Lesson: Linear Equations in One Variable

Exercise – 2.1

Question 1

Solve:

27x

Answer:

x = 9

Solution:

We have,

27x

Adding 2 to both sides, we get

2 2 7 2x

9x

Question 2

Solve:

3 10y

Answer:

y = 7

Solution:

We have,

3 10y

10 3y

(Transposing 3 to RHS)

7y

Question 3

Solve:

62z

Answer:

z = 4

Solution:

We have,

62z

62 z

(Transposing 2 to LHS)

4 z

4z

Question 4

Solve:

3 17

77

x

Answer:

x = 2

Solution:

We have,

3 17

77

x

17 3

77

x

(Transposing

3

7

to RHS)

14

7

x

2x

Question 5

Solve:

6 12x

Answer:

x = 2

Solution:

We have,

6 12x

6 12

66

x

(Dividing both sides by 6)

2x

Question 6

Solve:

10

5

t

Answer:

t = 50

Solution:

We have,

10

5

t

Multiplying both sides by 5,

5 10 5

5

t

50t

Question 7

Solve:

2

18

3

x

Answer:

x = 27

Solution:

We have,

2

18

3

x

Multiplying both sides by 3,we get

2

3 18 3

3

x

2 54x

2 54

22

x

(Dividing both sides by 2)

27x

Question 8

Solve:

1.6

1.5

y

Answer:

y = 2.4

Solution:

We have,

1.6

1.5

y

Multiplying both sides by 1.5,we get

1.6 1.5 1.5

1.5

y

2.4 y

2.4y

Question 9

Solve:

7 9 16x

Answer:

25

7

x

Solution:

We have,

7 9 16x

Adding 9 to both sides,

7 9 9 16 9x

7 25x

7 25

77

x

(Dividing both sides by 7)

25

7

x

Question 10

Solve:

14 8 13y

Answer:

3

2

y

Solution:

We have,

14 8 13y

Adding 8 to both sides, we get

14 8 8 13 8y

14 21y

14 21

14 14

y

(Dividing both sides by 14)

3

2

y

Question 11

Solve:

17 6 9p

Answer:

4

3

p

Solution:

We have,

17 6 9p

6 9 17p

(Transposing

17

to RHS)

68p

8

6

p

(Dividing both sides by 6)

4

3

p

Question 12

Solve:

7

1

3 15

x

Answer:

8

5

x

Solution:

We have,

7

1

3 15

x

7

1

3 15

x

(Transposing 1 to RHS)

8

3 15

x

8

33

3 15

x

(Multiplying both sides by 3)

8

5

x

Exercise – 2.2

Question 1

If you subtract

1

2

from a number and multiply the result by

1

2

, you get

1

8

.

What is the number?

Answer:

3

4

Solution:

Let the number be

x

.

According to the question, we have

1 1 1

2 2 8

x

11

2

22

x

1

2

8

(Multiplying both sides by 2)

11

24

x

11

42

x

(Transposing

1

2

to RHS)

3

4

x

Question 2

The perimeter of a rectangular swimming pool is 154 m.

Its length is 2 m more than twice its breadth.

What is the length and the breadth of the pool?

Answer:

The breadth and length of the pool are 25 m and 52 m respectively

Solution:

Let the breadth of the swimming pool be x m, then its length =

2 2 mx

.

The perimeter of the swimming pool is

2 154 mlength breadth

Therefore,

154 2[(2 2) ]xx

154 2 2 2

154 2 3 2

154 6 4

xx

x

x

154 – 4

6 4 – 4x

(Subtracting 4 from both sides)

150 6x

150 6

66

x

(Dividing both sides by 6)

25 x

The length

(2 2) mx

=

(2 25 2) m

= (50 + 2) m = 52 m

Hence, the breadth and the length of the swimming pool are 25 mand 52 m respectively.

Question 3

The base of an isosceles triangle is

4

3

cm.

The perimeter of the triangle is

2

4

15

cm.

What is the length of either of the remaining equal sides?

Answer:

2

1 cm

5

Solution:

Let the length of each of the equal sides be

cmx

Perimeter of the triangle

x cm x cm base

2

4

15

cm

4 62

2

3 15

x

62 4

2

15 3

x

(Transposing

4

3

to R H S)

62 20

2

15

x

42

2

15

x

42

30

x

(Dividing both sides by 2)

7

5

x

2

1

5

Therefore, the length of each of the equal sides is

2

1 cm

5

.

