**NCERT Solutions for Class 8 Maths Chapter 2- Linear Equations in One Variable**

The NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable are provided here in PDF format which can be downloaded for free. All the solved questions of linear equations in one variable are with respect to the latest NCERT syllabus and guidelines to help students solve each and every practice question present in the book and prepare for the exam.

These solutions also serve as a reference tool for students to do their homework and assignments. The NCERT Solutions for Class 8 contain exercise-wise answers for all the chapters, thus being a very useful study material for the students studying in Class 8. Practicing these solutions helps the students to perform well in the final exam and perform well in the subject. These solutions are prepared on the basis of most updated NCERT syllabus, which covers all the important topics of the respective subjects.

**Solve the following equations.**

**1. x – 2 = 7**

**Solution to:**

x – 2 = 7

x = 7 + 2

x = 9

**2. Y + 3 = 10**

**Solution to:**

y + 3 = 10

y = 10 -3

y = 7

**3. 6 = Z + 2**

**Solution to:**

6 = Z + 2

Z + 2 = 6

Z = 6-2

Z = 4

**4. 3/7 + x = 17/7**

**Solution to:**

**5. 6x = 12**

**Solution to:**

6x = 12

x = 12/6

x = 2

**6. T/5 = 10**

**7. 2x/3 = 18**

**Solution to:**

⇒ 2x = 54

⇒ 2x ÷ 2 = 54 ÷ 2 (dividing both sides by 2)

⇒ x = 27

Thus, x = 27 is the required solution.

**8. 1.6 = Y/15**

**Solution to:**

⇒ 2.40 = y

Thus, y = 2.40 is the required solution.

**9. 7x – 9 = 16**

**Solution to:**

7x – 9 = 16

7x = 16+9

7x = 25

x = 25/7

**10. 14y – 8 = 13**

**Solution to:**

14 years – 8 = 13

14y = 13+8

14y = 21

Y = 21/14

y = 3/2

**11. 17 + 6 p = 9**

**Solution to:**

17 + 6 p = 9

6p = 9 – 17

6p = -8

p = -8/6

p = -4/3

**12. x/3 + 1 = 7/15**

**Solution to:**

**Exercise 2.2 Page: 28**

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

**Solution to:**

Let the number be x.

According to the question

(X – 1/2) × = 1/8

x/2 – = 1/8

x/2 = 1/8 +

x/2 = 1/8 + 2/8

x/2 = (1+ 2)/8

x/2 = 3/8

x = (3/8) × 2

x =

**2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 meters, which is twice its width. What is the length and width of the pool?**

**Solution to:**

let me tell you

Perimeter of rectangular swimming pool = 154 m. Let x be the width of the rectangle

According to the question

Length of rectangle = 2x + 2 We know that,

Perimeter = 2 (Length + Width)

2(2x + 2 + x) = 154 m

2(3x + 2) = 154

3x +2 = 154/2

3x = 77 – 2

3x = 75

⇒ X = 75/3

x = 25 m

Therefore, width = x = 25 cm

Length = 2x + 2

= (2 × 25) + 2

= 50+2

= 52 m

**3. The base of an isosceles triangle is 4/3 cm. is The perimeter of a triangle is NCERT Solutions for Class 8 Maths Chapter 2 Figure 1 cm. What is the length of any of the remaining equal sides?**

**Solution to:**

Base of isosceles triangle = 4/3 cm

Perimeter of triangle =

NCERT Solutions for Class 8 Mathematics Chapter 2 Diagram 2 Diagram cm = 62/15

Let x be the length of the equal sides of the triangle.

According to the question

4/3 + x + x = 62/15 cm

2x = (62/15 – 4/3) cm

⇒ 2x = (62 – 20)/15 cm

⇒ 2x = 42/15 cm

x = (42/30) × (½)

x = 42/30 cm

x = 7/5 cm

One of the remaining equal sides has length 7/5 cm.

**4. The sum of two numbers is 95. If one number is 15 more than the other, find the number.**

Solution to:

Let a number = x.

Then the second number becomes x + 15. According to the question

x + x + 15 = 95

2x + 15 = 95

2x = 95 – 15

⇒ 2x = 80

x = 80/2

x = 40

First number = x = 40

And second number = x + 15 = 40 + 15 = 55

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

**Solution to:**

Let two numbers be 5x and 3x. According to the question

5x – 3x = 18

⇒ 2x = 18

x = 18/2

x = 9

like this,

The numbers are 5x = 5 × 9 = 45

And 3x = 3 × 9 = 27.

