Maths Concept Map 8 CBSE

July 12, 2017 | Author: Nilesh Gupta | Category: Rational Number, Fraction (Mathematics), Division (Mathematics), Exponentiation, Area
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Concept Maps Class VII Triangle & It's Properties Lines & Angles Algebraic Expression Area Ratio Symbol Ratio & Proportion Percentage Profit & Loss Simple Interest Exponent Congruent Triangle Integers Fraction Decimals Rational Numbers Probability Statistics Symmetry Solid Shapes Perimeter

TRIANGLE & IT'S PROPERTIES C

(iii)

Z

(ii)

X

P

(iii)

(ii)

15º

90º

15º B

A

C

Classification

60º

60º

30º

(i)

(i)

R

150º

50º

Q

P Obtuse angle D : an angle :greater than 90º, less than 180º

Right angled D :an angle equals to 90º

70º C

B

A

Acute angle D : all angles :greater than 0º less than 90º

3cm

B

Scalene D : No two sides are equal

x x R 3cm Q Isosceles D : Two sides are equal, base angles opposite to equal sides, are equal

60º

60º Y

Z

Equilateral D : All sides are equal, each angle has measure 60º

A Side : AB, BC, CA Angle : BAC, ABC, BCA Closed curve made up of Vertices : three line segment. A, B, C B

C

A P

B Q

A

P

(i)

100º

30º

(ii)

X Exterior angle property : ÐPRS = ÐQPR + ÐPQR 150º

50º 70º B

D

C

R

Q

80º C

R

Sum of the length of any two sides of a triangle is greater than the length of the third side. 3 cm + 5 cm > 6 cm 6 cm + 5 cm > 3 cm 6 cm + 3 cm > 5 cm.

(iii)

S Y

Z

LINES AND ANGLES Acute (0º < q < 90º) l m q

p

P

Parallel lines : never intersects

Line : have infinite length

l

Right (q = 90º)

m

Obtuse (90º < q < 180º)

Intersecting lines

Straight (q = 180º)

Types of angles Reflex (180º < q < 180º)

Line x

2x

4x

Complete (q = 360º)

Sum of all angles on one side of line is 180º

5x

5x + 2x + x + 4x = 180º

2x 3x x 4x

5x

Sum of all angle around a point is 360º x + 2x + 3x + 4x + 5x = 360º

Complementary angles Two angles whose Sum is 90º

1

Angles made by transversal Corresponding angles (1, 5) (2, 6) (3, 7) (4, 8)

Alternate Int. angles (3, 6) (4, 5)

5 7

Co.int. Angles (4, 6) ) (3, 5)

2 4

3 6

Supplementary angles Two angles whose sum is 180º Vertically opp. angles 1

8

4

2 3

Alternat Ext. angle (1, 8) (2, 7)

If lines are ||el then

Corresponrding angles are equal

Alternate angles are equal x

x+y y

Sum of Co. Int. angles is 180º

Ð1 = Ð3 ; Ð4 =Ð 2 A

Adjacent angles Ð1andÐ2 are adjacent angles

1 2

O

Common vertex Linear pair two adjacent angles whose sum is 180º 150º P

O

B common arm C opp.arm R 30º Q

ALGEBRAIC EXPRESSION Variables

Constants

(x, y, z,a, b,t,......) Its value is not fixed

(2, 3, -4, 100.........) It has fixed value 2

2

Expression : 9x - 3yx + 10

2

2

2x + 3y x + 3x - 5x + 6

Mathematical Operation

Standerd form

[ +, - , × , ÷ ]

2

2

(i) 3xy + 5x - 5x + 6 2

2

(ii) 6 - 5x + 5x +3xy

Terms

9x

2

-3yx

10

9 x x -3 x y Factors Factors

Numerical coefficient (-3)

In -3x2y

Coefficient 2

xy is -3

-3x is 7

2

Unlike terms:

x + 3y ; x2 + 3

of x is -3xy

Like Terms :

Different algebraic factors 2 3x y = 3 × x × x × y

-3y is 2 x

2

3xy = 3 × x × y × y

Same algebraic factors 6xab = 6 × x ×a ×b -3abx = -3 × a × b × x addition 9ab + 5ab = (9 + 5) ab = 14ab Subtraction 9ab – 4ab = (9 – 4)ab = 5ab

Types of Algebraic Exprission Monomial

Binomial

Trinomial

Using Algebraic Expression

Value of Expression

One term : eg : 3x, -2t ; 3

Two terms : eg : 3x -2t ; 3S+2

at x = a [put x = a] 2 eg : value of x -5x at x = -1 2 (-1) -5(-1) = 1 + 5 = 6

Rules for no. patterns

Formulas

Successor of n

Three terms : eg : 3x + 2y - 5 Perimeter

Quadrinomials

Four terms : eg ab + 2pq + 3rs + t

Equilateral triangle 3× a ; where a is side Square 4× l ; where l is side Regular pentagon 5 × s ; where s is side

Area

n+1

Even no. =2n

Triangle = 1 × b × h 2

odd no (2n +1) or (2n-1)

Rectangle = l × b

3, 6, 9, .......3n.......

AREA S

eg.

R D

8 4

4+2+2 P

Area of cross road = (8 × 2 + 4 × 2 – 2× 2) unit2

4unit 2unit

8 unit

Q

8+2+2

A

r

h2

a unit

C B

A h3

Area of shaded region = p(5)2 - p (3)2

Area = pr r = Area ÷ p r=d 2

C

a unit

Triangle

P h1

R = 3 + 2 cm = 5 r = 3 cm 2

A

h1 B

Area of Path /verandah = Ar. PQRS - Ar. ABCD

2 B

A 2

eg.

