Maths Concept Map 8 CBSE

Triangle & It's Properties

Concept Maps

Class VII

Lines & Angles

Algebraic Expression

Area

Ratio Symbol

Ratio & Proportion

Percentage

Profit & Loss

Simple Interest

Exponent

Rational Numbers

Probability

Statistics

Symmetry

Solid Shapes

Perimeter

Congruent Triangle

Integers

Fraction

Decimals

A C B A C B Q P R x x X Z 60º 60º Y Z C B 70º 50º 60º C Q P 150º 15º 60º ₁₅ º R (ii) B A (iii) (i) Side Angle : AB, BC, CA

: BAC, ABC, BCA Closed

curve made up of three line segment.

Classification

ScaleneD:

No two sides are equal

Isosceles : , , D

Two sides are equal base angles opposite to equal sides are equal

Equilateral : , D

All sides are equal

each angle has measure 60º

Vertices : A, B, C Acute angleD: all angles :-greater than 0º less than 90º Obtuse angle : , D an angle :-greater than 90º less than 180º Right angledD :-an :-angle equals to 90º P Q R A B D C X Z Y A C 70º 80º B R 50º Q 150º S 100º P

Exterior angle property : PRS = QPR + PQR

Ð Ð Ð

30º

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. (i) (ii) (iii) 3cm 3cm 90º 30º

(i) (ii) (iii)

Acute (0º < q< 90º)

Complementary angles

Two angles whose Sum is 90º

Supplementary angles

Two angles whose sum is 180º

Types of angles

Line

Right (q = 90º) Obtuse (90º <q< 180º) Straight ( = 180q º) Reflex (180º <q<180º) Complete ( = 360q º) l m p q

Line : have infinite length

Parallel lines : never intersects

Intersecting lines

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

4x x 2x 5x 5x + 2x + x + 4x = 180º 3x x 2x 5x 4x

Sum of all angle around a point is 360º x + 2x + 3x + 4x + 5x = 360º 1 ₂ 4 3 6 8 5 7

Angles made

by transversal

Corresponding angles (1, 5) (2, 6) (3, 7) (4, 8)

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

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

Alternat Ext. angle

(1, 8) (2, 7) If lines are ||el then

Corresponrding angles are equal

Alternate angles are equal

x y x + y

common arm

Vertically opp. angles

1 2 3 4 Adjacent angles O A B C opp.arm Common vertex Ð1 =Ð3 ;Ð4 =Ð2 30º 150º P Q R O Linear pair

two adjacent angles whose sum is 180º

Sum of Co. Int. angles is 180º P m l 1 2 Ð1and 2 are adjacent angles Ð

Unlike terms:

Different algebraic factors 3x y = 32 × x × x × y 3xy = 32 × x × y × y

Same algebraic factors 6xab = 6× x ×a ×b -3abx = -3 × a × b × x

addition

Subtraction

9ab + 5ab = (9 + 5) ab = 14ab

9ab– 4ab = (9 – 4)ab = 5ab Value of Expression at x = a [put x = a] eg : value of x -5x2 at x = -1 (-1) -5(-1) = 1 + 5 = 62 Using Algebraic Expression

Formulas Rules for no.

patterns Perimeter Equilateral triangle 3× a ; where a is side Square 4× l ; where l is side Regular pentagon 5× s ; where s is side Triangle = × b × h Rectangle = l× b 1 2 Successor of n n+1 odd no (2n +1) or (2n-1) 3, 6, 9, ...3n... Types of Algebraic Exprission

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

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

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

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

x + 3y ; x + 3

2 Standerd form 2x + 3y x + 3x - 5x + 62 2 2 (i) 3xy + 5x - 5x + 62 2 (ii) 6 - 5x + 5x +3xy2 2 Mathematical Operation Constants Variables (x, y, z,a, b,t,...) Its value is not fixed

[ +, - ,× , ÷ ]

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

Expression : 9x - 3yx + 102 Terms 9x2 -3yx 10 x x -3 y Factors Factors Numerical coefficient (-3) In -3x y 2 Coefficient of x y2 -3x2 x _-3y is is is is -3 7 -3xy x2 Even no. =2n Area Like Terms : 9 x

ALGEBRAIC EXPRESSION

Rhombus : Trapezium Quadrilateral : Square Parallelogram Circle Triangle Rectangle h1 h2 A D C B d h h _a a a a d1 h1 h2 a unit h1 h2 h b a d a unit r R 2cm B A C a unit B A C a unit h1 B A C a unit h2 h3 h1 P R Q a 1 2a× h1 4unit 2unit 8 unit 2unit 8+2+2 P Q R S D C B A 2 2 4+2+2 eg.

