8 COMPARING QUANTITIES WORKSHEET-39 - …...... Cost of TV including VAT = Rs. 13500 Rate of VAT = 8% Let the original price = Rs. x 1 2 x + 8% of x = Rs. 13,500 or x + 8 100 x = Rs

P-80 M A T H E M A T I C S – VIII

1. (A) Rs. 100 1

2. (C) 12 years 1

3. 72% of 25 students are good in Mathematics.

(100 – 72)% of 25 students are not good in Mathematics. 12

or, 28% of 25 students are not good in Mathematics.

or, 28

25100

´ = 7 students are not good in Mathematics. 12

4. Increase in price this year = 20% [the last year’s price]

= 20% of Rs. 45,000

= ´20

45,000100

12

= Rs. 20 × 450= Rs. 9,000

This year’s price = [Last year’s price] + [Increase in price] 12

= Rs. 45,000 + Rs. 9,000= Rs. 54,000.

5. Let the original salary = Rs. xIncrease in salary = 10% of Rs. x

=10

Rs.100

x = Rs. 10

x1

New salary = Rs. Rs.10

xx+ =

11Rs

10x

11

Rs.10

x = Rs. 154000 (given)

or x =´154000 10

Rs.11

= Rs. 140000

Thus, the original salary = Rs. 140000 1

6. We have P = Rs. 8000, R = 5% p.a., T = 2 years

A = 1100

nR

Pé ù+ê úë û

A =25

8000 1100

=221

800020

1

=21 21

800020 20

= (20 × 21 × 21) = Rs. 8820 ½Now, compound interset

= A – P= Rs. 8820 – Rs. 8000= Rs. 820 ½

7. Here P = Rs. 10800, n = 3 years,

R = 1

122

% p.a. = 25

%2

p.a.

8 COMPARING QUANTITIES WORKSHEET-39

P-81C O M P A R I N G Q U A T I T I E SN

We have, A = 1100

nR

Pé ù+ê úë û

=3

25Rs. 10800 1

2 100

é ù+ê úë û´

[ Interest compounded annually, n = 3]

=3

225Rs. 10800

200

é ùê úë û 1

=225 225 225

Rs. 10800200 200 200

´ ´ ´

=675 9 9 9

Rs.4 8

´ ´ ´

´=

492075Rs.

32

Amount = Rs. 15377·34

= Rs. 15377·34 1Now, Compound interest

= Rs. 15377·34 – Rs. 10800= Rs. 4577·34 1

8. Here, we shall calculate the amount for 2 years using the C.I. formula. Then this amount

will become the principal for next 4 months, i.e., 4

12 years.

Here, P = Rs. 26400, n = 2 years

R = 15% p.a.

A = 1100

nR

Pé ù+ê úë û

= Rs. 26400 2

151

100

é ù+ê úë û

= Rs. 26400 × 2

23

20

é ùê úë û 1

= Rs. 26400 × 23 23

20 20´

= Rs. 66 × 23 × 23= Rs. 34914

Again, P = 34914, T = 4 months

=4

12years, R = 15% p.a. 1

Using S.I. =100

P R T´ ´, we have

S.I. =15 4

Rs. 34914100 12

´ ´

=5819 3

Rs.10

´= Rs.

17457

10= Rs. 1745·70

Amount = S.I. + P= Rs. (1745·70 + 34914)= Rs 36659·70 1

Thus, the required amount to be paid to the bank after 2 years 4 months = Rs 36659·70.


1. (C) 40% 1

2. (B) 10 1

3. Sale price of the pair of shoes = Rs. 900

Sales tax = 5% × Rs. 900

= ´5

(Rs. 900)100

12

= Rs. 45

Bill amount = Sale price + Sales tax

= Rs. 900 + Rs. 45

= Rs. 945. 12

4. Cost of the two soap bars

= (Rs. 35) × 2 = Rs. 70

Sales tax = 5% of Rs. 70

=5

Rs 70100

´ = Rs. 3·50 ½

Thus, the buying price

= Rs.70 + Rs. 3·50

= Rs. 73·50 12

5. The price of the calculator = Rs. 750

Sales tax = 5% of Rs. 750

=5

Rs. 750100

´

=75

Rs.2

= Rs. 37·50 ½

Bill amount = Sales Price + Sales Tax

= Rs. 750 + Rs. 37·50

= Rs. 787·50 ½

6. Total Cost Price of the two-wheeler (C.P.)

= Rs. 15000 + overhead expenses

= Rs. 15000 + Rs. 500

= Rs. 15500 1

Selling Price (S.P.)

= Rs. 18600 S.P. > C.P. She got profit

Profit = Rs. 18600 – Rs. 15500

= Rs. 3100 ½



Now, Profit percent =3100

100%15500

´

= 20% ½

Thus, she earned a profit of 20%.

7. At simple interest

SI on ` 12,000 at 6% per annum for 2 years

= 12,000 6 2

100 = ` 1,440

At compound interest

P = ` 12,000

R = 6% per annum

n = 2 years

A = R

P 1100

n

1

=

26

12,000 1100

=

23

12,000 150

=

253

12,00050

= 53 53

12,00050 50

= ` 13,483.20 1

CI = A – P

= ` 13,483.20 – ` 12,000

= ` 1,483.20

Excess amount = ` 1,483.20 – ` 1,440

= ` 43.20

Hence, I would have to pay to him as excess amount of ` 43.20. 1

8. Principal = Rs 10,000 ½

Time =1

12

years

Rate = 10% p.a.

Case I. Interest is compounded half yearly

We have R = 10% p.a. = 5% per half yearly

T = 12

1 years n = 3

A = 1100

nR

Pé ù+ê úë û

= Rs. 10000 3

51

100

é ù+ê úë û

= Rs. 10000 3

21

20

é ùê úë û


= Rs. 10,000 21 21 21

20 20 20´ ´ ´

=5 21 21 21

Rs.4

´ ´ ´

=46305

Rs.4

= Rs. 11576·25 1

Amount = Rs. 11576·25

Now, C.I. = Amount – Principal

= Rs. 11576·25 – Rs. 10000

= Rs. 1576·25 ½

Case II. Interest is compounded annually

We have R = 10% p.a.

T =1

12

year

Amount for 1 year

= 1100

nR

Pé ù+ê úë û

= Rs. 10000 1

101

100

é ù+ê úë û

=11

Rs. 1000010

´

= Rs. 11000 Interest for 1st year

= Rs. 11000 – Rs. 10000

= Rs 1000 1

Interest for next 1

2year on Rs. 11000

=100

P R T´ ´

=10 1

Rs. 11000100 2

´ ´

= Rs. 55 × 10 = Rs. 550

Total interest = Rs. 1000 + Rs. 550

= Rs. 1550 1

Since Rs. 1576.25 > Rs. 1550

Interest would be more in case of it is compounded half yearly. ½


1. (A) Rs. 51,000 1

2. (B) 4 years 1

3. Number of visitors on Sunday = 845

Number of visitors on Monday = 169

Decrease in the number of visitors

= 845 – 169 = 676 12

Percent decrease

=676

100%845

´ = (4 × 20%)

= 80% 12

4. Cost price= Rs. 2400

Profit = 16% of Rs. 2400

=16

Rs 2400100

´

= Rs.16 × 24 = Rs. 384

Selling price = Rs. 2400 + Rs. 384

= Rs. 2784

Now, selling price per article = Rs 2784 80 [ Number of articles = 80]

=348

Rs Rs.34·8010

= 1

5. (a) Cost of TV including VAT = Rs. 13500

Rate of VAT = 8%

Let the original price = Rs. x 12

x + 8% of x = Rs. 13,500

or x + 8

100x = Rs. 13500

or x8

1100

é ù+ê úë û= Rs. 13500

or x × 108

100= Rs. 13500

or x =13500 100

Rs.108

´

= Rs. 12500

Thus, the original price

= Rs 12500 1

(b) Let original price = Rs. x

Price including VAT

= Rs. x + 8% of x



=8

Rs. 1100

é ù+ê úë ûx

=108

Rs.100

x

But the original price + VAT

= Rs. 180

108

100x = Rs. 180 1

2

or x =100

Rs. 180108

´

=500

Rs.3

= 2

Rs. 1663

Thus, the original price

=2

Rs. 1663

1

6 People who like cricket = 60%

People who like footbal = 30%

People who like other games

= [100 – (60 + 30)%

= [100 – 90]%

= 10%

Now, Total number of people = 50,00,000 1

60% of 50,00,000

=60

50,00,000100

´

= 6 × 5,00,000

= 30,00,000 ½

30% of 50,00,000

=30

100× 50,00,000 1

= 3 × 5,00,000

= 15,00,000 ½

10% of 50,00,000

=10

100× 50,00,000

= 1 × 5,00,000

= 5,00,000 ½

Thus, number of people who like

Cricket 30,00,000Football 15,00,000

Other games 5,00,000

ü= ï= ý

ï= þ½


7. Total S.P. of the shoapkeeper =Rs. 16000/-S.P. of VCR = Rs. 8000/-

Let C.P. of the shopkeeper = Rs. xLoss = 4%

S.P. of VCR = C.P. of VCR – Loss

8000 = 4

100x x

x = 100 4

8000100

x ½

x = 8000 100 25000

96 3

½

C.P. of VCR = Rs. 25000

3

S.P. of TV = Rs. 8000/-

Let shopkeeper cost price by Rs. y

Profit = 8%

S.P. of TV = C.P. of TV + 8

100 of C.P. of TV ½

8000 = 8

100y y

y 8

1100

= 8000 ½

y = 8000 100

74074.40108

½

Total C.P. = x + y = Rs. 25000

0.7407.403

= 15740.740

Also total S.P. = Rs. 16000/- 1

S.P. > C.P.

Overall profit = S.P. – C.P.

= 259.26

Profit % = 259.26

10015740.740

= 1.65% 1



[A] Lab Activity

Object :

To verify experimentally that a single discount of some percentage say x% provided onan article is greater than the sum of two successive discounts of any percentage say y%and z% with x = y + z.

Let’s start :

The discount is a reduction given on the marked price. If the discount percentage isd%, then

Discount = ´Marked Price100

d½

EFGH is a rectangle (Fig. (i). Its opposite sides are equal and each angle is of measure90°.

Area (EFGH) = l × b sq. units.

H G

E F

b

l

½

(i)

Meterial Required :

(1) Thick white paper(2) Sketch pens(3) Geometry box(4) A pair of scissors(5) Fevicol ½

Procedure :

(1) Take the MP of an article as ` 100

(2) Discount (in `) on the article by a single discount of x%

= 100100

xx´ = ½

(3) Using scissors, cut a rectangle of length x and breadth unity. Name it as ABCD andcolour it (Fig. (ii))

D C

A Bx

1

1

(ii)

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(4) Discounts on the article by two successive discounts of y% and z% are respectively

100100

y´ ` = y and ` (100 – y)

100

z´ = `

100

yzzæ öç ÷-è ø. ½

Sum of these discounts (in `) = y + z 100

yz-

(5) Using scissors, cut another rectangle of length y + z and breadth unity. Colour it with adifferent colour and name it as PQRS (Fig. (iii). It is obvious that y + z = x. ½

S R

P Q

N

M

(y + z)

yz100

(iii)

(6) Cut a rectangle PMNS from the rectangle PQRS obtained in step (4) along the dotted

line such that PM = 100

yz (Fig. (iii))

(7) Now, paste the remaining rectangle MQRN on the rectangle ABCD such that the pointQ concides with B and R with C (Fig. (iv)). ½

D C, R

A B, Q

N

M

(iv)

Observation :

We observe that region AMND is clearly free from overlapping on the rectangle ABCD.