Question 4

Sum of two numbers is 95.

If one exceeds the other by 15, find the numbers.

Answer:

The first number = 40 and the second number = 55

Solution:

Let one of the numbers be ‘

a

’

Therefore, the second number will be

a

+ 15

According to the question,

a

+

a

+ 15 = 95

⇒ 2

a

+ 15 = 95

⇒2

a

+ 15 – 15 = 95 – 15(subtracting 15 from both sides)

⇒ 2

a

= 95 – 15

⇒ 2

a

= 80

⇒

2 80

22

a

(dividing both sides by 2)

⇒

40a

So, the second number =

a

+ 15 = 55

Thus, the first number is 40 and the second number is 55.

Question 5

Two numbers are in the ratio 5:3.

If they differ by 18, what are the numbers?

Answer:

45 and 27

Solution:

Let one of the numbers be x, then the other number is x+18.

According the question,

18 5

3

x

x

By cross multiplication, we get

3x + 54 = 5x

54 = 5x –3x

54 = 2x

27 x

(Dividing both sides by 2)

18 27 18 45x

So, the numbers are 27 and 45.

Question 6

Three consecutive integers add up to 51.

What are these integers?

Answer:

16, 17 and18

Solution:

Let the three consecutive integers be

1,aa

, and

1a

.

The sum of these numbers =

( 1) ( ) ( 1) 51a a a

⇒

a

+

a

+

a

= 51

⇒

a

+

a

+

a

= 51

⇒ 3

a

= 51

⇒

3 51

33

a

(dividing both sides by 3)

⇒

17a

⇒

1 16a

⇒

1 18a

Hence, the consecutive integers are 16, 17, and 18.

Question 7

The sum of three consecutive multiples of 8 is 888.

Find the multiples.

Answer:

288, 296 and 304

Solution:

Let the three consecutive multiples of 8 be 8a, 8 (a + 1) and 8 (a + 2) respectively.

The sum of the numbers, as given, is 888

Therefore, 8

a

+ [8 (

a

+ 1)] + [8 (

a

+ 2)] = 888

⇒ 8

a

+ (8

a

+ 8) + (8

a

+ 16) = 888

⇒ 8

a

+ 8

a

+ 8 + 8

a

+ 16 = 888

⇒ 8

a

+ 8

a

+ 8

a

+ 8 + 16 = 888

⇒ 24

a

+ 24 = 888

⇒ 24

a

= 888 – 24(transposing 24 to RHS)

⇒ 24

a

= 864

⇒

24 864

24 24

a

(dividing both sides by 24)

⇒

36a

8 288a

(The first multiple of 8)

8( 1) 296a

(The second multiple of 8)

8( 2) 304a

(The third multiple of 8)

Thus, the three consecutive multiples of 8 are288, 296 and 304 respectively.

Question 8

Three consecutive integers are such that when they are taken in increasing order and

multiplied by 2, 3 and 4 respectively, they add up to 74.

Find these numbers.

Answer:

7, 8 and 9

Solution:

Let the first integer be

a

Therefore, second consecutive integer is (

a

+ 1) and the third consecutive integer is (

a

+ 2)

As per the question,

(First integer

2) + (second consecutive integer

3) + (third consecutive integer

4) = 74

( 2) [( 1) 3]aa

[( 2) 4] 74a

2 (3 3)aa

(4 8) 74a

2 3 3aa

4 8 74a

2 3 4a a a

3 8 74

9 11 74a

⇒

9 11 11 74 11a

(Subtracting 11 from both sides)

⇒

9 63a

⇒

9 63

99

a

(dividing both sides by 9)

⇒

7a

a + 1 = 8

a + 2 = 9

Thus, the three consecutive integers are 7, 8 and 9.

Question 9

The ages of Rahul and Haroon are in the ratio 5:7.

Four years later the sum of their ages will be 56 years.

What are their present ages?

Answer:

Rahul’s present age is 20 years; Haroon’s present age is 28 years

Solution:

Let the present age of Rahul be 5x years, then the present age of Haroon is 7x years.

After 4 years, the age of Rahul and Haroon will be 5x+4 years and 7x+ 4 years respectively.

According to the question,

(5 4) (7 4) 56xx

5 4 7 4 56xx

5 7 4 4 56xx

12 8 56x

12 56 8x

(Transposing 8 to RHS)

12 48x

4x

(Dividing both sides by 12)

Therefore, the present ages of Rahul and Haroon, in years, are

5 5 4 20x

and

7 7 4 28x

respectively.

