# GSEB Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

Question 1.
If you subtract $$\frac { 1 }{ 2 }$$ from a number and multiply the result by $$\frac { 1 }{ 2 }$$ you get $$\frac { 1 }{ 2 }$$. What is the number?
Solution:
Let the required number be x.
∴According to the condition, we have
(- $$\frac { 1 }{ 2 }$$) × $$\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 8 }$$ × 2
(Multipling both sides by 2)
x – $$\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 4 }$$
Adding both sides $$\frac { 1 }{ 2 }$$ we have
x – $$\frac { 1 }{ 2 }$$ + $$\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 4 }$$ + $$\frac { 1 }{ 2 }$$ or x = $$\frac { 1 + 2 }{ 4 }$$ = $$\frac { 3 }{ 4 }$$
∴The required number = $$\frac { 3 }{ 4 }$$ Question 2.
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Perimeter of the rectangular pool =154 m Let breadth = x metres
∴Length = 2(breadth) + 2 metres.
= (2x + 2) metres.
Since, perimeter of a rectangle
∴2[(2x + 2) + x] = 154
or (2x + 2) + x = $$\frac { 154 }{ 2 }$$ = 77
(Dividing both sides by 2)
or 2x + 2 + x = 77 or 3x + 2 = 77
Transposing 2 to RHS, we have
3x = 77 – 2 = 75
Dividing both sides by 3, we have
x = $$\frac { 75 }{ 3 }$$ = 25
Length = 2(25) + 2 = 52 m Question 3.
The base of an isosceles triangle is $$\frac { 4 }{ 3 }$$ cm.
The perimeter of the triangle is 4$$\frac { 2 }{ 15 }$$ cm. What is the length of either of the remaining equal sides?
Solution:
Base of the isosceles triangle = $$\frac { 4 }{ 3 }$$ cm.
Let the length of either equal sides = x cm.
Perimeter of the triangle = $$\frac { 4 }{ 3 }$$ + x + x = $$\frac { 4 }{ 3 }$$ + 2x
But the perimeter of the triangle = 4$$\frac { 2 }{ 15 }$$ cm
∴$$\frac { 4 }{ 3 }$$ + 2x = 4$$\frac { 2 }{ 15 }$$ = $$\frac { 62 }{ 15 }$$
Transposing $$\frac { 4 }{ 3 }$$ to RHS, we have 2x = $$\frac { 62 }{ 15 }$$ – $$\frac { 4 }{ 3 }$$
or 2x = $$\frac { 62 }{ 15 }$$ – $$\frac { 4 }{ 3 }$$ or 2x = $$\frac { 62 – 20 }{ 15 }$$ = $$\frac { 42 }{ 15 }$$
Dividing both sides by 2, we have
x = $$\frac { 42 }{ 15 }$$ × $$\frac { 1 }{ 2 }$$ = $$\frac { 21 }{ 15 }$$ = $$\frac { 7 }{ 5 }$$
or x = $$\frac { 7 }{ 5 }$$ = 1 $$\frac { 2 }{ 5 }$$ cm

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the smaller number = x.
∴ The other number = x + 15.
According to the condition, we have
x + [x + 15] = 95
2x + 15 = 95
or 2x = 95 – 15 = 80
Dividing both sides by 2, we have
x = 80 ÷ 2 = 40
∴The smaller number = 40
The other number = 40 + 15 = 55 Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Let the two numbers be 5x and 3x. According to the condition, we have
5x – 3x = 18 or 2x = 18
Dividing both sides by 2, we have
$$\frac { 2x }{ 2 }$$ = $$\frac { 18 }{ 2 }$$ or x = 9
∴5x = 5 x 9 = 45 and 3x = 3 x 9 = 27
∴The required numbers are 45 and 27.

Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive numbers be x, x + 1 and x + 2.
According to the condition, we have
x + [x + 1] + [x + 2] = 51
or x + x + 1 + x + 2 = 51
or 3x + 3 = 51
Transposing 3 to RHS, we have
3x = 51 – 3 = 48
Dividing both sides by 3, we have:
$$\frac { 3x }{ 3 }$$ = $$\frac { 48 }{ 3 }$$ or x = 16
Now x + 1 = 16 + 1 = 17
and x + 2=16 + 2 = 18
Thus, the required three consecutive numbers
are: 16, 17 and 18. Question 7.
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Solution:
Let the three multiples of 8 are: x, x + 8 and x+8+8=x+ 16.
According to the condition, we have
[x] + [x + 8] + [x + 16] = 888
or x + x + 8+ x + 16 = 888
or 3x + 24 = 888
Transposing 24 to RHS, we have
3x = 888 – 24 = 864
Dividing both sides by 3, we have
x = 864 + 3 = 288
∴ x + 8 = 288 + 8 = 296
and x + 16 = 288 + 16 = 304
Thus, the required multiples of 8 are: 288, 296 and 304.

