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Linear Equations in One Variable : Exercise 2.2 (Mathematics NCERT Class 8th)


Q.1 If you subtract{1 \over 2}from a number and multiply the result by {1 \over 2}, you get {1 \over 8}. What is the number?
Sol. Let the number be x.
Given, \left( {x - {1 \over 2}} \right) \times {1 \over 2} = {1 \over 8}
Dividing both the sides by {1 \over 2}, we get,
\left( {x - {1 \over 2}} \right) \times {1 \over 2} \div {1 \over 2} = {1 \over 8} \div {1 \over 2}
\left( {x - {1 \over 2}} \right) \times {1 \over 2} \times {2 \over 1} = {1 \over 8} \times {2 \over 1}
\left( {x - {1 \over 2}} \right) = {2 \over 8}
\left( {x - {1 \over 2}} \right) = {1 \over 4}
Transposing  - {1 \over 2}to RHS, we get,
x = {1 \over 4} + {1 \over 2}
x = {{1 + 2} \over 4}
x = {3 \over 4}

Q.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?
Sol. Let the breadth be x m. Therefore, the length will be (2x + 2) m.
Now, given perimeter of swimming pool = 2(length + breadth) = 154 m.
2 (2x + 2 + x) = 154
2 (3x + 2) = 154
Dividing both the sides by 2, we get,
{{2(3x + 2)} \over 2} = {{154} \over 2}
(3x + 2) = 77
Transposing 2 from LHS to RHS, we get,
3x = 77 – 2
3x = 75
Dividing by 3 on both the sides, we get,
{{3x} \over 3} = {{75} \over 3}
x = 25
Therefore, (2x + 2) = (2×25 + 2) = 52
Hence, breadth and length of the pool are 25 m and 52 m respectively.

Q.3 The base of an isosceles triangle is cm. The perimeter of the triangle is 4 cm. What is the length of either of the remaining equal sides?
Sol. Let the length of equal sides be x cm.
Now, perimeter of triangle = x + x + base = 4{2 \over {15}}
2x + {4 \over 3} = {{62} \over {15}}
Transposing {4 \over 3}from LHS to RHS, we get,
2x = {{62} \over {15}} - {4 \over 3}
2x = {{62 - 4 \times 5} \over {15}}
2x = {{62 - 20} \over {15}}
2x = {{42} \over {15}}
Dividing both the sides by 2, we get,
{{2x} \over 2} = {{42} \over {15}} \times {1 \over 2}
x = {7 \over 5} = 1{2 \over 5}
Therefore, the length of equal sides is 1{2 \over 5} cm.

Q.4 Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Sol. Let one number be x. Thus, the other number will be x + 15.
Given, x + x + 15 = 95
2x + 15 = 95
Transposing 15 from LHS to RHS, we get,
2x = 95 – 15
2x = 80
Dividing both the sides by 2, we get,
{{2x} \over 2} = {{80} \over 2}
x = 40
x + 15 = 40 + 15 = 55
Thus, the numbers are 40 and 55.

Q.5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Sol. Let the first number be 5x and second number be 3x.
Given, 5x 3x = 18
2x = 18
Dividing by 2 on both the sides, we get,
{{2x} \over 2} = {{18} \over 2}
x = 9
Since, first number = 5x = 5 × 9 = 45
Second number = 3x = 3 × 9 = 27
Therefore, the numbers are 45 and 27.

Q.6 Three consecutive integers add up to 51. What are these integers?
Sol. Let the three consecutive integers be x, x + 1 and x + 2.
Given, x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 from LHS to RHS, we get,
3x = 51 – 3
3x = 48
Dividing by 3 on both the sides, we get,
{{3x} \over 3} = {{48} \over 3}
x = 16
Therefore, x + 1 = 17 and x + 2 = 18.
Hence, three consecutive integers are 16, 17 and 18.

Q.7 The sum of three consecutive multiples of 8 is 888. Find the multiples.
Sol. Let the three consecutive multiples of 8 be 8a, 8(a + 1) and 8(a + 2).
Given, 8a + 8(a + 1) + 8(a + 2) = 888
8(a + a + 1 + a + 2) = 888
8(3a + 3) = 888
Dividing by 8 on both the sides, we get,
{{8(3a + 3)} \over 8} = {{888} \over 8}
3a + 3 = 111
Transposing 3 from LHS to RHS, we get,
3a = 111 – 3
3a = 108
Dividing by 3 on both the sides, we get,
{{3a} \over 3} = {{108} \over 3}
a = 36
Therefore, 8a = 8 × 36 = 288,
8(a + 1) = 8(36 + 1) = 8 × 37 = 296
8(a + 2) = 8(36 + 2) = 8 × 38 = 304
Thus, the required numbers are 288, 296 and 304.

Q.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.
Sol. Let the three consecutive integers be a, a + 1 and a + 2.
Given, 2a + 3(a + 1) + 4(a + 2) = 74
2a + 3a + 3 + 4a + 8 = 74
9a + 11 = 74
Transposing 11 from LHS to RHS, we get,
9a = 74 – 11
9a = 63
Dividing by 9 on both the sides, we get,
{{9a} \over 9} = {{63} \over 3}
a = 7
Therefore, a + 1 = 7 + 1 = 8, a + 2 = 7 + 2 = 9
Therefore, the required numbers are 7, 8 and 9.