Question 10

The number of boys and girls in a class are in the ratio 7:5.

The number of boys is 8 more than the number of girls.

What is the total class strength?

Answer:

48

Solution:

Let the number of boys be

7x

,

then the number of girls =

5x

As per the question,

7 8 5xx

8 5 7xx

(Transposing

7x

to RHS)

82x

82x

82

22

x

(Dividing both sides by 2)

4 x

4x

The strength of the class = Number of boys + Number of girls

Strength of the class

7 5 12x x x

12 4 48

Question 11

Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than

Baichung.

The sum of the ages of all the three is 135 years.

What is the age of each one of them?

Answer:

Baichung’s = 17 years, Baichung’s father’s age = 46 years and Baichung’s grandfather’s age

= 72 years

Solution:

Let the age of Baichung’s father in years be m.

Then, the ages of Baichung’s grandfather and Baichung, in years, will be m + 26 and

m – 29 respectively.

According to the question,

Age of Baichung + Age of Baichung’s father + Age of Baichung’s grandfather = 135 years

⇒ (m – 29) + m + (m + 26) = 135

⇒m – 29 + m + m + 26 = 135

⇒m + m + m – 29 + 26 = 135

⇒ 3m – 3 = 135

⇒3m = 135 + 3(transposing – 3 to RHS)

⇒ 3m = 138

⇒

3m 138

33

(dividing both sides by 3)

m 46

The age of Baichung, in years, = m – 29 = 46 – 29 = 17

The age of Baichung’s grandfather, in years, = m + 26 = 46 + 26 = 72

Question 12

Fifteen years from now Ravi’s age will be four times his present age.

What is Ravi’s present age?

Answer:

Ravi’s present age is 5 years

Solution:

Let the present age of Ravi be m years.

According to question,

m + 15 = 4

m

⇒m + 15 = 4 m

⇒15 = 4m – m(transposing m to RHS)

⇒ 15 = 3m

⇒

15 3

33

m

(dividing both sides by 3)

⇒

5m

Thus, Ravi’s present age is 5 years.

Question 13

A rational number is such that when you multiply it by

5

2

and add

2

3

to the product, you get

7

12

. What is the number?

Answer:

1

2

Solution:

Let the number be x.

According to the given question,

5 2 7

2 3 12

x

Transposing

2

3

to R.H.S, we get

5 7 2

2 12 3

x

5

2

x

7 2 4

12

5 15

2 12

x

On multiplying both sides by

2

5

, we get

15 2 1

12 5 2

x

Hence, the rational number is

1

2

.

Question 14

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs.100,

Rs.50 and Rs.10, respectively.

The ratio of the number of these notes is 2:3:5.

The total cash with Lakshmi is Rs.4, 00,000.

How many notes of each denomination does she have?

Answer:

The number of currency notes of Rs.100, Rs. 50 and Rs. 10 are 2000, 3000 and 5000

respectively.

Solution:

Let the number of currency notes of denomination of Rs.100 be 2

x

,

then the number of

currency notes of denomination of Rs.50 and Rs. 10 would be 3

x

and 5x respectively.

The value of currency notes of denomination of Rs.100 = Rs 2

x

100 = Rs 200

x

,

The value of currency notes of denomination of Rs.50 = Rs 3

x

50 = Rs 150

x

and

The value of currency notes of denomination of Rs.10 = Rs 5

x

10 = Rs 50

x

According to the question,

200 x + 150 x + 50 x = 4, 00,000.

400 400000x

1000x

2 2000x

3 3000x

5 5000x

Thus, the number of currency notes of Rs.100,Rs.50 and Rs.10 are 2000, 3000 and 5000

respectively.

Question 15

I have a total of Rs.300 in coins of denomination Rs.1, Rs.2 and Rs.5.

The number of Rs.2 coins is 3 times the number of Rs.5 coins.

The total number of coins is 160.

How many coins of each denomination are with me?

Answer:

The number of coins of Rs.1, Rs. 2 and Rs. 5 are 80, 60 and 20 respectively.

Solution:

Let the number of coins of Rs.5 be m.