2unit

C

2cm R

Parallelogram

Circle

h2

h1

1 a×h 1 2

h1

a unit

B

C a Q a unit 1 1 Area = 2 a ×h1 = 2 b×h2 = 1 c×h3 Sq. units 2 base = 2Area ÷ altitude altitude = 2Area ÷ base 3 2 Area of Equilateral D = 4 (side)

R

h2

Area = a ×h1 sq. unit = b × h2 sq. units base = Area ÷ corresponding altitude altitude = Area ÷ corresponding base

Rhombus : d

Rectangle h

h

l unit

a Area = l × b sq. units length = (Area ÷ breadth) unit breadth = (Area ÷ length) unit d = l2 + b2

Quadrilateral : Trapezium D

a

d h

a unit A = a × a sq. unit Side = area unit d=2a

When to calculate Area : (a) Painting / white washing (b) levelling (c) Ploughing (d) grazing / watering

b

h1

A

C h2

1 A = h × (a +b) sq. units 2

B

1 =2

1 cm2 = 100 mm2 2 2 1 m = 10000 cm 2 1 hectare = 10000 m

d

a

a

Area = a × h sq. units = Square

d1 d2

a

Area × diagonal × sum of offsets = 12 × d × [h1 + h2] sq. units

1 d × d2 sq. units 2 1

RATIO SYMBOL

Convert ratio into fraction Age of A 7 eg : ! Age of B 5 age of A = 7x age of B = 5x

After 6 years

eg :

age of A ! 7x " 6 age of B ! 5x " 6 age of A 5 ! age of B 4 #

ages of A 7 x " 6 5 ! ! ages of B 5 x " 6 4

To compare two quantities, the units must be the same eg : ratio of 3 km to 3m 3km : 3m = 3000 m : 3m =1000 : 1

4 > 3 Compare numerator when 7 7 Denominator same 5 5 When Numerator same eg : < 4 3 Compare denominator eg :

1 5 : 15

Comparison of two ratios

a and b should be of same kind

¹

If a : b is in simplest form Þ H.C.F. (a , b)=1

÷9= 2 27 27 ÷ 9 3

eg :18 = 18

H.C.F. of 18 and 27 is 9

H.C.F of 2 and 3 is 1

Simplest form/lowest term

a : b = a ® anticedent b® consequent

Equivalent Ratio

1:2=2:4=3:6

eg : divide Rs. 5000 in Ratio 3 : 5

2 3 1 = 2 = 4 6 1 1×2 1×3 or. 2 = 2×2 = 2×3

Rs. 5000 × 3 = Rs. 5000 ×3 = Rs. 1875 8 3+5 Rs. 5000 × 5 = Rs. 5000 × 5 8 = Rs. 3125 3+5

eg : If 3 A = 2B = 5C the find A : B : C Let 3 A = 2B = 5C = K (K ¹ 0) \ 3 A = K ; 2 B = K ; 5C = K K K A=K 3 ; B = 2 ; C= 5

eg : If A : B = 2 : 3 and B : C = 5 : 4

L.C.M. (3,2,5) = 30

A 2 5 5 then A : C = B × B C= 3 ×4 =6 A:C=5:6

A =10 × 3 ; B = 15 × 2 ; C = 6 × 5

10 × K

15 × K

\ A : B : C = 10 : 15 : 6

6×K

RATIO & PROPORTION If a : b :: c : d Þ ad = bc or extreme terms

a c If b = d

fourth proportion

If a, b, c are in continued proportion

(i) a : b = b : c a b (ii) ! b c ( iii) b 2 ! ac $ b !

middle term/means

ac

mean proportion a b b= c

Condition of proportionity ad = bc eg : are 20 , 30, 40, 50, in proportion. 20 × 5 ¹ 30 × 40 Not in proportion

Third proportion

eg : 8, 12 and x in continued proportion then 8 12 12= x Þ

Proportion

x = 12 ×8 12= 18

Continued Proportion

eg : find fourth pupoltion to 15, 20 and 30 15 30 20 = x Þ

× 20 x = 3015 = 40

eg : No. of chair 9 7 #

eg : mean proportion b/w 2 and 8

Cost of chair Rs. 720 Rs. x

b ! 2% 8 !

2% 2% 2% 2 ! 4

9 Rs. 720 7 % 720 ! $ x! 7 Rs. x 9 Rs. x = Rs. 560

K K A=K 3 ; B = 2 ; C= 5

L.C.M. (3,2,5) = 30 ×K A =10 10 × 3

;B

15 × K = 15 × 2 ;

C

= 66 ×× K5

\ A : B : C = 10 : 15 : 6

If a, b, c, d are in continued proportion then a b= c = b c d

PERCENTAGE eg : Meeta saves Rs 400 from her salary. If this is 10% of her salary. What is her salary ; Sol. Let her salary be Rs x 10% of Rs x = Rs 400

10 % Rs. x ! Rs.400 100 400 % 100 Rs. x ! Rs ! Rs.4000 10

eg : 5% of 25 5 % 25 ! 1.25 100

eg : Total students in class = 25 Number of girls = 15

% of girls !

