NCERT Solutions for Class 8th: Ch 2 Linear Equations in One Variable Maths Part-I

NCERT Solutions for Class 8th: Ch 2 Linear Equations in One Variable Maths Part-I

Page No: 23

Exercise 2.1

Solve the following equations.
(1) x – 2 = 7           (2) y + 3 = 10         (3) 6 = z + 2
(4) 3/7 + x = 17/7        (5) 6x = 12        (6) t/5 = 10
(7) 2x/3 = 18         (8) 1.6 = y/1.5        (9) 7x – 9 = 16
(10) 14y – 8 = 13        (11) 17 + 6p = 9         (12) x/3 + 1 = 7/15

Answer

(1) x - 2 = 7
⇒ x = 7 + 2
⇒ x = 9

(2)  y + 3 = 10
⇒ y = 10 - 3
⇒ y = 7

(3) 6 = z + 2
⇒ z + 2 = 6
⇒ z = 6 - 2
⇒ z = 4

(4) 3/7 + x = 17/7
⇒ x = 17/7 - 3/7
⇒ x = (17 - 3)/7
⇒ x = 14/7
⇒ x = 2

(5) 6x = 12
⇒ x = 12/6
⇒ x = 2

(6) t/5 = 10
⇒ t = 10 × 5
⇒ t = 50

(7) 2x/3 = 18
⇒ 2x = 18 × 3
⇒ x = 54/2
⇒ x = 27

(8) 1.6 = y/1.5
⇒ y/1.5 = 1.6
⇒ y = 1.6 × 1.5
⇒ y = 2.4

(9) 7x – 9 = 16
⇒ 7x = 16 + 9
⇒ x = 25/7

(10) 14y – 8 = 13
⇒ 14y = 13 + 8
⇒ 14y = 21
⇒ y = 21/14
⇒ y = 3/2

(11) 17 + 6p = 9
⇒ 6p = 9 - 17
⇒ 6p = -8
⇒ p = -8/6
⇒ p = -4/3

(12) x/3 + 1 = 7/15
⇒ x/3 = 7/15 - 1
⇒ x/3 = 7/15 - 15/15
⇒ x/3 = (7-15)/15
⇒ x/3 = -8/15
⇒ x = -8/15 × 3
⇒ x = -8/5

Page No: 28

Exercise 2.2

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

Answer

Let the number be x.
A/q,
(x - 1/2) × 1/2 = 1/8
⇒ x/2 - 1/4 = 1/8
⇒ x/2 = 1/8 + 1/4
⇒ x/2 = 1/8 + 2/8
⇒ x/2 = (1+2)/8
⇒ x = 3/8 × 2
⇒ x = 6/8 = 3/4

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

Given,
Perimeter of rectangular swimming pool = 154 m
Let the breadth of rectangle be x.
A/q,
Length of the rectangle = 2x + 2
Perimeter = 2(length + breadth)
⇒ 2(2x + 2 + x) = 154 m
⇒ 2(3x + 2) = 154
⇒ 3x + 2 = 154/2
⇒ 3x = 77 - 2
⇒ x = 75/3
⇒ x = 25 m
Thus,
Breadth = x = 25 m
Length = 2x + 2 = 50 + 2 = 52 m

3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 62/15 cm. What is the length of either of the remaining equal sides?

Answer

Base of isosceles triangle = 4/3 cm
Perimeter of triangle = 62/15 cm
Let the length of equal sides of triangle be x.
A/q,
4/3 + x + x = 62/15 cm
⇒ 2x = (62/15 - 4/3) cm
⇒ 2x = (62 - 20)/15 cm
⇒ 2x = 42/15 cm
⇒ x = 42/15 × 1/2
⇒ x = 42/30 cm
⇒ x =7/5 cm
The length of either of the remaining equal sides are 7/5 cm.

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

Answer

Let the one of the number be x.
Then, other number will be x + 15
A/q,
x + x + 15 = 95
⇒ 2x + 15 = 95
⇒ 2x = 80
⇒ x = 40
First number = x = 40
Other number = x + 15 = 40 + 15 = 55

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

Answer

Let the two numbers be 5x and 3x.
A/q,
5x - 3x = 18
⇒ 2x = 18
⇒ x = 9
Thus, the numbers are 5x = 45 and 3x = 27.