Therefore, number of coins of Rs.2 = m

3 = 3m and the number of coins of Rs.1 =

= 160 – (m + 3m) = 160 – 4m

Total Amount= (Rs.1

Number of Rs.1 coins) + (Rs.2

Number of Rs.2 coins) + (Rs.5

Number of Rs.5 coins)

⇒ 300 = [1

(160 – 4m)] + (2

3m) + (5

m)

⇒ 300 = (160 – 4m) + 6m + 5m

⇒ 300 = 160 – 4m + 6m + 5m

⇒300 = 160 – 4m + 11m

⇒ 300 = 160 + 7m

⇒300 – 160 = 7m(transposing 160 to LHS)

⇒ 140 = 7 m

⇒

140 7

77

m

(dividing both sides by 7)

140

7

m

140

7

m

20m

Number of coins of Rs.1 = 160 – (4

20) = 160 – 80 = 80

Number of coins of Rs.2 = 3m = 3

20 = 60

Therefore, the number of coins of Rs1,Rs 2 and Rs 3 are 80, 60 and 20 respectively.

Question 16

The organisers of an essay competition decide that a winner in the competition gets

a prize of Rs.100 and a participant who does not win gets a prize of Rs.25.

The total prize money distributed is Rs.3, 000.

Find the number of winners, if the total number of participants is 63.

Answer:

19

Solution:

Let the number of winners be m

Number of losers = 63 – m

Total Prize money distributed to winners

= Number of winners

prize money distributed to each winner

= Rs m

100 = Rs100 m

Total prize money distributed to participants who are not winners

= Rs (63 – m)

25 = Rs ((63

25) – 25 m) = Rs (1575 – 25 m)

Total Prize money of winners + Total Prize money of losers = Total prize money

⇒100 m + 1575 – 25 m = 3000

⇒100 m – 25 m + 1575 = 3000

⇒100 m – 25 m = 3000 – 1575(transposing 1575 to RHS)

⇒75 m = 1425

⇒

75 1425

75 75

m

(dividing both sides by 75)

1425

19

75

m

Thus, number of winners is 19

Exercise – 2.3

Question 1

Solve the equation and check your result.

3 2 18xx

Answer:

x = 18

3 2 18xx

3 2 18xx

(Transposing

2x

to LHS)

18x

L.H.S

3 3 18 54x

R.H.S

2 18x

2 18 18

36 18 54

L.H.S = R.H.S

Question 2

Solve the equation and check your result.

5 3 3 5tt

Answer:

t =

1

Solution:

5 3 3 5tt

5 3 5 3tt

(Transposing

3t

to LHS and

3 to RHS)

22t

1t

(Dividing both sides by 2)

L.H.S

53t

5 ( 1) 3

5 3 8

R.H.S

3 5 3t

( 1) 5

3 5 8

L.H.S = R.H.S

Question 3

Solve the equation and check your result.

5 9 5 3xx

Answer:

x =

2

Solution:

5 9 5 3xx

5 3 5 9xx

(Transposing

3x

to LHS and 9 to RHS)

24x

2x

(Dividing both sides by 2)

L.H.S

5 9 5( 2) 9x

10 9 1

R.H.S

5 3 5 3( 2)x

5 6 1

L.H.S = R.H.S

Question 4

Solve the equation and check your result.

4 3 6 2zz

Answer:

3

2

z

Solution:

4 3 6 2zz

4 2 6 3zz

(Transposing

2z

to LHS and 3 to RHS)

23z

3

2

z

(Dividing both sides by 2)

L.H.S

3

4 3 4 3

2

z

6 3 9

R.H.S

3

6 2 6 2

2

z

6 3 9

L.H.S = R.H.S

Question 5

Solve the equation and check your result.

2 1 14xx

Answer:

x = 5

Solution:

2 1 14xx

2 14 1xx

(Transposing

x

to LHS and

1 to RHS)

3 15x

5x

(Dividing both sides by 3)

L.H.S

2 1 2(5) 1x

10 1 9

R.H.S

14 14 5 9x

L.H.S = R.H.S

Question 6

Solve the equation and check your result.

8 4 3 1 7xx

Answer:

x = 0

Solution:

8 4 3 1 7xx

8 4 3 3 7xx

8 4 3 4xx

8 3 4 4xx

(Transposing

3x

to LHS and 4 to RHS)

50x

0x

L.H.S

8 4 8(0) 4 4x

R.H.S

3 1 7x

3(0 1) 7

3 7 4

L.H.S = R.H.S

Question 7

Solve the equation and check your result.

4

10

5

xx

Answer:

x = 40

Solution:

4

10

5

xx

44

10

55

xx

4

8

5

xx

4

8

5

xx

(Transposing

4

5

x

to LHS)

54

8

5

xx

8

5

x

Multiplying both sides by 5, we get

40x

L.H.S

40x

R.H.S

4

10

5

x

4

(40 10)

5

4

(50) 40

5

L.H.S = R.H.S

Question 8

Solve the equation and check your result.

