 16kmph (D)18kmph

39.

Michael Faraday a book binder got an opportunity to work with a scientist and later succeededhim. Name of the scientist is (A) Hans Christian Oersted (B) Humphrey Davy (C) Heinrich Lenz (D) James Clerk Maxwell

40.

When a car turns on a curved road, you are pushed against one of the doors of the car because of (A) inertia (B) the centripetal force (C) the centrifugal force (D) the frictional force

41.

The number of oxygen atoms required to combine with 14g of nitrogen to form N2O3 when 80% of nitrogen is converted to N2O3 . (A) 2.4 N A

(B) 1.2 N A

(C) N A

(D) 12 N A

42.

Calculate the amount of NaOH containing 20% H 2O is neutralized completely by 0.05M & 6 litre sulfuric acid? (A) 15g (B) 30g (C) 60g (D) 24g

43.

The size of species ,  1,  1 decreases in the order ________ (A)  1   1  

44.

(B)  1     1

(C)  1   1  

(D)    1   1

Consider the following radioactive process. 



   P   Q    R   S Which statement is incorrect among the following? (A) P & S are isotopes (B) P & Q are isodiapheres (C) Q & R are isobars (D) P & Q are isotopes

45.

A hydrocarbon having the molecular weight 72 give only one monochloro compound on chlorination with chlorine in the presence of sunlight. Compound can be _______ (A) n-pentane (B) isopentane (C) neopentane (D) none

46.

1 mole of Ethyl alcohol is treated with 1 mole of acetic acid such that 2/3 of acid changes into ester. Calculate equilibrium constant (kc) (A) 0.25 (B) 4 (C) 0.5 (D) 2

47.

Solubility product of a salt A2B is 4  10 9 . Its solubility is _______ (A) 10 2 (B) 10 3 (C) 10 6

(D) 10 5

48.

10 mL of 0.45N H2SO4 is mixed with 40mL of 0.1 N NaOH. Calculate the pH of resulting solution? (A) 2 (B) 3 (C) 1 (D) 4

49.

Oxidation number of Cr in CrO5 is ………. (A) 10 (B) 6 (C) 5 Space For Rough Work

(D) 4

IJSO STAGE-I _MOCK TEST-1_PAGE # 7

50.

Equal weights of CH4 & SO3 are placed in two separated vessels at 27ºC & 1atm pressure. Then the ratio of number of molecules of CH4 & SO3 is………... (A) 4:1 (B) 1:5 (C) 2:3 (D) 5:1

51.

One of the following sets of quantum numbers represents an impossible arrangement

n



m

ms

(A)

3

2

2



(B)

4

0

0

(C)

3

2

3

(D)

5

3

0



1 2

1 2 



1 2

1 2

52.

Thermite reaction is one of the important reactions in the metallurgical industry. This reaction is best described as......... (A) Fe displacing aluminum from its ore where iron acts as reducing agent & aluminum as oxidizing agent (B) Aluminum is displacing iron from its ore where iron acts as oxidizing agent & aluminum reducing agent. (C) Aluminum is displacing iron from its ore where iron acts as reducing agent & aluminum as oxidizing agent. (D) Iron displacing aluminum from its ore where iron acts as oxidizing agent & aluminum as reducing agent

53.

A compound is not isomeric with diethyl ether is............. (A) Methyl-n-propyl ether. (B) Butane-1-ol (C) 2-Methylpropan-2-ol (D) Butane

54.

The pair having similar geometry is........... (A) BF3 , NH3

(B) H2O, C2 H2

(C) CO2 , SO2

(D) NH3 , PH3

55.

The ratio of rates of diffusion of A & B is 1:4. Ratio of their weights present in the mixture is 2:3. The ratio of their mole fractions is ? (A) 24:1 (B) 1:24 (C) 5:6 (D) 7:6

56.