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

l unit d

Area = a×h =1 b×h =2 c×h Sq. units3

A = h× (a +b) sq. units

Area = × diagonal × sum of offsets

= × d × [h + h ] sq. units1 2

Area = a× h sq. units = d × d sq. units1 2

Area = a×h sq. unit = b × h sq. units

base = Area÷ corresponding altitude

altitude = Area÷ corresponding base

1 2

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

(a) Painting / white washing (b) levelling

(c) Ploughing (d) grazing / watering

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

When to calculate Area :

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

Area = l× b sq. units

length = (Area÷ breadth) unit

breadth = (Area÷ length) unit

d = l + b2 2

base = 2Area÷ altitude

altitude = 2Area÷ base

1 2 1 2 1 2 3 4 eg. Area = r = Area÷ p p 2 =d 2 4 8

Area of Equilateral = (side)2

D

2 1

Area of cross road = (8× 2 + 4 × 2 – 2× 2) unit2 r r d2 2 1 2 1 2 1 1 cm = 100 mm 1 m = 10000 cm 1 hectare = 10000 m 2 2 2 2 2

AREA

Comparison of two ratios

Equivalent Ratio a and b should

be of same kind

Simplest form/lowest term Convert ratio into fraction

eg : > Compare numerator when Denominator same eg : < When Numerator same

Compare denominator 3 7 4 7 5 4 5 3 eg : < find L.C. M. of denominators L.C.M. of (3,5)=15 1×5 3×5 5 15 = 2×3 5×3 6 15 = ; 6 : 15 > 5 : 15 Þ 1 3 2 5 1 : 2 = 2 : 4 = 3 : 6 1 2 = 2 4 3 6 = or.1 2= 1×2 2×2= 1×3 2×3

To compare two quantities, the units must be the same eg : ratio of 3 km to 3m 3km : 3m = 3000 m : 3m =1000 : 1 eg : If A : B = 2 : 3 and B : C = 5 : 4 then A : C =A_B×B_{C =}2₃×5₄=5₆ A : C = 5 : 6 eg : divide Rs. 5000 in Ratio 3 : 5 Rs. 5000× = Rs. 5000 × = Rs. 1875 Rs. 5000× = Rs. 5000 × = Rs. 3125 3 3+5 3 8 5 8 5 3+5 5 7 B of Age A of Age : eg ! age of A = 7x age of B = 5x If a : b is in simplest form ÞH.C.F. (a , b)=1 eg : H.C.F of 2 and 3 is 1 H.C.F. of 18 and 27 is 9 18 27= 18÷ 9 27÷ 9 2 3 = a : b = anticedent b consequent a® ® 4 5 B of age A of age 6 x 5 B of age 6 x 7 A of age years 6 After ! " ! " ! 4 5 6 x 5 6 x 7 B of ages A of ages ! " " ! # eg : Let 3 A = 2B = 5C = K (K 0) 3 A = K ; 2 B = K ; 5C = K If 3 A = 2B = 5C the find A : B : C \ A =K₃; B =K₂ ; C=K₅ L.C.M. (3,2,5) = 30 A = ; B = ; C = A : B : C = 10 : 15 : 6 \ 10× K 10× 3 15× K 15× 2 6× K 6× 5 ¹ ¹

RATIO SYMBOL

Proportion

If a : b :: c : dÞ ad = bc or a b= c d If fourth proportion middle term/means extreme terms

Continued Proportion

If a, b, c are in continued proportion

ac b ac b ) iii ( 2 ! $ ! (i) a : b = b : c c b b a ) ii ( ! mean proportion a b= b c Third proportion Condition of proportionity ad = bc eg : are 20 , 30, 40, 50, in proportion. 20× 5¹ 30 × 40 Not in proportion

eg : find fourth pupoltion to 15, 20 and 30 = Þx = = 40 15 20 30 x 30× 20 15

eg : No. of chair Cost of chair

9 Rs. 720 7 Rs. x 9 720 7 x x . Rs 720 . Rs 7 9 % ! $ ! # Rs. x = Rs. 560 A =K₃ ; B =K₂ ; C=K₅ L.C.M. (3,2,5) = 30 A = ; B = ; C = A : B : C = 10 : 15 : 6 \ 10× K 10× 3 15× K 15× 2 6× K 6× 5 eg : 8, 12 and x in continued proportion then = Þx = =18 8 12 12 x 12× 128

eg : mean proportion b/w 2 and 8 4 2 2 2 2 8 2 b! % ! % % % ! If a, b, c, d are in

continued proportion then

a b= b c= c d

RATIO & PROPORTION

To convert

Fraction into Percent

Ram's Report Total : 320/400

Percent : 80%

Percent into Fraction

Percent into decimal

06 . 100 6 % 6 ) i ( ! ! 005 . 100 5 . 0 % 5 . 0 ) ii ( ! !