½

Result :

A single discount of x% is greater than the two successive discounts of y% and z% suchthat x = y + z. ½

[B] Fill in the Blanks 5 × ½ = 2½

1. 90

2. 120

3. Marked

4. Cost price

5. R


[C] True/False 4 × ½ = 2

1. False

2. False

3. False

4. True

[D] Viva-Voce

1. The price reduces . 1

2. The first discount is applicable on the marked price, the second discount is applicableon the reduced price after the first discount and so on. 1

3. Rs. 250. 1

4. A single discount of 10%. 1

5. Discount = Marked Price – Sale Price. 1

[E] Quiz

1. Value Added Tax. 1

2. Discount percent = Discount

MarkedPrice× 100 1

3. Increase or decreases in population. 1

4. Cost Price. 1

5. Two successive discounts of 12% and 8%. 1

P-91ALGEBRAIC EXPRESSIONS AND IDENTITIES

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1. (D) 2 2 2 11p q pq 1

2. (A) 2 7 5 10xy yz zx xyz 1

3. (a) 25 3xyz zy

25 3

5 3

xyz zy-

-

Terms

Their coefficients12

(b) 1 + x + x2

2 1

1 1 1

x xTerms

Their coefficients12

4. (a) ab – bc

+ bc – ca

– ab + ca

0 + 0 + 0 = 0 12

(b) a – b + ab 12

b – c + bc – a c + ac

0 + 0 + ab + 0 + bc + ac = ab + bc + ca 12

5.æ ö æ ö- ´ç ÷ ç ÷è ø è ø

3 310 6

3 5pq p q = 3 310 6

3 5pq p q

é ùæ ö æ ö- ´ ´ ´ç ÷ ç ÷ê úè ø è øë û

1

=4 410 6

3 5p q

æ ö æ ö- ´ ´ç ÷ ç ÷è ø è ø

= 4 44p q . 1

6. 2( 1) 5a a a = 2 1 5a a a a a

= 3 2 5a a a 12

(i) For a = 0,

a3 + a2 + a + 5 = (0)3 + (0)2 + 0 + 5 = 5 12

(ii) For a = 1,

3 2 5a a a = 3 2(1) (1) 1 5 = 1 + 1 + 1 + 5

= 8 12

(iii) For a = – 1

3 2 5a a a = 3 2( 1) ( 1) 1 5

= – 1 + 1 – 1 + 5

= 4. 12

9ALGEBRAIC EXPRESSIONSAND IDENTITIES WORKSHEET-43


7. (a + b)(a – 2b + 3c) = a × (a – 2b + 3c) + b(a – 2b + 3c)

= 2 22 3 2 3a ab ac ab b bc

=2 2( 2 ) 3 2 3a ab ab ac b bc 1

= 2 22 3 3a b ab bc ac

and (2a – 3b) × c = 2ac – 3bc 1

+ - + - - ´( )( 2 3 ) (2 3 )a b a b c a b c = 2 2( 2 3 3 ) (2 3 )a b ab bc ac ac bc- - + + - -

= 2 22 3 3 2 3a b ab bc ac ac bc- - + + - +

= 2 22 (3 3 ) (3 2 )a b ab bc bc ac ac

[Combining the like terms bc and ac]

= 2 22 6a b ab bc ac . 1

8. (1.5x – 4y)(1·5x + 4y + 3) – 4·5x + 12y

= + + - + +1·5 (1·5 4 3) 4 (1·5 4 3)x x y y x y – 4·5x 12y

= (1·5 1·5 ) (1·5 4 ) (1·5 3)x x x y x - ´ - ´ - ´(4 1·5 ) (4 4 ) (4 3)y x y y y - +4·5 12x y 1

= + + - - -2 22·25 6 4·5 6 16 12x xy x xy y y - +4·5 12x y

= 2 22·25 (6 6) (4·5 4·5) 16x xy x y (12 12)y 1

= 2 22·25 (0) (0) 16 (0)x xy x y y

= 2 22·25 0 0 16 0x y

= 2 22·25 16x y . 1

Another Method :

(1·5x – 4y)(1·5x + 4y + 3) – 4·5x + 12y

= (1·5x – 4y)(1·5x + 4y) + (1·5x – 4y) × 3 – 4·5x + 12y 2

= (1·5x)2 – (4y)2 + (4·5x – 12y) – (4·5x + 12y)

= 2·25x2 – 16y2. 1


1. (A) 2xy 1

2. (B) a6 1

3. 12a – 9ab + 5b – 3

– 4a – 7ab + 3b + 12 12

(–) (+) (–) (–) Subtract

8a – 2ab + 2b – 15 12

4. We have,

(– 4p) × (7pq) = (– 4 × 7) × p × pq

= – 28p2q 1

5. 2 3( 5)( 3) 5a b = 2 3 3( 3) 5( 3) 5a b b 1

= 2 3 2 3( ) ( 3) (5 )a b a b (5 3) 5

= 2 3 2 33 5 15 5a b a b

= 2 3 2 33 5 20a b a b . 1

6. 2 2(2 5) (2 5)x x

= 2 2[(2 ) (5) 2 2 5]x x 2 2[(2 ) (5) 2 2 5]x x 1

= 2 2[4 25 20 ] [4 25 20 ]x x x x

= 2 24 25 20 4 25 20x x x x

= 20x + 20x

= 40x. 1

Alternative Method :

Since, a2 – b2 = (a + b)(a – b)

(2x + 5)2 – (2x – 5)2

= [(2x + 5) + (2x – 5)][(2x + 5) – (2x – 5)] 1

= (4x)[2x + 5 – 2x + 5]

= (4x)(10) = 40x. 1

7. L.H.S. =

24 3

23 4

m n mn

=

2 24 3 4 3

2 23 4 3 4

m n m n mn

1

=2 216 9

2 29 16

m mn n mn



=2 216 9

( 2 2)9 16

m mn n 1

=2 216 9

(0)9 16

m mn n

=2 216 9

9 16m n 1

= R.H.S. Hence Proved.

8. L.H.S. = 2 2(4 3 ) (4 3 )pq q pq q

= [(4pq + 3q) + (4pq – 3q)] × [(4pq + 3q) – (4pq – 3q)]

(... x2 – y2 = (x + y) (x – y)) 1

= (8pq) (4pq + 3q – 4pq + 3q) 1

= (8pq) (6q)

= 248 = R.H.S.pq Hence Proved. 1


1. (B) x + 1 1

2. (C) 2 2a b 1

3. 3x(4x – 5) + 3 = (3x)(4x) – (3x)(5) + 3

= (3 × 4) × (x × x) – 15x + 3

= 12x2 – 15x + 3 ½

4. 71 = 70 + 1

2(71) = 2(70 1)

= (70)2 + 2 × 70 × 1 + (1)2 1

[·.· (a + b)2 = a2 + 2ab + b2]

= 4900 + 140 + 1

= 5041. 12

5. (i) Using the Identity,

(a + b)(a – b) = 2 2a b

We have,

(6x + 7)(6x – 7) = 2 2(6 ) (7)x

= 236 49.x 1

(ii) Using the identity, 2( )a b = a2 + b2 + 2ab, we have

3 3

2 4 2 4

x y x y

=

23

2 4

x y

=

2 23 3

22 4 2 4

x y x yæ ö æ ö æ ö æ ö+ + ´ ´ç ÷ ç ÷ ç ÷ ç ÷

è ø è ø è ø è ø

=2 29 3

4 16 4

x y xy . 1

6. L.H.S. = 2(3 7) 84x x

= + + ´ ´ -2 2(3 ) (7) 2 3 7 84x x x

= 29 49 42 84x x x

= 29 49 42x x 1

R.H.S. = 2(3 7)x

= 2 2(3 ) (7) 2 3 7x x

= 29 49 42x x 1



Since L.H.S. = R.H.S.

2(3 7) 84x x = 2(3 7)x

7. 1·05 × 9·95 = (1 + 0·05)(1 – 0·05) 1

= (1)2 – (0·05)2 1

= 1 – 0·0025

= 1·0000 – 0·0025

= 0·9975. 1

8. We have

- + - - +2 25 2 5 11 3 28p q pq pq q p

2 24 5 3 7 8 10p q pq pq q p 1

(–) (–) (+) (–) (+) (+)

2 27 8 18 5 38p q pq pq q p 2



1. (A) A binomial 1

2. (B) 225 30 9c c 1

3. 2 251 49 = (51 + 49) (51 – 49)

= (100) × (2)

= 200. 1

4. (x2 – 5)(x + 5) + 25 = x2(x + 5) – 5(x + 5) + 25

= 2 2( ) ( 5) (5 ) (5 5) 25x x x x 1

= 3 25 5 25 25x x x

= + -3 25 5 .x x x 1

5. (x + 7y) × (7x – y) = x(7x – y) + 7y(7x – y)

= ´ + ´- + ´ + ´-( 7 ) ( ) (7 7 ) (7 )x x x y y x y y 1

= - + -2 27 49 7x xy xy y

= + -2 27 48 7x xy y . 1

6. First expression

= 3a(a + b + c) – 2b(a – b + c)

= (3a) × (a) + (3a) × (b) + (3a) × (c) – (2b) × (a)

+ (2b) × (b) – (2b) × (c)

= 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc

= 3a2 + 2b2 + 3ab – 2ab – 2bc + 3ac

= 3a2 + 2b2 + ab – 2bc + 3ac 12

Second expression

= 4c (– a + b + c)

= 4c × (– a) + 4c × b + 4c × c

= – 4ac + 4bc + 4c2 12


Subtraction

– 4ac + 4bc + 4c2

3a2 + 2b2 + ab + 3ac – 2bc

– – – – +————————————————

– 3a2 – 2b2 – ab – 7ac + 6bc + 4c2 1

7. Area of the new rectangle = (l + 5) × (b – 3) 2

8. Volume of a cuboid = Length × Breadth × Height

= (x2 – 2) [(2x + 2) (x – 1)]

= (x2 – 2) [2x2 – 2x + 2x – 2] 1

= (x2 – 2) [2x2 – 2]

= 2(x2 – 2) (x2 – 1)

= 2[x4 – x2 – 2x2 + 2] 1

= 2[x4 – 3x2 + 2]

= 2x4 – 6x2 + 4. 1


9 WORKSHEET-47

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ALGEBRAIC EXPRESSIONSAND IDENTITIES

[A] Lab Activity

Object :

To verify the following identity by paper cutting and pasting.

(a + b)2 = a2 + 2ab + b2

Pre-requisite Knowledge :

(i) Area of a square

(ii) Area of a rectangle

(iii) Addition of algebraic expressions

Materials Required :

(i) Thick chart paper

(ii) A pair of scissors

(iii) White papers

(iv) Sketch pens

(v) Fevicol. 1

Procedure :

(1) Draw a square of side ‘a’ units on a thick sheet of paper.

(2) Draw another square of side ‘b’ units on the thick sheet of paper.

a2 b2

a

a b

b

½

(3) Draw two rectangles each of length ‘a’ units and breadth ‘b’ units as shownbelow :

ab ab

a a

b b½


(4) Draw a square of side (a + b) units.

(a + b)2 (a + b)

(a + b)

½

(5) Take the cutouts of all the above figures.

(6) Arrange and paste the cutouts 2 2, ,a b ab and ab on the cutout of the square (a + b)2.

ab

abb2

a2

½

Observations :

The square of side (a + b) units is completely covered by the cutouts of a2, b2, ab and

ab. ½

Conclusion :2( )a b = 2 2a ab ab b

Or2( )a b = 2 22a ab b . ½

[B] Fill in the Blanks 5 × ½ = 2½

1. 2

2. Coefficient

3. 7

4. Monomial

5. Trinomial

[C] True/False 5 × ½ = 2½

1. True

2. False


3. True

4. False

5. True

[D] Viva-Voce ½ × 10 = 5

1. (b) 1

2. 2

3. Yes

4. Two

5. Binomial

6. An expression is a group of terms, separated by + or – signs whereas a polynomial is a

group of terms separated by + or – or × signs.

7. (c) – 3

8. ‘Variable’ is a symbol which can have different values whereas a ‘constant’ has a fixed

value.

9. If an equality is true for all values of variables in it, then it is called an identity.

10. 0.

