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

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Exercise Questions and Answers.

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

Question 1.
Amina thinks of a number and subtracts 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?
Answer:
Let the required number be x.
According to question,
8(x - \(\frac{5}{2}\)) = 3x
or 8x - 20 = 3x
or 8x - 3x = 20
or 5x = 20
or x = \(\frac{20}{5}\) = 4    
∴ 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?
Answer:
Let the required numbers be x and 5x.
According to question,
 2(x + 21) = 5x + 21
or 2x + 42 = 5x +21
or 2x - 5x = 21 - 42
or - 3x = - 21
or \(\frac{-3 x}{-3}=\frac{-21}{-3}\)
or x = 7
Hence, 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 num-ber is greater than the original number by 27. What is the two-digit number?
Answer:
Let the digit in the units place be x.
Then the digit in the tens place = (9 - x)
Two digit number = 10 × (9 - x) + x
= 90 - 10x + x = 90 - 9x

With interchange of digits, the resulting two digit number
= 10x + (9 - x) = 9x + 9

According to question,
(90 - 9x) + 27 = 9x + 9
or 117 - 9x= 9x + 9
or - 9x - 9x= 9 - 117
or - 18x= - 108
or x = \(\frac{-108}{-18}\) = 6
∴ Digit in units place = 6
and Digit in tens place 9 - x = 9 - 6 = 3
Hence two digit number = 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?
Answer:
Let the digit in unit place be x.
Then digit in tens place be 3x.
∴ Number = 10 x 3x + x
= 30x + x = 31x ....(1)

With interchange of digits, the resulting two digit number = 10x + 3x = 13x
The sum of resulting number and original number is 88.
∴ 31x + 13x= 88
or 44x = 88
or x = \(\frac{88}{44}\)= 2
Putting the value of x in equation (1)
Number = 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?
Answer:
Let the Shobo’s present age be x years.
According to question,
Shobo’s mother’s present age = 6x years
After 5 years : Shobo’s age = (x + 5) years
Hence (x + 5) = \(\frac{1}{3}\) × (6x) 
3 × (x + 5)= 3 × \(\frac{1}{3}\) × (6x)
(Multiplying both sides by 3)
or 3x + 15= 6x
or 3x - 6x = -15
or -3x = - 15
or x = \(\frac{-15}{-3}\) = 5
∴ Shobo’s present age = 5 years 
Shobo’s Mother’s present age= 30 years

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuii 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?
Answer:
Let the length and breadth of rectangular plot be 11x meter and 4x meter re-spectively, then
Perimeter = 2 (11x + 4x) meter
= 30x meter

∵ Perimeter of rectangular plot = 2 (L + B)
The total cost of fencing the plot at the rate of ₹ 100
= ₹ (100 × 30x)
= ₹ 3000x
But, Actual Cost = ₹ 75000
∴ 3000x = 75000
x = \(\frac{75000}{3000}\) = 25
∴ Length = 11x = 11 × 25 meters
= 275 meters

and Breadth = 4x = 4 × 25 meters
= 100 meters

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per meter and trouser material that costs him Rs 90 per meter. For every 3 meters of the shirt material he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?
Answer:
Let Hasan buys 3x meters of shirt Cloth material and 2x, meters of trouser cloth material
Cost of shirt material = ₹ 50 per meter
Total cost of shirt material = 50 × 3x
= ₹150x
Cost of trouser Material = ₹ 90 per meter Total cost of trouser material
= 90 × 2x = ₹ 180x

Profit on shirt material =12%
= \(\frac{150 x \times 12}{100}\) = 18x

SP of Shirt Material =150x + 18x = 168x
Profit on trouser Material = 10%
= \(\frac{180 x \times 10}{100}\) =18x

SP of trouser material
= 180x + 18x = 198x
Hence Total Selling price = 168x + 198x
= 366x
According to question SP = 36,600
Or 366x = ₹ 36,600
= 200 Meters

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.
Answer:
Let the number of deer in herd = x
According to question,
Number of deer grazing in the field = \(\frac{x}{2}\)
Number of deer playing nearby
= \(\frac{3}{4} \times\left(x-\frac{x}{2}\right)\)
= \(\frac{3}{4} \times \frac{x}{2}=\frac{3 x}{8}\)

Number of deer drinking water from the pond = 9
∴ Total deer = \(\frac{x}{2}+\frac{3 x}{8}\) + 9 = x
or 8 × \( \frac{x}{2}\) + 8 × \(\frac{3 x}{8}\) + 8 × 9 = 8 × x
(Multiplying both sides by 8)
or 4x + 3x + 72 = 8x
or 7x - 8x = - 72
or - x= - 72
or x = 12
Hence, the total 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.
Answer:
Let granddaughter’s age be x years,
then grandfather’s age = 10x years

According to question,
10x = x + 54
or 10x - x= 54
or 9x = 54
or x = \(\frac{54}{9}\) = 6
∴ Granddaughter’s age = 6 years and, Grandfather’s age = 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.
Answer:
Let the present son’s age be x, then Aman’s age 3x years Son’s age 10 years ago was (x - 10) years.
Aman’s age 10 years ago was (3x - 10) years According to question,
3x - 10 = 5(x - 10)
or 3x - 10 = 5x - 50
or 5x - 3x= - 10 + 50
or - 2x = 40
or x = 20
∴ Son’s age = 20 years
and Aman’s age = 20 × 3 = 60 years