eg : 20% less than 70 20 20% of 70 ! % 70 ! 14 100

15 % 100 ! 60% 25

% of boys = 100 - 60 = 40%

20% less then 70 = 70 - 14 = 56 eg : 10% more of 90 10

In c re a s e d Va lu e & O rig in a l v a lu e % 100 O rig in a l v a lu e

10% of 90 = 100 × 90 = 9 10% more of 90 = 90 + 9 = 99

eg : School team won 6 games this year; 4 games last year; increase = 6 - 4 = 2 2 Increase% ! % 100 ! 50 % 4

eg :

Ram's Report Shyam's Report Total : 320/400 Total : 300/360 Percent : 80% Percent : 63.33%

eg : A's salary is 50% more than B's salary If B's salary = Rs. 100 50 $ 50% of B' s salary ! % 100 ! Rs.50 100 A's salary = Rs 100 + Rs.50 = Rs 150 B's salary is less then by A's

salary by

50 1 % 100 ! 33 % 150 3

Original value & Decreased value % 100 Original value

In 10 years number of illitrate persons in a countery decreased from 150 lakh to 100 lakhs decrease = 150 - 100 lakh = 50 lakh 50 1 decrease% ! % 100 ! 33 % 150 3

Percent into Ratio

25% $

25 1 ! ! 1: 4 100 4

To convert Percent into Fraction

30% $

Multiply by 100 Fraction into Percent 3 3 (i) $ % 100 ! 60% 5 5 3 3 300 6 (ii) $ % 100 ! % ! 42 % 7 7 7 7

Ratio into Percent 3 3 3:8$ $ % 100 ! 37.5% 8 8

30 3 ! 100 100

Decimal into Percent 67 6. 7 $ % 100 ! 670% 10

(i) 6% ! Percent into decimal

6 ! .06 100

(ii) 0.5% !

0 .5 ! .005 100

PROFIT & LOSS

If C.P. < S. P. gain = S.P. -C.P. gain%!

Profit Loss

S.P. & C.P. gain %100 ! %100 C.P. C.P.

loss% !

Cost Price (C.P.) :Buying price of any item Selling price (S.P.) : Price at which you sell

eg : A toy bought for Rs. 72 Sold for Rs. 80 Since S.P. > C.P. gain = Rs.80 - Rs.72 = Rs. 8 gain% !

If C.P. > S.P loss = C.P. - S.P

eg : Cost . Price of T.V. = Rs. 12500 S.P. of T.V. = Rs 12000 Since C.P. > S.P. loss = Rs. 12500 - Rs.12000= Rs.500

Gain % ; Profit% always calculated on C.P.

8 100 1 % 100 ! % ! 11 % 72 9 9

loss% !

C.P. % (100 " gain%) 100

eg :C.P. = Rs. 540 gain % = 10% Rs. 540 % (100 " 10) 100 540 % 110 ! Rs. ! Rs.594 100

S.P. !

Rs. 500 % 100 ! 4% Rs. 25000

Loss% Given

Gain% Given

S.P. !

C.P. & S.P. loss ! % 100 C.P. C.P.

C.P. !

S.P. % 100 (100 " gain)

S.P. !

S.P.= Rs 4025 gain% = 15

eg . C.P. of a cooler = Rs. 6200 loss% = 15%

Rs.4025 % 100 C.P. ! 100 " 15

S.P. Rs.

4025 % 100 ! Rs. 115

Rs.

= Rs.3500

Also S.P. = C.P. + gain C.P= S.P. - gain If C.P. = Rs. 72 gain =Rs. 8 S.P. = Rs. 72 + 8 = Rs. 80

If S.P. = Rs 80 gain = Rs. 8 C.P. = Rs.80 - Rs. 8 = Rs. 72

C.P. % (100 & loss%) 100

gain% !

2 % 100 ! 20% 10

S.P. % 100 (100 & loss%)

S.P. of a washing machine = Rs.13500 loss % = 20%

6200 % (100 & 15 ) 100

6200 % 85 ! Rs. 5270 100

eg : S.P.of 10 pens = C.P. of 12 pens C.P. of one pen = Rs. 1 C.P. of 12 pens = Rs.12 S.P. of 10 pens = Rs. 12 C.P. of 10 pens = Rs. 10 S.P. of 10 Pens > C.P. of 10 pens gain- Rs. 12 - Rs.10= Rs. 2 (on 10 pens)

C.P. !

C.P. ! Rs.

Rs.13500 % 100 100 & 20

13500 % 100 ! Rs.16875 80

also S.P. =C.P. - loss C.P. = S.P. + loss If C.P. = Rs. 12500 loss = Rs.500 S.P. =Rs. 12500-Rs.500 =Rs 12000 If S.P. = Rs.12000 loss = Rs. 500 C.P. = Rs. 12000 + Rs.500 = Rs. 12500

SIMPLE INTEREST

P = Rs. 7200 R = 18% P.A. T = 3 years

eg : A sum of money doubles itself in 8 years what is the rate of interest. Sol. Principal = P ; A = 2P T = 8 years. S.I. = 2P - P = P

S.I. !

eg : S.I. = Rs. 450 Time = 3years Rate = 5% P.A.

Rs.7200 % 18 % 3 100

= Rs. 3888 Amount = P + I = Rs. 7200 + Rs. 3888 = Rs. 11088

P % 100 R! ! 12.5% P% 8

A = P + S. I.

P!

Simple Interest S.I.

Interest S.I.

