# 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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