NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.4

Question 1.
Amina thinks of numbers and subtracts 52 from it. She multiplies the results 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
By subtracting 52, we get x – 52
According to the given question
8(x−52)=3x
8x – 8×52 = 3x
8x – 20 = 3x
By transposing 3x to L.H.S. and -20 to R.H.S., we get
8x – 3x = 20
5x = 20
Dividing both sides by 5, we get
5x/5=20/5
x = 4
∴ The required number 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 number be x
The other positive number is 5x
on adding 21 to both numbers, we get (x + 21) and (5x + 21)
According to the question, we get
2(x + 21) = 5x + 21
2x + 42 = 5x + 21
Transposing 42 to R.H.S and 5x to L.H.S., we get
2x – 5x = 21 – 42
-3x = -21
Dividing both sides by -3, we get
−3x/−3=−21/−3
x = 7
the other 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 units digit be ‘x’.
the tens digits = 9 – x(sum of the digits is 9)
The original two digit number = 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
On interchanging the digits,
the new number = 10x + 9 – x = 9x + 9
According to the given question, we get
New number = Original number + 27
9x + 9 = 90 – 9x + 27
9x + 9 = 117 – 9x
Transposing 9 to R.H.S. and -9x to L.H.S., we get
9x + 9x = 117 – 9
18x = 108
Dividing both sides by 18, we get
18x/18=108/18
x = 6
∴ The original number = 90 – 9x = 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 in the unit place be ‘x’
Then, the digit at tens place = 3x
The number = 10(3x) + x
= 30x + x
= 31x
On interchanging the digits,
The new number = 10x + 3x = 13x
According to the question
31x + 13x = 88
44x = 88
Dividing both sides by 44
44x/44=88/44
x = 2
∴ The number = 31x = 31 × 2 = 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?
Solution:
Let Shobos present age be ‘x’ years
Mother’s present age = 6x years
After 5 years
Shobos age = (x + 5) years
Shobos mothers age = (6x + 5) years
According to the question, we get
1/3 (mothers present age) = Shobos age after 5 years
1/3 × 6x = x + 5
2x = x + 5
Transposing x to LHS
2x – x = 5
x = 5
∴ Shobo’s present age = 5 years
Mothers present age = (6 × 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 of ₹ 100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the length of the rectangular plot be 11x metres
and the breadth of the rectangular plot be 4x metres
Perimeter of the plot = 2(l + b)
= 2(11x + 4x)
= 2 × 15x
= 30x
Cost of fencing = ₹ 75000
100 × 30x = 75000
3000x = 75000
Dividing both sides by 3000, we get
3000x/3000=75000/3000
x = 25
Length of the plot = 11 × 25 = 275 metres
Breadth of the plot = 4 × 25 = 100 metres

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 of ₹ 36,600. How much trouser material did he buy?
Solution:
Let the length of cloth for shirts be ‘3x’ metres
and the length of cloth for trousers be ‘2x’ metres
Cost of shirts cloth = 3x × 50 = ₹ 150x
Cost of trouser cloth = 2x × 90 = ₹ 180x
S.P of shirts cloth at 12% profit

Material bought for trousers (2 × 100) = 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 be ‘x’

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
Present age of grandfather = 10x years
According to the given question, we get
10x – x = 54
9x = 54
Dividing both sides by 9, we get
9x/9=54/9
x = 6
Present age of granddaughter = 6 years.
Present age of grandfather = 10 × 6 = 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
Present age of Aman = 3x
Ten years ago Son’s age = (x – 10) years
Aman’s age = (3x – 10) years
According to the given question, we get
3x – 10 = 5(x – 10)
3x – 10 = 5x – 50
Transposing -10 to R.H.S. and 5x to L.H.S
3x – 5x = 10 – 50
-2x = -40
Dividing both sides by -2
−2x/2=−40/−2
x = 20
∴ Sons present age = 20 years
Aman’s present age = (3 × 20) = 60 years.