# PSEB 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for 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 x.
Subtracting $$\frac {5}{2}$$ from it, we get, x – $$\frac {5}{2}$$
By multiplying this result by 8,
we get 8 (x – $$\frac {5}{2}$$)
According to the condition,
8(x – $$\frac {5}{2}$$) = 3x
∴ 8x – 20 = 3x
∴ 8x = 3x + 20 [Transposing (-20) to RHS]
∴ 8x – 3x = 20 (Transposing 3x to LHS)
∴ 5x = 20
∴ $$\frac{5 x}{5}=\frac{20}{5}$$ (Dividing both the sides by 5)
∴ x = 4
Thus, the number though of by Amina is 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 smaller number be x
∴ The greater number is 5x
On adding 21 to both the numbers,
we get (5x + 21) and (x + 21)
According to the condition,
5x + 21 = 2 (x + 21)
∴ 5x + 21 = 2x + 42
∴ 5x = 2x + 42 – 21 (Transposing 21 to RHS)
∴ 5x = 2x + 21
∴ 5x – 2x = 21 (Transposing 2x to LHS)
∴ 3x = 21
∴ $$\frac{3 x}{3}=\frac{21}{3}$$ (Dividing both the sides by 3)
∴ x = 7
The smaller number x = 7
The greater number = 5x = 5 × 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 digit at the units place be x.
Then, the digit at the tens place = (9 – x)
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,
New number = Original number + 27
∴ 9x + 9 = (90 – 9x) + 27
∴ 9x + 9 = 90 – 9x + 27
∴ 9x + 9 = 117 – 9x
∴ 9x = 117 – 9x – 9 (Transposing 9 to RHS)
∴ 9x = 108 – 9x
∴ 9x + 9x = 108 [Transposing (-9x) to LHS]
∴ 18x = 108
∴ $$\frac{18 x}{18}=\frac{108}{18}$$ (Dividing both the sides by 18)
∴ x = 6
∴ Original number = 90 – 9x
= 90 – 9(6)
= 90 – 54 = 36
Thus, the original number is 36.

Question 4.
One of the two digits of a two-digit number is three times the other digit. I 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 units place be x and the digit at tens place be 3x.
The number = 10 (3x) + x
= 30x + x
= 31x
On interchanging the digits, the number – 10x + 3x = 13x
According to the condition,
31x + 13x = 88
44x = 88
∴ $$\frac{44 x}{44}=\frac{88}{44}$$ (Dividing both the sides by 44)
∴x = 2
The number = 31x = 31 × 2 = 62
Thus, original number is either 62 or 26.

Question 5.
Saroj’s mother’s present age is six times ; Saroj’s present age. Saroj’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution:
Let Saroj’s present age be x years and mother’s present age be 6x years
After 5 years:
Saroj’s age will be x + 5 years
Mother’s age will be 6x + 5 years
According to the condition,
$$\frac {1}{3}$$ (Mother’s present age) = Saroj’s age after 5 years
∴ $$\frac {1}{3}$$(6x) = x + 5
∴ 2x = x + 5
∴ 2x – x = 5 (Transposing x to LHS)
∴ x = 5
Saroj’s present age = x = 5 years
Mother’s present age = 6x = 6 × 5
= 30 years
Thus, present ages of Saroj and her mother are 5 years and 30 years respectively. 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 ₹ 100 per metre it will cost the village panchayat ₹ 75,000 to fence the plot. What are the dimensions of the plot ?
Solution :
Length : Breadth = 11 : 4
Let the length be 11x metres.
Then, the breadth = 4x metres.
Perimetre = 2 (length + breadth)
= 2 (11x + 4x)
= 2 (15x) = 30x
Cost of fencing = ₹ 100 × 30x
= ₹3000 x
But, the cost of fencing = ₹ 75,000 (Given)
∴ 3000 x = 75000
∴ $$\frac{3000 x}{3000}=\frac{75000}{3000}$$ (Dividing both the sides by 3000)
∴ x = 25
Length = 11x
= 11 × 25
= 275 metres
= 4 × 25
= 100 metres
Thus, the length and breadth of the rectangular plot are 275 metres and 100 metres respectively.