[E] Quiz

1. 2 2 sq. cmx 1

2. 2 2 2( ) 2a b a ab b 1

3. 2( )( ) ( )x a x b x a b x ab 1

4. ( )( )a b a b 1

5. No 1

6. An equation is not true for every value of variable in it, but an identity is. 1


1. (B) 6 1

2. (A) 3 faces 1

3. A polyhedron has a minimum number of your faces. 1

4. No, not always, because it can be a cuboid also. 1

5. Since, a prism is a polyhedron having two of its faces congruent and parallel, where as

other faces are parallelogram.

(i) No, a nail is not a prism. 12

(ii) Yes, unsharpened pencil is a prism. 12

(iii) No, table weight shown is not a prism. 12

(iv) Yes, box is a prism. 12

6. A solid is a polyhedron if Eulers formula F + V – E = 2 is satisfied. ½

Here, F = 10, E = 20 and V = 15

10 + 15 – 20 = 2

or 25 – 20 = 2 ½

or 5 = 2, which is not true

i.e. F + V – E 2 1

Thus, such a polyhedron is not possible.

7. Here :

Number of vertices (V) = 20

Number of edges (E) = 30

Let the number of faces = F. If do decahedron is to exist.

Using Euler’s formula, we have

F + V = E + 2 ....(1) 1

Substituting the values of V and E in (1), we get

F + 20 = 30 + 2 1

F + 20 = 32

F = 32 – 20

F = 12

Thus the required number of faces = 12. 1

8. The (bases) of the given solid are congruent polygons each of six sides.

It is a hexagonal prism. ½

In a hexagonal prism, we have :

The number of faces = 8 ½

The number of edges = 18 ½

The number of vertices = 12 ½

10 VISUALISING SOLID SHAPES WORKSHEET-48

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P-103V I S U A L I S I N G S O L I D S H A P E S

1. (A) triangles 1

2. (D) 12 1

3. The plural of a polyhedron is polyhedra. 1

4. If the base of a pyramid is a triangle, then we call it as a tetrahedron. 1

5. A polyhedron is bounded by four or more than four polygonal faces. 12

(i) No, it is not possible that a polyhedron has 3 triangles for its faces.

(ii) Yes, 4 triangles can be the faces of a polyhedron. 12

(iii)Yes, a square and 4 triangles can be the faces of a polyhedron. 1

6. (i)If we look at the given solid structure from the top, we would see just a square. 1

(ii)If we look at it from a side, i.e. left or right, then we would se a figure as shown here.

1

7. (i) A cylinder is not a polyhedron. 12

(ii) A cuboid is a polyhedron. 12

(iii) A cube is a polyhedron. 12

(iv) A cone is not a polyhedron. 12

(v) A sphere is not a polyhedron. 12

(vi) A pyramid is a polyhedron. 12

8. (i) Polyhedron : A solid shape bound by polygons which are called its faces is called a

polyhedron. The faces meet at line segments called edges and edjes meet at points

called vertices. For example, cube, cuboid, etc. 1

(ii) Prism : A prism is a solid, whose faces are parallelograms and whose ends (or

bases) are congruent parallel rectilinear figures. 1

(iii) Pyramid : A pyramid is a polyhedron whose base is a polygon of any number of

sides and whose other faces are triangles with a common vertex. 1


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1. (C) 5 1

2. (B) 6 1

3. When the bases of a prism are parallelograms, then it is called a parallelopiped. It is

formed by six parallelogram. 1

4. Both of the prisms and cylinders have their base and top as congruent faces and parallel

to each other. Also, a prism becomes a cylinder as the number of sides of its base

becomes larger and larger. i.e. A cylinder is a limiting case of a prism as the number of

sides of the base of a regular prism approach infinity. 1

5. The front view and the top view of the given solid are given below : 1

(i) (ii)

Front view Top view

2

6. Here, number of faces (F) = 20


Let the number of edges be E.

Using Euler’s formula, we have 1

F + V = E + 2

20 + 12 = E + 2

32 = E + 2

E = 32 – 2 = 30

Thus, the required number of edges = 30. 1

7. If a polyhedron is having number of faces as F, number of edges as E and the number of

vertices as V, then the relationship

F + V = E + 2

is known as Euler’s formula. 1

Following figure is a solid pentagonal prism.


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It has : Number of faces (F) = 7

Number of edges (E) = 15 1


Substituting the values of F, E and V in the relation,

F + V = E + 2

we have 7 + 10 = 15 + 2

17 = 17

Which is true, the Euler’s formula is verified. 1

8. (a)1 1

12 3

æ öç ÷+ +è økm 1

(b) Meena’s. 1

(c) Letter box, Petrol pump, Hospital. 1


1. (C) Cuboid 1

2. (C) Sphere 1

3. The pyramid and cones are alike because their lateral faces meet at a vertex. Also a

pyramid becomes a cone as the number of sides of its base becomes larger and larger. 1

4. 6 vertices 1

5. (i) F + V = E + 2

F + 6 = 12 + 2

F + 6 = 14

F = 14 – 6 = 8 1

(ii) F + V = E + 2

20 + 12 = E + 2

32 = E + 2

E = 32 – 2 1

E = 30

6. In figure (i), we have

F = 7, V = 10 and E = 15 1

F + V – E = 7 + 10 – 15

= 17 – 15 = 2, 1

which is true.

Thus, Euler’s formula is verified.

(ii) In figure (ii), we have

F = 9, V = 9 and E = 16 1

F + V – E = 9 + 9 – 16

= 18 – 16 = 2,

which is true.

Thus, Euler’s formula is verified. 1

7. At least 4 planes can enclose a solid. Tetrahedron is the simple polyhedron. Following

figure represent a simplest regular polyhedron called tetrahedron. 2


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A tetrahedron has :

4 triangular faces, i.e. F = 4 1

4 vertices, i.e., V = 4

6 edges, i.e., E = 6

Now, substituting the values of F, V and E in Euler’s formula, i.e.,

F + V = E + 2

we have, 4 + 4 = 6 + 2

8 = 8, which is true.

Thus, Euler’s formula is verified for a tetrahedron. 1



[A] Lab Activity

Make a prism and verify the Euler’s formula for it.

Object :

To make a prism say triangular prism and to verify the Euler’s formula for it.

Materials Required

(1) Cardboard (2) Ruler (3) A pair of scissors (4) Cellophane tape (5) Fevicol (6) Glazedpapers ½

Procedure :

(1) Cut two dimensional equilateral triangles of sides 15 cm each from the cardboard as

shown in fig. (i).

(i)

(2) Cut out the three identical rectangular surfaces each of length 20 cm and breadth 15 cm

leaving tab, on sides as shown in fig. (ii).

(3) Put one of the triangles horizontally. Then put the three rectangular faces vertically on

the three edges of the triangular surface such that breadth of one rectangular face lies

exactly on each edge of the triangle.

(4) Now, apply fevicol on the tabs and use the cellophane tape to fix these pairs of edges as

shown in fig. (iii).

(ii) (iii) (iv)

(5) In the same way, you have to fix the other triangular surfaces on the top of the figure

obtained in step 4.

(6) Paste Red glazed paper on the triangular faces and three other different colours on

rectangular faces.

(7) The three dimensional figure thus obtained is a triangular prism as shown in the fig.

(iv). 1½

Observation :

We observe that

Number of faces, F = 5

Number of vertices, V = 6

Number of edges, E = 9 ½

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Result :

Verification of Euler’s formula :

F + V – E = 2

Here, F + V – E = 5 + 6 – 9

= 11 – 9 = 2

Hence, Euler’s formula is verified. ½

[B] Fill in the Blanks 4 × ½ = 2

1. 8, 12, 6

2. Vertex

3. Prism

4. Tetrahedron

[C] True/False 4 × ½ = 2

1. True

2. False

3. True

4. False

[D] Match the column 1 × 5 = 5

(1) (C)

(2) (D)

(3) (E)

(4) (B)

(5) (A)

[E] Viva-Voice

1. F + V – E = 2 1

2. 6 1

3. Parallelogram 1

4. Hexagonal prism 1

5. 6 1

[F] Quiz 6 × ½ = 3

1. 4

2. 4

3. 8

4. 12

5. No

6. Dice


1. (C) 7 cm 12. (D) 15 cm 1

3. Area of a rhombus =1

2 × Product of diagonals

Area of given rhombus=1

7·5 cm 12 cm2 1

2

=1 75 12

2 10 1 cm2

= 15 × 3 cm2

= 45 cm2.12

4. Area of a quadrilateral =1

2 × Diagonal × [Sum of the perpendiculars

on the diagonal from opposite vertices] 12

=21

24 (8 13) m2

= ´ ´ 2124 21 m

2

= 212 21 m

= 2252 m . 12

5. Area of a parallelogram = Base × Corresponding height

Area of a tile =224 10

m100 100

(converting cm into m)

=2240

m10000

1

= 20·024 m

Now area of the floor = 1080 m2

Number of tiles =Total area

Area of one tile

=1080

0·024

= 45000 tiles. 1

6. A rhombus is a parallelogram

Area of a parallelogram = Base × Height

Area of the rhombus = Base × Height

= 6 × 4 cm2

= 24 cm2 1

Let the other diagonal be d.

Also area of the rhombus =1

8 42

d d [ One of the diagonals = 8 cm]

11 MENSURATION WORKSHEET-53

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P-111M E N S U R A T I O N

So, 4d = 24

or d =24

4 = 6 cm

Thus, the required other diagonal is 6 cm. 1

7. (a) Side of the square = 60 m

Its perimeter = 4 × side

= 4 × 60 m = 240 m ½

Area of the square = Side × Side

= 60 m × 60 m

= 3600 m2 12

(b) Perimeter of the rectangle = Perimeter of the given square

Perimeter of the rectangle = 240 m

But the perimeter of a rectangle = 2 × [Length + Breadth] = 240 m

or 2 × [80 m + Breadth] = 240 m

or 80 m + Breadth =240

2 m

= 120 m

Breadth = (120 – 80) m 1

= 40 m

Now, Area of the rectangle

= Length × Breadth

= 80 × 40 m2

= 3200 m2

Since, 3600 m2 > 3200 m2 ½

Area of the square field (a) is greater. ½

8. Perimeter of food-piece (a)

= (r + 1·5 + 2·8 + 1·5) cm

=22 2·8

5·8 cm7 2

æ ö´ +ç ÷

è ø

= (22 × 0·2 + 5·8) cm

= (4·4 + 5·8) cm = 10·2 cm 1

Perimeter of food-piece (b)

= (r + 2 + 2) cm

=22 2·8

4 cm7 2

æ ö´ +ç ÷

è ø

= (22 × 0·2 + 4) cm

= (4·4 + 4) cm = 8·4 cm 1

Since 10·2 > 8·4 > 7·2

The ant will take a longer round in case of food-piece (b). 1


1. (C) 2rh 1

2. (B) 7 cm 1

3. Let the height of the cuboid = h cm

Now base area × Height = Volume

or 180 × h = 900 12

or h =900

180 = 5 cm

Hence, the required height of the cuboid = 5 cm. 12

4. (i) Area of quadrilateral ABCD

= 1

2AC [Sum of perpendiculars on AC from opposite vertices] ½

= 1

6 cm [3 cm + 5 cm]2

= 1

6 cm 8 cm = 3 cm 8 cm2

= 24 cm2 ½

5. Tiles are rhombus shaped, having d1 = 45 cm and d2 = 30 cm.

Area of a tile (rhombus) = 1 2

1

2d d

=21

45 30 cm2

= 45 × 15 cm2 = 675 cm2 12

Total number of tiles = 3000

Area of floor = 675 × 3000 cm2

= 2025000 cm2

=22025000

m100 100´

2 2[ 1 m = 100 100 cm ]

=22025

m10

12

Cost of polishing the floor= Rs. 4 per sq. m

Cost of polishing the floor = Rs. 4 × 2025

10

= Rs. 2 × 405

= Rs. 810. 1


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6. Let the length of the side along the river

= 2x cm

Other side (parallel to the road) = x cm 12

Area of the trapezium shaped field

= 212 100 m

2x x

= 2 21 3003 100 m m

2 2x 1

2

But area of the field = 10500 m2

300

2x = 10500 1

2

or x =10500 2

300

=105 2

35 23

= 70 m.