4

:

1

4

1

100

25

%

25

$

!

!

100 3 100 30 % 30 $ !

Ratio into Percent Decimal into Percent

% 670 100 10 67 7 . 6 $ % ! % 5 . 37 100 8 3 8 3 8 : 3 $ $ % ! % 60 100 5 3 5 3 ) i ( $ % ! % 7 6 42 % 7 300 100 7 3 7 3 ) ii ( $ % ! ! Shyam's Report Total : 300/360 Percent : 63.33% eg : 5% of 25 25 . 1 25 100 5 ! %

100

value

Original

value

Decreased

value

Original

%

&

1 0 0 v a lu e O rig in a l v a lu e O rig in a l Va lu e In c re a s e d % &

eg : School team won 6 games this year; 4 games last year; increase = 6 - 4 = 2

% 50 100 4 2 % Increase ! % !

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

400

.

Rs

x

.

Rs

100

10

!

%

4000

.

Rs

10

100

400

Rs

x

.

Rs

!

%

!

50 . Rs 100 100 50 salary s ' B of % 50 ! % ! $

eg : Total students in class = 25 Number of girls = 15 % 60 100 25 15 girls of % ! % ! % of boys = 100 - 60 = 40%

eg : In 10 years number of illitrate persons in a countery decreased from

150 lakh to 100 lakhs

decrease = 150 - 100 lakh = 50 lakh % 3 1 33 00 1 150 50 % decrease ! % ! 20% less then 70 = 70 - 14 = 56

Percent into Ratio

Multiply by 100 eg : 10% more of 90 10% of 90 =₁₀₀× 90 = 9 10 10% more of 90 = 90 + 9 = 99 eg : 20% less than 70 14 70 100 20 70 of % 20 ! % !

eg : A's salary is 50% more than B's salary If B's salary = Rs. 100

A's salary = Rs 100 + Rs.50 = Rs 150 B's salary is less then by A's

%

3

1

33

100

150

50

by

salary

%

!

PERCENTAGE

If C.P. < S. P. gain = S.P. -C.P. 100 . P . C gain 100 . P . C . P . C . P . S % gain ! & % ! % Also S.P. = C.P. + gain C.P= S.P. - gain

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

also S.P. =C.P. - loss C.P. = S.P. + loss If C.P. > S.P loss = C.P. - S.P 100 . P . C loss . P . C . P . S . P . C % loss ! & ! % Gain% Given 100 %) gain 100 ( . P . C . P . S ! % " _{(100 gain)} 100 . P . S . P . C " % ! S.P.= Rs 4025 gain% = 15 15 100 100 4025 . Rs . P . C " % ! 115 100 4025 . Rs % ! = Rs.3500 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 eg :C.P. = Rs. 540 gain % = 10% 100 ) 10 100 ( 540 . Rs . P . S ! % " 594 . Rs 100 110 540 . Rs % ! !

eg : A toy bought for Rs. 72 Sold for Rs. 80 Since S.P. > C.P. gain = Rs.80 - Rs.72 = Rs. 8 % 9 1 11 % 9 100 100 72 8 % gain ! % ! !

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)

% 20 100 10 2 % gain ! % ! 100 %) loss 100 ( . P . C . P . S ! % & eg . C.P. of a cooler = Rs. 6200 loss% = 15% 100 ) 15 100 ( 6200 . Rs . P . S % & 5270 . Rs 100 85 6200 . Rs % ! 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 % 4 100 25000 . Rs 500 . Rs % 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

Profit Loss

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

Loss% Given %) loss 100 ( 100 . P . S . P . C & % ! S.P. of a washing machine = Rs.13500 loss % = 20% 20 100 100 13500 . Rs . P . C & % ! 16875 . Rs 80 100 13500 . Rs % !

Simple Interest S.I.

Sum paid for use of Rs. 100

S.I. = Rs. 280 P = Rs. 56000 T = 2 years T P 100 .I . S % R % % ! % 4 1 2 56000 100 280 . Rs % R ! % % !

Interest S.I.

