Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1.

Amina thinks of a number and subtracts \(\frac { 5 }{ 2 }\) 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?

Solution:

Let the number be āxā.

Subtracting \(\frac { 5 }{ 2 }\), we get x – \(\frac { 5 }{ 2 }\).

According to the condition, we have

8 Ć (x – \(\frac { 5 }{ 2 }\)) = 3 Ć x or 8x – 20 = 3x

Transposing -20 to RHS, we have

8x = 3x + 20

Again transposing 3x to LHS, we have

8x – 3x = 20 or 5x = 20 or x = \(\frac { 20 }{ 5 }\) = 4

Thus, the required number = 4

Question 2.

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution:

Let the positive number = 5x

The other number = x

On adding 21 to both numbers, we get (5x + 21) and (x + 21)

According to the condition, we have

2 Ć (x + 21) = 5x + 21 or 2x + 42= 5x + 21

Transposing 42 to RHS, we have

2x = 5x + 21 – 42

Transposing 5x to LHS, we have

2x – 5x = -21 or – 3x = -21

Dividing both sides by (-3), we have

\(\frac { -3x }{ -3 }\) = \(\frac { -21 }{ -3 }\) or x = 7 and 5x = 5 Ć 7 = 35

Thus, the required numbers are 7 and 35.

Question 3.

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?

Solution:

Let the units digit = x

ā“ The tens digits = (9 – x)

ā“ The original number = 10(9 – x) + x

= 90 – 10x + x = 90 – 9x

On interchanging the digits, the new number

= 10x + (9 – x) = 10x + 9- x = 9x + 9

According to the condition, we have

[New number] = [Original number] + 27

or 9x + 9 = 90 – 9x + 27

or 9x + 9 = 117 – 9x

or 9x = 117 – 9 – 9x [Transposing 9 to RHS]

or 9x + 9x = 108 [Transposing (-9x) to LHS]

or 18x = 108

Dividing both sides by 18, we have

x = \(\frac { 108 }{ 18 }\) = 6

ā“The Original number = 90 – (9Ć6)

= 90 – 54 = 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?

Solution:

Let the digit at unit place = x

The digit at tens place = 3x

The number = 10(3x) + x = 30x + x = 31x

On interchanging the digits, the number

= 10x + 3x = 13x

According to the condition, we have

31x + 13x = 88 or 44x = 88

Dividing both sides by 44, we have

\(\frac { 44x }{ 44 }\) = \(\frac { 88 }{ 44 }\) or x = 2

ā“The number = 31 x 2 = 62

Question 5.

Shoboās motherās present age is six times Shoboās present age. Shobo s age five years from now will be one-third of his mother s present age. What are their present ages?

Solution:

Let Shoboās present age = x years /. Motherās present age = 6x years After 5 years: Shoboās age = (x + 5) years Motherās age = (6x + 5) years According to the condition, we have

(Motherās present age) = (Shoboās age after 5 years)

ie. \(\frac { 1 }{ 3 }\) (6x) = x + 5 or \(\frac { 1 }{ 3 }\) Ć 6x = x + 5

or 2x = x + 5 or 2x – x = 5

(Transporting x to LHS)

or x = 5

ā“ Shobo’s present age = 6 Ć 5 = 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 ā¹ 100 per metre it will cost the village panchayat ā¹ 75000 to fence the plot. What are the dimensions of the plot?

Solution:

Let the length = 11x metres and breadth = 4x metres

ā“Perimeter = 2(Length + Breadth)

= 2(11x + 4x) = 2 x 15x = 30x

Cost of fencing = ā¹ 100 x 30x = ā¹ 3000x

But the cost of fencing is ā¹ 75000.

3000x = 75000. or x = \(\frac { 75000 }{ 3000 }\) = 25

Length = 11 x 25 = 275 metres

Breadth = 4 x 25 = 100 metres

Question 7.

Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him 750 per metre and trouser material that costs himā¹ 90 per metre. For every 3 metres 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 ā¹ 36,600. How much trouser material did he buy?

Solution:

Let the length of cloth for trousers = 2x metres

ā“ The length of cloth for shirts = 3x metres

Cost of trouserās cloth = 2x Ć ā¹90 = ā¹ 180x

Cost of shirtās cloth = 3x Ć ā¹ 50 = ā¹ 150x

S.P. of trouserās cloth at 10% profit =

= ā¹ \(\frac { 110 }{ 100 }\) Ć108x = ā¹ 198x

S.p of shirtās clth at 12% profit = ā¹ \(\frac { 112 }{ 100 }\) Ć 150x = ā¹ 168x

ā“ Total S.P = ā¹ 198x + ā¹ 198x = ā¹366x

But the total S.P = ā¹36,600

ā“ 366x = 36,600 or x = \(\frac { 36600 }{ 366 }\)

[Dividing both sides by 366] = 100

ā“ 2x = 2 Ć 100 = 200

Thus, he bougth cloth for trouser = 200 metres

Question 8.

Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Solution:

Let the total number of deer = x

ā“ Number of deer: Grazing in the field = \(\frac { x }{ 2 }\)

Playing nearly = \(\frac { 3 }{ 4 }\) Ć (Remaining no. of deer)

Thus, number of deer in herd = 72

Question 9.

A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Solution:

Let the present age of grandfather = x years

ā“Present age of granddaughter = \(\frac { 1 }{ 2 }\) years

According to the condition, we have [Present age of granddaughter] + 54 = x or \(\frac { x }{ 10 }\) + 54 = x

Transposing 54 to RHS and x to LHS, we have

Present age of grandfather = 60 years

ā“Present age of granddaughter = 6 years

Question 10.

Atnanās age is three times his son s age. Ten years ago he was five times his sonās age. Find their present ages.

Solution:

Let present age of son = x years Present age of Aman = 3x years Ten years ago sonās age = (x – 10) years Amanās age = (3x – 10) years According to the condition, we have [Sonās age (10 years ago)] x 5 = [Amanās age (10 years ago)] or (x – 10) x 5 = (3x – 10) or 5x – 50 = 3x – 10

Transposing (-50) to RHS and 3x to LHS, we have

5x – 3x = -10 + 50

or 2x = 40 or x = \(\frac { 40 }{ 2 }\) = 20

[Dividing both sides by 2]

and 3x = 3 Ć 20 = 60

Sonās present age = 20 years

Amanās present age = 60 years