Additional money paid by borrower to the lender for the money used

P = Rs. 2000 R% = 14% P.A. T = 5 years Rs. 2000 % 14 % 15 S.I. ! 100 = Rs. 1400 P = Rs. 3700

1 5 T ! 2 years ! years 2 2 1 11 R ! 5 %P.A. ! % 2 2 S.I. !

Rs.3700 11 5 % % 100 2 2

= Rs. 508.75

S.I. % 100 R% T

Principal (P)

Amount (A)

P%R% T 100

Rs.450 % 100 ! Rs.3000 5% 3

Cash borrowed by you from the lander is called principal/ sum borrowed.

Sum of Interest and the principal

If after certain years sum of money : (A) Doubles Þ A= 2P Þ S.I. = 2P-P = P (B) Triples Þ A = 3P Þ S.I. = 3P-P = 2P

S.I. !

P!

Time for which money is barrowed (calculated in years) S.I. % 100 T! R% T

Rate (R) Percent

Sum paid for use of Rs. 100

S.I. % 100 R% ! P% T

Time (T)

If T is given in days then T !

no. of days years 365

If T is in m o n th s , n o . o f m o n th s ) Then T ! * 'y e a rs 12 + (

P = Rs. 2400 R% = 5% P. A. S.I. = Rs. 240 Rs.240 % 100 Time ! Rs.2400 % 5 = 2 years

S.I. = Rs. 280 P = Rs. 56000 T = 2 years R% !

Rs.280 % 100 1 ! % 56000 % 2 4

eg : P = Rs. 1200 R = 6% P. A.

T ! 146 days ! S.I. ! Rs.

146 2 ! years 365 5

1200 % 6 2 % 100 5

= Rs. 28.80

eg : P = Rs 2400 R = 8% P.A. T = 6 month = 6 year 12

Rs.2400 % 8 6 % 100 12 ! Rs. 96

S.I. !

Solve :

2 % 3 4 % 25 9% 4

4

!

5

2

3

10 m= 1km

5 "1

4

5

2% 3 % 2 2 %3 ! 2 3 2 % (2 2 ) 2 3 % 24 = 26-4×34-2 = 22×32 = 4 × 9 = 36 4

10 cm= 1km

a × a × a....×a = an base a multiplied n times exponential notation/power notation

3

2

an®'a' raised to power n or th n power of 'a' 102®10 squared or 10 raised to power 2 3 10 ®10 cubed or 10 raised to power 3

,2 ) ,2 ) * ' * ' 2 3( 3 ,2 ) ,5 ) + ! ! + (1 ! * ' - * ' 4& 3 3 ,5 ) ,5 ) + ( +6 ( * ' * ' 6 + ( +6 ( !

To make vary large number's Easy to read, understand and compare ,we use exponents a = a1 a × a = a2 a × a × a = a3 exponent/inden/power

,2 ) ,5 ) * ' % * ' 3 ( + +6 ( eg : 4 2 ,5 ) ,2 ) * ' % * ' +6 ( +3 ( 4& 2

Exponent

22 6 4 6 8 % ! % ! 3 2 5 9 5 15

eg : (i) 3x = 81 = 3 × 3 × 3 × 3 = 34 Þ x=4 (ii) (-5) x = -125 = (-5) × (-5) × (-5) = (-5)3 Þ x = 3

Express 432 as product of powers of prime no. Prime factorisation 2 432 2 216 2 108 2 54 3 27 3 3

9 3 1

= 2 × 2 × 2 × 2 × 3 × 3 × 3 = 24 × 33

multiplication powers subtracted

division m

n

m–n

(2) a ÷ a = a

[m > n]

(1) am × an

=

am+n

Powers added

base same

eg : 23 × 25 = 23 + 5 = 28

base same eg : 35 ÷ 32 = 35-2 = 33

(6) aº = 1 m m m-m 1= a ÷ a = a = aº eg : 2º + 3º + 4º = 1 + 1 + 1 = 3

m

(3) (am)n

am×n

=

Low's of Exponents

am ,a ) (5) * ' ! m ! am - bm b b + ( 4

24 ,2 ) eg :. * ' ! 4 ! 2 4 - 3 4 3 +3 (

eg : (53)7 = 53×7 = 521

base different eg : (-2)4 × (-3)4 = (-2 × -3)3 = (+6)4

eg : Compare 8

(a) 2.7 × 10

; 1.5 × 10

8

Same Compare 8 8 2.7 × 10 ; 1.5 × 10 different 8

(b) 2.7 × 10

12

; 1.5 × 10

Compare 8

2.7 × 10

12

< 1.5 × 10

power same (4) am × bm = (a × b)m

eg : Express in exponential notation 4 (i) 625 = 5 × 5 ×5×5 = 5 3 (ii) -27 = (-3) × (-3) × (-3)= (-3) 3 2 (iii) a × a × a × b × b × y × y = a b y2 Note : b3a2 ¹ b2a3

Standard form/Scientfic notation Þ A × 10n 1.0 < A < 10.0 ; n ® integers eg : 59 = 5.9 × 101 2 590 = 5.9 × 10 3 5900 = 5.9 × 10 Expand form : 47561 4 3 2 0 = 4 × 10 + 7 ×10 + 6 ×10 + 1 × 10 eg : 1000= 10 ×100 = 10 × 10 × 10 = (2×5)×(2×5)×(2×5) = 23 × 53

(corresponding) 2 sides and the included = 2 sides and the included angle of a triangle angle of other triangle