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

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

Time for which money is barrowed (calculated in years) T R 100 .I . S T % % ! If T is given in days years 365 days of . no T then ! eg : P = Rs. 1200 R = 6% P. A. years 5 2 365 146 days 146 T! ! ! 5 2 100 6 1200 . Rs .I . S ! % % = Rs. 28.80

Cash borrowed by you from the lander is called principal/ sum borrowed. eg : S.I. = Rs. 450 Time = 3years Rate = 5% P.A. 3000 . Rs 3 5 100 450 . Rs P ! % % ! T R 100 .I . S P % % ! P = Rs. 2400 R% = 5% P. A. S.I. = Rs. 240 5 2400 . Rs 100 240 . Rs Time % % ! = 2 years y e a rs 1 2 m o n th s o f . n o T T h e n m o n th s in is T If ' ( ) * + , ! eg : P = Rs 2400 R = 8% P.A. T = 6 month =6 year 12 96 . Rs 12 6 100 8 2400 . Rs .I . S ! % % ! Sum of Interest and the principal

A = P + S. I. P = Rs. 7200 R = 18% P.A. T = 3 years 100 3 18 7200 . Rs .I . S ! % % = Rs. 3888 Amount = P + I = Rs. 7200 + Rs. 3888 = Rs. 11088

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

B) Triples A 3P S.I. = 3P-P = 2P

Þ Þ

Þ Þ

( =

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 % 5 . 12 8 P 100 P R ! % % ! 100 15 14 2000 . Rs .I . S ! % % P = Rs. 3700 years 2 5 years 2 1 2 T! ! % 2 11 . A . P % 2 1 5 R! ! 2 5 2 11 100 3700 . Rs .I . S ! % % = Rs. 508.75

Rate (R) Percent

Time (T)

Amount (A)

Principal (P)

Exponent

To make vary large number's Easy to read, understand and compare ,we use exponents 10 m= 1km3

10 cm= 1km5

n

th

a 'a' raised to power n or

n power of 'a'

®

10 10 squared or 10 raised to power 2 10 10 cubed or 10 raised to power 3

2 3® ® 2 4 3 4 3 2 6 5 6 5 3 2 : e g ' ( ) * + , % ' ( ) * + , ' ( ) * + , % ' ( ) * + , ' ( ) * + , -' ( ) * + , ! ' ( ) * + , ' ( ) * + , ! ' ( ) * + , ' ( ) * + , ! & & 6 5 3 2 6 5 3 2 6 5 3 2 2 1 2 3 4 2 4 15 8 5 6 9 4 5 6 3 2 2 2 ! % ! % ! 2 5 4 4 9 2 3 2 : Solve % % % 4 2 4 1 5 2 2 2 5 4 2 3 3 2 2 3 2 3 2 % % ! % % % ! " ) ( = 26-4×34-2 = 22×32 = 4× 9 = 36 Express 432 as product of powers of prime no.

1 3 3 9 3 27 3 54 2 108 2 216 2 432 2 = 2× 2 × 2 × 2 × 3 × 3 × 3 = 2 × 34 3 Prime factorisation eg : 2º + 3º + 4º = 1 + 1 + 1 = 3 eg : (i) 3 = 81 = 3x × 3 × 3 × 3 = 34Þx=4 (ii) (-5) = -125 = (-5)× (-5) × (-5) = (-5) x = 3 x 3 Þ (6) aº = 1 1= am÷ a = am m-m= aº m m m m m b a b a b a 5 ' ! ! -( ) * + , ) ( 4 4 4 4 4 3 2 3 2 3 2 eg ' ! ! -( ) * + , . : (4) am × bm = (a × b)m base different power same eg : (-2)4× (-3) = (-2 × -3) = (+6)4 3 4

Low's of Exponents

multiplication Powers added (1) am× an = am+n base same eg : 23× 2 = 2 + 5 = 25 3 8

division powers subtracted (2) am÷ a = an m–n[m > n]

base same eg : 35÷ 3 = 32 5-2= 33

(3) (a )m n = am×n

eg : (5 ) = 53 7 3×7= 521

exponential notation/power notation

base exponent/inden/power a = a1 a× a = a2 a× a × a = a3 a× a × a....×a = an a multiplied n times eg : Compare (a) 2.7× 108 ; 1.5 × 108 Same Compare 2.7× 108 ; 1.5 × 108 different (b) 2.7× 108 ; 1.5 × 1012 Compare 2.7× 108 < 1.5 × 1012

eg : Express in exponential notation (i) 625 = 5 × 5 ×5×5 = 5 (ii) -27 = (-3) × (-3) × (-3)= (-3) 4 3 (iii) a× a × a × b × b × y × y = a b y3 2 2 Note : b a3 2 b a2 3 ¹

Standard form/Scientfic notation

A× 10 Þ n 1.0 < A < 10.0 ; n®integers eg : 59 = 5.9× 10 590 = 5.9× 10 5900 = 5.9× 10 1 2 3 eg : 1000= 10×100 = 10 × 10 × 10 = (2×5)×(2×5)×(2×5) Expand form : 47561 = 4× 10 + 7 ×10 + 6 ×10 + 1 × 104 3 2 0 = 23× 53