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

Answer

Let the three consecutive integers are x, x+1 and x+2.
A/q,
x + (x+1) + (x+2) = 51
⇒ 3x + 3 = 51
⇒ 3x = 51 - 3
⇒ 3x = 48
⇒ x = 16
Thus, the integers are x = 16, x+1 = 17 and x+2 = 18

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

Answer

Let the three consecutive multiples of 8 are 8x, 8(x+1) and 8(x+2).
A/q,
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 = 108
⇒ x = 36
Thus, the three consecutive multiples of 8 are 8x = 8 × 36 = 288,
8(x+1) = 8 × (36+1) = 8 × 37 = 296 and
8(x+2) = 8 × (36+2) = 8 × 38 = 304

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 are x, x+1 and x+2.
A/q,
2x + 3(x+1) + 4(x+2) = 74
⇒ 2x + 3x +3 + 4x + 8 = 74
⇒ 9x + 11 = 74
⇒ 9x = 74 - 11
⇒ x = 63/9
⇒ x = 7
Thus, the numbers are x = 7, x+1 = 8 and x+2 = 9

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 the ages of Rahul and Haroon be 5x and 7x.
Four years later their age will be (5x + 4) and (7x + 4) respectively.
A/q,
(5x + 4) + (7x + 4) = 56
⇒ 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 - 8
⇒ x = 48/12
⇒ x = 4
Present age of Rahul = 5x = 5×4 = 20
Present age of Haroon = 7x = 7×4 = 28

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 be 7x and girls be 5x.
A/q,
7x = 5x + 8
⇒ 7x - 5x = 8
⇒ 2x = 8
⇒ x = 4
Number of boys = 7×4 = 28
Number of girls = 5×4 = 20

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 father be x.
Therefor, Age of Baichung’s grandfather = (x+26)
and, Age of Baichung = (x-29)
A/q,
x + (x+26) + (x-29) = 135
⇒ 3x + 26 - 29 = 135
⇒ 3x = 135 + 3
⇒ 3x = 138
⇒ x = 138/3
⇒ x = 46
Age of Baichung’s father = x = 46
Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72
Age of Baichung = (x-29) = 46 - 29 = 17

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Answer

Let the present age of Ravi be x.
Fifteen years later, Ravi age will be x+15 years.
A/q,
x + 15 = 4x
⇒ 4x - x = 15
⇒ 3x = 15
⇒ x = 5
Present age of Ravi = 5 years.

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 be x.
A/q,
x × (5/2) + 2/3 = -7/12
⇒ 5x/2 + 2/3 = -7/12
⇒ 5x/2 = -7/12 - 2/3
⇒ 5x/2 = (-7 - 8)/12
⇒ 5x/2 = -15/12
⇒ 5x/2 = -5/4
⇒ x = (-5/4) × 2/5
⇒ x = -10/20
⇒ x = -1/2
Thus, the rational number is -1/2

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

Let the numbers of notes of  ₹100, ₹50 and ₹10 be 2x, 3x and 5x respectively.
Value of ₹100 = 2x × 100 = 200x
Value of ₹50 = 3x × 50 = 150x
Value of ₹10 = 5x × 10 = 50x
A/q,
200x + 150x + 50x = 4,00,000
⇒ 400x = 4,00,000
⇒ x = 4,00,000/400
⇒ x = 1000
Numbers of ₹100 notes = 2x = 2000
Numbers of ₹50 notes = 3x = 3000
Numbers of ₹10 notes = 5x = 5000

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

Let the number of ₹5 coins be x.
Therefor, 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)
A/q,
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

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.
Therefor, number of participant didn't win = 63 - x
Total money given to the winner = x × 100 = 100x
Total money given to participant didn't win = 25×(63-x)
A/q,
100x + 25×(63-x) = 3,000
⇒ 100x + 1575 - 25x = 3,000
⇒ 75x = 3,000 - 1575
⇒ 75x = 1425
⇒ x = 1425/75
⇒ x = 19
Thus, the numbers of winners are 19.