Thus, the length of the side along the river

= 2x = 140 m. 12

7. Area of trapezium I :

Parallel sides are 24 cm and 16 cm.

Height =28 20

4 cm2

Area = ´ + ´ 21[16 24] 4 cm

2

= 40 × 2 cm2

= 80 cm2 1

Area of trapezium II :

Parallel sides are 20 cm and 28 cm.

Height =24 16

2

= 4 cm

Area =21

[20 28] 4 cm2 1

= 48 × 2 cm2 = 96 cm2

Area of trapezium III :

Area of trapezium III = Area of trapezium I

= 80 cm2 12


Area of trapezium IV :

Area of trapezium IV = Area of trapezium II

= 96 cm2. 12

8. Total surface area of the cuboid (a)

= 2[lb + bh + hl]

= 22[(60 40) (40 50) (50 60)] cm

= 22[2400 2000 3000] cm 1

= 2 22[7400 cm ] 14800 cm

Total surface area of the cube (b)

= 6l2 = 6 (50 × 50)

= 6(2500)

= 15000 cm2 1

Since the total surface area of the second (b) is more. 1

Cuboid (a) will require lesser material.


1. (B) 13860 litres ( 1 l = 0.001 m3) 1

2. (C) 1 litre ( 1 l = 1000 cm3) 1

3. Let d1 and d2 be the diagonals of the rhombus.

d1 = 12 cm and d2 = 9·2 cm

Area of rhombus = 1 2

1

2d d 1

2

Area of given rhombus =21

12 9·2 cm2

= 6 × 9·2 cm2 = 55·2 cm2. 12

4. (i) l = 6 cm, b = 4 cm and h = 2 cm

Total surface area of the cuboid

= 2( )lb bh hl

= 2(6 × 4 + 4 × 2 + 2 × 6) cm2

= 2(24 + 8 + 12) cm2 1

= 2 × 44 cm2 = 88 cm2

5. Here, r = 7 m and h = 3 m

Total surface area of the cylindrical tank

= 2r(r + h)

= 2222 7 (7 3) m

7´ ´ ´ + 1

= 2 × 22 × 10 m2

= 440 m2

Thus, 440 m2 metal is required. 1

6. Lateral surface area of a cylinder

= 4224 cm2 = 2rh

4224 = (2r) (33)

and length of the rectangular sheet = l cm = circumference of circular base = 2r

33 × l = 4224

l =4224

33 = 128 cm 1

Now perimeter of the rectangular sheet

= 2[Length + Breadth]

= 2[128 cm + 33 cm]

= 2 × 161 cm

= 322 cm 1


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7. Total surface area of a suitcase

= 2(lb + bh + hl)

= 2(80 48 48 24 24 80) 1

= 22(3840 1152 1920) cm

= 22(6912) cm

= 213824 cm

Total S. A. of 100 suitcases

= 2 2100 13824 cm 1382400 cm 1

Surface area of 1 m of tarpaulin

= 100 cm 96 cm´

= 9600 cm2

i.e. 9600 cm2 surface covered by 1 m of tarpaulin.

1382400 cm2 surface covered by tarpaulin 1382400

9600 metre or 144 m.

Thus, 144 metres of tarpaulin will be requird to cover 100 suitcases. 1

8. Let the side be x cm.

Total S. A. of the cube = 2 26 cmx 1

But the S. A. of the cube = 2600 cm

6x2 = 600 1

or x2 =600

6 = 100

or x2 = 102 x = 10 1

The required side of the cube = 10 cm.


1. (D) 4·62 litres 1

2. ... 1 cubic cm = 0.001 1

(b) Side (edge) of the cube = 1·5 cm

Volume of the cube = (edge)3

= 3(1·5 m)

=315 15 15

m10 10 10

=3 33375

m 3·375 m1000

. 1

3. (i) Radius (r) = 7 cm

Height (h) = 10 cm

Volume of the cylinder = r2h

=2 322

7 10 cm7

=322

7 7 10 cm7

= 322 7 10 cm

= 1540 cm3 12

(ii) Base area = 250 m2

Height = 2 m

Volume of the cylinder = Base area × Height

= 250 m2 × 2 m

= 500 m3. 12

4. The road roller is a cylinder

Radius =84

cm = 42 cm2

Length (h) = 1 m = 100 cm

Lateral surface area = 2rh

= 2222 42 100 cm

7

= 2 × 22 × 6 × 100 cm2

= 26400 cm2 1

Area levelled by the roller in 1 revolution = 26400 cm2

or Area levelled in 750 revolutions

= 2750 26400 cm

= 2750 26400 m

100 100

= 15 × 132 m2 = 1980 m2 1Thus, the required area of the road

= 1980 m2.


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5. Volume of the cuboid = 60 cm × 54 cm × 30 cm ½= (60 × 54 × 30) cm3

Volume of the small cube = (6 × 6 × 6) cm3 ½

Required Number of cubes =Volume of the cuboid

Volume of the small cube½

=3

3

(60 54 30) cm

(6 6 6) cm

=60 54 30

6 6 6

= 10 × 9 × 5 = 450 ½

Thus, 450 small cubes can be placed in the given cuboid.

6. Volume of the reservoir = 108 m3

1 m 3 = 1000 litres

Capacity of the reservoir = 108 × 1000 litres

= 108000 litres 1

Amount of water poured in 1 minute = 60 litres

Amount of water to be poured in 1 hour

= 60 × 60 litres 1

Thus, number of hours required to fill the reservoir = 108000

60 60 = 30

The required number of hours = 30. 1

7. (i) Total length of the fencing surrounding it

= Perimeter of the park

= 30 m + 20 m + 30 m + 20 m

= 100 m 1

(b) Land occupied by the park = Area of the park

= 30 × 20 m2

= 600 m2 1

(c) Area of cemented path = Area of park – Area of park left after cementing the path.

Now, since path is 1 m wide, so, the rectangular area left after cementing the path

= {(30 – 2) × (20 – 2)} m2

= (28 × 18) m2 = 504 m2 12

Area of cemented path= 600 m2 – 504 m2 = 96 m2

Number of cement bags used

=area of the path

area cemented by 1 bagIf 1 bag of cement is required to cement 4 m2 area, then the number of cement bagsused

=96

4 = 24 1

2

(d) Area of rectangular beds = 2 × (1·5 × 2) m2 = 6 m2

Area of the park left after cementing the path= 504 m2

Area covered by the grass

= 504 m2 – 6 m2 = 498 m2 . 1



[A] Lab Activity

Objective :To derive the formula for the total surface area of a right circular closed cylinder.

Materials Required :1. A closed cylindrical box2. A pair of scissors3. A pen4. Graph papers ½

Procedure :

(1) Take the cylindrical box. Let its radius be r and height be h.(2) Trace one of its bases on a graph paper and cut it (Fig. (i)). ½

r

h

½

(i)(3) You get a circular strip of radius r (Fig. (ii))(4) The circumference of this strip is 2r and area is r2 (Fig. (ii)).(5) Thus, the surface area of each base of the cylindrical box is r2. ½(6) Take another graph paper in such a way that its width is equal to the height of the box,

i.e. h.(7) Wrap this paper around the box such that it just fits around the box (cut out the excess

paper) (Fig. (iii)). ½

(ii)

½

(iii)

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(8) The shape of this paper obtained is a rectangular strip as shown in Fig. (iv).(9) The length of this strip is same as the circumference of the circular strip, obtained in the

step 3, i.e., 2r, and height h.(10) Area of this (rectangular) strip = 2rh. 1

This represent the lateral surface area of the cylindrical box, i.e., lateral surface area of thebox = 2rh.

h

2r

1

(iv)

Result :

Total surface area of the cylindrical box= 2 × surface area of each base + lateral surface area= 2 × r2 + 2rh= 2r (r + h).

Hence, we obtain that total surface area of a right circular closed cylinder = 2r(r + h)sq. units. 1

[B] Fill in the blanks ½ × 4 = 2

1. Base, height

2. Base, height

3. Sum of parallel sides, distance between them

4. 128

[C] True/False : ½ × 4 = 2

1. True

2. False

3. True

4. False

[D] Viva-Voce 1 × 5 = 5

1. 4 cm

2. 2rh

3. No vertex

4. 3, one curved face and two circular faces

5. They are identical.

[E] Quiz 1 × 5 = 5

1. a3 cubic units

2. 1000

3. 24 cm2

4. 1 litre

5. 1 : 1

P-121E X P O N E N T A N D P O W E R SS

1. (B) 1

101

2. (D) 16 1

3. The multiplicative inverse of 2– 4 is 24. 1

4. 22

1 1 1( 4) ·

( 4) ( 4) 16( 4)

1

5. (i) 1 = 13

3

1

2 =

33

3

1 1

22

Now

2

3

1

2

=

23 3 21 1

2 2

[Using (am)n = amn]

=

6

6

1 1

2 2

1

(ii)4

4 5( 3)

3

=

45

( 3)3

= 4[( 1) 5]- ´

= 4 4[( 1) (5) ]- ´

= 4 41 (5) (5) . 1

6. (i) 0 1 2(3 4 ) 2 0 1 1

1 and a aa

0 1 2(3 4 ) 2 =21

1 24

=5

4 54 1

(ii)

2 2 21 1 1

2 3 4

= (2)2 + (3)2 + (4)2

= 4 + 9 + 16

= 29 1

7. (i) m ma b = ( )mab

1 1 2(2 4 ) 2 = 1 2(2 4) 2

= 1 2 1 2(2 2 ) 2 1

12 EXPONENTS AND POWERS WORKSHEET-58

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= 1 2 1 2(2 ) 2

= 3 1 2(2 ) 2

= 3 22 2

= 3 ( 2) 1 12 (2)

2 1

2

(ii) 13 = 1 11 1 1, 4 and 5

3 4 5- -= =

1 1 1 0[3 4 5 ] =

01 1 1

3 4 5

1

=

020 15 12

60

=

047

160

12

8. m na a = m na

35 5m = 55

( 3)5m- - = 55 1

35m = 55

Since, bases 5 are equal, therefore exponents are also equal. 1

i.e., m + 3 = 5

or m = 5 – 3 = 2

Thus, the required value of m is 2. 1


1. (D) 1 1

2. (D) 7

31

3.