The rate law equation for a certain reaction is given below: Rate = k  A   B 

x

1

y

C 

x

y

. The order of

reaction is............. (A) x  y 

x y

 1 x   (B) x    y 

(C) x 

1 1  y z

(D) 2  x  y 

Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 8

57.

A mixture of 2 moles of CO & 1 mole O2 in a closed vessel is ignited to convert to CO into CO2 then.......... (A) H  E (B) H  E (C) H  E (D) The relation depends on the capacity of the vessel.

58.

Which d-orbital has different shape from rest of all d-orbital’s? (A) d x 2  y 2

59.

60.

61.

(B) d z2

Froth floatation process is based on ............ (A) Specific gravity of ore particles (C) Wetting property of ore particles The IUPAC name of

is...........

(A) But-2-ene

(B) Buta-1, 3 – diene

(C) d xy

(D) d xz

(B) magnetic properties of ore (D) electrical property of ore particles

(C) Butene

(D) But-2, 3 - diene

When a cell is fully turgid, which of the following will be zero? (A) turger pressure (B) wall pressure (C) suction pressure

(D) osmotic pressure

62.

When a cell fails to communicate with other cells in multicellular organism, it (A) becomes cancerous (B) enters mitotic phase (C) chooses to die (D) is eaten up by other cells

63.

Which of the following organelle is the site for ribosome synthesis? (A) Ribosome (B) Cytoplasm (C) Nucleus

64.

65.

66.

(D) Nucleolus

Lateral meristem includes : (A) fascicular cambium and cork cambium (C) procambium and interfascicular cambium

(B) phellogen and protoderm (D) dermatogen and ground meristem

Which one of the following is a true fruit ? (A) Pear (B) Coconut

(C) Apple

Cycas is classified as gymnosperm because of (i) presence of naked seeds (iii) presence of sieve tubes. The correct reason is (A) (i) and (iv) (B) (i) and (iii)

(D) Cashewnut

(ii) lack of vessels in the xylem (iv) fruit formation. (C) (i) and (ii)

(D) (ii) and (iii)

67.

With reference to human beings, the correct order of taxonomical classification is (A) chordata, primata, mammalia, hominidae. (B) chordata, mammalia, primata, hominidae. (C) chordata, primata, hominidae, mammalia. (D) chordata, mammalia, hominidae, primate

68.

“Green fuel” means (A) fuel obtained from plant leaves (B) green coloured fuel (C) chemicals used for the growth of plants (D) fuel obtained from plastic waste Space For Rough Work

IJSO STAGE-I _MOCK TEST-1_PAGE # 9

69.

Which of the folowing is an ex situ method of conservation ? (A) Agro-forestry (B) Sanctuary (C) Cryopreservation

(D) Biosphere reserve

70.

Bacteria cannot survive in a highly salted pickle because (A) they become plasmolysed and consequently die. (B) they do anaerobic respiration. (C) water is not available to them. (D) of all the reasons mentioned above.

71.

Why do antibiotics kill bacteria but not viruses ? (A) Antibiotics stimulate the immune system against bacteria but not viruses (B) Viruses have a way of blocking antibiotics (C) Viruses are too small to be affected by antibiotics (D) Viruses do not have a metabolism

72.

The part of the digestive system that digests lipids in the food is : (A) stomach (B) duodenum (C) ilium

(D) large intestine

Human body cannot digest carbohydrate in the form of (A) sugars. (B) starch (C) cellulose

(D) glycogen

Which organ secretes enzymes that are active at a low pH ? (A) Mouth (B) Pancreas (C) Stomach

(D) Liver

73.

74.

75.

Which of teh following does not happen during the Calvin cycle ? (A) Regeneration of the CO2 acceptor (B) Oxidation of NADPH (C) Release of oxygen (C) Consumption of ATP

76.