Question 8.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let the integers be x, (x + 1) and (x + 2).
∴ According to the condition, we have
or 2x + 3x + 3 + 4x + 8 = 74
or 9x+11 = 63
Dividing both sides by 9, we have
x = 63 ÷ 9 = 7
The required integers are:

Question 9
The ages Of Rahul and Haroon are in the
ratio 5 : 7. Four years later the sum of their ages
will be 56 years. What are their present ages?
Solution:
Since, ages of Rahul and Haroon are in the
ratio of 5 : 7.
Let their present ages are: 5x and 7x.
4 years later: Age of Rahul = (5x + 4) years
Age of Haroon = (7x + 4) years
According to the condition, we have
(5x + 4) + (7x + 4) = 56
or 5x + 4 + 7x + 4 = 56
or + 8 56
Transposing 8 to RHS, we have
12K = 56 – 8 – 48
Dividing both sides by 12, we have x = $$\frac { 48 }{ 12 }$$ = 4
∴Present age of Rahul = 5x 5 x 4 = 20 years
Present age Of Haroon 7x = 7 x 4 = 28 years Question 10.
The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Solution:
Since, [number of boys] : [number of girls] = 7:5
Let the number of boys = lx and the number of girls = 5x
According to the condition, we have
7x = 5x + 8
Transposing 5x to LHS, we have
7x – 5x = 8 or 2x = 8
Dividing both sides by 2,
x = 8 = 2 = 4
∴Number of boys = 7x = 7 x 4 = 28
Number of girls = 5x = 5 x 4 = 20
∴Total class strength = 28 + 20 = 48 students

Question 11.
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let the age of Baichung = x years
His father’s age = (x + 29) years and
His grandfather’s age = (x + 29 + 26)
years = (x + 55) years
According to the condition, we have
[x] + [x + 29] + [x + 55]= 135 years
or x + x + 29 + x + 55 = 135
or 3x + 84 = 135
Transposing 84 to RHS, we have
3x = 135 – 84 or 3x = 51
Dividing both sides by 3, we have
x = 51 + 3 = 17
Now, Baichung’s age = 17 years
His father’s age = 17 + 29 = 46 years
His grandfather’s age = 46 + 26 = 72 years

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi s present age?
Solution:
Let Ravi’s present age = x years
4 times the present age= 4 x x = 4x years
15 years from now, Ravi’s age = (x + 15) years
According to condition, we have
x + 15 = 4x
Transposing 15 and 4x, we have
x – 4x = -15 or -3x = -15
Dividing both sides by (-3), we have
x = (-15) ÷ (-3) = 5
∴ Ravi’s present age = 5 years.

Question 13.
A rational number is such that when you multiply it by $$\frac { 5 }{ 2 }$$ and add $$\frac { 2 }{ 3 }$$ to the product, you 7 get . What is the number?
Solution:
Let the required rational number be x.
∴ According to the condition, we have Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2:3: 5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of:
₹ 100-notes = 2x
₹ 50-notes = 3x
₹ 10-notes = 5x
Value of ₹ 100-notes = 2x x ₹ 100 = ₹ 200x Value of ₹ 50-notes = 3x × ₹ 50 = ₹ 150x Value of ₹ 10 we have 10-notes = 5x × ₹ 10 = ₹ 50x
According to the condition, we have
₹ 200x + ₹ 150x + ₹ 50x = ₹ 4,00,000
or 200x + 150x + 50x = 400000
or 400x = 400000

Dividing both sides by 400, we have
x = 400000 – ÷ 400 = 1000
2x = 2 x 1000 = 2000
3x = 3 x 1000 = 3000
5x = 5 x 1000 = 5000
Thus, the number of
₹ 100-notes = 2000
₹ 50-notes
₹ 10-notes Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number off 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of ₹ 5 coins = x
The number of ₹ 2 coins = 3x
Total number of coins = 160
Number of ₹ 1 coins = 160 – 3x – x = (160 – 4x)
Now, value of:
₹ 5-coins = ₹ 5 x x = ₹ 5x
₹ 2-coins = ₹ 2 x 3x = ₹ 6x
₹ 1-coins = ₹ 1 x (160 – 4x) = ₹ (160 – 4x)
According to the condition, we have
5x + 6x + (160 – 4x) = 300
or 5x – 6x +160 – 4x = 300
or 7x + 160 = 300
Transposing 160 to RHS, we have
7x = 300 – 160 = 140
Dividing both sides by 7, we have
x = $$\frac { 140 }{ 7 }$$ = 20
Number of: ₹ 5-coins = 20
₹ 2-coins = 3 x 20 = 60
₹ 1-coins = 160 – (20 x 4) = 80

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners = x
Number of participants who are not winners = (63 – x)
∴ Prize money given to:
Winners = x × ₹ 100 = ₹ 100x
Non-winner participants = ₹ 25 x (63 – x) = ₹ 25 x 63 – ₹ 25x = ₹ 1575 – ₹ 25x
According to the condition, we have
100x + 1575 – 25x = 3000
or 75x = 3000 – 1575
[Transposing 1575 to RHS]
or 75x = 1425
or x = 1425 – 75 = 19
Thus, the number of winners = 19 