27

13

3 15

xx

Answer:

x = 10

Solution:

27

13

3 15

xx

27

31

3 15

xx

(Transposing

7

15

x

to LHS and 1 to RHS)

3

2

15

x

Multiplying both sides by 15,

we get,

3 30x

10x

(Dividing both sides by 3)

L.H.S

2

1

3

x

2(10)

1

3

20

1

3

23

3

R.H.S

7

3

15

x

7(10)

3

15

70

3

15

70 45

15

115 23

15 3

L.H.S = R.H.S

Question 9

Solve the equation and check your result.

5 26

2

33

yy

Answer:

7

3

y

Solution:

5 26

2

33

yy

26 5

2

33

yy

(Transposing –

y

to LHS and

5

3

to RHS)

21

3

3

y

37y

7

3

y

(Dividing both sides by 3)

L.H.S

5

2

3

y

75

2

33

14 5

33

19

3

R.H.S

26

3

y

26 7

33

19

3

L.H.S = R.H.S

Question 10

Solve the equation and check your result.

8

35

5

mm

Answer:

4

5

m

Solution:

8

35

5

mm

8

35

5

mm

(Transposing

5m

to LHS)

8

2

5

m

8

10

m

(Dividing both sides by 2)

4

5

m

L.H.S

4 12

33

55

m

R.H.S

8

5

5

m

48

5

55

20 8

5

12

5

L.H.S = R.H.S

Exercise – 2.4

Question 1

Amina thinks of a number and subtracts

5

2

from it.

She multiplies the result by 8.

The result now obtained is 3 times the same number she thought of.

What is the number?

Answer:

4

Solution:

Let the number thought by Amina be

a

According to the question,

5

83

2

The number The number

5

83

2

aa

5

83

2

aa

5

8 8 3

2

aa

40

83

2

aa

8 20 3aa

8 20 3 0aa

(Transposing

3a

to LHS)

8 3 20 0aa

8 3 20aa

(Transposing

20 to RHS)

5 20a

5 20

55

a

(Dividing both sides by 5)

20

4

5

a

Question 2

A positive number is 5 times another number.

If 21is added to both the numbers, then one of the new numbers becomes twice the other new

number.

What are the numbers?

Answer:

7 and 35

Solution:

Let the given positive number be

a

. Then

the other number will be 5

a

.

According to the question,

5

a

+ 21 = 2 (

a

+ 21)

⇒ 5

a

+ 21 = 2

a

+ 42

⇒ 5

a

+ 21 – 2

a

= 42(transposing 2

a

to LHS)

⇒ 5

a

– 2

a

= 42 – 21(transposing 21 to RHS)

⇒ 3

a

= 21

3 21

33

a

(dividing both sides by 3)

7a

Therefore, the other number is 5

a

= 5

7 = 35

Thus, the numbers are 7 and 35

Question 3

The sum of the digits of a two-digit number is 9.

When we interchange the digits, it is found that the resulting new number is greater

than the original number by 27.

What is the two-digit number?

Answer:

36

Solution:

Let the number at the ones place of the two digit number be a, then the digit at the

ten’s place = 9 –

a

Thus, the number = 10(9 –

a

) +

a

After interchange of digits, the new number =10

a

+ (9 –

a

)

As per the question, the new number – 27 = the original number

⇒ 10

a

+ (9 –

a

) – 27 = 10 (9 –

a

) +

a

⇒ 10

a

+ 9 –

a

– 27 = 90 – 10

a

+

a

⇒ 10

a

–

a

+ 9 – 27 = 90 – 9

a

⇒ 9

a

– 18 = 90 – 9

a

⇒9

a

= 90 – 9

a

+ 18(transposing

18 to RHS)

⇒9

a

+ 9

a

= 90 + 18(transposing – 9a to LHS)

⇒18

a

= 108

⇒

18 108

18 18

a

(dividing both sides by 18)

108

18

a

6a

9 –

a

= 9 – 6 = 3

Thus, the digit at the ten’s place = 3 and the digit at the ones place = 6

The number is 36.

Question 4

One of the two digits of a two digit number is three times the other digit.

If you interchange the digits of this two-digit number and add the resulting number

to the original number, you get 88.

What is the original number?