When one glucose molecule undergoes one turn of aerobic respiration, 38 ATP molecules are produced. Cellular respiration takes partly in cytoplasm and partly in mitochondria. During the process, some ATP molecules are produced in the cytoplasm, some in the mitochondrial matrix and some in the oxysomes on cristae. Maximum number of these ATP molecules is produced in.... (A) cytoplasm (B) mitochondrial matrix (C) cytoplasm and mitochondrial matrix together (D) oxysomes on cristae

77.

Which is the correct sequence of blood flow in normal human circulation ? (A) pulmonary vein ; right atrium ; aorta ; vena cava (B) vena cava ; pulmonary vein ; aorta ; right atrium (C) vena cava ; right atrium ; pulmonary vein ; arota (D) pulmonary vein ; vena cava ; aorta ; right atrium

78.

Where in the kidney does ultra filtration take place ? (A) Glomerulus (B) Loop of Henle (C) Proximal tubule

(D) Collecting ducts

79.

Like sickle cell anaemia, the other genetic disorder related to blood is (A) thyroiditis (B) phenylketoneuria (C) thalassemia (D) xeroderma pigmentosis

80.

Down’s syndrome is a result of (A) XO genotype (C) Trisomy (chromosome-12)

(B) XXY genotype Space For Rough Work (D) Trisomy (chromosome-21)

IJSO STAGE-I _MOCK TEST-1_PAGE # 10

ALL INDIA IJSO(STAGE-I) TEST SERIES

MOCK TEST PAPER # 1 ONLINE MODE

HINTS & SOLUTIONS ANSWER KEY Que s.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

Ans.

A

B

D

B

A

A

C

A

B

B

B

A

A

C

D

A

A

B

B

A

Que s. 21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

B

A

D

B

B

A

B

A

D

C

A

D

B

C

C

A

B

A

B

C

Que s. 41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

B

B

B

D

C

B

B

A

B

D

C

B

D

D

B

B

C

B

C

B

Que s. 61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

C

C

D

A

B

C

B

A

C

A

D

B

C

C

C

D

C

A

C

D

Ans.

Ans.

Ans.

5.

pq – qr = (p + r)r–q Let 11 > p > r > q  prime number 73 – 35 = (7 + 3)5–3 100 = 100 (p + q) = 7 + 3 = 10 = 5(5 – 3) = r(r – q)

1.

x

f(1) sum of all its coefficients.

3.

a b c b c d c d e             = b c d c d e d e f 

6.

1 x2  x  1 =x+ +1 =  3 x x 33333 70 + 71 + 72 + 73 + ........ + 7365 =

a b c   d e f

7 366  1 1( 7 366  1) = 7 1 6

It is divisible since (an-bn) is divisible by (a – b)

1 1 1 1   1 =   2    2   3    3   = 2 4 2 3   2

now

abc def



abc 1 3 3 = 3 = def 3 8 8

7.

a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 = 0 a2 + b2 + c2 = –2(ab + bc + ca)

a

2

 b2  c 2

2



a 2b 2  b 2 c 2  c 2 a 2

=

=

1 =2 x



2.

4.

for any positive value of x the minimum value of

=

[ 2(ab  bc  ca)] 2 a 2b 2  b 2 c 2  c 2 a 2

4(ab  bc  ca) 2 a 2b 2  b 2 c 2  c 2 a 2

4(a 2 b 2  b 2 c 2  c 2 a 2  2abc (a b  c )) a 2b 2  b 2c 2  c 2a 2

7 366  1 5

4 1 3 2 366  1 ( 4)183  1    5 5 5 5

Since the reminder is 3 Hence krishna will receive 3 hens. let |x – 2| = m m2 + m – 2 = 0 (m – 1) (m + 2) = 0 m=1 m = –2 (rejected) |x – 2| = 1 x–2=1 x= 3 x = 1,3 2–x=1 sum of roots = 1 + 3 =4 x =1

=4

SOL. IJSO STAGE-I _MOCK TEST-1_PAGE # 1

8.

x1 = a – d x2 = a x3 = a + d d = common difference x3 – x2 + bx + y = 0 x1 + x2 + x3 = 1 x 1x 2 + x 2x 3 + x 3x 1 = b a=

1 3

since

12.