5

5

1 1 1 1

1 1 1 1 1 12 12 2 2 2 2 322

-æ ö= = =ç ÷

æ ö æ öè ø æ ö ´ ´ ´ ´ç ÷ ç ÷ç ÷ è ø è øè ø

= 32 1

4. (i) 0·000000564 = 564

1000000000

= 2

9 7

5·64 5·6410

10 10 = 5·64 × 10– 7 1

0·000000564 = 5·64 × 10– 7

(ii) 0·0000021 = 21

10000000 =

2·1 10

10000000

= 6

2·1

10 = 2·1 × 10– 6

0·0000021 = 2·1 × 10– 6 1

5.1 3

4

8 5

2

=

1

4

8 (5 5 5)

2

=41

2 1258 1

=1

2 2 2 2 1258

= 2 × 125 = 250. 1

6. 1 1 1(5 2 ) 6 = 1 1(5 2) 6 1

= 1 1(10) 6

= 1 1(10 6) (60)- -´ =

=1

601

7. (i) (a)– 1 =1

a

11 11 1

3 4

=1

1 1

1 1

3 4

= 1 1(3 4) ( 1)

=1

11

1


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(ii) a– m =1ma

48

5

=

45

8

Now,

7 45 8

8 5

=

7 45 5

8 8

=

7 45

8

- +æ öç ÷è ø

=

35

8

-æ öç ÷è ø

[ ]m n m na a a

=

38

5

æ öç ÷è ø

=8 8 8

5 5 5

´ ´

´ ´

=512

125. 1

8. 125 = 5 × 5 × 5 = 53

10– 5 = (2 × 5)– 5 = 2– 5 × 5– 5

6– 5 = 5 5 5(2 3) 2 3- - -´ = ´ 1

5 5

7 5

3 10 125

5 6

=

5 5 5 3

7 5 5

3 2 5 5

5 2 3

- - -

- - -

´ ´ ´

´ ´

=

5 5 5 3

5 5 7

3 2 5 5

3 2 5

1

= 5 ( 5) 5 ( 5) 5 3 ( 7)3 2 5- -- - -- - + - -´ ´

= 5 5 5 5 5 103 2 5

= 0 0 53 2 5

= 51 1 5

= 55. 1


1. (C) – 1 1

2. (C) 3 1

3. (i) 63·02 10 = 2 6302 10 10- -´ ´

= 8302 10-´

=302

100000000

= 0·00000302 12

(ii) 4·5 × 104 =45

1000010

= 45 × 1000

= 45000 12

4. (i) 1

1000000 =

66

11 10

10

1 micron = 1 × 10– 6 m12

(ii) 0·07 =2

2

7 77 10

100 10

Thickness of a thick paper is 7 × 10– 2 mm. 12

5. ( )m n mna a=

222

3

=

( 2) 22

3

1

=

4 4

4

( 2)2

3 (3)

m m

m

a a

b b

=4

4

3

( 2)

=3 3 3 3

( 2) ( 2) ( 2) ( 2)

=81

.16

1

6. Since, 25 = 52

4

3 8

25

5 10

t

t

=

2 ( 4) ( 8)

3

5

5 5 2

t - - -

-

´

´ ´

=2 3 1 4 85

2

t 1

=4 45

2

t´


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= 45 5 5 5

2t

´ ´ ´´

=4625.

2t 1

7. Since

21

4 =

24

1 = (4)2 = 16

31

3 = (3)3 = 3 × 3 × 3 = 27 1

41

2= (2)4 = 2 × 2 × 2 × 2 = 16

2 3 41 1 1

4 3 2 = 16 + 27 + 16 = 59 1

8. (i) 0·00000000000942 =942

100000000000000

=2

14

9·42 10

10

= 9·42 × 102 – 14

= 9·42 × 10– 12 1

(ii) 6020000000000000 = 602 × 10000000000000

= 13602 10

= 2 136·02 10 10

= 156·02 10 . 1

9. Thickness of one book = 20 mm

Thickness of 5 books = 5 × 20 mm

= 100 mm

Again,

Thickness of 1 paper sheet = 0·016 mm

Thickness of 5 paper sheets

= 5 × 0·016 mm

= 0·080 mm 1

Total thickness = 100 mm + 0·080 mm

= 100·08 mm

= 1·0008 × 102 mm. 1



1. (B) 6 1

2. (A) (ab)m 1

3. (i) 3 10p p = 3 ( 10) 7( ) ( )p p [... am × an = am+n] 12

= 7

1

p

(ii) 2 5 63 3 3 = 2 ( 5) 63 + - +

= 8 5 33 3- = . 12

4. 33

15

5

--´ = 3 3 05 5 1.- + = = 1

5. 0·000,000,000,000,000,000,16

=16

100000000000000000000

= 20 19

1·6 10 1·6

10 10

= 191·6 10

Charge on an electron is 1·6 × 10– 19 coulomb. 1

6. 34500000 = 345 × 100000

= 3.45 × 100 × 100000

= 3.45 × 102 × 105 12

= 3.45 × 102+5

= 3.45 × 107

Thus, 34500000 = 3.45 × 107 12

7. We have (80 + 5– 1)– 2 × 32

=

221

1 35

0 1 1

1 and x xx

=

226

35

=

225

36

1

=5 5

3 36 6

=5 5

2 2

=25

4. 1

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8. Since 125 = 5 × 5 × 5 = 53

6– 5 = (2 × 3)– 5 = 5 52 3

5 5

4 5

2 3 125

5 6

=

5 5 3

4 5 5

2 3 5

5 2 3

1

=5 5

3 45 5

2 35

2 3

1

= 1 × 1 × 57

= 5 × 5 × 5 × 5 × 5 × 5 × 5

= 78125. 1

9. Distance between Sun and Moon

= 1·496 × 1011 – 3·84 × 108 1

= 1·496 × 1000 × 108 – 3·84 × 108

= (1496 – 3·84) × 108 m 1

= 1492·16 × 108 m

= 1·49216 × 1011 m. 1



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[A] Lab Activity

Object :

To verify law of exponents using paper folding.

Let’s Start :

We know that, 5 2 32 2 2 8. Let us verify it by paper folding.


A butter paper. 1

Procedure :

(1) Take a butter paper or a very thin paper. Fold it as many number of times as power ofdividend. Here, we have folded it 5 times (see Fig. i). 1

1

(i)

(2) Open up the fold and count the number of boxes you get.

(3) Then make groups of boxes equal to divisor, that is, 4 in this case (see Fig. ii). 1

1

(ii)

(4) Now, count the number of groups you get.

(5) The number of groups you get will be equal to the quotient.

Observation

We observe that as we go on increasing the number of folds the number of boxes multiplies.

Conclusion

We conclude that the total boxes we obtain is equal to dividend and the number ofgroups gives quotient. 1

Result

25 22 = 8, i.e., 23

Hence, am an = am – n 1

[B] Fill in the Blanks ½ × 5 = 2½

1. 3·6 × 105

2. 1·23 × 10– 5


3.9

4

4. 0·003

5. 0·000532

[C] True/False ½ × 5 = 2½

1. True

2. False

3. True

4. True

5. False

[D] Viva-Voce ½ × 6 = 3

1. First number is 5 times the second number.

2. 1·275 × 10–5

3. (– 5)–9

4. 1/729

5. 24

6. 22

[E] Quiz 1 × 5 = 5

1. 24

2. No

3. x = 2

4. – 1

5. 7·25 × 10– 4

P-131D I R E C T A N D I N V E R S E P R O P O R T I O N S

1. (D) 720 1

2. (A) 17·6 1

3. Clearly, 3 5 7 9 12 1

6 10 14 18 24 2

x

y= = = = = = 1

(Constant)

x, y are directly proportional.

4. Let the required distance be x km. Then; we have

36 25

432 x

Quantity of petrol (in litres)

Distance (in km)

Clearly, less is the quantity of petrol consumed, less is the distance covered. 12

So, it is a case of direct proportion

36 25 1 25

432 12x x x = 12 × 25 = 300

required distance is 300 km. 12

5. We have :

60 154 hours Rs. 60

4 1100 25

8 hours Rs. 1008 2

140 3512 hours Rs. 140

12 3180 15

24 hours Rs. 18024 2

Parking Time Parking Charges Parking charges/Parking time

1

Since15

1

15 35 15

2 3 2

The parking charges are not in direct proportion with the parking time. 1

6. Let the required number of sugar crystals be x in 5 kg. of sugar. We have :

(i) 62 9 10

5 y

Weight of sugar Number of sugar

(in kg) crystals

13 DIRECT AND INVERSE PROPORTIONS WORKSHEET-63

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Since, more the amount of sugar, more would be the number of sugar crystals. ½

it is a case of direct proportion.

2

5 =

69 10

x

½

or 2 × x = 6[9 10 ] 5

or x =6 69 10 5 45 10

2 2

½

= 22·5 × 106 = 2·25 × 107 [Standard form]

Thus, the required number of sugar crystals = 2·25 × 107. ½

7. Let the red pigment be represented by x1, x2, x3, ..... and the base represented by y1, y2,

y3, ..... .

As the base increases, the required number of red pigments will also increase.

The quantities are in direct proportion. ½

i.e. 1

1

x

y = 32

2 3

.....xx

y y ½

For x1 = 1, y1 = 8

1

1

x

y =

1

8

Now, 2

2

x

y =

2

1 4 1

8 8y ½

or y2 = 4 8 32

3

3

x

y =

3

1 7 1

8 8yÞ = ½

or y3 = 7 × 8 = 56

4

4

x

y =

4

1 12 1

8 8y ½

or y4 = 12 × 8 = 96

5

5

x

y =

5

1 20 1

8 8y

or y5 = 20 × 8 = 160

Thus, the required parts of base are : 32, 56, 96 and 160. ½


8. Let the required distance covered in the map be x cm. We have :

18 1

72 x

Actual distance Distance

covered on the road represented

(in km) on the map (in cm)1

Since, it is a case of direct proportion,

18

72 =

1

x1

or 18 × x = 72

or x =72

418

Thus, the required distance on the map is 4 cm. 1


1. (A) 2700 12. (C) 29·5 kg 13. Let the required number of days be x. Then we have

45 35

49 x

Number of men

Number of days

Clearly, less men will take more days to finish the work. 12

So, it is a case of inverse proportion. 45 × 49 = 35 × x

x =45 49

6335

The required number of days = 63. 12

4. Let the required number of workers be x. Then we have

14

45 35

xNumber of workers

Number of hours

For finishing the work in less hours, more workers will be needed. 12

So, it is a case of inverse proportion. 14 × 45 = x × 35

x =14 45

1835

Hence, the required number of workers = 18. 12

5. Since the speed is constant.

for longer distance, more time will be required.

So, it is a case of direct proportion.

Let the required distance to be travelled in 5 hours (= 300 minutes) be x km. 1We have

14 25

300x

Distance (in km.) Time in (minutes)

12

14

x =

25

300or 25 × x = 14 × 300

or x =14 300

25

´

= 14 × 12 = 168

The required distance = 168 km. 12

6. Number of animals added = 10

now, the total number of animals

= 10 +20 = 30

For more number of animals, the food will last less number of days. 1

It is a case of inverse proportion.


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Thus, we have 30 × x = 20 × 6

or x =20 6

430

Therefore, the food will now last for 4 days. 17. (i) Obviously, more the number of spokes, less is the measure of angle between a pair of

consecuitve spokes. It is a case of inverse proportion. 1

Thus, 4 × 90° = 8 × x1 x1 = 4 90

458

´ °= °

4 × 90° = 10 × x2 x2 = 4 90

3610

´ °= °

4 × 90° = 12 × x3 x3 = 4 90

3012

´ °= °

Thus, the table is completed as below :

4 6 8 10 12

90 60 45 36 30° ° ° ° °

Number of spokes

Angle beween a pair of

consecutive spokes1

(ii) Let required number of spokes be n.

n × 40° = 4 × 90°

or n =4 90

40

= 9

The required number of spokes = 9. 1

8. (i) Let the time taken by the remaining persons to complete the job be x. 2 persons – 1 person = 1 personand lesser the number of person, more will be the number of days to complete the job.

it is a case of inverse proportion. ½

We have

2 3

1 x

Number of persons Number of days

2 × 3 = 1 × x

or x =2 3

61

1

1 person will complete the job in 6 days.(ii) Let the number of persons required to finish the job in 1 day be y.

2 3

1y

Number of persons Number of days

2 × 3 = 1 × y ½

or y =2 3

61

6 persons will be required to complete the job in 1 day. 1


1. (B) 5 : 3 12. (B) xy = k, being a constant 13. Let x bottles be filled in 5 hours.

840 6

5x

Number of bottles filled Number of hours

For more number of hours, more number of bottles would be filled. Thus given quantities

vary directly. 12

840

x =

6

5

or 6x = 5 × 840

or x =5 840

6

= 5 × 140 = 700 12

Thus, the required number of bottles = 700.

