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NCERT Solutions(Part- 2)- Linear Equations in One Variable Class 8 Notes | EduRev

Mathematics (Maths) Class 8

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Class 8 : NCERT Solutions(Part- 2)- Linear Equations in One Variable Class 8 Notes | EduRev

The document NCERT Solutions(Part- 2)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.
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EXERCISE 2.2

Ques 1. If you subtract 1/2 from a number and multiply the results by 1/2, you get 1/8. What is the number?

Ans: Let the required number be x.

âˆ´ According to the condition, we have

Multiplying both sides by 2

â‡’

â‡’
Adding 1/2 both sides, we have

â‡’

â‡’
âˆ´ The required number = 3/4

Ques 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?

Ans: Perimeter of the rectangular pool = 154 m
âˆ´ Length = 2(breadth) + 2 metres = (2x + 2) metres.
Since, perimeter of a rectangle = 2(length + breadth)

âˆ´ 2[(2x + 2) + x] = 154
Dividing both sides by 2

â‡’ (2x + 2) + x = 154/2 = 77
â‡’ 3x + 2 = 77

Transposing 2 to RHS, we have
3x = 77 â€“ 2 = 75
Dividing both sides by 3, we have

x = 75/3 =  25
Length = 2(25) + 2 = 52 m.

Ques 3. 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?

Ans: Base of the isosceles triangle = 4/3 cm.
Let the length of either equal sides = x cm.

âˆ´ Perimeter of the triangle

But the perimeter of the triangle =

Transposing 4/3 to RHS, we have

â‡’

Dividing both sides by 2, we have

â‡’

Thus, the required length =

Ques 4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Ans: Let the smaller number = x.

The other number = x + 15.
According to the condition, we have

x + [x + 15] = 95
2x + 15 = 95
or  2x = 95 â€“ 15 = 80

Dividing both sides by 2, we have

x = 80/2 = 40
Therefore, The smaller number = 40
and the other number = 40+15 = 65

Ques 5. Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?

Ans: Let the two numbers be 5x and 3x.

According to the condition, we have

5x â€“ 3x = 18
â‡’ 2x = 18

Dividing both sides by 2, we have

2x/2 = 18/2
â‡’ x = 9
âˆ´  5x = 5 Ã— 9 = 45 and 3x = 3 Ã— 9 = 27
Thus, the required numbers are 45 and 27.

Ques 6. Three consecutive integers add up to 51. What are these integers?

Ans: Let 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 â€“ 3 = 48

Dividing both sides by 3, we have

Now x + 1 = 16 + 1 = 17
and x + 2 = 16 + 2 = 18

Thus, the required three consecutive numbers are: 16, 17 and 18.

Ques 7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Ans: 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 â€“ 24 = 864

Dividing both sides by 3, we have

x = 864/3 = 288

âˆ´ x + 8 = 288 + 8 = 296 and x + 16 = 288 + 16 = 304

Thus, the required multiples of 8 are: 288, 296 and 304.

Ques 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.

Ans: Let the integers be x, (x + 1) and (x + 2).
âˆ´ According to the condition, we have

2 Ã— (x) + 3 Ã— (x + 1) + 4 Ã— (x + 2) = 74
â‡’ 2x + 3x + 3 + 4x + 8 = 74
â‡’ 9x + 11 = 74
â‡’ 9x = 74 â€“ 11 = 63

Dividing both sides by 9, we have

x = 63/9 = 7
âˆ´ The required integers are: 7, (7 + 1) and (7 + 2) or 7, 8, and 9.

Ques 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?

Ans: Since, ages of Rahul and Haroon are in the ratio of 5 : 7.
Let their present ages are: 5x and 7x.
âˆ´ 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  12x + 8 = 56

Transposing 8 to RHS, we have

12x = 56 â€“ 8 = 48

Dividing both sides by 12, we have
x = 48/12 = 4
âˆ´ Present age of Rahul = 5x = 5 Ã— 4 = 20 years

Present age of Haroon = 7x = 7 Ã— 4 = 28 years

Ques 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?
Ans: Since, [number of boys] : [number of girls] = 7 : 5
Let the number of boys = 7x
and the number of girls = 5x
According to the condition, we have
7x = 5x + 8
Transposing 5x to LHS, we have

7x â€“ 5x = 8

or  2x = 8

Dividing both sides by 2,

x = 8/2 = 4

âˆ´ Number of boys = 7x = 7 Ã— 4 = 28
Number of girls = 5x = 5 Ã— 4 = 20

âˆ´ Total class strength = 28 + 20 = 48 students.

