Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Exercise Questions and Answers.
Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 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 Amina thinks the number be x.
She subtracts \(\frac{5}{2}\) from x.
∴ Now, the result will be x – \(\frac{5}{2}\)
The result is multiply by 8
Therefore, (x – \(\frac{5}{2}\)) × 8 = 8x – 20
This number is three times the same number, therefore
8x – 20 = 3x
or, 8x – 3x = 20
or, 5x = 20
or, x = 4
∴ The number will be 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 another number be x.
The first number is 5x
21 is added to both the numbers then one of the new numbers becomes twice the other new number.
∴ 2(x + 21) = 5x + 21
or, 2x + 42 = 5x + 21
or, 2x – 5x = 21 – 42
or, -3x = -21
∴ x = 7
Other number = 7 × 5 = 35.
∴ The numbers are 7, 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 ones place the number be x.
∴ Ten’s place = 9 – x
The number will be = 10 (9 – x) + x
(Two digit number = 10, tens place + one’s place)
After interchanging the digit we get 10x + 9 – x
This number is 27 more than original number
According to the question
10x + 9 – x = 10 (9 – x) + x + 27
or, 9x + 9 = 90 – 10x + x + 27
or, 9x = 117 – 9x – 9
or, 9x + 9x = 108
or, 18x = 108
or x = \(\frac{108}{18}\) = 6
∴ Ten’s place = 9 – 6 = 3
∴ The number will be = 10 × 3 + 6 = 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 one’s place of a two digit number be x
Ten’s place = 3x
The number will be 10.3x + x
After interchanging the digits we get
10x + 3x
According to the question,
10.3x + x + 10x + 3x = 88
or, 30x + x + 10x + 3x = 88
or, 44× = 88
or, x = 2
The original number = 10 × 3 × 2 + 2 = 62
[Note : According to the question, If you let one’s place is 3 times of ten’s place digit, the answer will be 26.]
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 Shobo’s present age be x years.
Shobo’s mother’s present age 6x years
5 years after Shobo’s age = x + 5 years
According to the question
x + 5 = \(\frac{6x}{3}\)
or, x + 5 = 2x
or, 2x – x = 5
or, x = 5
∴ Shobo’s present age = 5 year’s
Shobo’s mother’s 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 Rs. 100 per metre it will cost the village panchayat Rs. 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the ratio be x
Length = llx and breadth = 4x
Rs. 100 cost 1 metre to fence the plot.
Rs. 7500 cost \(\frac{1}{100}\) × 75000 metre to fence the plot = 750 metre
∴ Perimeter of the plot
= 750 m
2 (11x + 4x) = 750
[∵ Perimeter -2(l + b)]
2 × 15x = 750
or, 30x = 750
or, x = 25
Therefore length = 11 × 25 = 275 m
breadth = 4 × 25 = 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 2 metres of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs. 36,660. How much trouser material did he buy?
Solution:
Hasan bpys shirt and trouser in ratio 2 : 3
Let the ratio be x
He buys 2x m of the trouser and 3cc m of the shirt material
The cost of the shirt material
= Rs. 3x × 50 = Rs. 150x
The cost of the trouser meterial =Rs. 2x × 90 = Rs. 180x
The S.P of the shirt material
= Rs. 150x + 12% of 150x
= 150x + \(\frac{12}{100}\) × 150x = Rs. 168x
The S.P of the trouser material
= Rs. 180x + 10% of 180x
= 180x + \(\frac{10}{100}\) × 180x = 198x
According to, the question,
168x + 198x = 36,660
or, 366x = 36,660
or, x = \(\frac{36,660}{366}\) = 100.16
Then trouser material is bought by him = 200 m.
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 number of deer in the herd be x
Number of deer who grazing in the field
= \(\frac{1}{2}\) of x = \(\frac{x}{2}\)
Rest Number of deer
= x – \(\frac{x}{2}\) = \(\frac{x}{2}\)
Three fourths of the remaining
= \(\frac{3}{4}\) of \(\frac{x}{2}\) = \(\frac{3x}{8}\)
Rest = \(\frac{x}{2}\) – \(\frac{3x}{8}\)
9 = \(\frac{4x-3x}{8}\)
or, 9 = \(\frac{x}{8}\)
∴ x = 9 × 8 = 72
2nd Method
Total number of deer in the herd = x
Hints :
∴ x = \(\frac{x}{2}\) + \(\frac{3x}{8}\) + 9
= \(\frac{4x+3x+72}{8}\)
∴ x = \(\frac{7x+72}{8}\)
8x = 7x + 72
8x – 7x = 72
x = 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
The present age of grandfather = 10x years
According to the question
10x – x = 54
or, 9x = 54
∴ x = 6
∴ The present age of granddaughter = 6 years
The present age of grandfather = 10 × 6 years = 60 years
Question 10.
A man’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 son’s present age be x years
Father’s present age = 3x years
Ten years ago son’s age = x – 10 years
Ten years ago father’s age = 3x – 10 years
According to the question
3x – 10 = 5 (x – 10)
or, 3x – 10 = 5x – 50
or, 3x – 5x = -50 + 10
or, -2x = -40
∴ x = 20
Therefore, Son’s present age = 20 years.
Father’s present age = 3 x 20 = 60 years.