NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable PDF Download
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Study Materials for Class 8 Maths Chapter 2 Linear Equations in One Variable 


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 = (715)/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 = (x29)
A/q,
x + (x+26) + (x29) = 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 = (x29) = 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×(63x)
A/q,
100x + 25×(63x) = 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
Page No: 31
Exercise 2.4
1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Answer
Let the number be x.
A/q,
(x  5/2) × 8 = 3x
⇒ 8x  40/2 = 3x
⇒ 8x  3x = 40/2
⇒ 5x = 20
⇒ x = 4
Thus, the number is 4.
2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Answer
Let one of the positive number be x then other number will be 5x.
A/q,
5x + 21 = 2(x + 21)
⇒ 5x + 21 = 2x + 42
⇒ 5x  2x = 42  21
⇒ 3x = 21
⇒ x = 7
One number = x = 7
Other number = 5x = 5×7 = 35
The two numbers are 5 and 35.
3. Sum of the digits of a twodigit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the twodigit number?
Answer
Let the digit at tens place be x then digit at ones place will be (9x).
Original two digit number = 10x + (9x)
After interchanging the digits, the new number = 10(9x) + x
A/q,
10x + (9x) + 27 = 10(9x) + x
⇒ 10x + 9  x + 27 = 90  10x + x
⇒ 9x + 36 = 90  9x
⇒ 9x + 9x = 90  36
⇒ 18x = 54
⇒ x = 3
Original number = 10x + (9x) = (10×3) + (93) = 30 + 6 = 36
Thus, the number is 36.
4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this twodigit number and add the resulting number to the original number, you get 88. What is the original number?
Answer
Let the digit at tens place be x then digit at ones place will be 3x.
Original two digit number = 10x + 3x
After interchanging the digits, the new number = 30x + x
A/q,
(30x + x) + (10x + 3x) = 88
⇒ 31x + 13x = 88
⇒ 44x = 88
⇒ x = 2
Original number = 10x + 3x = 13x = 13×2 = 26
5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Answer
Let the present age of Shobo be x then age of her mother will be 6x.
Shobo's age after 5 years = x + 5
A/q,
(x + 5) = 1/3 × 6x
⇒ x + 5 = 2x
⇒ 2x  x = 5
⇒ x = 5
Present age of Shobo = x = 5 years
Present age of Shobo's mother = 6x = 30 years
6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre it will cost the village panchayat ₹75000 to fence the plot. What are the dimensions of the plot?
Answer
Let the length of the rectangular plot be 11x and breadth be 4x.
Rate of fencing per metre = ₹100
Total cost of fencing = ₹75000
Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 2×15x = 30x
Total amount of fencing = (30x × 100)
A/q,
(30x × 100) = 75000
⇒ 3000x = 75000
⇒ x = 75000/3000
⇒ x = 25
Length of the plot = 11x = 11×25 = 275
Breadth of the plot = 4x = 4×25 = 100
7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹50 per metre and trouser material that costs him ₹90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹36,600. How much trouser material did he buy?
Answer
Let 2x m of trouser material and 3x m of shirt material be bought by him.
Selling price of shirt material per metre = ₹ 50 + 50×(12/100) = ₹ 56
Selling price of trouser material per metre = ₹ 90 + 90×(10/100) = ₹99
Total amount of sale = ₹36,600
A/q,
(2x × 99) + (3x × 56) = 36600
⇒ 198x + 168x = 36600
⇒ 366x = 36600
⇒ x = 36600/366
⇒ x =100
Total trouser material he bought = 2x = 2×100 = 200 m.
Page No: 32
8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Answer
Let the total number of deer be x.
Deer grazing in the field = x/2
Deer playing nearby = 3/4(x  x/2) = 3/4×x/2 = 3x/8
Deer drinking water = 9
A/q,
x/2 + 3x/8 + 9 = x
⇒ (4x + 3x)/8 + 9 = x
⇒ 7x/8 + 9 = x
⇒ x  7x/8 = 9
⇒ (8x  7x)/8 = 9
⇒ x = 9×8
⇒ x = 72
9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Answer
Let the age of granddaughter be x and grandfather be 10x.
Also, he is 54 years older than her.
A/q,
10x = x + 54
⇒ 10x  x = 54
⇒ 9x = 54
⇒ x = 6
Age of grandfather = 10x = 10×6 = 60 years.
Age of granddaughter = x = 6 years.
10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Answer
Let the age of Aman's son be x then age of Aman will be 3x.