Congruent Fig

A

P

(Corresponding) L

P

Equal Shape and Equal Size

80º

60º 60º Q

R

C

B

In DPQR and DABC PQ = BC = 5cm ÐP = ÐB = 60º PR = AB = 4cm

QR = AC ÐR = ÐA ÐQ = ÐC

R

30º DPQR

DBCA [by S.A.S property]

3 sides of a = triangle

Q

Corresponding three sides of other triangle X

A

30º

Criteria for Congruence of Triangle

R

30º A P B mÐPQR = mÐ ABC ÐPQR Ð ABC

N

M

In DPQR and DLMN ÐQ = ÐM = 30º QR = LM = 5cm ÐR = ÐL = 80º DQRP DMLN by A. S. A. Property

C

P

30º

80º

Q

then by c.p.c.t. ÐP = ÐN RP = LN QP = MN

In rt. angle triangle's The hypotenuse = and one side of a triangle X

A

L

(cosrresponding) hypotenuse and one side of other triangle

5cm B

6cm

C Y

In D ABC & DXYZ AB = XY = 3cm BC = XZ = 6cm AC = YZ = 4cm D ABC DYXZ [by S.S.S property]

z

80º

60º

c 8cm

R

B

C

90º D PQR

D ABC

Y

Corresponding vertex : A«P ; B«Q ; R«C Corresponding sides : AB = PQ ; BC = QR ; AC = PR Corresponding angles :Ð A«ÐP ; ÐB«ÐQ ; ÐC«ÐR

4cm Z In rt. DXYZ and rt. DLMN ÐY = ÐN = 90º hyp. XZ = hyp. LM = 5cm YZ = LN = 4cm DYZX DNLM by R.H.S. propery.

90º M Then by c.p.c.t ÐX = ÐM ÐZ = ÐL XY = MN

P

a

b

Q

then by c.p.c.t. ÐA = Ð Y ÐB = Ð X ÐC = ÐZ

A

B

Z

4cm

C

Q

x

40º y

Given : DABC DPQR to Find : x,y,z, a, b,c. So, (i) AB = PQ (iv) ÐA = ÐP 6cm = z a = 80º (ii) BC = QR (v) ÐB = ÐQ 8cm = y b = 60º (iii) AC = PR (vi) ÐC = ÐR 7cm = x c = 40º

R

A

eg : In DAOB and DQOP AO = OQ ÐAOB = ÐQOP P A OB = OP

O

B

Q DAOB DQOP by S.A.S. property

C

A

B

If AB = AC C Þ ÐB = ÐC

Q B In DACQ and DABQ ÐC = ÐB = 90º hyp. QA = hyp. QA CQ = BQ DACQ DABQ by R.H.S. property

N

Integers

Addition

(ii) –27 + (–3) = – (27 + 3) = –30 (b) unlike sign

- - - - - - - - - - - - – 4, –3, –2, –1, 0 , 1, 2, 3, 4, - - - - - - - - -

(i) –27+ (3) = – 24 (sign of bigger absolute value) (ii) 27 + (–3) = 24

Negative integers

Positive integers Neither positive Nor negative Multiplication

Subtraction b - (a) = b + (-a) b - (-a) = b + (+a) eg : (i) 27-3= 24 (ii) -27 - (-3) = - 27 + 3 = – 24 (iii) 27 –(-3) = 27 + 3 = 30 (iv) –3 –(-27) = -3 + 27 = 24 (v) –27 –(3) = –27–3 = –30 Properties of addition and subtraction of Integers (i) Closer property : a and b are two integers :then (a + b) and (a – b) are also integers (ii) Commutative property (a) a + b = b + a eg : 3 + 4 = 4 + 3 = 7 ; -3 + 4 = 4 + (-3) = 1 (b) a – b = b – a eg : –3 –(4) = –7 ; 4 –(–3) = 4 + 3 = 7 (iii) Associative property (a) (a + b) + c = a + (b+c) eg :[(–3) + (–9)] + 17 = (–12) + 17 = 5 (–3) + [(–9) + 17] = (–3) + 8 = 5 (b) (a – b) – c / a – (b – c) eg :[(–3) – (–9)] – 17 = 6 – 17 = – 11 (–3) – [(–9) –(17)] = –3 – [–26] = –3 + 26 = 23 (iv) Additive identity a + 0 =0 + a =a [o is additive identify] eg : 3 + 0 = 0 + 3 = 3 (v) Additive inverse a + (–a) = (–a) + a = 0 [ a additive inverse (–a) ] eg : 3 + (–3) = (–3) + 3 = 0

Division

[All natural numbers + 0 + negative of counting numbers]

When integers are of (a) like sign (i) 27 + 3 = 30

When integers are (a) of unlike sign: product is negative a × (– b) = –(a × b) eg . 2 × (– 6) = –12 (–a) × (b) = –(a × b) eg .(– 2) × ( 6) = –12 (b) of like sign: product is positive a × b = a × b eg : 2 × 5 = 10 (–a) × (–b) = a × b (–2) × (–5) = 10

Properties : If a, b and c are integers (i) Closure property a × b = integer eg : (-3) × 10 = –30

When the dividend and divisor are of unlike signs , the quotient is negative other wise positive , If a and b are ( + ve) then (i) (-a) ÷ b = – (a ÷ b) (–6) ÷ 3 = –(6 ÷ 3) = – 2 (ii) a ÷ (–b) = – (a ÷ b) 15 ÷ (–5)= –(15 ÷ 5) = ÷ 3 (iii) (–a) ÷ (–b) = – (a ÷ b) (–4) ÷ (–2)= (4 ÷ 2) = 2