Page No: 30

Exercise 2.3

Solve the following equations and check your results.
(1) 3x = 2x + 18           (2) 5t – 3 = 3t – 5          (3) 5x + 9 = 5 + 3x
(4) 4z + 3 = 6 + 2z        (5) 2x – 1 = 14 – x       (6) 8x + 4 = 3 (x – 1) + 7
(7) x = 4/5(x + 10)           (8) 2x/3 + 1 = 7x/15 + 3
(9) 2y + 5/3 = 26/3 - y           (10)  3m = 5 m – 8/5

Answer

(1) 3x = 2x + 18
⇒ 3x - 2x = 18
⇒ x = 18
Putting the value of x in RHS and LHS we get,
3 × 18 = (2 × 18)+18
⇒ 54 = 54
⇒ LHS = RHS

(2) 5t – 3 = 3t – 5
⇒ 5t - 3t = -5 + 3
⇒ 2t = -2
⇒ t = -1
Putting the value of t in RHS and LHS we get,
5×(-1) - 3 = 3×(-1) - 5
⇒ -5 - 3 = -3 - 5
⇒ -8 = -8
⇒ LHS = RHS

(3) 5x + 9 = 5 + 3x
⇒ 5x - 3x = 5 - 9
⇒ 2x = -4
⇒ x = -2
Putting the value of x in RHS and LHS we get,
5×(-2) + 9 = 5 + 3×(-2)
⇒ -10 + 9 = 5 + (-6)
⇒ -1 = -1
⇒ LHS = RHS

(4) 4z + 3 = 6 + 2z
⇒ 4z - 2z = 6 - 3
⇒ 2z = 3
⇒ z = 3/2
Putting the value of z in RHS and LHS we get,
(4 × 3/2) + 3 = 6 + (2 × 3/2)
⇒ 6 + 3 = 6 + 3
⇒ 9 = 9
⇒ LHS = RHS

(5) 2x – 1 = 14 – x
⇒ 2x + x = 14 + 1
⇒ 3x = 15
⇒ x = 5
Putting the value of x in RHS and LHS we get,
(2×5) - 1 = 14 - 5
⇒ 10 - 1 = 9
⇒ 9 = 9
⇒ LHS = RHS

(6) 8x + 4 = 3 (x – 1) + 7
⇒ 8x + 4 = 3x – 3 + 7
⇒ 8x + 4 = 3x + 4
⇒ 8x - 3x = 4 - 4
⇒ 5x = 0
⇒ x = 0
Putting the value of x in RHS and LHS we get,
(8×0) + 4 = 3 (0 – 1) + 7
⇒ 0 + 4 = 0 - 3 + 7
⇒ 4 = 4
⇒ LHS = RHS

(7) x = 4/5(x + 10)
⇒ x = 4x/5 + 40/5
⇒ x - 4x/5 = 8
⇒ (5x - 4x)/5 = 8
⇒ x = 8 × 5
⇒ x = 40
Putting the value of x in RHS and LHS we get,
40 = 4/5(40 + 10)
⇒ 40 = 4/5 × 50
⇒ 40 = 200/5
⇒ 40 = 40
⇒ LHS = RHS

(8) 2x/3 + 1 = 7x/15 + 3
2x/3 - 7x/15 = 3 - 1
⇒ (10x - 7x)/15 = 2
⇒ 3x = 2 × 15
⇒ 3x = 30
⇒ x = 10
Putting the value of x in RHS and LHS we get,
(2×10)/3 + 1 = (7×10)/15 + 3
⇒ 20/3 + 1 = 70/15 + 3
⇒ (20 + 3)/3 = (70 + 45)/15
⇒ 23/3 = 115/15
⇒ 23/3 = 23/3
⇒ LHS = RHS

(10) 3m = 5 m – 8/5
⇒ 3m - 5m = -8/5
⇒ -2m = -8/5

⇒-2m × 5 = -8
⇒ -10m = -8
⇒ m = -8/-10
⇒ m = 4/5
Putting the value of m in RHS and LHS we get,
3 × 4/5 = (5 × 4/5) – 8/5
⇒ 12/5 = 4 - 8/5
⇒ 12/5 = (20 - 8)/5
⇒ 12/5 = 12/5
⇒ LHS = RHS

Part-II (Exercise 2.4 to 2.6)

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