A/q,
5(x  10) = 3x  10
⇒ 5x  50 = 3x  10
⇒ 5x  3x = 10 + 50
⇒ 2x = 40
⇒ x = 20
Aman's son age = x = 20 years
Aman age = 3x = 3×20 = 60 years
Page No: 33
Exercise 2.5
Solve the following linear equations.
(1) x/2  1/5 = x/3 + 1/4
(2) n/2  3n/4 + 5n/6 = 21
(3) x + 7  8x/3 = 17/6  5x/2
(4) (x  5)/3 = (x  3)/5
(5) (3t  2)/4  (2t + 3)/3 = 2/3  t
(6) m  (m  1)/2 = 1  (m  2)/3
Answer
(1) x/2  1/5 = x/3 + 1/4
⇒ x/2  x/3 = 1/4 + 1/5
⇒ (3x  2x)/6 = (5 + 4)/20
⇒ 3x  2x = 9/20 × 6
⇒ x = 54/20
⇒ x = 27/10
(2) n/2  3n/4 + 5n/6 = 21
⇒ (6n  9n + 10n)/12 = 21
⇒ 7n/12 = 21
⇒ 7n = 21×12
⇒ n = 252/7
⇒ n = 36
(3) x + 7  8x/3 = 17/6  5x/2
⇒ x  8x/3 + 5x/2 = 17/6  7
⇒ (6x  16x + 15x)/6 = (17  42)/6
⇒ 5x/6 = 25/6
⇒ 5x = 25
⇒ x = 5
(4) (x  5)/3 = (x  3)/5
⇒ x/3  15 = x/5  15
⇒ x/3  x/5 = 15 + 15
⇒ (5x  3x)/15 = 0
⇒ 2x/15 = 0
⇒ x = 0
(5) (3t  2)/4  (2t + 3)/3 = 2/3  t
⇒ 3t/4  1/2  (2t/3 + 1) = 2/3  t
⇒ 3t/4  1/2  2t/3  1 = 2/3  t
⇒ 3t/4  2t/3 + t = 2/3 + 1 + 1/2
⇒ (9t  8t + 12t)/12 = 2/3 + 3/2
⇒ 13t/12 = (4 + 9)/6
⇒ 13t/12 = 13/6
⇒ t = 13/6 × 12/13
⇒ t = 12/6 = 2
(6) m  (m  1)/2 = 1  (m  2)/3
⇒ m  (m/2  1/2) = 1  (m/3  2/3)
⇒ m  m/2 + 1/2 = 1  m/3 + 2/3
⇒ m  m/2 + m/3 = 1 + 2/3  1/2
⇒ m/2 + m/3 = 1/2 + 2/3
⇒ (3m + 2m)/6 = (3 + 4)/6
⇒ 5m/6 = 7/6
⇒ m = 7/6 × 6/5
⇒ m = 7/5
Page No: 34
Simplify and solve the following linear equations.
(7) 3(t – 3) = 5(2t + 1)
(8) 15(y – 4) –2(y – 9) + 5(y + 6) = 0
(9) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
(10) 0.25(4f – 3) = 0.05(10f – 9)
Answer
(7) 3(t – 3) = 5(2t + 1)
⇒ 3t  9 = 10t + 5
⇒ 3t  10t = 5 + 9
⇒ 7t = 14
⇒ t = 14/7
⇒ t = 2
(8) 15(y – 4) –2(y – 9) + 5(y + 6) = 0
⇒ 15y  60 2y + 18 + 5y + 30 = 0
⇒ 15y  2y + 5y = 60  18  30
⇒ 18y = 12
⇒ y = 12/18
⇒ y = 2/3
(9) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
⇒ 15z  21  18z + 22 = 32z  52  17
⇒ 15z  18z  32z = 52  17 + 21  22
⇒ 35z = 70
⇒ z = 70/35
⇒ z = 2
(10) 0.25(4f – 3) = 0.05(10f – 9)
⇒ f  0.75 = 0.5f  0.45
⇒ f  0.5f = 0.45 + 0.75
⇒ 0.5f = 0.30
⇒ f = 0.30/0.5
⇒ f = 30/5 = 6
Page No: 35
Exercise 2.6
Solve the following equations.