Answer:

26 or 62

Solution:

Let one of the digits, which is at ones place, of the two digit number be

a

Therefore, the other digit of the two digit number = 3

a

Therefore, the number = (10

3

a

) +

a

= 30

a

+

a

= 31

a

After interchange of the digits, the new number = 10

a

+ 3

a

= 13

a

As per the question,

31

a

+ 13

a

= 88

⇒ 44

a

= 88

44 88

44 44

a

(Dividing both sides by 44)

2a

31

a

= 31

2 = 62

13

a

= 13

2 = 26

Therefore, the two-digit number is either 26 or 62.

Question 5

Shobo’s mother’s present age is six times Shobo’s present age.

Shobo’s age five years from now will be one third of his mother’s present age.

What are their present ages?

Answer:

Shobo’s age: 5 years; Shobo’s mother’s age: 30 years

Solution:

Let the present age of Shobo be

a

years

Therefore, Shobo’s mother’s present age = 6

a

years

After five years, the age of Shobo = (

a

+ 5) years

As per the question,

1/3 of the present age of Shobo’s mother = Age of Shobo after 5 years.

6

5

3

a

a

25aa

25aa

(Transposing a to LHS)

5a

6 30a

Therefore, the present age of Shobo is 5 years and the present age of Shobo’s

mother is 30 years.

Question 6

There is a narrow rectangular plot, reserved for a school, in Mahuli village.

The length and breadth of the plot are in the ratio 11:4.

At the rate Rs.100 per metre it will cost the village panchayat Rs.75000 to fence

the plot.

What are the dimensions of the plot?

Answer:

Shob Length = 275 m; breadth = 100 m

Solution:

Let the length and breadth of the rectangular plot be

11 x

m and

4 x

m respectively.

The perimeter of the plot = 2 (length + breadth)

2 11 4 m 30 mx x x

It is given that the cost of fencing the plot, at the rate of Rs 100 per metre, is Rs 75000.

100 Perimeter = Rs 75000

100 30 75000x

3000 75000x

Dividing both sides by 3000, we get

25x

The length

11 m 11 25 m 275 mx

and the breadth

4 m 4 25 m 100 mx

Hence, the dimensions of the plot are, length = 275 m and breadth = 100 m.

Question 7

Hasan buys two kinds of cloth materials for school uniforms, shirt material that

costs him Rs. 50 per metre and trouser material that costs him Rs. 90 per metre.

For every 3 meters of the shirt material he buys 2 metres of the trouser material.

He sells the materials at 12% and 10% profit respectively.

His total sale is Rs.36, 600. How much trouser material did he buy?

Answer:

200 m

Solution:

Given:

The rate of the shirt material = Rs.50 per meter

The rate of the trouser material = Rs.90 per meter

The profit on the shirt material = 12%

The profit on the trouser material = 10%

The total sale price = Rs.36600.00

Since the profit on the cost price of the shirt material = 12%, therefore, the sale

price of the material = cost price + 12% of cost price

12

.50.000 .50.00

100

Rs Rs

.50.00 .6.00Rs Rs

.56.00Rs

Since, the profit on the cost price of the trouser material = 10%, therefore, sale

price of the trouser material = cost price of trouser material + 10% of cost price

10

.90.000 .90.00

100

Rs Rs

.90.00 .9.00Rs Rs

.99.00Rs

Let the shirt material and the trouser material being purchased by Hasan be

3mx

and

2mx

respectively.

Total sale price = Total sale price of the shirt material + Total sale price of

the trouser material

.36600Rs

3 .56 2 .99x Rs x Rs

.36600Rs

.168 .198Rs x Rs x

.36600 .366Rs Rs x

36600

366

x

366

366

x

(Dividing both sides by 366)

100x

2 200x

Thus, Hasan bought 200 m of trouser material.

Question 8

Half of a herd of deer is grazing in the field and three fourths of the remaining is

playing nearby.

The rest 9 are drinking water from the pond.

Find the number of deer in the herd.

Answer:

72

Solution:

Let the total number of deer be

x

Since, half of the herd is grazing in the field, the number of deer grazing =

2

x

Since,

3

4

th

of remaining are playing nearby,the number of deer playing

near b

33

2 4 8

xx

Number of deer drinking water = 9

(Given by the question)

Now, the total number of deer = the number of deer grazing + the number of

deer playing + the number of deer drinking water.

