Chapter 2 Linear Equations in One Variable Exercise 2.2
Question 1: If you subtract 1/2 from a number and multiply the result by 1/2, you will get 1/8. What is the number?Answer:
Let the number be x.
= (x - 1/2) × 1/2 = 1/8
= (x - 1/2) = 1/8 ÷ 1/2
= (x - 1/2) = 1/8 × 2/1
= x - 1/2 = 1/4
= x = 1/4 + 1/2
= x = 3/4
Therefore, the number is 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 breadth of the
pool?
Answer:
Given
Perimeter of the rectangular swimming pool = 154 m
Length is 2 m more than twice its breadth.
Let the breadth of the swimming pool be x metre.
Length of swimming pool = (2x + 2) metre
Perimeter of rectangle = 2(l + b)
= 154 m = 2[(2x + 2) + x]
= 154 m = 2[3x + 2]
= 154 m = 6x + 4
= 154 - 4 = 6x
= 150 = 6x
= x = 150/6
= x = 25
Length = 2x + 2 = 2 × 25 + 2 = 52
Breadth = x = 25
Therefore, the length of swimming pool is 52 m while the breadth of swimming pool is 25 m.
Question
3: The base of an isosceles triangle is 4/3 cm. The perimeter of the
triangle is 4 2/15 cm. What is the length of either of the remaining
sides?
Answer:
Given
Base of isosceles triangle = 4/3 cm
Perimeter of isosceles triangle = 4 2/15 cm (62/15 cm)
Length of 2 remaining sides = ?
Let the sides be a.
= 62/15 = a + a + 4/3
= 62/15 = 2a + 4/3
= 62/15 - 4/3 = 2a
= 42/15 = 2a
= a = 42/15 ÷ 2/1
= a = 42/15 × 1/2
= a = 42/30 = 1 12/30 cm
Therefore, the length of one side of isosceles triangle is 1 12/30 cm.
Question 4: Sum of 2 numbers is 95. If one of them exceeds the other by 15. Find the numbers.
Answer:
Let the smaller number be x.
= x + x + 15 = 95
= 2x + 15 = 95
= 2x = 95 - 15
= 2x = 80
= x = 80/2
= x = 40
x = 40
x + 15 = 40 + 15 = 55
Therefore, the 2 numbers are 40 and 55.
Question 5: Two numbers are in the ratio of 5:3. If they differ by 18, what are the numbers.
Answer:
Let the numbers be 3x and 5x.
= 5x - 3x = 18
= 2x = 18
= x = 18/2
= x = 9
= 5x = 5 × 9 = 45
= 3x = 3 × 9 = 27
Therefore, the 2 number are 27 and 45.
Question 6: Three consecutive integers add up to 51. What are these integers?
Answer:
Let the three consecutive integers be a, a + 1, a + 2.
= a + (a + 1) + (a + 2) = 51
= 3a + 3 = 51
= 3a = 51- 3
= 3a = 48
= a = 48/3
= a = 16
= a = 16
= a + 1 = 16 + 1 = 17
= a + 2 = 16 + 2 = 18
Therefore, the 3 consecutive integers that add up to 51 are 16, 17 and 18.
Question 7: The sum of 3 consecutive multiples of 8 is 888. Find the multiples.
Answer:
Let the 3 consecutive multiples of 8 be x, x + 1, x + 2.
= 8x + 8(x + 1) + 8(x + 2) = 888
= 8x + 8x + 8 + 8x + 16 = 888
= 24x + 24 = 888
= 24x = 888 - 24
= 24x = 864
= x = 864/24
= x = 36
8x = 36 × 8 = 288
8x + 8 = 36 × 8 + 8 = 296
8x + 16 = 36 × 8 + 16 = 304
Therefore, the three consecutive multiples of 8 that sum up to 888 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.
Answer:
Let the three consecutive integers be x, x + 1 and x + 2.
= 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
x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
Question 9: The age of Rahul and Haroon are in the ratio 5:7. Four years later sum of their ages will be 56 years. What are their present ages?