(1) (8x  3)/3x = 2
(2) 9x/(7  6x) = 15
(3) z/(z + 15) = 4/9
(4) (3y + 4)/(2  6y) = 2/5
(5) (7y + 4)/(y + 2) = 4/3
Answer
(1) (8x  3)/3x = 2
⇒ 8x/3x  3/3x = 2
⇒ 8/3  1/x = 2
⇒ 8/3  2 = 1/x
⇒ (8  6)/3 = 1/x
⇒ 2/3 = 1/x
⇒ x = 3/2
(2) 9x/(7  6x) = 15
⇒ 9x = 15(7  6x)
⇒ 9x = 105  90x
⇒ 9x + 90x = 105
⇒ 99x = 105
⇒ x = 105/99 = 35/33
(3) z/(z + 15) = 4/9
⇒ z = 4/9(z + 15)
⇒ 9z = 4(z + 15)
⇒ 9z = 4z + 60
⇒ 9z  4z = 60
⇒ 5z = 60
⇒ z = 12
(4) (3y + 4)/(2  6y) = 2/5
⇒ 3y + 4 = 2/5(2  6y)
⇒ 5(3y + 4) = 2(2  6y)
⇒ 15y + 20 = 4 + 12y
⇒ 15y  12y = 4  20
⇒ 3y = 24
⇒ y = 8
(5) (7y + 4)/(y + 2) = 4/3
⇒ 7y + 4 = 4/3(y + 2)
⇒ 3(7y + 4) = 4(y + 2)
⇒ 21y + 12 = 4y  8
⇒ 21y + 4y = 8  12
⇒ 25y = 20
⇒ y = 20/25 = 4/5
6. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Answer
Let the age of Hari be 5x and Hari be 7x.
After 4years,
Age of Hari = 5x + 4
Age of Harry = 7x + 4
A/q,
(5x + 4)/(7x + 4) = 3/4
⇒ 4(5x + 4) = 3(7x + 4)
⇒ 20x + 16 = 21x + 12
⇒ 21x  20x = 16  12
⇒ x = 4
Hari age = 5x = 5×4 = 20 years
Harry age = 7x = 7×4 = 28 years
7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.
Answer
Let the numerator be x then denominator will be (x + 8).
A/q,
(x + 17)/(x + 8  1) = 3/2
⇒ (x + 17)/(x + 7) = 3/2
⇒ 2(x + 17) = 3(x + 7)
⇒ 2x + 34 = 3x + 21
⇒ 34  21 = 3x  2x
⇒ 13 = x
The rational number is x/(x + 8) = 13/21
Go Back To NCERT Solutions for Class 8 Maths
NCERT Solutions for Class 8 Maths Ch 2 Linear Equations in One Variable
Chapter 2 Linear Equations in One Variable Class 8 Maths is very important chapter from the examinations point of view as a lot of questions appear in the question paper from this chapter. The NCERT Solutions for Class 8 Maths presented here detailed and accurate so you can always clear your doubts when in need.
We will discuss those equations with linear expressions in one variable only. Mainly, we are dealing with word problems that are about finding a number. We will first start with solving equations which have linear expressions on one side and numbers on the other side and after that we will solving equations having the variable on both side. We will also reduce equations to simpler form equation reducible to the linear form. These Class 8 Maths NCERT Solutions are very effective in gaining basic concepts in an easy way.
Studyrankers experts have prepare detailed step by step NCERT Solutions of every exercise which you find below.
NCERT Solutions for Class 8 Maths Chapters:
FAQ on Chapter 2 Linear Equations in One Variable
How many exercises in Chapter 2 Linear Equations in One Variable
There are 6 exercise in the Chapter 2 Linear Equations in One Variable that will give good experience and provide opportunities to learn new things.
What are Algebra Equations?
Expressions that involve variables in equality are called algebraic equations. These expressions are formed by two equations involving expressions on left and right with an equality sign in between the two. The expression on the left is called Left Hand Side (L.H.S) and the expression on the right is the Right Hand Side expression (R.H.S).
In the last three months Mr.Sharma lost 5 and a half kg, gained 2 and onefourth kg and then lost 3 and three fourth kg weight. If he now weighs 95 kg, then how much did Mr.Sharma weighs in beginning?
Let x be the Mr.Sharma’s weight in beginning.
According to given in instructions,
x  5.5 + 2.25  3.75 = 95
x  7 = 95
Transposing 7 to R.H.S, we get
x = 102 kg
A number is such that it is as much greater than 65 as it is less than 91. Find the number.
Let the number be x.
Since, we have [The number] – 65 = 91 – [The number]
⇒ x – 65 = 91 – x
⇒ x + x = 91 + 65
⇒ 2x = 156
⇒ x = 156/2 = 78
Thus, the required number is 78.