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

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

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 two-digit 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 two-digit number?

Answer

Let the digit at tens place be x then digit at ones place will be (9-x).
Original two digit number = 10x + (9-x)
After interchanging the digits, the new number = 10(9-x) + x
A/q,
10x + (9-x) + 27 = 10(9-x) + x
⇒ 10x + 9 - x + 27 = 90 - 10x + x
⇒ 9x + 36 = 90 - 9x
⇒ 9x + 9x = 90 - 36
⇒ 18x = 54
⇒ x = 3
Original number = 10x + (9-x) = (10×3) + (9-3) = 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 two-digit 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

Part-I (Exercise 2.1 to 2.3)

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