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

Gujarat BoardĀ 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.
ā“According to the condition, we have
(- $$\frac { 1 }{ 2 }$$) Ć $$\frac { 1 }{ 2 }$$ = $$\frac { 1 }{ 8 }$$ Ć 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 = $$\frac { 1 + 2 }{ 4 }$$ = $$\frac { 3 }{ 4 }$$
ā“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
ā“Length = 2(breadth) + 2 metres.
= (2x + 2) metres.
Since, perimeter of a rectangle
ā“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
ā“$$\frac { 4 }{ 3 }$$ + 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 = $$\frac { 42 }{ 15 }$$ Ć $$\frac { 1 }{ 2 }$$ = $$\frac { 21 }{ 15 }$$ = $$\frac { 7 }{ 5 }$$
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.
ā“ 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 Ć· 2 = 40
ā“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
ā“5x = 5 x 9 = 45 and 3x = 3 x 9 = 27
ā“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
ā“ 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).
ā“ 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 Ć· 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
ā“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
ā“Number of boys = 7x = 7 x 4 = 28
Number of girls = 5x = 5 x 4 = 20
ā“Total class strength = 28 + 20 = 48 students

Question 11.
Baichungās father is 26 years younger than Baichungā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 fatherās age = (x + 29) years and
His grandfatherā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, Baichungās age = 17 years
His fatherās age = 17 + 29 = 46 years
His grandfatherās age = 46 + 26 = 72 years

Question 12.
Fifteen years from now Raviās age will be four times his present age. What is Ravi s present age?
Solution:
Let Raviās present age = x years
4 times the present age= 4 x x = 4x years
15 years from now, Raviā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) Ć· (-3) = 5
ā“ Raviā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.
ā“ According to the condition, we have

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations ā¹ 100, ā¹ 50 and ā¹ 10, respectively. The ratio of the number of these notes is 2:3: 5. The total cash with Lakshmi is ā¹ 4,00,000. How many notes of each denomination does she have?
Solution:
Let the number of:
ā¹ 100-notes = 2x
ā¹ 50-notes = 3x
ā¹ 10-notes = 5x
Value of ā¹ 100-notes = 2x x ā¹ 100 = ā¹ 200x Value of ā¹ 50-notes = 3x Ć ā¹ 50 = ā¹ 150x Value of ā¹ 10 we have 10-notes = 5x Ć ā¹ 10 = ā¹ 50x
According to the condition, we have
ā¹ 200x + ā¹ 150x + ā¹ 50x = ā¹ 4,00,000
or 200x + 150x + 50x = 400000
or 400x = 400000

Dividing both sides by 400, we have
x = 400000 – Ć· 400 = 1000
2x = 2 x 1000 = 2000
3x = 3 x 1000 = 3000
5x = 5 x 1000 = 5000
Thus, the number of
ā¹ 100-notes = 2000
ā¹ 50-notes
ā¹ 10-notes

Question 15.
I have a total of ā¹ 300 in coins of denomination ā¹ 1, ā¹ 2 and ā¹ 5. The number of ā¹ 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 ā¹ 5 coins = x
The number of ā¹ 2 coins = 3x
Total number of coins = 160
Number of ā¹ 1 coins = 160 – 3x – x = (160 – 4x)
Now, value of:
ā¹ 5-coins = ā¹ 5 x x = ā¹ 5x
ā¹ 2-coins = ā¹ 2 x 3x = ā¹ 6x
ā¹ 1-coins = ā¹ 1 x (160 – 4x) = ā¹ (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: ā¹ 5-coins = 20
ā¹ 2-coins = 3 x 20 = 60
ā¹ 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 ā¹ 100 and a participant who does not win gets a prize of ā¹ 25. The total prize money distributed is ā¹ 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)
ā“ Prize money given to:
Winners = x Ć ā¹ 100 = ā¹ 100x
Non-winner participants = ā¹ 25 x (63 – x) = ā¹ 25 x 63 – ā¹ 25x = ā¹ 1575 – ā¹ 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