4. Let the number of workers employed to build the wall in 30 hours be y. We have thefollowing table.

48 30

15 y

Number of hours

Number of workers

Obviously more the number of workers, faster will they build the wall. So, the number

of hours and number of workers vary in inverse proportion. 12

So 48 × 15 = 30 × y

Therefore,48 15

30

= y

or y = 24

i.e., to finish the work in 30 hours, 24 workers will be required. 12

5. More is the number of persons, less is the time to complete job.

it is a case of inverse proportion. ½

3 4

4 x

Number of days

Number of persons to Complete the

wiring job½

3 × 4 = 4 × x

or x =3 4

34

the required number of days = 3. 1


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6. Suppose that x machines would be required. We have the following table.

Number of machines 42

Number of days 63 54

x

Lesser the number of machines, more will be the number of days to produce the samenumber of articles. 1

So, this is a case of inverse proportion.

Hence, 42 × 63 = x × 54 1

x =42 63

54

x = 49

Hence, 49 machines would be required. 1

7. Let x1 = p1, x2 = 200, x3 = 300

are y1 = 60, y2 = 30, y3 = p2

Since x and y are in inverse variation.

(i) x1y1 = x2y2

p1 × 60 = 200 × 30 1

p1 =200 30

60

´ = 100. 1

(ii) Also x2y2 = x3y3

200 × 30 = 300 × p2

p2 =200 30

300

´ = 20. 1

8. Actual length of the bacteria

=5

50000 cm

=1

10000 cm = 10–4 cm 1

More the number of times a photograph of a bacteria is enlarged, more the lengthattained. So, the number of times a photograph of a bacteria is enlarged and the lengthattained are directly proportional to each other.

So,1

1

x

y =2

2

x

y

50000

5 =

2

20000

y 1

50000y2 = 5 × 20000

y2 =5 20000

50000

y2 = 2

Hence, its enlarged length would be 2 cm. 1


1. (A) Inverse proportion 1

2. (C) 20 days 1

3. Reduced number of children = 24 – 4 = 20

Since, more the number of children, less is the quantity of sweets, that each child gets.

It is a case of inverse variation.

24 × 5 = 20 × x 12

or x =24 5

20

= 6

Each student will get 6 sweets. 12

4. Let the required length of the model of the ship = x m.

We have :

28 12

9x

Length of the ship Height of the mast

Since, more the length of the ship, more would be the height of its mast.

It is a case of direct proportion.

28

x =

12

91

or x =28 9

12

= 7 × 3 = 21

Thus, the required length of the model ship = 21 m. 2

5. Present duration of each period = 50 minutes

= 45 minutes 1

Duration of each period after reduction.

Let the present number of periods after reduction in time be ‘x’.

Since, more the number of periods, less is the duration of a period.

It is a case of inverse variation. 1

we have

9 50

45x

Duration of ech periodNumber of periods

(in minutes)

1

9 × 50 = x × 45

x =9 50

45

= 10

Thus, the required number of periods = 10. 1


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6. ·.· Quantity of water required for 2 persons

= 300 mL

Quantity of water required for 5 persons

=300

52 mL = 750 mL 1

Quantity of sugar required for 2 persons

= 2 spoons

Quantity of sugar required for 5 persons

=25

2 = 5 spoons 1

Quantity of tea leaves required for 2 persons

= 1 spoon

Quantity of tea leaves required for 5 persons

=15

2 =

122

spoons 1

Quantity of milk required for 2 persons

= 50 mL

Quantity of milk required for 5 persons

=50

52 mL = 125 mL 1

Thus, Mohan will require 750 mL of water, 5 spoons of suagar, 122

spoons of tea leaves

and 125 mL of milk of 5 persons. 1



[A] Lab Activity

Objective :

To show relation between two things in direct and indirect proportion by plotting the

data of both on a graph paper.

Let’s Start :

Observe the following tables :

(A)

1 2 3 4

25 50 75 100

No. of items

purchased

Total Cost

(in Rs.)

½

(B)

5 10 15 20

300 150 100 75

No. of children

No. of days food

will last in pantry½


(i) Data (ii) Graph papers (iii) Pencil (iv) Ruler.

Procedure :

(1) Taking suitable scale plot the points of Data 1 (Table A) on the graph paper and join the

points (see Fig. (i)).

(2) Take suitable scale plot the points of Data 2 (Table B) on a different graph paper and

join the points (see Fig. (ii)).

(3) Observe and compare both graphs by moving from left to right.

1

(i)

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1

(ii)

Observations :

For 1st graph : We observe that as we go from left to right the number of items

purchased increases, so the total cost also increases.

For 2nd graph : We observe that as number of children increases the number of days

for which the food will last decreases.

Conclusion :

We conclude that 1st graph is of direct proportion and 2nd graph is of indirect proportion.

2

Result :

In direct proportion, the two things given are in direct relation, i.e., if one increases the

other also increases and if one decreases the other also decreases. In indicrect proportion,

the result is inverse that is the two given things are in inverse relation, i.e., if one

increases other will decrease and vice-versa. ½


1. 10– 4 cm

2. 3

3. 120

4. 2

5. 13

[C] True/False ½ × 4 = 2

1. True

2. True

3. False

4. False


[D] Viva-Voce

1. If the number of items purchased will be reduced, then the total cost will also reduce.1

2. This happens beause items purchsed and total cost are in direct proportion. 1

3. In graph (ii), we observe downward sloping curve because as the number of children is

increasing, the no. of days the food will last is decreasing. 1

4. This happens because number of children and the number of days food will last are in

indirect proportion. 1

5. For a given job, more the number of workers, less will be the time taken to complete thework. 1

[E] Quiz

1. Rs. 80 1

2. For the given table 1

·3

x

y 1

3. k is constant of proportionality. 1

4. Simple interest. 1

5. The required number of sheets is 375. 1

P-143F A C T O R I S A T I O N

1. (A) 10xy(4x + 3y)(4x– 3y) 1

2. (D) 5(z + 4)(z – 4) 1

3. (a) 14pq = 2 7 p q = (2 7 )p q

28p2q2 = 2 2 7 p p q q´ ´ ´ ´ ´ ´

= (2 7 ) 2p q p q

The common factor = (2 7 )p q

= 14pq 12

(b) 2x = 1 2 1 2x x

3x2 = 1 3 1 3x x x x´ ´ ´ = ´ ´ ´

and 4 = 1 2 2 1 2 2´ ´ = ´ ´

The common factor = 1. 12

4. (a) 7x = 7 (7)x x

42 = 2 7 3 (7) 2 3

7x – 42 = 7[( ) (2 3)]x

= 7[ 6]x 12

(b) 6p = 2 3 p

= (2 3) p´ ´

12q = 2 × 2 × 3 × q

= (2 × 3) × 2 × q

6p – 12q = (2 3)[( ) (2 )]p q

= 6[ 2 ]p q . 12

5. 2 33x y = 3 x x y y y

= ( ) 3x x y y y

3 210x y = 2 5 x x x y y

= ( ) 2 5x x y y x 1

2 26x y z = 2 3 x x y y z

= ( ) 2 3x x y y z

The common factor = (x × x × y × y)

= 2 2x y . 1

6. 2 249 84 36y yz z+ + = 2 2(7 ) 2(7 )(6 ) (6 )y y z z 1

= 2(7 6 )y z

= (7 6 )(7 6 )y z y z 1

14 FACTORISATION WORKSHEET-68

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7. 2 7 10y y = 2 2 5 10y y y [Splitting 7y in 2y + 5y such that

2y × 5y = 10y2] 1

= ( 2) 5( 2)y y y

= ( 2)( 5)y y 1

2 7 10

( 5)

y y

y

=

( 2)( 5)( 2)

( 5)

y yy

y

+ += +

+

2( 7 10) ( 5)y y y = (y + 2). 1

8. 25 25 20p p = 25( 5 4)p p

= 5(p2 – p – 4p + 4)

= 5[ ( 1) 4( 1)]p p p- - - 1

= 5[( 1)( 4)]p p

25 25 20

1

p p

p

=

5( 1)( 4)

( 1)

p p

p

- -

-1

=5 ( 4)

1

p´ -

Thus, 2(5 25 20) ( 1) 5( 4)p p p p 1


1. (D) q3 – p3 1

2. (C) 162 1

3. 2 8 8x xy x y

= ( ) 8( )x x y x y+ + + 12

= ( )( 8)x y x . 12

4. 2 24 9p q = 2 2(2 ) (3 )p q 12

= (2 3 )(2 3 )p q p q . 12

[Using a2 – b2 = (a + b)(a – b)]

5. (i) 4 4a b = 2 2 2 2( ) ( )a b

= 2 2 2 2( )( )a b a b 12

2 2[using ( )( )]a b a b a b

= 2 2( )( )( )a b a b a b 12

(ii) p4 – 81 = 2 2 2( ) (9)p

= 2 2( 9)( 9)p p 12

= 2 2 2( 9)[( ) (3) ]p p

= 2( 9)( 3)( 3)p p p+ + - 12

6. 2 4 6 2 2 363 7a b c a b c =2 4 6

2 2 3

63

7

a b c

a b c

=2 4 6

2 2 3

3 3 7

7

a b c

a b c

=2 4 6

2 2 33 3

a b c

a b c

= 2 2 4 2 6 39 a b c 1

= 0 2 39 a b c

= 2 3 2 39 1 9b c b c . 1

7. 2 29 16x y = 2 2(3 ) (4 )x y

= (3 4 )(3 4 )x y x y 1


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2 212 (9 16 )

4 (3 4 )

xy x y

xy x y

=

2 2 3 (3 4 ) (3 4 )

2 2 (3 4 )

x y x y x y

x y x y

1

=3 (3 4 )

1

x y

2 212 (9 16 ) 4 (3 4 )xy x y xy x y- ¸ + = 3(3x – 4y) 1

8. 3a – 12 = 3(a – 4)

5b – 30 = 5(b – 6)

96 = 2 × 2 × 2 × 2 × 2 × 3

and 144 = 2 × 2 × 2 × 2 × 3 × 3 1

96 (3 12)(5 30)

144( 4)( 6)

abc a b

a b

- -

- -

=2 2 2 2 2 3 3 ( 4) 5 ( 6)

2 2 2 2 3 3 ( 4) ( 6)

a b c a b

a b

´ ´ ´ ´ ´ ´ ´ ´ ´ ´ - ´ ´ -

´ ´ ´ ´ ´ ´ - ´ -1

= 2 × 5 × a × b × c = 10abc

Thus, 96abc(3a – 12)(5b – 36) 144(a – 4)(b – 6) = 10abc. 1


1. (C) (x + 2) 1

2. (C) (x – 1)(x2 + 1) 1

3. 428 56x x =428

56

x

x =

42 2 7

2 2 2 7

x

x

12

=4 1 31 1

2 2x x . 1

2

4. 2(5 6 ) 3x x x =25 6 (5 6)

3 3

x x x x

x x

- -=

=5 6

3

x-. 1

5. Taking out 2x as common from each term, we have

3 2 22 2 2x xy xz = 2 2 22 ( )x x y z . 1

6. 3z – 24 = 3(z – 8)

2 29 (3 24)

27 ( 8)

x y z

xy z

=

2 23 3 3 ( 8)

3 3 3 ( 8)

x y z

x y z

´ ´ ´ ´ ´ -

´ ´ ´ ´ ´ -

=

2 22 1 2 1x y

x yx y

= xy. 1

7. 10x – 25 = 5(2x – 5)

10 25

5

x =

5(2 5)(2 5)

5

xx

1

8. 4 4( )x y z = 2 2 2 2[ ] [( ) ]x y z

= 2 2[( ) ( ) ]x y z 2 2[( ) ( ) ]x y z 1

2 2[Using ( )( )]a b a b a b

We can factorise [x2 – (y + z)2] further as

2 2( ) ( )x y z = [( ) ( )][( ) ( )]x y z x y z

= ( )( )x y z x y z 1

4 4( )x y z = ( )( )x y z x y z 2 2[ ( ) ]x y z 1


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9. We have

4 4( )x x z = 2 2 2 2[ ] [( ) ]x x z- -

= 2 2 2 2[ ( ) ][ ( ) ]x x z x x z 1

Now, factorising x2 – (x – z)2 further, we have

2 2( )x x z = [ ( )][ ( )]x x z x x z

= ( )( )x x z x x z

= (2 )( )x z z 1

4 4( )x y z = 2 2(2 )[ ( ) ]z x z x x z

= 2 2 2(2 )[ ( 2 )]z x z x x xz z 1

= 2 2 2(2 )[ 2 ]z x z x x xz z

= 2 2(2 )(2 2 )z x z x xz z 1



1. (A) 12x 1

2. (B) a + b and a + c 1

3. 5 316 144x x = 3 216 ( 9)x x 12

= 3 2 216 [( ) (3) ]x x

= 316 ( 3)( 3)x x x . 12

[Using a2 – b2 = (a – b)(a + b)]

4. 2 27 21p q = 7 7 3p p q q

= 7[ 3 ]p p q q 12

= 2 27( 3 )p q .