3

9

28

xx

x

43

9

8

xx

7

9

8

x

7

9

8

x

x

(Transposing

7

8

x

to LHS)

87

9

8

xx

9

8

x

72x

Hence, the total number of deer in the herd is 72.

Question 9

A grandfather is ten times older than his granddaughter.

He is also 54 years older than her.

Find their present ages.

Answer:

Granddaughter’s age = 6 years; Grandfather’s age = 60 years

Solution:

Let the age of the granddaughter be

x

years.

Therefore, the age of the grandfather =

10x

As per the question,

10 54xx

9 54x

9 54

99

x

(Dividing both sides by 9)

6x

Thus, the ages of the granddaughter and the grandfather are 6 years and 60 years

respectively.

Question 10

Aman’s age is three times his son’s age.

Ten years ago he was five times his son’s age.

Find their present ages.

Answer:

Aman’s age: 60 years; Aman’s son’s age: 20 years

Solution:

Let the age of Aman’s son be

x

years

Therefore, Aman’s age = 3x years

As per the question, the present age of Aman –10 years

(the present age of his son

10 years)

5

3 5 50 10xx

2 40x

2 40x

2 40

22

x

20x

3 60x

Hence the present age of Aman is 60 years and the present age of his son is 20 years.

Exercise – 2.5

Question 1

Solve the following linear equations.

11

2 5 3 4

xx

Answer:

27

10

x

Solution:

11

2 5 3 4

xx

L.C.M. of the denominators 2, 3, 4, and 5 is 60.

On multiplying both sides by 60, we get

1

60

25

x

1

60

34

x

⇒30x − 12 = 20x + 15 (opening the brackets)

⇒ 30x − 20x = 15 + 12(transposing

20x

to LHS and

12

to RHS)

⇒ 10x = 27

27

10

x

Question 2

Solve the following linear equation.

35

2 4 6

n n n

21

Answer:

n = 36

Solution:

35

21

2 4 6

n n n

L.C.M. of the denominators 2, 4, and 6, is 12.

On multiplying both sides by 12, get

6n − 9n + 10n = 252

7n = 252

252

7

n

36n

Question 3

Solve the following linear equations.

8 17 5

7

3 6 2

xx

x

Answer:

x =

5

Solution:

8 17 5

7

3 6 2

xx

x

L.C.M. of the denominators 2, 3, and 6 is 6.

Multiplying both sides by 6, we get

6x + 42 − 16x = 17 − 15x

6x − 16x + 15x = 17 – 42(transposing

15x

to LHS and 42 to RHS)

5x = −25

25

5

x

5x

Question 4

Solve the following linear equation.

53

35

xx

Answer:

x =

8

Solution:

53

35

xx

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

On multiplying both sides by 15, we get

5(x − 5) = 3(x − 3)

5x − 25 = 3x – 9 (Opening the brackets)

5x − 3x = 25 − 9

2x = 16

16

2

x

8x

Question 5

Solve the following linear equation.

3 2 2 3

43

tt

2

3

t

Answer:

t = 2

Solution:

3 2 2 3

43

tt

2

3

t

L.C.M. of the denominators, 3 and 4, is 12.

Multiplying both sides by 12, we get

3(3t − 2) − 4(2t + 3) = 8 − 12t

9t − 6 − 8t − 12 = 8 − 12t (opening the brackets)

t − 18 = 8 − 12t

t + 12t = 8 + 18 (transposing

18 to RHS and

12t to LHS)

13t = 26

26

13

t

2t

Question 6

Solve the following linear equation.

1

2

m

m

2

1

3

m

Answer:

7

5

m

Solution:

1

2

m

m

2

1

3

m

L.C.M. of the denominators, 2 and 3, is 6.

Multiplying both sides by 6, we get

6m − 3(m − 1) = 6 − 2(m − 2)

6m − 3m + 3 = 6 − 2m + 4 (Opening the brackets)

3m + 3 = 10 − 2m

3m + 2m = 10– 3(transposing 3 to RHS and

2m to LHS)

5m = 7

7

5

m

Question 7

Solve the following linear equation.

3( 3) 5(2 1)tt

Answer:

2t

Solution:

3(t − 3) = 5(2t + 1)

3t − 9 = 10t + 5(opening the brackets)

−9 − 5 = 10t − 3t(transposing 3t to RHS and 5 to LHS)

−14 = 7t

14

7

t

2t

Question 8

Solve the following linear equation.