Congruent Fig

Equal Shape and Equal Size

Criteria for Congruence of

Triangle

In PQR then by c.p.c.t. QR = LM = 5cm P = N R = L = 80º RP = LN QRP MLN QP = MN D D D and LMN Q = M = 30º D Ð Ð Ð Ð Ð Ð by A. S. A. Property 30º R P Q B m PQR = m ABC PQR ABC Ð Ð Ð Ð _{In rt. angle triangle's}

The hypotenuse = (cosrresponding) and one side hypotenuse of a triangle and one side of

other triangle C A B _{If AB = AC} B = C Þ Ð Ð P A

In AOB and QOP AO = OQ AOB = QOP OB = OP D D Ð Ð eg : B C b c a A 8cm Q R P x y 60º 40º 80º z Given : ABC PQR to Find : x,y,z, a, b,c.

So, (i) AB = PQ (iv) A = P 6cm = z a = 80º (ii) BC = QR (v) B = Q 8cm = y b = 60º (iii) AC = PR (vi) C = R 7cm = x c = 40º D D Ð Ð Ð Ð Ð Ð 6cm 4cm In AB = XY = 3cm then by c.p.c.t. BC = XZ = 6cm A = Y AC = YZ = 4cm B = X ABC YXZ C = Z D D Ð Ð Ð Ð Ð .

ABC & XYZ

[ ] Ð D D by S S.S property M N 30º R Q 30º 80º B C A Y Z X

3 sides of a = Corresponding three triangle sides of other triangle

P L

80º

(Corresponding)

P A

2 sides and the included angle of a triangle

(corresponding) = 2 sides and the included

angle of other triangle

B C 60º R Q 60º In PQR PQ = BC = 5cm P = B = 60º [ ] D and ABC QR = AC R = A PR = AB = 4cm Q = C PQR BCA D Ð Ð Ð Ð Ð Ð D D by S.A.S proper yt Y X Z 90º 4cm 5cm 90º M N L In t. XYZ and t LMN Y = N = 90º hyp. XZ = hyp. LM = 5cm YZ = LN = 4cm YZX NLM r D r .D Ð Ð D D by R.H.S. propery. Then by c.p.c.t X = M Z = L XY = MN Ð Ð Ð Ð 30º C A R B C A DPQR DABC Corresponding vertex : A P ; B Q ; R C Corresponding sides : AB = PQ ; BC = QR ; AC = PR Corresponding angles : A P ; B Q ; C R « « « Ð «Ð Ð «Ð Ð «Ð AOB QOP by S.A.S. property D D

In ACQ and ABQ C = B = 90º hyp QA = h p QA CQ = BQ ACQ ABQ by R.H.S. property D D Ð D Ð D . y . Q A B C Q P O Q B

Integers

[All natural numbers + 0 + negative

of counting numbers]

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

-Negative integers Positive integers Neither positive Nor negative (ii) -27 - (-3) = - 27 + 3 =– 24 (iii) 27–(-3) = 27 + 3 = 30 (iv)–3 –(-27) = -3 + 27 = 24 (v)–27 –(3) = –27–3 = –30

When the dividend and divisor

are of unlike signs , the quotient is negative other wise positive If a and b are ( + v ) then (-a)÷ b = – (a ÷ b) (–6) ÷ 3 = –(6 ÷ 3) = – 2 , e (i) Properties :

(i) a÷ o = not integer [Not closed]

(ii) If a 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 ¹ ; Addition

When integers are of (a) like sign

(i) 27 + 3 = 30

(ii)–27 + (–3) = – (27 + 3) = –30

(b) unlike sign

(i)–27+ (3) = – 24 (sign of bigger

absolute value) (ii) 27 + (–3) = 24 Subtraction b - (a) = b + (-a) b - (-a) = b + (+a) eg : (i) 27-3= 24 Multiplication 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

(a)of unlike sign: product is negative

(–a) × (b) = –(a × b) eg .(– 2) × ( 6) = –12 (ii) a÷ (–b) = – (a ÷ b) 15÷ (–5)= –(15 ÷ 5) = ÷ 3 (iii) (–a) ÷ (–b) = – (a ÷ b) (–4) ÷ (–2)= (4 ÷ 2) = 2 (iv) 8÷ 4 = 2