5. 2 3 266 11pq r qr¸ =2 3

2

66

11

pq r

qr1

=2 3 11

11

p q q r r r

q r r

12

=2 3

1

p q r = 6pqr. 1

2

6. Using the identity a2 – b2 = (a + b)(a – b), we have

2 2( ) ( )l m l m = [( ) ( )]l m l m [( ) ( )]l m l m

= [ ][ ]l m l m l m l m 1

= (2 )(2 ) 2 2( )l m l m

= 4lm. 1

7. 2 2 23 (50 98) 26 (5 7)ay y y y

=

2 2

2

3 (50 98)

26 (5 7)

ay y

y y

=

2 2

2

3 2 (25 49)

26 (5 7)

ay y

y y12

=

2 2 2

2

3 2 [(5 ) (7) ]

26 (5 7)

ay y

y y12

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=

2

2

3 2 (5 7)(5 7)

26 (5 7)

ay y y

y y =

2

2

3 2(5 7)

26

ay y

y

=3 3

(5 7) (5 7)13 13

a ay y 1

9. (m2 – 14m – 32) (m + 2) =

2 14 32

2

m m

m

=

2 16 2 32

2

m m m

m1

=

( 16) 2( 16)

2

m m m

m

=

( 16)( 2)

2

m m

m

= (m – 16) 1

10. Joseph and Sirish are not correct while Suman is correct as she has followed the rule

mentioned. ½+½+½+½



[A] Lab Activity

Objective :

To verify experimentally that the factors of 3xy + 2y + 3x + 2 are (3x + 2)(y + 1) such that3xy + 2y + 3x + 2 = (3x + 2)(y + 1).

Let’s Start :

We know that area of a rectangle is the product of its length and breadth. ½Area of rectangle ABCD = a × b = ab (Fig. (i))We have

3xy = 3x × y2y = 2 × y3x = 3x × 1

and 2 = 2 × 1

A B

D C

a

b

½

(i)Materials Required :

(i) White thick paper sheet (ii) Pen (iii) Ruler (iv) A pair of scissors (v) Fevicol (vi)Colours 1

Procedure :(1) Take suitable values of x and y.(2) Cut out two rectangles of lengths 3x units and 2 units and breadth y units each from the

paper sheet.

D1 C1

A1 B13x

y

C2D2

B2A22

y

(ii) (iii)Colour these rectangles with different colours and name them as A1B1C1D1 and A2B2C2D2

respectively (Fig. (ii) and (iii)). ½

(3) Area of rectangle A1B1C1D1 = 3x × y

= 3xy.

(4) Area of rectangle A2B2C2D2 = 2 × y = 2y. ½

(5) Join and paste these two rectangles as shown in Fig. (iv)

D1 C2

C1D2

A1 B2B1A2

3 + 2x

y y

½

(iv)

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(6) Area of rectangle 1 2 2 1A B C D

= Area of rectangle 1 1 1 1A B C D + Area of rectangle 2 2 2 2 (Fig. iv)A B C D

= 3xy + 2y (Use steps 3 and 4).

(7) Cut out two other rectangles of lengths 3x units and 2 units and breadth 1 unit eachfrom the paper sheet.Colour these rectangles with different colours and name them as

A3B3C3D3 and 4 4 4 4A B C D (Fig. (v) and (vi)). ½

D3 C3

A3 B3

1

3x(v)

D4 C4

A4 B4

1

2

½

(vi)

(8) Area of rectangle 3 3 3 3 3 1 3 .A B C D x x= ´ =

(9) Area of rectangle 4 4 4 4 2 1 2.A B C D

(10)Join and paste these two rectangles as shown in (Fig. (vii)).

D3 C4

A3 B4

C3 D4

B3 A4

3 + 2x

1 1½

(vii)

(11) Area of rectangle 3 4 4 3A B C D

= Area of rectangle 3 3 3 3A B C D + Area of rectangle 4 4 4 4 (Fig. (vii))A B C D

= 3x + 2 (Use steps 8 and 9).

(12)Now, join and paste the rectangles 1 2 2 1A B C D and 3 4 4 3A B C D as shown in (Fig. (viii)). 1

y+1

3 + 2x

D , A1 3 C , B2 4

C4D3

B2A1

(viii)

(13) Area of the rectangle 1 2 4 3A B C D

= Area of rectangle 1 2 2 1A B C D + Area of rectangle 3 4 4 3A B C D (Fig. (viii))

= 3 2 3 2.xy y x+ + + (Use steps 6 and 11)

(14)Length of the rectangle 1 2 4 3 2 3A B C D x and breadth of it = y + 1 (Fig. (viii)).

(15)Therefore, area of the rectangle 1 2 4 3A B C D

= (3 2) ( 1)x y

= (3 2)( 1)x y

Result :

We obtain that the factors of 3xy + 2y + 3x + 2 are 3x + 2 and y + 1 such that

3xy + 2y + 3x + 2 = (3x + 2)(y + 1). 1



1. 6

2. 6(x – 7)

3. a(a2 + ab + b2)

4. (x + 4)2

5. 9801, (... 992 = (100 – 1)2 = (100)2 – 2 × 100 × 1 + 12

= 10000 – 200 + 1

= 9801)

[C] True/False ½ × 5 = 2½

1. True

2. False

3. True

4. True

5. False

[D] Viva-Voce ½ × 8 = 4

1. 1, b and c

2. 1, 2 and 4

3. 1

4. 2y

5. 1, 2, 3, 4, 6 and 12

6. No

7. (x + 3)(x – 3)

8. xy

[E] Quiz ½ × 8 = 4

1. No

2. 1

3. 2( 4)( 2)( 2)x x x

4. Every natural number

5. 5 × 3 × x × y

6. – 2x

7. 2 × x

8. No


1. (B) 2 axes 1

2. (A) 0 1

3. (i) On y-axis 1

(ii) On x-axis 1

4. In case (iii), there are an infinite number of temperatures at the same time which is not

possible. 1

Case (iii) does not represent a time-temperature graph. 1

5. (a) The patient’s temperature at 1 p.m. was 36·5º. 1

(b) The patient’s temperatre 38·5°C was at 12 noon. 1

(c) The patient’s temperature was same (i.e., 36·5°C) at 1 p.m. and 2 p.m.

(d) The patient’s temperature at 1:30 p.m. was 36·5°C [because the temperature of thepatient was constant (i.e. 36·5°C) from 1 p.m. to 2 p.m.] 1

(e) The temprature of patient showed an upward trend during 9 a.m. to 11 a.m. and 2

p.m. to 3 p.m. 1

6. From the figure, we have

The co-ordinates of P are (0, 4). 1

The co-ordinates of Q are (5, 5).

The co-ordinates of R are (10, 3). 1

The co-ordinates of S are (10, 2). 1

The co-ordinates of T are (8, 0). 1

15 INTRODUCTION TO GRAPHS WORKSHEET-73

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P-155I N T R O D U C T I O N T O G R A P H S

1. (B) 1 1

2. (D) Its perpendicular distance from the y-axis. 1

3. At the point on intersection of the x and y-axis. 1

4. (i) A, x = – 3, y = 2 ½

(ii) C, x = 0, y = – 7 ½

5. (a) (i) Company’s sales in 2003 were Rs. 4 crores.

(b) Difference between the sales in 2002 and 2006 = [Rs. 8 crores] – [Rs. 4 crores]

= Rs. 4 crores 1

(c) The greatest difference between the sales of two consecutive years 2004 and 2005.

16. (i) The forecast temperature was the same as the actual temperature on Tuesday,

Friday and Sunday. 1

(ii) The maximum forecast temperature during the week was 35°C. 1

(iii) The minimum actual temperature during the week was 15°C. 1

7. (i) A (5, 100), D (12, 500). 1

(ii) The car started from the town A at 5 a.m. 1

(iii) The car stopped at town B for 1 hour (6 a.m. to 7. a.m.) 1

(iv) The car reached at town D at 12 noon.

The car look 2 hours to reach from town C to town D. 1


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1. (B) A straight line 1

2. (A) (0, 0) 1

3. (i) 6 (ii) 0 (iii) 2

(iv) – 3 1

4. (a) The time is taken along the x-axis. The scale along x-axis is 4 units = 1 hour. 1

(b) Total travel time = 8 a.m. to 11·30 a.m. = 1

32

hours. 1

(c) Distance of the merchant from the town = 22 km. 1

(d) Yes, the stoppage time = 10·00 a.m. to 10·30 a.m. 1

5. Linear graph to show, snow fall in different years

12

11

10

9

8

7

6

5

4

3

2

1

Da

ys

for

wh

ich

rain

fall

rec

eived

Years2003 2004 2005 2006

X

Y

3

6. From the graph, the vertices of ABC are : A, B and C. We find that

The co-ordinates of A are (2, 5). ½

The co-ordinates of B are (2, 1). ½

The co-ordinates of C are (6, 1). ½


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The required triangle with vertices as A(5, 2), B(1, 2) and C(1, 5) is given in the following

graph.

1½+1


1. (B) (3, 4) 1

2. (A) (1, 0) 1

3. (i) A, x = 0, y = 5, (ii) B, x = – 6, y = – 4. 1

4. On plotting the points A(2, 3) and B(3, 2) and joining them, draw a line. On extending

the line meets the x-axis at C(5, 0)and the y-axis at D(0, 5). 1

D (0, 5)

4

3

2

1

O(0, 0) 1 2 3 4 5 6 7

A(2, 3)

B(3, 2)

C(5,0)

X

Y

1

5. (i) The co-ordinates of :

O are (0, 0)

A are (2, 0)

B are (2, 3)

C are (0, 3) 1

(ii) The co-ordinates of :

P are (4, 3)

Q are (6, 1)

R are (6, 5)

S are (4, 7) 12

(iii) The co-ordinate of :

K are (10, 5)

L are (7, 7)

M are (10, 8) 12

6. (i) Taking the side of the square along the x-axis and the perimeter along the y-axis

and plotting the points (2, 8), (3, 12), (3·5, 14), (5, 20) and (6, 24), we get the required

graph as a straight line.


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This graph is a linear graph. 1

Y

24

20

16

12

8

4

Per

imet

er (

in c

m)

1 2 3 3·54 5 6

Side (in cm)

X(0, 0)

1

(b) Taking the side of the square along the x-axis and area (in cm2) along the y-axis, we

can draw the required graph by plotting the points (2, 4), (3, 9), (4, 16), (5, 25) and

(6, 36) as shown in the following figure.

Y

40

30

20

10Are

a (

in c

m)

2

(0,0) 1 2 3 4 5 6 7 8 9

Side (in cm)

1

the graph in not a straight line.

it is not a linear graph.

7. (a) The horizontal axis (or the x-axis) indicates the matches played during the

year 2007. The vertical axis (or the y-axis) shows the total runs scored in each

match. 1

(b) The dotted line shows the runs scored by Batsman A. (This is already indicated at

the top of the graph). 1

(c) During the 4th match, both have scored the same number of 60 runs. (This is

indicated by the point at which both graphs meet). 1

(d) Batsman A has one great ‘‘peak’’ but many deep ‘‘valleys’’. He does not appear to be

consistent. B, on the other hand has never scored below a total of 40 runs, even

though his highest score is only 100 in comparison to 115 of A. Also A has scored a

zero in two matches and in a total of 5 matches he has scored less than 40 runs.

