RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions 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 \(\frac{1}{8}\) What is the number?
Answer:
Let the required number be x,
∴ (x - \(\frac{1}{2}\)) × \(\frac{1}{2}=\frac{1}{8}\)
or x - \(\frac{1}{2}=\frac{1}{8} \div \frac{1}{2}\)
or x - \(\frac{1}{2}=\frac{1}{8} \times \frac{2}{1}\)
or x - \(\frac{1}{2}=\frac{1}{4}\)
or x = \(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)
∴ \(\frac{3}{4}\) is the required number. 

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?
Answer:
Let x be the breadth of swimming pool, then length
= (2x + 2) meter Perimeter = 2(Length + Breadth)
= 2(2x + 2 + x)
= 4x + 4 + 2x = 6x + 4

But, according to question, perimeter is 154 meter.
∴ 6x + 4 = 154
or 6x= 154 - 4
or 6x= 150
or x = \(\frac{150}{6}\) = 25
Length = (2 × 25 + 2) meter = 52 meter
and Breath = 25 meter

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?
Answer:

Let the length of the side of isosceles triangle by x cm, then perimeter = AB + BC + CA
= (x + \(\frac{4}{3}\) + x)cm
= (2x + \(\frac{4}{3}\))cm
But, According to question perimeter
= 4\(\frac{2}{15}\)cm = \(\frac{62}{15}\)cm
∴ 2x + \(\frac{4}{3}=\frac{62}{15}\).
or (2x + \(\frac{4}{3}\)) × 15 = \(\frac{62}{15}\) × 15
(Multiplying both sides by 15)
or 30x + 20 = 62
or 30x = 62 - 20
or 30x = 42
or x = \(\frac{62}{15}\)
Hence required length of side = 1\(\frac{2}{5}\)cm

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer:
Let one number be x, then other number is x + 15
According to question,
x + (x + 15) = 95
or 2x + 15= 95
or 2x + 15 - 15 = 95 - 15
(Subtracting 15 from both sides)
or 2x = 80
or \(\frac{2 x}{2}=\frac{80}{2}\)
(Dividing both sides by 2)
or x =40
Hence required numbers are 40 and (40 + 15) = 55

Question 5.
Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer:
Let the numbers be 5x and 3x.
According to question,
5x - 3x = 18
or 2x = 18
or x = \(\frac{18}{2}\) = 9
Using x = 9,
First Number 5x = 5 × 9 = 45
Second Number 3x = 3 × 9 = 27

Question 6.
Three consecutive integers add up to 51. What are these integers?
Answer:
Let three numbers be x, (x + 1) and (x + 2).
According to question,
x + (x + 1) + (x + 2) = 51
or 3x + 3 = 51
or 3x = 51 - 3 = 48
or x = \(\frac{48}{3}\) = 16
Hence, multiples axe 8 × 36, 8 × (36 + 1) and 8 × (36 + 2) i.e. 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 respec¬tively, they add up to 74. Find these numbers.
Answer:
Let three consecutive integers be x, (x + 1) and (x + 2).
According to question,
2 × x + 3 × (x + 1) + 4 × (x + 2) = 74
or 2x + 3x + 3 + 4x + 8 = 74
or 9x + 11 = 74
or 9x = 74 - 11
or 9x = 63
x = \(\frac{63}{9}\) = 7
Hence, the required numbers are 7, 8 and 9.

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?
Answer:
Let x ages of Rahul and Haroon be 5x and 7x respectively.
After 4 years - Age of Rahul = 5x + 4
and age of Haroon = lx + 4

According to question,
(5x + 4) + (7x+ 4)= 56
or 12x + 8 = 56
or 12x = 56 - 8
or 12x = 48
or x = \(\frac{48}{12}\) = 4
∴ Age of Rahul = 5 × 4 = 20 year
and Age of Haroon = 7 × 4 = 28 year
 
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?
Answer:
Let the number of boys and girls in class be 7x and 5x respectively. According to question,
Hence, total students in class = 28 + 20 = 48
7x = 5x + 8
7x - 5x = 8
2x = 8
or x = 4
∴ Number of boys = 7 × 4 = 28
and Number of girls = 5 × 4 = 20
Hence, total students in class = 28 + 20 = 48