Q.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?
Sol. Let the age of Rahul be 5x and the age of Haroon be 7x.
After 4 years, the age of Rahul will be (5x + 4) and that of Haroon will be (7x + 4)
Given, (5x + 4) + (7x + 4) = 56
12x + 8 = 56
Transposing 8 from LHS to RHS, we get,
12x = 56 – 8
12x = 48
Dividing by 12 on both the sides, we get,
{{12x} \over {12}} = {{48} \over {12}}
x = 4
Therefore, age of Rahul = 5x = 5 × 4 = 20 years
Age of Haroon = 7x = 7 × 4 = 28 years

Q.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?
Sol. Let the number of boys be 7x and number of girls be 5x.
Given, 7x – 8 = 5x
Transposing 7x from LHS to RHS, we get,
- 8 = 5x – 7x
- 8 = – 2x
8 = 2x
Dividing both the sides by 2, we get,
{8 \over 2} = {{2x} \over 2}
4 = x
x = 4
Therefore, number of boys = 7x = 7 × 4 = 28
Number of boys = 5x = 5 × 4 = 20
Hence, total class strength = 28 + 20 = 48 students.

Q.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?
Sol. Let Baichung’s father’s age be a years.
Hence, age of Baichung’s father and Baichung’s grandfather will be (a – 29) and (a + 26) respectively.
Given, a + (a – 29) + (a + 26) = 135
3a – 3 = 135
Transposing 3 from LHS to RHS, we get,
3a = 135 + 3
3a = 138
Dividing by 3 on both the sides, we get,
{{3a} \over 3} = {{138} \over 3}
a = 46
Thus, age of Baichung’s father = a years = 46 years
Age of Baichung = (a – 29) = (46 – 29) = 17 years
Age of Baichung’s Grandfather = (a + 26) = (46 + 26) = 72 years.

Q.13 A rational number is such that when you multiply it by {5 \over 2}and add {2 \over 3} to the product, you get - {7 \over {12}}. What is the number?
Sol. Let the number be a.
Given, {5 \over 2}a + {2 \over 3} = - {7 \over {12}}
Transposing from LHS to RHS, we get,
{5 \over 2}a = - {7 \over {12}} - {2 \over 3}
{5 \over 2}a = - {{7 - (2 \times 4)} \over {12}}
{5 \over 2}a = {{ - 15} \over {12}}
Multiplying by on both the sides, we get,
a = {{ - 15} \over {12}} \times {2 \over 5}
a = - {1 \over 2}
Thus, the required rational number is  - {1 \over 2}.

Q.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?
Sol. Let the number of Rs 100 notes, Rs 50 notes and Rs 10 notes be 2x, 3x, and 5x respectively.
Therefore, amount of Rs 100 notes = Rs (100 × 2x) = Rs 200x
Amount of Rs 100 notes = Rs (50 × 3x) = Rs 150x
Amount of Rs 100 notes = Rs (10 × 5x) = Rs 50x
Given, 200x + 150x + 50x = 400000
400x = 400000
Dividing by 400 on both the sides, we get,
x = 1000
Thus, number of Rs 100 notes = 2x = 2 × 1000 = 2000
Number of Rs 50 notes = 3x = 3 × 1000 = 3000
Number of Rs 10 notes = 5x = 5 × 1000 = 5000

Q.15 I have a total of Rs 300 in coins of denomination 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?
Sol. Let the number of Rs 5 coins be a.
Number of Rs 2 coins = 3 x number of Rs 5 coins = 3a
Number of Rs 1 coins = 160 – (number of coins of Rs 5 and of Rs 2) = 160 – (3a + a) = 160 – 4a.
Now, amount of Rs 1 coins = Rs [1 x (160 – 4a)] = Rs (160 – 4a)
Amount of Rs 2 coins = Rs (2 x 3a) = Rs 6a
Amount of Rs 5 coins = Rs (5 x a) = Rs 5a
Given, 160 – 4a + 6a + 5a = 300
160 + 7a = 300
Transposing 160 from LHS to RHS, we get,
7a = 300 – 160
7a = 140
Dividing by 7 on both the sides, we get,
{{7a} \over 7} = {{140} \over 7}
a = 20
Thus, number of Rs 5 coins = a = 20
Number of Rs 2 coins = 3a = 3 x 20 = 60
Number of Rs 1 coins = 160 – 4a = 160 - 4 x 20 = 160 – 80 = 80

Q.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.
Sol. Let the number of winners be y.
Hence, the number of participants who did not win will be 63 – y.
Amount given to the winners = Rs (100 x y) = Rs 100y
Amount given to the participants who did not win = Rs [25 (63 – y)] = Rs (1575 – 25y)
Given, 100y + 1575 – 25y = 3000
Transposing 1575 from LHS to RHS, we get,
75y = 3000 – 1575
75y = 1425
Dividing by 75 on both the sides, we get,
{{75y} \over {75}} = {{1425} \over {75}}
y = 19.
Thus, number of winners are 19.



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