(iv) 8÷ 4 = 2 Properties :

(i) a ÷ o = not integer [Not closed] (ii) If a ¹ o then a ÷ a = 1 ; 3 ÷ 3 = 1 a÷1=a;3÷1=3 a ÷ (–a) = (–a) ÷ a = –1 a ÷ (–1) = –a a÷a=0

(ii) Commutative a × b = b ×a eg : (-50) × 2 = 2 × (-50) = 100 (iii) Associative (a × b) × c = a × (b × c) eg : - 1 × (3 × 2) = -6 (-1 ×3) × 2 = -6 (iv) 1 is multiplicative identity ie 1 × a = a × a 1 × a = a × 1 = a : eg : 6 × 1 = 1 × 6 = 6 (v) for integer a : a × 0 = 0 × a = 0 eg : 500 × 0 = 0 (vi) a × -1 = ( - 1) × a = – a eg : (–3) × 1 = (–3) = 3 (vi) Distributive property a × ( b × c) = a × b + a × c ; a × (b – c) = a × b – a × c 3 × [4 + 5] = 3 × 4 + 3 × 5 3 × [4 – 5] = 3 × 4 – 3 × 5

Operation precedence [BODMAS] I. (bar) I. – B®Brackets II. ( ) (small) II. – O®Of III. { } (curly) III. – D®Division Vi [ ] (big) IV. – M®Multiplication V. – A®Addition done VI. – S®Subatract simultaneously

1 = 4

1 4

4 16

FRACTION

5 1 !1 4 4

+

Decimal fraction Denominator is 10,100 , 1000, 10000 - - - 1 3 eg : ; e.t.c. 10 100

Fraction as an operator 1 of 1 2 1 1 % 1! 2 2

1 of 2 2

eg :

converted into

7 2 ! 1 5 5

; 14

4 3 0 7 7

eg :

5 5 1 4 3

eg :

1 2 1 3 5

Proper fraction numerator < denominator

Improper fraction denominator > numerator 7 eg : 5

3

1 1 kg " 1 kg 4 6

L.C.M. (4,6) = 12 5 (13 % 3 " 7 % 2)kg 53 ! kg ! 4 kg 12 12 12 9 16 1 1 1 1 ! % % eg : 4 % 3 % 2 5 3 2 5 3 9 % 16 % 1 24 4 ! ! ! 4 2% 5% 3 5 5

Find L.C.M. of denominators L.C.M. of (3,5) = 15 1% 5 5 2% 3 6 ! ; ! 3 % 5 15 5 % 3 15

eg : After spending 4 of her money, Nidhi 9 had Rs. 45 left. How much did she spend

2 14 % 3 " 2 44 ! ! 3 3 3

4/9

Operation on Fraction

5/9

Rs. 45

Rs. 36

Multiplication

Subtraction 13 7 kg " kg 4 6

Compare denominator

3 3% 2 6 ! ! 5 5 % 2 10 3 6 9 ! ! 3 3 % 2 9 = 5 10 15 ! ! 5 5 % 3 15

1 % 2!1 2

Addition

Compare numerator Denominator same Numerator same

Equivalent fractions

5 7

Mixed fraction [combination of whole no. and a proper fraction] 1 of 1 2 1 1 % 1! 2 2

eg :

2 1 12 5 2 – 1 ! – 5 4 5 4

(i) Fraction by fraction Product of numerator ! Product of denomin ator

L.C.M . (5,4) = 20

3 5 3 % 5 15 eg : % ! ! 4 7 4 % 7 28

12 % 4 " 5 % 5 48 – 25 23 3 ! ! !1 20 20 20 20 14 2 2 1 eg : 1 % 14 – 2 % 2 11 3 5 4 15 % 44 12 % 9 – 11 % 3 5% 4

15 44 12 9 % – % 11 3 5 4 20 27 – 1 5

20 % 5 – 27 % 1 73 – 5 5

(ii) Fraction by whole number 3 3 2 3%1 3 %2! % ! ! 4 4 1 2%1 2 Product of two proper fraction is less than each of the fraction. *Value of ,product of two improper fraction is more then each of the two factions

*

Division (i) 3 -

5 3 4 12 ! % ! 4 1 5 5

5 5 3 5 1 5 (ii) - 3 ! - ! % ! 4 4 1 4 3 12 1 5 1 6 2 (ii) - ! % ! 3 6 3 5 5 * Reciprocal or multiplicative inverse of a non-zero faction a b a b is ie % !1 b a b a * Multiplicative inverse of 1 is 1 Reciprocal of 0 does not exist.