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?
Answer:
Let the age of Baichung’s be x years then age of father of Baichung = (x + 29) years
and age of Baichung’s grandfather = x + 29 + 26
= (x + 55) years

According to question,
x + (x + 29) + (x + 55) = 135
or 3x + 84 = 135
or 3x = 135 - 84
or 3x = 51
or x = \(\frac{51}{3}\) = 17
∴ Age of. Baichung = 17 years
Age of. Baichung’s father
= 17 + 29 = 46 years

Age of Baichung’s grandfather
= 17 + 55 = 72 years. 

Question 12.
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Answer:
Let die present age of Ravi be x years.
After 15 years, Ravi’s age = (x + 15) years

According to question, fifteen years from now Ravi’s age will be four times his present age.
∴ 4x= x + 15
or 4x - x = 15
or 3x = 15
or x = \(\frac{15}{3}\) = 5
Hence, the Ravi’s present age is 5 years.

Question 13.
A rational number is such that when you multiply it by \(\frac{5}{2}\) and \(\frac{2}{3}\) add to the product, you get \(\frac{-7}{12}\). What is the number?
Answer:
Let the required number be x.

According to question,
x × \(\frac{5}{2}+\frac{2}{3}=-\frac{7}{12}\)
or \(\frac{5}{2}\)x = \(\frac{-7}{12}-\frac{2}{3}\)
or \(\frac{5}{2}\)x = \(\frac{-7-8}{12}\)
or \(\frac{5}{2}\)x = \(\frac{-15}{12}\)
or x = \(\frac{-15}{12} \times \frac{2}{5}=-\frac{1}{2}\)
Hence, the required number is \(-\frac{1}{2}\)

Question 14.
Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?
Answer:
Let the currency notes of ₹ 100, ₹ 50 and ₹ 10 be 2x, 3x and 5x respectively. Now the amount of 2x notes of ₹ 100
= ₹ (2x × 100)
= ₹ 200x

Amount of 3x notes of ₹ 50 = ₹ (3x × 50)
= ₹ 150x
and amount of 5x notes of ₹ 10 = ₹ (5x × 10)
= ₹ 50x

But die total cash is ₹ 4,00,000
200x + 150x + 50x = 400000
or 400x = 400000
or x = 1000
Hence, number of notes of ₹ 100
= 2x = 2 × 1000 = 2000

number of notes of ₹ 50
= 3x = 3 × 1000 = 3000

and number of notes of ₹ 10
= 5x = 5 × 1000 = 5000

Question 15.
I have a total of Rs 300 in coins of denominations Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer:
Let the number of coins of ₹ 5 be x,
then number of coins of ₹ 2 = 3x
And number of coins of ₹ 1 = 160 - x - 3x
= 160 - 4x

Now, Amount of ₹ 5 coins = ₹ (5 x x)
= ₹ 5x

Amount of ₹ 2 coins = ₹ (2 × 3x)
= ₹ 6x

Amount of ₹ 1 coins = ₹ 1 × (160 - 4x) = ₹ (160 - 4x)
But Total amount = ₹ 300
5x + 6x + (160 - 4x)= 300
or 7x - 300 - 160
or 7x = 140
or x = 20

Hence, Number of coins of ₹ 5
= x = 20
Number of coins of ₹ 2= 3x
= 3 × 20 = 60
and, Number of coins of ₹ 1= 160 - 4x
= 160 - (4 × 20)
= 160 - 80 = 80

Question 16.
The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63. 
Answer:
Let the number of winners be x,
then number of participants who does not win gets a prize = (63 - x)
amount of winners = (100 × x)
= ₹ 100x

amount of participants who does not win gets a prize = ₹ 25(63 - x)
But the total prize money distributed
= ₹ 3,000

According to question
100x + 25(63 - x)= ₹ 3000
or 4x + (63 - x) = 120
(Dividing both sides by 25)
or 3x + 63 = 120
or 3x= 120 - 63
or 3x= 57
or x = \(\frac{57}{3}\) = 19
∴ Number of winners = 19