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

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KSEEB 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 Amina think a number = x
She subtracts \(\frac{5}{2}\) from x
∴ According to equation
8(x – \(\frac{5}{2}\)) = 3x
⇒ 8x – \(\frac{5}{2}\) × 8 = 3x
⇒ 8x – 20 = 3x
⇒ 5x = 20
⇒ x = 4
∴ 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 another number = x
The first number = 5x
According to question,
2(x + 21) = 5x + 21
⇒ 2x + 42 = 5x + 21
⇒ 42 – 21 = 5x – 2x
⇒ 21 = 3x
⇒ 7 = x
First number = 5 × 7 = 35
Hence, the numbers are 7 and 35.

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

Question 3.
The 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 ones place digit = x
Tens place digit = y
Number = 10y + x
By interchanging the digits, number = 10x + y
x + y = 9 ……… (1)
⇒ (10x + y) – (10y + x) = 27
⇒ 10x + y – 10y – x = 27
⇒ 9x – 9y = 27
⇒ 9(x – y) = 27
⇒ x – y = \(\frac{27}{9}\)
⇒ x – y = 3
⇒ x = 3 + y ………(2)
Substituting the value of x in 1
3 + y + y = 9
⇒ 3 + 2y = 9
⇒ 2y = 6
⇒ y = 3
x = 3 + 3 = 6
Hence, number is = 10(3) + 6 = 30 + 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 unit digit = x
Tens digit = 3x
Number = 10(3x) + x = 30x + x = 31x
On interchanging the digits, number = 10x + 3x = 13x
According to question,
31x + 13x = 88
⇒ 44x = 88
⇒ x = 2
unit digit = 2
Tens digit = 2 × 3 = 6
Number = 10 × 6 + 2 = 62

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

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 = x years
Shobo’s mother’s present age = 6x years
After 5 years,
Shobo’s age = (x + 5) years
According to question,
(x + 5) = \(\frac{6 x}{3}\)
⇒ (x + 5) = 2x
⇒ 5 = 2x – x
⇒ x = 5
Shobo’s present age = 5 years
Shobo’s mother age = 5 × 6 = 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 Rs. 100 per meter it will cost the village panchayat Rs. 75,000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the ratio = x
Length of rectangular plot = 11x
Breadth of rectangular plot = 4x
Fencing costs Rs. 100 for = 1 m
Fencing costs Re. 1 for = \(\frac{1}{100}\) mts
Fencing costs Rs. 75000 for = \(\frac{1}{100} \times 75000\) = 750 m
Hence, perimeter of plot = 750 m
But perimeter = 2(l + b)
⇒ 2(11x + 4x) = 750
⇒ 2 × 15x = 750
⇒ 30x = 750
⇒ x = 25
Length of plot = 11 × 25 = 275 m
Breadth of plot = 4 × 25 = 100 m

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

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:
Ratio of cloth for trouser and shirt = 2 : 3
Let ratio = x
Length of cloth for shirt = 3x m
Length of cloth for trouser = 2x m
Cost of shirt material = 3x × 50 = Rs. 150x
Cost of trouser material = 2x × 90 = Rs. 180x
Profit on shirt material = 12%
S.P. of the shirt material = 150x + 12% of 150x
= 150x + \(\frac{12}{100} \times 150 x\)
= 150x + 18x
= 168x
Profit on trouser material = 10%
S.P. of trouser material = 180x + \(\frac{10}{100} \times 180 x\)
= 180x + 18x
= 198x
According to queastion,
168x + 198x = 36,660
⇒ 366x = 36,660
⇒ x = \(\frac{36,660}{366}\) = 100.16 or 100 (approx.)
∴ Cost of trouser material = 2 × 100 = Rs. 200

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 = x
Number of deer grazing in the field = \(\frac{1}{2}\) of x = \(\frac{x}{2}\)
Remaining Deer = x – \(\frac{x}{2}\) = \(\frac{x}{2}\)
Three-fourth of the remaining = \(\frac{3}{4} \times \frac{x}{2}=\frac{3 x}{8}\)
Remaining = \(\frac{x}{2}-\frac{3 x}{8}=\frac{4 x-3 x}{8}\)
⇒ 9 = \(\frac{x}{8}\)
⇒ x = 72

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

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 = x years
The present age of grandfather = 10x years
According to the question,
10x – x = 54
⇒ 9x = 54
⇒ x = 6
Age of granddaughter = 6 years
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 = x years
Present age of father = 3x years
10 years ago,
Age of son = (x – 10) years
Age of father = (3x – 10) years
According to question,
(3x – 10) = 5(x – 10)
⇒ 3x – 10 = 5x – 50
⇒ -10 + 50 = 5x – 3x
⇒ 2x = 40
⇒ x = 20
Son’s present age = 20 years
Father’s age = 3 × 20 = 60 years.