Comparing Decimals

Convert

Decimals

eg : Arrange in ascanding order 2.01 , 1.9, 0.95, 3.2 and 2.758 I Change to like decimals 2.010 , 1.900,0.950, 3.200, 3.200, 2.758 II write corresponding numbers 2010, 1900, 950,3200,2758 III Now are angle 950, 1900, 2010, 2753, 3200 IV 0.95, 1.9, 2.01, 2.750, 3.2

(i) decimal into fraction eg : 3.75 ! Decimals are fractions whose denominators are 10,100,100 e.t.c. eg : 23 . 005 whole number part

(ii) fraction into decimal 1 1% 25 25 (i) 9 ! 9 " ! 9" ! 9.25 4 4 % 25 100

Decimal part

(ii)

Addition like decimals eg : Add 3.01 , 2.58 and 6.9

unlike decimals

same decimal place 5.25 , 6.89

different decimal place 5.25 , 6.893 Subtract : 8.96 from 25.1 25 . 10 & 8 .96 16 . 14

Multiplication

decimal by decimal (i) First multiply as whole no (ii) Count the no. of digits to the right of the decimal point (iii) Add the no. of digits counted (iv) Put decimal point in the product by counting the digits from the right most place eg : 256. 7 × 0.005 = 3 decimal place

7 7% 8 56 ! ! ! 0.0056 125 125 % 8 1000

Subtraction

3 . 01 Þ Keep decimal points under each other 2 . 58 Þ Write digts in the correct place value columns 6 . 9 Þ Add like whole numbers. 12 . 49

1 decimal place

375 15 ! 100 4

by 10,100, 1000 move decimal point in the number to the right by as many places as there are zeros. over 1 eg 0.53 × 10 = 5.3 0.53 × 100 = 53 0.53 × 1000 = 530

by 10, 100 or 1000 -[shift the decimal point] 15.6 ÷ 10 = 1.56 [1 place to the left] 30.9 ÷ 100 = 0. 309 [2 place to the left]

by decimal no. 0.165 by 1.5

1+3 = 4 decimal place

eg :- cost of 2.5m cloth = Rs. 41.25 cost of 1m cloth = Rs. 41.25 ÷ 2.5 = Rs. 16.50

like decimals , then subtract like whole no.

Division.

= 1.2835

eg :- cost of 1m cloth = Rs. 15.5 cost of 1.5m cloth = Rs. 15.5 × 1.5 = Rs. 23.25

Þ change to

!

165 100

!

1 [1 6 5 - 1 5 ] 100

!

11 100

15 10

!

0 .1 1

by whole number 1.575 ÷ 35 =

1 [1575 ÷ 35] 1000

= 1 [45] = 0.045 1000

Rational Numbers

Rational number on number line

A number that can be expressed in the form 17 7

(i) To represent ( i ) 0 A 0

P

17 7

eg :

divide into 7 equal parts

(ii)

&9 4

Q

B

O

-9

&9 4

0

divide into 4 equal parts

&3 &4 ; 5 &5

(iii) If p ! 0 the

eg :

&1 3

;

&

&3 5 & 3 (& 5) " ! " 8 & 12 8 12

Add the additive inverse ie eg

:

a c a ,& c ) & ! "* ' b d b + a (

S u b tr a c t

& 11 24 & 11 24

& 5 , & * 36 + & 10 " 33 23 ! ! 72 72 !

&5 fr o m 36 & 5 11 ) " ' ! 3 6 2 4 (

eg : Additive inverse of (i)

& 15 15 is 2 2

(ii)

4 &4 is 9 9

*Two positive rational numbers can be compared as fractions *(i) Compare the rational no.s ignoring their negative signs (ii) Then reverse the order 7 5 &7 &5 eg : 1 $ 0 5 3 5 3 * A negative rational no. is always less then positive rational no. &2 1 eg : 1 7 2 Division

Product of Rational No. product of numerators ! product of deno min ators

&3 % 5 (& 3 ) % 2 % ! 8 % 7 %

eg :

a c a d 7 c 4 - ! % such that / 02 5 b d b c 6 d 3

2 (& 5 ) % 7 8 (& 5 ) 3 ! 8 28

Simplest (Lowest) or standard form

1 33 52 & 34 ! % 17 & 34 17 33 & 104 5 ! ! 3 33 33

eg : 3

Multiplicative inverse of

p q is where p, q / 0 q p 3 5 eg : Re ciprocal of is 5 3

( & 3) % 3 " ( & 5) % (2) & 19 ! 24 24

p &p p (&p) is [ ie " ! 0] q q q q

Comparison

Multiplication

L. C. M. of 8 and 12 = 24

Additive Inverse of

&3 4 or 5 &5

3 &9 &1 &5 &9 &7 &6 &5 ! ; ! # 1 1 1 5 15 3 15 15 15 15 15

Subtraction

a c " [Find L. C. M. of b and d] b a

eg :

* A fraction is a rational number, but a rational number may or may not be a fraction 3 3 eg :® rational no. + fraction ; & ®rational no but not a fraction 5 5 * There are infinite rational no's between two rational no's :

Addition

(ii)

p 0 0 ! eg , q 8 &6

1 0 5 &9 * Natural numbers + whole numbers + integers + fractions are rational no.s eg : 1, 1 , 6 , 1000 e.t.c.

eg : Rational No. b/w &53 and

Convert each of them into number with a positive denominator. a b a " b [when denominator (i) " ! are same] c c c & 11 7 & 11 " 7 & 4 eg : " ! ! 5 5 5 5

but q / o

(ii) Negative rational number : p and q both are have opposite sign

(i) Positive rational number : p and q both are positive or negative

17

p integer q integer

Equivalent Rational no.

p p%m p p-m ! also ! w here m ! non & zero int eger q q% m q q-m

If denominator is a positive integer and

2 2 % 5 10 ! ! 3 3 % 5 15

denominator and numerator have no common factor other then 1 6 6 - ( & 3) &2 is ! eg : Standard form of & 15 & 15 - ( & 3) 5

eg

divide by H.C.F of 6 and 15

eg : &

equivalent Rational no.