Ques 11. Baichungâ€™s father is 26 years younger than Baichungâ€™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?
Ans: Let the age of Baichung = x years
âˆ´  His fatherâ€™s age = (x + 29) years
â‡’ His grandfather â€™s age = (x + 29 + 26) = (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 â€“ 84

or 3x = 51

Dividing both sides by 3, we have
x = 51/3 = 17

Now, Baichungâ€™s age = 17 years

His fatherâ€™s age = 17 + 29 = 46 years

His grandfather â€™s age = 46 + 26 = 72 years.

Ques 12. Fifteen years from now Raviâ€™s age will be four times his present age. What is Raviâ€™s present age?

Ans: Let Raviâ€™s present age = x years
4 times the present age = 4 Ã— x = 4x years
15 years from now, Raviâ€™s age = (x + 15) years

According to condition, we have

x + 15 = 4x

Transposing 15 and 4x, we have

x â€“ 4x = â€“15
or  â€“3x = â€“15

Dividing both sides by (â€“3), we have

x = (â€“15)/(â€“3) = 5

âˆ´ Raviâ€™s present age = 5 years.

Ques 13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product,you get . What is the number?

Ans: Let the required rational number be x.

âˆ´ According to the condition, we have

â‡’
Transposing 2/3 to RHS, we have

Dividing both sides by 5/2, we have

âˆ´ The required integer

Ques 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?

Ans: Let the number of:
Rs 100-notes = 2x
Rs 50-notes = 3x
Rs 10-notes = 5x
âˆ´ Value of Rs 100-notes = 2x Ã— Rs 100 = Rs 200x
Value of Rs 50-notes = 3x Ã— Rs 50 = Rs 150x
Value of Rs 10 we have 10-notes = 5x Ã— Rs 10 = Rs 50x

According to the condition, we have
Rs 200x + Rs 150x + Rs 50x = Rs 4,00,000
or 200x + 150x + 50x = 400000
or 400x = 400000

Dividing both sides by 400, we have

x = 400000/400 = 1000
Therefore, 2x = 2000 = Rs 100-notes
3x = 3000 = Rs-50 notes
5x = 50000 = Rs-10 notes

Ques 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?
Ans:
Let the number of Rs 5 coins = x
âˆ´  The number of Rs 2 coins = 3x
âˆµ Total number of coins = 160
âˆ´  Number of Re 1 coins = 160 â€“ 3x â€“ x
= (160 â€“ 4x)

Now, value of:
Rs 5 coins = Rs 5 Ã— x = Rs 5x
Rs 2 coins = Rs 2 Ã— 3x = Rs 6x
Rs 1 coins = Rs 1 Ã— (160 â€“ 4x) = Rs (160 â€“ 4x)

According to the condition, we have

5x + 6x + (160 â€“ 4x) = 300
or 5x + 6x + 160 â€“ 4x = 300
or 7x + 160 = 300

Transposing 160 to RHS, we have
7x = 300 â€“ 160 = 140

Dividing both sides by 7, we have

x = 140/7 = 20
âˆ´ Number of Rs 5-coins = 20
Rs 2-coins = 3 Ã— 20 = 60
Rs 1-coins = 160 â€“ (20 Ã— 4) = 80

Ques 16. The organizers 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.

Ans: Let the number of winners = x
âˆ´ Number of participants who are not winners = (63 â€“ x)
âˆ´ Prize money given to: Winners = x Ã— Rs 100 = Rs 100x
Non-winner participants
= Rs 25 Ã— (63 â€“ x)
= Rs 25 Ã— 63 â€“ Rs 25x
= Rs 1575 â€“ Rs 25x

According to the condition, we have

100x + 1575 â€“ 25x = 3000
Transposing 1575 to RHS
â‡’ 75x = 3000 â€“ 1575

â‡’ 75x = 1425
Dividing both sides by 75

â‡’ x = 1425/75 = 19

Thus, the number of winners = 19

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