# KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2

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## Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.2

Question 1.
If 4 is added to a number and the sum is multiplied 3, the result is 30. Find the number.
Let the number be ‘x’.
If 4 is added to it will be x + 4.
Sum is multiplied by 3 the result is 30.
∴ (x + 4)3 = 30
3x +12 = 30
3x = 30 – 12
3x = 18
x = $$\frac { 18 }{ 3 }$$
x = 6
∴The number is 6

Question 2.
Find three consecutive odd numbers whose sum is 219.
Let the odd number be ‘x’.
The next two consecutive numbers are x + 2 and x + 4
x + (x + 2) + (x + 4) = 219
3x + 6 = 219
3x = 219 – 6
3x = 213
x = $$\frac { 213 }{ 3 }$$
x = 71
x + 2 = 71 + 2 = 73
x + 4 = 71 + 4 = 75
Three consecutive odd numbers are 71, 73, 75

Question 3.
A number subtracted by 30 gives 14 subtracted by 3 times the number. Find the number.
Let the number be x
Number subtracted by 30 = 30 – x
14 subtracted by 3 times the number 3x – 14
∴30 – x = 3x – 14
30+ 14 = 3x + x
44 = 4x
x = $$\frac { 44 }{ 4 }$$
x =11
∴ The number is 11

Question 4.
If 5 is subtracted from three times a number the result is 16. Find the number.
Let the number be x, 5 is subtracted from 3 times the number the result is 16.
3x – 5 = 16
3x= 16 + 5
3x = 21
x = $$\frac { 21 }{ 3 }$$
x = 7
∴ The number is 7

Question 5.
Find two numbers such that one of them exceeds the other by 9 and their sum is 81.
Let the number is x. The other number is x + 9.
Their sum is 81
∴ x + (x + 9) = 81
2x = 81 – 9
2x = 72
x = $$\frac{72}{2}$$
x = 36
x + 9 = 36 + 9 = 45
∴ The number are 36 and 45

Question 6.
Prakruthi’s age is 6 time Sahil’s age. After 15 years prakruthi will be 3 times as old as Sahil. Find their age.
Let Sahil’s present age be x. Prakruthi’s present age is 6x, 15 years later Sahil age will be (x + 15) years and Prakruthi age will be (6x + 15) years.
Given that the Prakruthi age will be 3 times as old as Sahil.
∴6x +15 = 3(x + 15)
6x + 15 = 3x + 45
6x – 3x = 45 – 15
3x = 30
x = $$\frac{30}{3}$$ = 10
Sahils age = x = 10 years Prakruthis age = 6x = 6 x 10 = 60 years

Question 7.
Ahmed’s father is thrice as old as Ahmed. After 12 years, his age will be twice that of his son. Find their present age.
Ahmed’s presents age be x
Fathers present age = 3x
12 years later Ahmed’s age = x + 12
and father’s age = 3x + 12
Given 3x + 12 = 2(x + 12)
3x+ 12 = 2x + 24
3x – 2x = 24 – 12
x = 12 years
∴ Ahmed’s age = 12 years
Fathers age = 3x = 3 × 12 = 36 years.

Question 8.
Sanju is 6 years older than his brother Nishu. If the sum of their ages is 28 years what are their present ages.
Let Nishu’s age be ‘x’ Sanju’s age = x + 6
Sum of their ages = 28 x + (x + 6) = 28
2x + 6 = 28
2x = 28 – 6
2x = 22
x = $$\frac{22}{2}$$
x = 11
Nishu’sage = x = 11 years
Sanju’ s age = x + 6 = 11 + 6 = 17 years

Question 9.
Viji is twice as old as his brother Deepu. If the difference of their ages is 11 years, find their present age.
Let Deep’s age be ‘x’, Viji’s age is 2x
Difference of their age = 11
2x – x = 11
x = 11
∴ Deepu’s age = x = 11 years
Viji’s age = 2x = 2 x 11 = 22 years

Question 10.
Mrs. Joseph is 27 years older than her daughter Bindu. After 8 years she will be twice as old as Bindu. Find their present ages.
Let Bindu’s present age be ‘x’ years.
Mrs. Joseph’s present age = x + 27 years.
After 8 years Bindu’s age = x + 8 and Mrs.
Josephs age = x + 27 + 8
= x + 35 years.
Given that x + 35 = 2(x + 8)
x + 35 = 2x + 16
35 – 16 = 2x – x
19 = x
x= 19
Bindu’s present age = 19 years
Mrs.Joseph’s age = x + 27 = 19 + 27 = 46 years

Question 11.
After 16 years Leena will be three times as old as she is now. Find her present age.
Let Leena’s present age be ‘x’ years
16 years later she will be (x + 16) years
Given x + 16 = 3x
16 = 2x
x = $$\frac { 16 }{ 2 }$$
x = 8 years
Leena’ s present age = 8 years

Question 12.
A rectangle has a length which is 5 cm less than twice its breadth. If the length is decreased by 5 cm and breadth is increased by 2 cm the perimeter of the resulting rectangle will be 74 cm. Find the length and breadth of the original rectangle.
Let the breadth of the original rectangle be ‘b’ twice the breadth is 2b.
The length of the rectangle is 5 cm less than twice the breadth.
∴ Length = 2b – 5
If the length is decreased by 5 then length
is 2b – 5 – 5 = 2b – 10
If the breadth is increased by 2cm then breadth is b + 2 cm.
Perimeter of new rectangle = 2(length + breadth)
74 = 2(2b – 10 + b + 2)
74 = 2(3b – 8)
74 + 16 = 6b 90 = 6b
b = $$\frac{90}{6}$$
b = 15cm.
Breadth of the original rectangle = 15 cm
Length of the original rectangle = 2b – 5
= 2 × 15 – 5
= 30 – 5
= 25 cm

Question 13.
The length of a rectangular field is twice its breadth. If the perimeter of the field is 288m. Find the dimensions of the field.
Let breadth of the rectangular field bee ‘x’m
∴ Its length = 2x
Its perimeter = 288
2(1 + b) = 288
2(2x + x) = 288
2(3x) = 288
6x = 288
x = $$\frac { 288 }{ 6 }$$ = 48
Its length = 2x = 2 × 48 = 96m
x = $$\frac{3750}{5}$$