Since A has a lot of ups and downs, B is a more consistent and reliable batsman. 1



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[A] Lab Activity

Objective :To prepare a graph paper and hence draw a linear graph of x + y = 3 on it.

Let’s Start :A graph paper is a plane surface consisting horizontal and vertical lines, on whichnumerical facts/change of one or more quantities are shown. A graph is a two-dimensionalfigure. A linear graph is of a line. ½

Materials Required :(i) A white paper (ii) Pencil (iii) Pen (iv) Sketch pens (v) Geometry box.

Procedure :(a) To prepare a graph paper :

(1) Take a point O in the middle of a paper [Fig. (i)](2) Draw a horizontal, semi-dark straight line passing through O. Name it as XOX [Fig. (ii)].

This line is called the x-axis. ½(3) Draw another semi-dark straight line which is perpendicular to the x-axis and passes

through O. Name it as YOY (Fig. iii). This line is called the y-axis.

O

X' X

O ½

(i) (ii)(4) The point O is the common point on both the axes, i.e., x-and y-axes. The point O is

called the origin. Its coordinates are (0, 0).(5) Draw many more light lines of equal separation, parallel to x-axis (Fig. (iv)).(6) Again draw many more light lines of equal separation parallel to the y-axis (Fig. (v)). ½

Y

X' X

Y'

O

Y

X' X

Y'

O

(iii) (iv) ½

OX' X

Y

Y'

(v) ½


ResultWe get a graph paper.

(b) To draw a linear graph of the line x + y = 3 :(1) First find out the coordinates of at least two points on the line.(2) Put x = 0 in x + y = 3 and it will give the value of y, i.e., y = 3. ½(3) Put y = 0 in x + y = 3 and it will give the value of x, i.e., x = 3.(4) Thus, you get the following table for the values of x and y.

0 3

3 0

x

y ½

(5) Plot two points A(0, 3) and B(3, 0) on the graph paper (Fig. vi)(6) Joint AB and extend to both the sides (Fig. vii). ½

Result :

AB

is the graph of the required line.

Y

A

X' X

Y'

O

(0, 3)

B

(3, 0)

Y

A

XX'

Y'

O

(0, 3)

B(3, 0)

x+y=3

½

(vi) (vii)

[B] Fill in the Blanks ½ × 1 = 2

1. y-axis2. A line parallel to the x-axis3. x = 04. Line graph

[C] True/False 1 × 3 = 3

1. True2. False3. True

[D] Viva-Voce

1. (0, 0) 12. 90° 13. Yes, x = y 14. Yes 15. A graph is a visual base to show the numerical facts/change in one or more quantities. 1

[E] Quiz

1. Second quadrant 12. 5 units 13. No 14. x-coordinate and y-coordinate 15. 2. 1

P-161P L A Y I N G W I T H N U M B E R S

1. (D) 100 × 4 + 10 × 2 + 1 × 1 12. (D) 804 1

3. The one’s digit, when divided by 5, must leave a remainder of 3 so the one’s digit mustbe either 3 or 8. 1

4. If remainder = 1, then the one’s digit of ‘N’ must be either 1 or 6. 1

5. 616

We have, 6 + 1 + 6 = 13

( A number is divisibile by 9 if sum of its digits is divisible by 9) ½

and 13 is not divisible by 9

616 is also not divisible by 9. ½

6. For A + S to be 2 or number whose one’s digit is 2 we must have A = 7. ½

For A + 8 to have one’s digit as 3 we must have A = 5

1 4 9 = B

or B = 1 4 12

4 5

9 8

1 4 3

+

Clearly C = 1 1

Thus A = 5, B = 4 and C = 1

7. We have the sum of the digits of 51x3

= 5 + 1 + x + 3 = 9 + x 1

Since, 51x3 is divisible by 9.

(9 + x) must be disible by 9.

(9 + x) must be equal to 0 or 9 or 18 or 27 or ..... 1

But x is a digit, then

9 + x = 9 x = 0

9 + x = 18 x = 9

x = 27 x = 18, which is not possible.

The required value of x = 0 or 9. 1

8. Since the ones digit of A × 3 is A.

We must have A = 0 or A = 5 ½

Now, let us look for B,

If B =1, then BA × B3 would at most be equal to 19 × 19

i.e. at the most it must be equal to 361.

But product is 57A which is more than 500.

So we cannot have B = 1 ½

If B = 3, then BA × A3 would be more than 30 × 30 i.e. more than 900.

16 PLAYING WITH NUMBERS WORKSHEET-78

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But 57A is less than 600

B 3 ½

B can be only equal to 2.

It is either 20 × 23 or 25 × 23.

But 20 × 23 = 460

So not possible 25 × 23 = 575 ½

A = 5; B = 2.

9. ·.· The value of a or b cannot exceed 9.

The sum a + b cannot exceed 18. ½

Yes ! dad is a multiple of 11.

dad is less than or equal to 198. ½

All the 3-digit numbers which one multiples of 11 upto 198 are

110, 121, 132, 143, 154, 165, 176, 187 and 198. 1

Clearly dad = 121

d = 1, a = 2 1


1. (A) 1 1

2. (C) 28 1

3. If remainder = 4, then the one’s digit of ‘N’ must be either 4 or 9. 1

4. (i) 108

We have 1 + 0 + 8 = 9

and 9 is divisible by 3.

108 is also divisible by 3. 1

(ii) 616

We have 6 + 1 + 6 = 13

and 13 is not divisible by 3.

Thus 616 is also not divisible by 3. 1

5. Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; Since x is a digit, it

can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have

any of four different values. [1+1+1]

6. We have 3 + 1 + z + 5 = 9 + z

31z5 is divisible by 3.

(9 + z) must be divisible by 3. 1

since, z is a digit.

If 9 + z = 9, then z = 0

If 9 + z = 12, then z = 3,

If 9 + z = 15, then z = 6,

If 9 + z = 18, then z = 9,

If 9 + z = 21, then z = 12, 1

Possible values of z are 0, 3, 6 or 9.

7. (i) 73 = 70 + 3

= 10 × 7 + 3 × 1 = 10 × 2 + 3 1

(ii) 129 = 100 + 29 + 9

= 100 × 1 + 10 × 2 + 1 × 9 = 100 × 1 + 10 × 20 + 9 1

(iii) 302 = 100 × 3 + 10 × 0 + 1 × 2 = 300 + 0 + 2 1


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1. (B) 54 1

2. (D) a = 1 1

3. N is odd; so its one’s digit is odd. Therefore, the one’s digit must be 1, 3, 5, 7, or 9. 1

4. Chosen number = 27

Number with reversed digits = 72

Sum of the two numbers = 27 + 72 = 99

Now, 99 = 11[9]

= 11[2 + 7]

= 11 [Sum of the digits of the chosen number] 1

This works always as if we choose the number say ab which is actually 10a + b. In

reversing the digits we have 10b + a as the number.

Adding the two we get

10a + b + 10b + a

= 11a + 11b

= 11 (a + b)

which is a multiple of 11. 1

5. We have 2 + 1 + y + 5 = 8 + y

21y5 is multiple of 9

(8 + y) must be divisible by 9 1

(8 + y) will be divisible by 9 if 8 + y is either 0, 9, 18, 27, ....., etc. 1

Since, y is a digit

So, 8 + y = 0 is not possible. The only possibility is 8 + y = 9 1

8 + y = 9 y = 9 – 8 or y = 1.

6. We have 3 + 1 + z + 5 = 9 + z

31z5 is divisible by 9. 1

(9 + z) must be equal to 0, or 9 or 18 or 27, .....


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But z is a digit. 1

9 + z = 9 or 9 + z = 18 1

If 9 + z = 9, then z = 0 and if 9 + z = 18, then z = 9.

7. (i) 10 × 5 + 6 = 50 + 6 = 56 1

(ii) 100 × 7 + 10 × 1 + 8 = 700 + 10 + 8 = 718 1

(iii) 100 × a + 10 × c + b = 100a + 10c + b = a c b 1


1. (B) 0, 3, 6 or 9 1

2. (B) 440 1

3. For N 5, remainder = 4

One’s digit can be 4 or 9 ....(i)

Again for N 2, remainder = 1

N must be an odd number.

So, one’s digit of N must be

1, 3, 5, 7 or 9 ....(ii) 1

From (i) and (ii), the one’s digit of N must be 9.

4. (i) 25 = 20 + 1 × 5

= 10 × 2 + 5 12

(ii) 73 = 70 + 3

= 10 × 7 + 1 × 3 12

5. 329

We have 329 = 300 + 20 + 9

= 100 × 3 + 10 × 2 + 1 × 9 1

6. We have

100 × 7 + 10 × 0 + 9 × 1

= 700 + 0 + 9

= 709

7. (i) 294

We have 2 + 9 + 4 = 15 2

and 15 is divisible by 3.

Thus, 294 is also divisible by 3. 1

8. (i) B can be 2, 4, 6 or 8.

We need product 111 or 222, or 333 or 444 or 888 out of them 111 and 333 are rejected.

Possible products are 222, 444 or 888

To obtain

The possible value is B = 4

6

A B

B B B

´

4

6

4

A

‘A’ can be either 2 or 7.


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A × 6 means 2 × 6 = 1 2 or 7 × 6 = 4 2 + 2 + 2

1 4 4 4 1 A × 6 = 7 × 6 is the accepted valueNow, 2 7 4 × 6

4 4 4 12

Thus, A = 7 and B = 4(ii) 10 – 1 = 9 B = 9

Also 9 – 1 = 8 – 1 = 7 12

A = 7

We have 7 1 + 1 9 9 0 1

Thus, A = 7 and B = 9

9. (a) Yes ! it is necessary that (a – b + c) should be divisible by 11. 1

(b) [(b + d) – (a + c)] is divisible by 11. 1

(c) Yes ! we can say that a number will be divisible by 11 if the difference between thesum of digits at its odd places and that of digits at the even places is divisible by 11. 1



[A] Lab ActivityObject :

To test the divisibility of a number by grouping method.

Let’s Start

Recall all the divisibility rules.


Math box. 1

Procedure :

(1) Throw out some match sticks from the match box on the table (Fig. (i)). (say 30 sticks)

Scatteredpencils

Pencil box

(i)(2) Make groups of two sticks each out of scattered sticks (see Fig. (ii)) and record your

observation in the observation table.

1

(ii)(3) Now make groups of three sticks each and record your observation [see Fig. iii(a)].

(a)

(b)

½

(iii)Keep repeating the grouping for different numbers (4, 5, 6, 7 and so on) and record theobservations. ½

Observations :

2 3 4 5 6 7 8 9 10 11

1. 30

S. Total No. of Grouping complete with numbers or not Number is

No. Sticks thrown divisible by

× × × × × 2, 3, 5, 6, 10

2.

3.

4.

5.

Conclusion :

From the above activity it can be concluded that if the grouping is completed then thenumber is divisible by the number used for grouping and if grouping is not completedthen the number is not divisibly by number used for grouping 1

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Result :

A given number is divisible by a number only when it completely divides the number,leaving no remainder. 1

[B] Fill in the Blanks ½ × 4 = 2

1. 10 × 5 + 2

2. 75

3. 100 × 1 + 10 × 2 + 3

4. 345

[C] True/False 1 × 4 = 4

1. True

2. False

3. True

4. True

[D] Viva-Voce 1 × 3 = 3

1. A number is divisible by 3 when sum of its digits is divisible by 3.

2. 9

3. Yes

[E] Quiz

1. No 1

2. A = 5 1

3. (i) A = 7, B = 4

(ii) A = 7, B = 9

(iii) A = 4, B = 7 1

4. 2 1

5. 4 1

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