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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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
breadth = x = 48m

Question 14.
Srishti’s salary is the same as 4 times Azar’s salary. If together they earn Rs 3750 a month find their individual salaries.
Answer:
Let Axar’s salary be x. Sristi’s salary is 4x together they earn Rs. 3750
∴ x + 4x = 3750
5x = 3750
x = \(\frac{3750}{5}\)
x = 750
4x = 4 × 750 = 3000
∴ Azar’s salary is Rs.750 and Sristi’s salary is Rs.3000

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