When integers are

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

(b) a– b = b – a eg : –3 –(4) = –7 ; 4 –(–3) = 4 + 3 = 7

(a) a + b = b + a eg : 3 + 4 = 4 + 3 = 7 ; -3 + 4 = 4 + (-3) = 1

(b) (a– b) – c a – (b – c)

eg :[(–3) – (–9)] – 17 = 6 – 17 = – 11

(–3) – [(–9) –(17)] = –3 – [–26] = –3 + 26 = 23

(iii) Associative property

(a) (a + b) + c = a + (b+c)

eg :[(–3) + (–9)] + 17 = (–12) + 17 = 5

(–3) + [(–9) + 17] = (–3) + 8 = 5

(iv) Additive identity

a + 0 =0 + a =a [o is additive identify] eg : 3 + 0 = 0 + 3 = 3

(v) Additive inverse

eg : 3 + (–3) = (–3) + 3 = 0

a + (–a) = (–a) + a = 0 [ a additive inverse (–a) ]

Properties : If a, b and c are integers (i) Closure property

a× b = integer eg : (-3) × 10 = –30 (ii) Commutative a× b = b ×a eg : (-50)× 2 = 2 × (-50) = 100 (a× b) × c = a × (b × c) eg : - 1× (3 × 2) = -6 (-1 ×3) × 2 = -6 (iii) Associative

(iv)1 is multiplicative identityie 1× a = a × a

1× a = a × 1 = a : eg : 6 × 1 = 1 × 6 = 6

(v) for integer a : a× 0 = 0 × a = 0eg : 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

Division

Operation precedence [BODMAS] I. – B Brackets

II. – O f

III.– D ivision

IV.– M ultiplication

V. – A dd

VI.– S Subatract simultaneously ® ® ® ® ® O D M A ition ® done I. (bar) II. ( ) (small) III. { } (curly) Vi [ ] (big) /

Division

FRACTION

4 1 1 4 5 ! 4 1 + 4 1 1 of 2 1 2 1 1 2 1 ! % 2 1 1 2 1 ! % 1 2 2 1 ! % 1 of 2 1 2 of 2 1 =

Fraction as an operator

Decimal fraction Denominator is 10,100 , 1000, 10000 -Proper fraction numerator < denominator . c . t . e 100 3 ; 10 1 : eg 7 5 : eg Improper fraction denominator > numerator eg :₅7 converted into Mixed fraction

[combination of whole no. and a proper fraction]

3 44 3 2 3 14 3 2 14 ; 5 2 1 5 7 ! " % ! ! Equivalent fractions 10 6 2 5 2 3 5 3 ! % % ! 15 9 3 5 2 3 5 3 ! % % ! 15 9 10 6 5 3 ! ! =

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

9 4 4/9 5/9 Rs. 45

Operation on Fraction

5 12 5 4 1 3 4 5 3 ) i ( - ! % ! 12 5 3 1 4 5 1 3 4 5 3 4 5 ) ii ( - ! - ! % ! 5 2 5 6 3 1 6 5 3 1 ) ii ( - ! % ! * Reciprocal or multiplicative inverse of a non-zero faction

b a * Multiplicative inverse of 1 is 1 1 a b b a ie a b ! %

Reciprocal of 0 does not exist.

Subtraction

4 1 1 – 5 2 2 4 5 – 5 12 ! L.C.M . (5,4) = 20 20 5 5 4 12% " % 20 3 1 20 23 20 25 – 48 ! ! !

Multiplication

(i) Fraction by fraction

ator min deno numerator of oduct Pr of oduct Pr !

(ii) Fraction by whole number

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

Product of two proper fraction is less than each of the fraction.

2 3 1 2 1 3 1 2 4 3 2 4 3 ! % % ! % ! %

*Value of ,product of two improper fraction is more then each of the two factions 4 1 2 5 2 2 – 3 2 14 11 14 1 : eg % % 4 9 5 12 – 3 44 11 15 % % 4 5 9 12 – 3 11 44 15 % % % % 5 27 – 1 20 5 73 – 5 1 27 – 5 20% %

Addition

kg 6 1 1 kg 4 1 3 " kg 6 7 kg 4 13 " L.C.M. (4,6) = 12 12 kg ) 2 7 3 13 ( % " % kg 12 5 4 kg 12 53 ! ! 3 1 5 1 3 2 1 4 : eg % % 3 1 5 16 2 9 % % ! 3 5 2 1 16 9 % % % % ! 5 4 4 5 24 ! ! is * 16 4 Rs. 36 Compare numerator Denominator same Numerator same Compare denominator Find L.C.M. of denominators L.C.M. of (3,5) = 15 15 6 3 5 3 2 ; 15 5 5 3 5 1 ! % % ! % % 7 3 7 4 : eg 0 3 5 4 5 : eg 1 5 2 3 1 : eg 1

Comparing 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 Division.