1 – d2 = b 3 1 1 – d2  3 3

1 b  3

b  (–  ,

1 ] a(a2 – d2) = – y 3

13.

a3 – ad2 = – y

  d2 1  +  3 27  y=

  =–y  

d2 1 – 3 27

14.

 (d2  0 )

No. of fruits in II half life = 11x –

 1  y   – 27 ,    

9.

10.

11.

Let (x – a) be the common factor f(a) = g (a) a2 + 6a + p = a2 + 7a + q p – q = a f(a) = 0 (p – q)2 + 6(p – q) + p = 0 p2 + q2 – 2pq + 6p – 6q + p = 0 p2 + q2 – 2pq + 7p – 6q = 0 f(x) = (x + 6) Q(x) – 6 f(x) + 6 = (x + 6)Q(x) ..(1) f(x) = [Q(x) + 1] (x + 6) + R(x) f(x) = (x + 6) Q(x) + x + 6 + R(x) – x – 12 = R(x) x33 = (x2 – 3x – 4) Q(x) + R(x)  mx + n x33 = (x – 4) (x + 1)Q(x) + mx + n Put x = 4 433 = 4m + n ..(1) n=m–1 Put x = - 1 =

4 33  1 –1 5

–1 = –m + n .....(2) n=m–1 From (1) 433 = 4m + m – 1  m=

433  1 5

Now, n = m – 1 n=

433  1 –1 5

n=

4 33  4 5

 4 33  1   4 33  4      Thus, R(x) =  5  x +   5     Given polynomials are quadratic and LCM is cubic  f(x) = (x + 8) (x – 2) = x2 + 6x – 16  a = – 16 g(x) = (x – 2) (x + 6) = x2 + 9x – 12 b = –12 h(x) = (x + 8) (x + 6) c = 48 a + b + c = – 16 – 12 + 48 = 20 If (P – q) is a solution of ax + by + c = 0 and bx + ay + c = 0 then p - q  sin = cos fruit that can’t ripen in first half = y  No of fruits in xth year = 11x – y In 4th year = 36 11  4 – y = 36 y=8

In 9th year = 11  9 – 15.

16.

y 2

8 = 95 2 |y|=b

|x|=a (3a + 5b = 8)  3 (7a – 3b = 48)  5 a=6 b=–2 |x| = 6 |y| = – 2 x + y = does not exist 1 Area of ABC = × base × height 2

1 × 16 × 14 = 112 cm2 2 1 Area of shaded semi-circle = r2 2 =

1 22 × × 7 × 7 = 77 cm2 2 7 Total area of shaded portion = 112 + 77 = 189 cm2. Area of unshaded part = Area of minor segment EC = Area of sector CE – Area (EOC) =

=

90 º 22 1 × ×7×7– ×7×7 360 º 7 2

77 49 28 – = = 14 cm2. 2 2 2 Ratio of shaded part to unshaded part = 189 : 14 = 27 : 2 =

SOL. IJSO STAGE-I _MOCK TEST-1_PAGE # 2

17.

In OBD

tan 30º =

xy = – (x+y) + 1 x + y + xy – 1 = 0 (x–1) + y (1+x) = 0 y (1 + x) = 1 – x

   

1 x

y=



1 3

x=

=

1 x

1 x . 1 x

A 3y F

D

3

19.