;

5 5 % &2 & 10 ! ! 9 9 % &2 & 18

equivalent Rational no.

3 9 9% 11 ! ;(&3) % x ! 9 % 11 ; x ! ; x ! &33 11 x (&3)

PROBABILITY Collection of all possible out comes eg : In a throw of a die S. S. = (1,2,3,4,5,6)

Sample Space

different possibilities which can occure eg : In tossing a coin outcome are Head and Tail

Any kind of activity eg : Tossing a coin

Probability

Deals with the measurement of uncertinity of the occurrence of some event in terms of percent or ratio

Outcome

Event

Experiment

A bag contains 5 red balls, 8 white balls, 4 green balls and 7 blackballs. If one ball is drawn at random, then probability that it is :

Equally likely outcomes Sort of an experiment eg : getting six in a throw of a die

There is equal uncerteinty of each outcome of an experment

Probability of an event A = P (A) !

Probability of getting a number greater than 7 0

0 < 1/4

Poor No Chance Chance

Num ber of outcom es in favour of A Total num ber of possible outcom es

Probability of getting Tuesday after Monday

P (A) < 1 1/2 Even Chance

3/4 Good Chance

1 Sure Chance

7 24 4 1 ! (b) Green P(G) = 24 6 (a) Black i.e. P(B) =

(c) P(Not red) = 1–P (Red) ! 1 &

5 24 & 5 19 ! ! 24 24 24

When a coin is tossed Total number of outcomes = 2 ie. T,H 1 Probability of getting head P(H) = 2 1 Probability of getting tail P(T) = 2

Statistics

Range Difference of the largest and the smallest observation eg : 15, 33, 24, 47, 91, 82 range = 91 – 15 = 76

Raw Data /Crude data Unorganised data Each entry in row data is called observation

Frequency Distribuation Table

Array/arranged data (organised data)

Measure of central tendency

Mean

9x

X!

i

n

8f % x 8f

xi

f

f×x

25

5

125

27

4

108

28

5

140

30

6

180

32

3

96

20

33

2

66

15

sun

3 " 6 " 9 " 12 " 15 x ! 9 x! 5 Quartarly Half yearly 25

8 f ! 25

10

8 fix ! 715

[divides the distribution into two equal parts] If n is number of observation

eg :

(i) When n is odd median th

= ,* n " 1 )' observation.

Number

7

10

11

12

13

17

Frequency

2

1

1

3

2

1

Mode = 12 [Highest frequency = 3]

+ 2 (

Favourate Colour No. of Students

3,1,4,3,6,5,9,5,3 arreange in ascending order 1,2,3,3,4,5,5,6,9 th

, n " 1) median ! * ' observatio n + 2 (

Ashish Arun Kavish Maya Rita

th

Ashish

Arun

Kavish

Maya

Rita

Quartely

10

15

12

20

9

Half yearly

15

18

16

21

15

Red 43 60 50 40

th

,n ) (ii) n is even median = , n ) * ' ob. " * " 1' ob. 2 + ( +2 ( 2 eg : 11, 10, 12, 9, 8, 16, 15, 14 arreange : 8, 9, 10, 11, 12, 14, 15,16 th

6 6 4 3 1

[The value which occurs the most or has highest frequency]

th

Students

0-10 10-20 20-30 30-40 40-50

Mode

, 9 " 1) !* ' observatio n ! 4 + 2 (

5

Grouped Number Tally Frequency Marks

Ungrouped Number Tally Frequency Marks 1 2 2 4 3 2 4 6 5 3 6 3

Median

x " x 2 " x 3 ......... " x n x! 1 ! n Mean of 3, 6, 9, 12, 15

Number of times an observation occurs in eg : 3,5,4,9,9,19 , frequency of 9 is 3

Frequency

Science of collection presentation , analysis and interpretation of numerical data

th

,8 ) ,8 ) * ' ob. " * " 1' ob 4 th ob. " 5 th ob. 11 " 12 2( + +2 ( ! 11 . 5 = Median ! .= 2 2 2

Green 19

Blue 55

Yellow Orange 49

34

Scale : 1 unit : 10 Students

30 20 10 Red Green Blue yellow orange

Symmetrical Figures

P a Q

a

Q

Rotation terns an object about a fixed point called Center of rotation

a R Clock wise Rotation (i)

The angle of turning during rotation is called angle of rotation 120º P

R

R

Order of rotational symmetry = 3

(ii)

Q

P 120º (iii)

Line of symmetry

Rotational Symmetry

A Line which divides the figure into two congruent parts is called line (axis) of symmetry

Symmetry

equilatral triangle has rotational as well as line of symmety

60º

60º

60º

Equilateal Triangle

Square

Regular Pentagon

Regalar Hexagon

Number of lines of symmetry = Number of sides of that polygon of a regular polygon

Infinite lines of Symmetry

SOLID SHAPES NETS

Solid Shapes Objects that occupy space and have three dimensions [length, breath and heigh or depth]

height Breadth Length Cuboid

h

h

Pyramis Sprere

Cone

Euler's formula V+F–E=2 Vertex Face Edges eg : For Triangular pyramid V=4;E=6;F=4 \ 4+4-6=2 Vertex Vertex Face

Face

Edge Edge

2 - D Repesentations of a 3- D figure

Perimeter Sum of all sides. 1 ( 2: r ) " 2r 2 22 P! % 2 " 2 % 2 units 7 44 72 P! " 4! units 7 7 P!

1 2 1 2

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