Chapter 2 Linear Equations in One Variable Ex 2.2 Solutions
Question - 1 : - The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is  cm. What is the length of either of the remaining equal sides?
  cm. What is the length of either of the remaining equal sides?
Answer - 1 : - 
Base of isosceles triangle = 4/3 cm
Perimeter of triangle =
  image cm = 62/15
image cm = 62/15Let the length of equal sides of triangle be x.
According to the question,
4/3 + x + x = 62/15 cm
⇒ 2x = (62/15 – 4/3) cm
⇒ 2x = (62 – 20)/15 cm
⇒ 2x = 42/15 cm
⇒ x = (42/30) × (½)
⇒ x = 42/30 cm
⇒ x = 7/5 cm
The length of either of the remaining equal sides are 7/5 cm.
Question - 2 : - I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer - 2 : - 
Let the number of ₹5 coins be x.
Then,
number ₹2 coins = 3x
and, number of ₹1 coins = (160 – 4x) Now,
Value of ₹5 coins = x × 5 = 5x
Value of ₹2 coins = 3x × 2 = 6x
Value of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)
According to the question,
5x + 6x + (160 – 4x) = 300
⇒ 11x + 160 – 4x = 300
⇒ 7x = 140
⇒ x = 140/7
⇒ x = 20
Number of ₹5 coins = x = 20
Number of ₹2 coins = 3x = 60
Number of ₹1 coins = (160 – 4x) = 160 – 80 = 80
Question - 3 : - 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.
Answer - 3 : - 
Let the numbers of winner be x.
Then, the number of participants who didn’t win = 63 – x
Total money given to the winner = x × 100 = 100x
Total money given to participant who didn’t win = 25×(63-x)
According to the question,
100x + 25×(63-x) = 3,000
⇒ 100x + 1575 – 25x = 3,000
⇒ 75x = 3,000 – 1575
⇒ 75x = 1425
⇒ x = 1425/75
⇒ x = 19
Therefore, the numbers of winners are 19.
Question - 4 : - If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8 what is the number?
Answer - 4 : - 
Let the number be x.
According to the question,
(x – 1/2) × ½ = 1/8
x/2 – ¼ = 1/8
x/2 = 1/8 + ¼
x/2 = 1/8 + 2/8
x/2 = (1+ 2)/8
x/2 = 3/8
x = (3/8) × 2
x = ¾
Question - 5 : - 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 - 5 : - 
Given that,
Perimeter of rectangular swimming pool = 154 m Let the breadth of rectangle be = x
According to the question,
Length of the rectangle = 2x + 2 We know that,
Perimeter = 2(length + breadth)
⇒ 2(2x + 2 + x) = 154 m
⇒ 2(3x + 2) = 154
⇒ 3x +2 = 154/2
⇒ 3x = 77 – 2
⇒ 3x = 75
⇒ x = 75/3
⇒ x = 25 m
Therefore, Breadth = x = 25 cm
Length = 2x + 2
= (2 × 25) + 2
= 50 + 2
= 52 m
Question - 6 : - Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer - 6 : - 
Let one of the numbers be= x.
Then, the other number becomes x + 15 According to the question,
x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 95 – 15
⇒ 2x = 80
⇒ x = 80/2
⇒ x = 40
First number = x = 40
And, other number = x + 15 = 40 + 15 = 55
Question - 7 : - Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer - 7 : - 
Let the two numbers be 5x and 3x. According to the question,
5x – 3x = 18
⇒ 2x = 18
⇒ x = 18/2
⇒ x = 19
Thus,
The numbers are 5x = 5 × 9 = 45
And 3x = 3 × 9 = 27.
Question - 8 : - Three consecutive integers add up to 51. What are these integers?
Answer - 8 : - 
Let the three consecutive integers be x, x+1 and x+2. According to the question,
x + (x+1) + (x+2) = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 – 3
⇒ 3x = 48
⇒ x = 48/3
⇒ x = 16
Thus, the integers are
x = 16
x + 1 = 17
x + 2 = 18
Question - 9 : - The sum of three consecutive multiples of 8 is 888. Find the multiples.
Answer - 9 : - 
Let the three consecutive multiples of 8 be 8x, 8(x+1) and 8(x+2). According to the question,
8x + 8(x+1) + 8(x+2) = 888
⇒ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)
⇒ 8 (3x + 3) = 888
⇒ 3x + 3 = 888/8
⇒ 3x + 3 = 111
⇒ 3x = 111 – 3
⇒ 3x = 108
⇒ x = 108/3
⇒ x = 36
Thus, the three consecutive multiples of 8 are:
8x = 8 × 36 = 288
8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296
8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304
Question - 10 : - 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.
Answer - 10 : - 
Let the three consecutive integers are x, x+1 and x+2. According to the question,
2x + 3(x+1) + 4(x+2) = 74
⇒ 2x + 3x +3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 – 11
⇒ 9x = 63
⇒ x = 63/9
⇒ x = 7
Thus, the numbers are:
x = 7
x + 1 = 8
x + 2 = 9