So, side of equilateral triangle = 2 + 2x = 2(1 + Area of ABC =

y B

3)

x

2x

E

area DCB =

3 (side)2 4

3 [2(1 + 4

=

3 (1 +

=

3 (1 + 3 + 2 3 )

3 )]2 =

3 )2

area AEC =

sec – tan = x

=

1 sin   =x cos  cos 

And

or ADF =

cosec  – cot  = y

1 cos  – =y sin  sin 

So,

x+y =

1  sin  1  cos  + cos  sin 

=

sin   sin 2   cos   cos 2  cos  sin 

=

sin   cos   1 cos  sin 

And

xy =

1  sin  1  cos  × cos  sin 



xy =

1  sin   cos   sin  cos  sin   cos 



xy =

1  sin   cos  1 sin  cos 

20.

1 1 area  ABC = × 120 = 60 2 2

3 3 area ADC = × 60 = 45 4 4

area EFC =

1  cos  =y sin 

2  area ABC 3

2  120 = 80 3

area ADC =

1  sin  =x cos 

1 area DBC 3

1  60 = 20 3

= 4 3 + 6. 18.

1 area ABC = 60 2

[D is mid point of AB] area DBE =

=

C

1  80 = 20 4

area DEF = area ABC – area DBE – area EFC – area ADF = 120 – 20 – 20 – 45 = 35 For a+1,b+1,c+1 AM  GM a  1 b  1 c  1 3  (a  1)(b  1)(c  1) 3 8 3  (a  1)(b  1)(c  1) 3 3

8    (a  1)(b  1)(c  1) 3

SOL. IJSO STAGE-I _MOCK TEST-1_PAGE # 3

43.

Atomic radius  =

44.

Z

anion of same elements cation of same element

 A–4 P A   2–1 R A – 4  z S A – 4 z–z Q

W W : S = (1 : 5) × 5  5 : 1 16 80 51.

A A–4 z P and z S same atomic no. different atomic mass so isotope.

 = 2 mean d orbital m = –3 m = –2 –1 0 +1 +2 not possible Oxidation

CH3

45.

CH3 Cl2

CH3 – C – CH3

52. CH3 – C – CH3 – Cl

CH3

53.

(same compoumnd will be obtained by chlorination at every carbon.) CH3 – CH2 – OH + CH3COOH Initial

1 mole

CH3 COOC2H5 + H2O

54.

1 mole

1–2/3 After dissociation

1–2/3

2/3

Mole at 1/3 equilibrum

1/3

2/3

47.

55.

2/32/3 4 1/ 3  1/ 3

48.

1/ 3

 4  10 – 9 =  4 

= 10–3

M2 M1

m 2 w1 m1 × w 2

1 2 × 16 3 1 = = 1 : 24 24 56.

Order of a reaction is the sum of pomers to which each concentration term is raised in a rate law.

57.

2CO(g) + O2(g)  2CO2(g)

+1

O Cr

+1 O

O+1 oxidation number = + 6

50.

1 = 4

=

+2

49.

M2 M1

=

0 .5 = Nmix 50 Nmax = 10–2 [H+] = 10–2 pH = 2

O

r1 r2 =

w1 m1 w1 m2 1 n1 Now, × = = w2  m1 × w 2 2 n2 m2

1/ 3

   

N1V1 – N2V2 = Nmix Vmix 45 × 10 – 40 ×0.1 = Nmax 50 4.5 – 4 = Nmax50

+1 O

3 NH3  sp  pyramidal

M2 1 = M1 16

KSP = 4s3

 K SP   S=   4 

CH3 – O – CH3 Molecules formula  C2H6O Butane  CH3 – CH2 – CH2 i.e.C4H10 different molecular formula

3 PH3  sp  pyramidal

[CH3COOC 2H5 ] KC = [CH – CH – OH][CH – COOH] 3 2 3

=

Fe + Al2 O3

Reduction

CH3

Neo-pentane

46.

Fe2O3 + Al

W W : MWCH4 MWSO3

n = 2 – 3 = –1 n = negative H < E 1

60.

2

3 4

CH2 = CH – HC = CH2 buta-1,3-dine SOL. IJSO STAGE-I _MOCK TEST-1_PAGE # 4