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 \(\frac{5}{2}\) 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 \(\frac{5}{2}\), we get x – \(\frac{5}{2}\)
According to the given question
\(8\left(x-\frac{5}{2}\right)=3 x\)
8x – \(\frac{8 \times 5}{2}\) = 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
\(\frac{5 \mathrm{x}}{5}=\frac{20}{5}\)
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
\(\frac{-3 x}{-3}=\frac{-21}{-3}\)
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
\(\frac{18 \mathrm{x}}{18}=\frac{108}{18}\)
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
\(\frac{44 x}{44}=\frac{88}{44}\)
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
\(\frac{1}{3}\) (mothers present age) = Shobos age after 5 years
\(\frac{1}{3}\) × 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
\(\frac{3000 \mathrm{x}}{3000}=\frac{75000}{3000}\)
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
= ₹ \(\frac{112}{100}\) × 150x
= ₹ 168x
S.P. of trousers cloth at 10% profit
= \(\frac{110}{100}\) × 180x
= ₹ 198x
Total S.P = ₹ 36,600
168x + 198x = ₹ 36,600
366x = 36600
Dividing both sides by 366, we get
\(\frac{366 \mathrm{x}}{366}=\frac{36600}{366}\)
x = 100
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’
Number of deer, grazing in the field = \(\frac{\mathrm{x}}{2}\)
Number of deer playing near by = \(\frac{3}{4}\left(x-\frac{x}{2}\right)=\frac{3 x}{4}-\frac{3 x}{8}=\frac{6 x-3 x}{8}=\frac{3 x}{8}\)
Number of deer drinking water = 9
\(\frac{x}{2}+\frac{3 x}{8}\) + 9 = x
Transposing 9 to RHS and x to LHS we get
\(\frac{x}{2}+\frac{3 x}{8}-x=-9\)
\(\frac{4 x+3 x-8 x}{8}=-9\)
\(\frac{-x}{8}=-9\)
Multiplying both sides by -8
\(\frac{-\mathrm{x}}{8}\) × (-8) = -9 × (-8)
x = 72
∴ 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 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
\(\frac{9 \mathrm{x}}{9}=\frac{54}{9}\)
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
\(\frac{-2 x}{2}=\frac{-40}{-2}\)
x = 20
∴ Sons present age = 20 years
Aman’s present age = (3 × 20) = 60 years.