Decimals

Decimals fractions e are

whose denominators are 10,100,100 .t.c. eg : 23 . 005

whole number Decimal part part

like decimals unlike decimals

same decimal place

5.25 , 6.89 different decimal place5.25 , 6.893

Multiplication

decimal by decimal

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

(i) First multiply as whole no (ii) Count the no. of digits to the

right of the decimal point

(iv) Put decimal point in the product by

counting the digits from the right most place (iii) Add the no. of digits counted

eg : 256. 7× 0.005 = 1 decimal 3 decimal place place = 1.2835 1+3 = 4 decimal place eg :- cost of 1m cloth = Rs. 15.5 cost of 1.5m cloth = Rs. 15.5× 1.5 = Rs. 23.25 eg :- cost of 2.5m cloth = Rs. 41.25 cost of 1m cloth = Rs. 41.25÷ 2.5 = Rs. 16.50 by 10,100, 1000

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 whole number 1.575÷ 35 = [1575 ÷ 35] 1000 1 = [45] = 0.045 1000 1 by decimal no. 0.165 by 1.5 1 0 1 5 1 0 0 1 6 5 -! ] 1 5 1 6 5 [ 1 0 0 1 -! 1 1 . 0 1 0 0 1 1 ! !

(i) decimal into fraction

eg : 3.75

(ii) fraction into decimal 4 1 5 1 0 0 3 7 5 ! ! 25 . 9 100 25 9 25 4 25 1 9 4 1 9 ) i ( ! " ! % % " ! 0056 . 0 1000 56 8 125 8 7 125 7 ) ii ( ! ! % % ! Subtraction Subtract : 8.96 from 25.1 change to Þ 14 . 16 96 . 8 10 . 25

& like decimals , then subtract like whole no.

Addition eg : Add 3.01 , 2.58 and 6.9 49 . 12 9 . 6 58 . 2 01 .

3 ÞKeep decimal points under each other

ÞWrite digts in the correct place value columns

ÞAdd like whole numbers.

Rational Numbers

A number that can be expressed in the form q

p integer

integer but q/ o

Addition

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

eg & " ! & " ! &

] d and b of . M . C . L Find [ a c b a ) ii ( " 12 ) 5 ( 8 3 12 5 8 3 : eg ! & " & & " & L. C. M. of 8 and 12 = 24 24 19 24 ) 2 ( ) 5 ( 3 ) 3 ( & ! % & " % & Additive Inverse of eg : Additive inverse of ] 0 q ) p ( q p ie [ q p is q p ! & " & 9 4 is 9 4 ) ii ( 2 15 is 2 15 ) i ( & & divide into 4 equal parts

(i) Positive rational number :

pandqboth are positive or negative

5 4 ; 5 3 : eg & & &

(ii) Negative rational number :

pandqboth are have opposite sign

5 4 or 5 3 : eg & & 6 0 , 8 0 eg q p the 0 p If ) iii ( & ! ! e ge r int zero non m w he re m q m p q p also m q m p q p & ! -! % % ! Division 2 3 4 5 6 7 / % ! - 0 d c that such c d b a d c b a 33 5 3 3 3 1 0 4 3 3 3 4 1 7 5 2 34 3 3 1 7 1 3 : e g ! & ! & % ! & -Subtraction

Add the additive inverse

' ( ) * + ,& " ! & a c b a d c b a ie 7 2 2 3 7 2 3 3 1 0 2 4 1 1 3 6 5 2 4 1 1 3 6 5 3 6 5 f r o m 2 4 1 1 S u b t r a c t : e g ! " & ! " & ! ' ( ) * + , & & & ! & & Multiplication

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

2 8

3

8

7

8

)

5

(

2

)

3

(

8

)

5

(

7

2

5

3

:

e g

!

%

%

&

%

%

&

!

&

%

%

&

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

Simplest (Lowest) or standard form

If denominator is a positive integer and denominator and numerator have no common factor other then 1 eg : Standard form of 5 2 ) 3 ( 15 ) 3 ( 6 is 15 6 & ! & -& & -& divide by H.C.F of 6 and 15

Equivalent Rational no.

18 10 2 9 2 5 9 5 ; 15 10 5 3 5 2 3 2 eg & & ! & % & % ! ! % % ! equivalent Rational no. equivalent Rational no. Comparison

*Two positive rational numbers can be compared as fractions

*(i) Compare the rational no.s ignoring their negative signs

(ii) Then reverse the order 3 5 5 7 3 5 5 7 : eg 1 $ & 0 & 2 1 7 2 : eg & 1

* A negative rational no. is always less then positive rational no.