Answer:
Let the three consecutive integers be x, x + 1 and x + 2.
= 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
x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
Question 9: The age of Rahul and Haroon are in the ratio 5:7. Four years later sum of their ages will be 56 years. What are their present ages?
Answer:
|
|
Present Age |
Age after 4 years |
|
Rahul |
5x |
5x + 4 |
|
Haroon |
7x |
7x + 4 |
= 5x + 4 + 7x + 4 = 56
= 12x + 8 = 56
= 12x = 56 - 8
= 12x = 48
= x = 48/12
= x = 4
= 5x = 5 × 4 = 20
= 7x = 7 × 4 = 28
Therefore, the age of Rahul is 20 years while the age of Haroon is 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?
Answer:
Let the number of boys and girls be 7x and 5x.
= 7x = 5x + 8
= 7x - 5x = 8
= 2x = 8
= x = 8/2
= x = 4
7x = 7 × 4 = 28
5x = 5 × 4 = 20
= Total strength = Number of Boys + Number of Girls
= 28 + 20
= 48
Therefore, 48 is the total class strength.
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:
Baichung age = x years
Baichung's father age = x + 29 years
Baichung's grandfather age = x + 29 + 26 years
= x + x + 29 + x + 29 + 26 = 135
= 3x + 84 = 135
= 3x = 135 - 84
= 3x = 51
= x = 51/3
= x = 17
Age
Baichung = x = 17 years
Baichung's father age = 17 + 29 = 46 years
Baichung's grandfather age = 17 + 29 + 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?
Answer:
|
|
Present Age |
Age after 15 years |
|
Ravi |
x |
x + 15 |
= x + 15 = 4x
= 15 = 4x - x
= 15 = 3x
= 15/3 = x
= 5 = x
= x = 5
Therefore, Ravi's present age is 5 years.
Question 13: A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get -7/12. What is the number?
Answer:
Let the rational number be x.
Given
5x/2 + 2/3 = -7/12
= 5x/2 = -712 - 2/3
= 5x/2 = -15/12
= x = -15/12 × 2/5
= x = -30/60
= x = -1/2
Therefore, the rational number is -1/2.
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?
Answer:
|
Denomination |
Number |
Amount |
|
₹100 |
2x |
200x |
|
₹50 |
3x |
150x |
|
₹10 |
5x |
50x |
= 200x + 150x + 50x = 400000
= 400x = 400000
= x = 400000/400
= x = 1000
|
Denomination |
Number |
Amount |
|
₹100 |
2x (2000 notes) |
200x (₹200000) |
|
₹50 |
3x (3000 notes) |
150x (₹150000) |
|
₹10 |
5x (5000 notes) |
50x (₹50000) |
Therefore, she has 2000 notes of ₹100, 3000 notes of ₹50 and 5000 notes of ₹10.
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 of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Answer:
|
Denomination |
Number |
Amount |
|
₹1 |
160 - 4x |
160 - 4x |
|
₹2 |
3x |
6x |
|
₹5 |
x |
5x |
|
|
160 |
₹300 |
= 160 - 4x + 6x + 5x = 300
= 160 + 7x = 300
= 7x = 300 - 160
= 7x = 140
= x = 140/7
= x = 20
|
Denomination |
Number |
Amount |
|
₹1 |
160 - 4x (80 coins) |
160 - 4x (₹80) |
|
₹2 |
3x (60 coins) |
6x (₹120) |
|
₹5 |
x (20 coins) |
5x (₹100) |
|
|
160 |
₹300 |
Therefore, I have 80 coins of ₹1, 60 coins of ₹2 and 20 coins of ₹5.
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.
Answer:
Let the numbers of winner be x.
The number of participants who didn’t win will be 63 - x.
Amount given to the winner = ₹(100 × x) = ₹100x
Amount given to participant who didn’t win = ₹25(63 - x)
= ₹(1575 - 25x)
= 100x + 1575 - 25x = 3000
= 75x = 3000 - 1575
= 75x = 1425
= x = 1425/75
= x = 19
Therefore, the numbers of winners are 19.
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