15( 4) 2( 9)yy

5( 6) 0y

Answer:

2

3

y

Solution:

15(y − 4) − 2(y − 9) + 5(y + 6) = 0

15y − 60 − 2y + 18 + 5y + 30 = 0(Opening the brackets)

18y − 12 = 0

18y = 12(transposing

12 to RHS)

12

18

y

2

3

y

Question 9

Solve the following linear equation.

3(5 7) 2(9 11)zz

4 8 13 17z

Answer:

z = 2

Solution:

3(5z − 7) − 2(9z − 11) = 4(8z − 13)−17

15z − 21 − 18z + 22 = 32z − 52 – 17 (Opening the brackets)

−3z + 1 = 32z – 69 (transposing 32z to LHS and 1 to RHS)

−3z − 32z = −69 − 1

−35z = −70

70

35

z

2z

Question 10

Solve the following linear equation.

0.25(4 3)f

0.05 10 9f

Answer:

f = 0.6

Solution:

0.25(4 3)f

0.05 10 9f

Multiplying both sides by 20,

5(4f − 3) = 10f − 9

20f − 15 = 10f – 9(Opening the brackets)

20f − 10f = − 9 + 15(transposing 10f to LHS and

15 RHS)

10f = 6

3

5

f

0.6f

Exercise – 2.6

Question 1

Solve:

83

2

3

x

x

Answer:

3

2

x

Solution:

83

2

3

x

x

On multiplying both sides by3x, we obtain

8x − 3 = 6x (transposing

6x

to LHS and

3

to RHS)

8x − 6x = 3

2x = 3

3

2

x

Question 2

Solve:

9

15

76

x

x

Answer:

35

33

x

Solution:

9

15

76

x

x

On multiplying both sides by (7 − 6x), we obtain

9x = 15(7 − 6x)

9x = 105 − 90x

9x + 90x = 105(transposing

90x

to LHS)

99x = 105

105

99

x

35

33

x

Question 3

Solve:

4

15 9

z

z

Answer:

z= 12

Solution:

4

15 9

z

z

On multiplying both sides by 9(z + 15), we obtain

9z = 4(z + 15)

⇒9z = 4z + 60

⇒9z − 4z = 60 (transposing

4z

to LHS)

⇒ 5z = 60

⇒z = 12

Question 4

Solve:

3 4 2

2 6 5

y

y

Answer:

y = −8

Solution:

3 4 2

2 6 5

y

y

On multiplying both sides by 5(2 − 6y), we obtain

5(3y + 4) = −2(2 − 6y)

⇒15y + 20 = − 4 + 12y

⇒ 15y − 12y = − 4 – 20 (transposing 12

y

to LHS and 20 to RHS)

⇒ 3y = −24

⇒ y =−8

Question 5

Solve:

7 4 4

23

y

y

Answer:

4

5

y

Solution:

7 4 4

23

y

y

On multiplying both sides by 3(y +2),we obtain

3(7y + 4) = −4(y + 2)

⇒ 21y + 12 = − 4y − 8

⇒ 21y + 4y = − 8 – 12(transposing

4y

to LHS and 12 to RHS)

⇒ 25y = −20

4

5

y

Question 6

The ages of Hari and Harry are in the ratio 5:7.

Four years from now the ratio of their ages will be 3:4.

Find their present ages.

Answer:

Hari’s age = 20 years; Harry’s age = 28 years

Solution:

Let the ages of Hari and Harry be 5x years and 7x years respectively.

Four years later, their ages will be (5x + 4) years and (7x + 4) years respectively.

According to the question,

5 4 3

7 4 4

x

x

By cross-multiplication,

4(5 4) 3(7 4)xx

20 16 21 12xx

16 12 21 20xx

4x

Hari’sage =5x years = (5 × 4) years = 20 years

Harry’s age = 7x years = (7 × 4) years = 28 years

Question 7

The denominator of a rational number is greater than its numerator by 8.

If the numerator is increased by 17 and the denominator is decreased by 1, the

Number obtained is

3

2

.

Find the rational number.

Answer:

13

21

Solution:

Let the numerator of the rational number be x.

Therefore, its denominator will be x + 8.

The rational number will be

8

x

x

.

According to the question,

17 3

8 1 2

x

x

17 3

72

x

x

By cross-multiplication,

⇒ 2(x +17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21 (transposing

2x

to RHS and 21 to LHS)

⇒34 − 21 = 3x − 2x

⇒13 = x

Numerator of the rational number = x = 13

Denominator of the rational number = x + 8 = 13 + 8 = 21

Hence, the rational number is

13

21