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 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 ₹ 36,600. How much trouser material did he buy?
Solution:
Let the length of cloth for trousers be 2x metres
Then, the length of cloth for shirts = 3x metres
Cost of trouser’s cloth = 2x × ₹ 90
= ₹ 180x
Cost of shirt’s cloth = 3x × ₹ 50
= ₹ 150x
10 % profit is made on trouser’s cloth.
If C.E of trouser’s cloth is ₹ 100, then S.E is ₹ 110.
S.E of trouser’s cloth at 10 % profit = ₹$$\frac {110}{100}$$ × 180x
= ₹ 198x
12 % profit is made on shirt’s cloth. If C.P of shirt’s cloth is ₹ 100, then S.P is ₹ 112.
S.P of shirt’s cloth at 12 % profit = ₹$$\frac {112}{100}$$ × 150x
= ₹ 168x
∴ Total S.P = ₹ 198x + ₹ 168x
= ₹ 366x
But the total S.P. = ₹ 36,600 (Given)
366x = 36600
∴ $$\frac {112}{100}$$ (Dividing both the sides by 366)
∴ x = 100
Cloth for trousers = 2x = 2 × 100 = 200
Thus, he bought 200 metres of cloth for trousers.

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 be x.
Number of deer grazing in the field = $$\frac{x}{2}$$
Remaining deer = x – $$\frac{x}{2}$$ = $$\frac{x}{2}$$
Deer playing near by = $$\frac {3}{2}$$ × (Remaining no. of deer)
= $$\frac {3}{4}$$ × $$\frac{x}{2}$$
= $$\frac{3 x}{8}$$
Number of deer drinking water = 9
∴ Total number of deer = $$\frac{x}{2}$$ + $$\frac{3 x}{8}$$ + 9
∴ $$\frac{x}{2}$$ + $$\frac{3 x}{8}$$ + 9 = x
$$\frac{x}{2}$$ + $$\frac{3 x}{8}$$ = x – 9 (Transposing 9 to RHS)
$$\frac{x}{2}$$ + $$\frac{3 x}{8}$$ -x = -9 (Transposing x to LHS)
$$\frac{4 x+3 x-8 x}{8}$$ = -9 (LCM = 8)
∴ $$\frac{-x}{8}$$ = -9
∴ $$\frac{-x}{8}$$ × 8 = -9 × 8 (Multiplying both the sides by 8)
∴ -x = – 72
∴ x = 72 [∵ Multiplying both sides by (- 1)]
Thus, total number of deer in 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.
Solution:
Let the present age of granddaughter be x years.
Grandfather’s age is 10x years.
Fresent age of daughter + 54 = Grandfather’s age
∴ x + 54 = 10x
∴ 10x = x + 54 (Interchanging the sides)
∴ 10x – x = 54 (Transposing x to LHS)
∴ 9x = 54
∴ $$\frac{9 x}{9}=\frac{54}{9}$$ (Dividing both the sides by 9)
∴ x = 6
Granddaughter’s age = x = 6 years
Grandfather’s age = 10x = 10 × 6 = 60 years
Thus, granddaughter’s age is 6 years and grandfather’s age is 60 years.

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.
Solution :
Let the present age of son be x years
Then, the present age of Aman = 3x years
Ten years ago their ages:
Son’s age was (x – 10) years
Aman’s age was (3x – 10) years
5 × (Son’s age 10 years ago) = Aman’s age 10 years ago
∴ 5 (x – 10) = (3x – 10)
∴ 5x – 50 = 3x – 10
∴ 5x = 3x – 10 + 50 [Transposing (-50) to RHS]
∴ 5x = 3x + 40
∴ 5x – 3x = 40 (Transposing 3x to LHS)
∴ 2x = 40
∴ $$\frac{2 x}{2}=\frac{40}{2}$$(Dividing both the sides by 2)
∴ x = 20
Son’s present age = x = 20 years
Aman’s present age = 3x = 3 × 20
= 60 years
Thus, present ages of Aman and his son are 60 years and 20 years respectively.