7 17 0 A P 17 0 divide into 7 equal parts 4 9 ) ii ( & -9 Q B O 0 4 9 &

Rational number on number line

(i) To represent 7 1 7 ) i ( 33 x ; ) 3 ( 11 9 x ; 11 9 x ) 3 ( ; x 9 11 3 : eg !& & % ! % ! % & ! &

* Natural numbers + whole numbers +

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

* A fraction is a rational number, but a rational number may or may not be a fraction eg :- ®rational no. + fraction; ®rational no but not a fraction

5 3

5 3

&

* There are infinite rational no's between two rational no's :

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

eg : Rational No. b/w and

5 3 & 3 1 &

;

There is equal uncerteinty of each outcome of an experment

Probability of an event A = P (A)

outcom es possible of num ber Total A of favour in outcom es of N um ber ! 0 < P (A) < 1 Probability of getting a number greater than 7

Probability of getting Tuesday after Monday

Sure Chance Good Chance Even Chance Poor Chance No Chance 0 1/4 1/2 3/4

Probability

Equally likely outcomes

Experiment

Sample Space

Outcome

Deals with the measurement

of uncertinity of the occurrence of some event in terms of

percent or ratio

Event

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 :

(a) Black i.e. P(B) =

24 7 (b) Green P(G) = 6 1 24 4 !

(c) P(Not red) = 1–P (Red)

24 19 24 5 24 24 5 1& ! & ! !

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

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

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

1

Any kind of activity eg : Tossing a coin

When a coin is tossed

Total number of outcomes = 2 ie. T,H Probability of getting head P(H) =

2 1

Probability of getting tail P(T) =

2 1

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

Number of times an observation occurs in

eg : 3,5,4,9,9,19 , frequency of 9 is 3 Frequency Distribuation Table

Measure of central tendency

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

(i) When n is odd median = observation.

3,1,4,3,6,5,9,5,3

arreange in ascending order 1,2,3,3,4,5,5,6,9 th 2 1 n ' ( ) * + , " 4 n observatio 2 1 9 n observatio 2 1 n median th th ! ' ( ) * + , " ! ' ( ) * + , " !

(ii) n is even median =

eg : 11, 10, 12, 9, 8, 16, 15, 14 arreange : 8, 9, 10, 11, 12, 14, 15,16 . 2 ob 1 2 8 . ob 2 8 Median th th ' ( ) * + , " " ' ( ) * + , ! 4 ob.₂5 ob. th th " sun 8f!25 8fix!715 xi f f× x 25 27 28 30 32 33 5 4 6 3 2 125 108 140 180 96 66 5 f x f X 8 % 8 ! Students Quartely Half yearly

Ashish Arun Kavish Maya Rita

10 15 12 20 9 15 21 16 18 15 25 20 15 10 5

Ashish Arun Kavish Maya Rita

Ungrouped Grouped Number Tally Marks Frequency 1 2 3 4 5 6 2 4 2 6 3 3 Number Tally Marks Frequency 0-10 10-20 20-30 30-40 40-50 6 6 4 3 1

[The value which occurs the most or has highest frequency] eg : Number Frequency 7 2 10 1 11 1 12 3 13 2 17 1 Mode = 12 [Highest frequency = 3]

Array/arranged data

(organised data)

Statistics

Science of collection presentation , analysis and interpretation of numerical data n x n x ... x x x x 1 2 3 n

9

i ! " " " ! Mean of 3, 6, 9, 12, 15 5 15 12 9 6 3 x! " " " " x!9

Mean

Quartarly Half yearly

Median

2 . ob 1 2 n . ob 2 n th th ' ( ) * + , " " ' ( ) * + ,

Mode

Favourate Colour No. of Students 43

Red Green Blue Yellow

19 55 49 Orange 34 50 40 30 20 10

Red Green Blue yellow orange 60 Scale : 1 unit : 10 Students

5 . 11 2 12 11 ! " = =

Rotation terns an object about a fixed point called

a a Q R P Center of rotation Clock wise Rotation (i) P Q R 120º (iii) a R P Q (ii) Order of rotational symmetry = 3

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

equilatral triangle has

rotational as well as line of symmety A Line which divides the figure into two

congruent parts is called line (axis) of symmetry Line of symmetry

60º

60º 60º

Infinite lines of Symmetry Regalar Hexagon

Regular Pentagon Square

Equilateal Triangle

Number of lines of symmetry of a regular polygon

= Number of sides of that polygon

Symmetry

Rotational Symmetry Symmetrical Figures

Length Cuboid Euler's formula V + F– E = 2 Face Edge Vertex

eg : For Triangular pyramid V = 4 ; E = 6 ; F = 4 4 + 4 - 6 = 2 \ h h Cone Sprere Pyramis Breadth height

Solid Shapes

Vertex Face Edges

Face

Vertex

Edge

2 - D Repesentations of a 3- D figure NETS

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

SOLID SHAPES

Perimeter

Sum of all sides.

1 2 1 2 r 2 ) r 2 ( 2 1 P! : " units 2 2 2 7 22 P! % " % units 7 72 4 7 44 P! " !