**6. The sum of three consecutive integers is 51. What are these integers?**

**Solution to:**

Let x, x+1 and x+2 be three consecutive integers. According to the question

x + (x+1) + (x+2) = 51

3x + 3 = 51

3x = 51 – 3

3x = 48

⇒ x = 48/3

x = 16

Thus are the integers

x = 16

x + 1 = 17

x + 2 = 18

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

**Solution to:**

Let the three consecutive multiples of 8 be 8x, 8(x+1) and 8(x+2). According to the question

8x + 8(x+1) + 8(x+2) = 888

8 (x + x+1 + x+2) = 888 (dividing 8)

8 (3x + 3) = 888

3x + 3 = 888/8

3x + 3 = 111

3x = 111 – 3

3x = 108

⇒ x = 108/3

x = 36

Thus, there are three consecutive multiples of 8:

8x = 8 × 36 = 288

8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296

8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

**8. There are three consecutive integers such that taking them in ascending order and multiplying them by 2, 3 and 4 respectively, their sum is 74. Find this number.**

Solution to:

Let x, x+1 and x+2 be three consecutive integers. According to the question

2x + 3(x+1) + 4(x+2) = 74

2x + 3x +3 + 4x + 8 = 74

9x + 11 = 74

9x = 74 – 11

9x = 63

x = 63/9

x = 7

Thus, the numbers are:

x = 7

x + 1 = 8

x + 2 = 9

**9. Ages of Rahul and Haroon are in the ratio of 5 : 7. After four years, the sum of their ages will be 56 years. What is his current age?**

**Solution to:**

Let us tell that the age of Rahul and Harun is 5 times and 7 times respectively. Four years later,

The ages of Rahul and Harun will be (5x + 4) and (7x + 4) respectively. According to the question

(5x + 4) + (7x + 4) = 56

5x + 4 + 7x + 4 = 56

12x + 8 = 56

12x = 56 – 8

12x = 48

x = 48/12

x = 4

So, present age of Rahul = 5x = 5×4 = 20

And, present age of Aaron = 7x = 7×4 = 28

**10. The ratio of the number of boys and girls in a class is 7 : 5. The number of boys is 8 more than the number of girls. What is the total square power?**

**Solutions to:**

Let the number of boys be 7 times and that of girls 5 times.

according to the question

7x = 5x + 8

⇒ 7x – 5x = 8

2x = 8

x = 8/2

x = 4

Thus, number of boys = 7×4 = 28

And, number of girls = 5×4 = 20

Total number of students = 20+28 = 48

**11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of all the three ages is 135 years. What is the age of each of them?**

**Solutions to:**

Let the age of Baichung’s father be x.

So, age of Baichung’s grandfather = (x+26)

And, Baichung’s age = (x-29). according to the question

x + (x+26) + (x-29) = 135

3x + 26 – 29 = 135

⇒ 3x – 3 = 135

3x = 135 + 3

3x = 138

x = 138/3

x = 46

Baichung’s father’s age = x = 46

Age of Baichung’s grandfather = (x+26) = 46 + 26 = 72

Baichung’s age = (x-29) = 46 – 29 = 17

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

**Solutions to:**

Let the present age of Ravi be x.

Ravi’s age after fifteen years will be x+15 years. according to the question

x + 15 = 4x

⇒ 4x – x = 15

3x = 15

⇒ x = 15/3

x = 5

Hence, the 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?**

Solutions to:

Let x be rational.

according to the question

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 = -½

Hence the rational number is -½.

**14. Lakshmi is a cashier in a bank. They have currency notes in the denominations of ₹100, ₹50 and ₹10 respectively. The number of these notes are in the ratio 2:3:5. Lakshmi has total cash of ₹ 4,00,000. How many notes of each denomination does he have?**

**Solutions to:**

Let the number of ₹100, ₹50 and ₹10 notes be 2x, 3x and 5x respectively.

Value of ₹100 = 2x × 100 = 200x

Value of ₹50 = 3x × 50 = 150x

Cost of ₹10 = 5x × 10 = 50x

according to the question

200x + 150x + 50x = 4,00,000

400x = 4,00,000

x = 400000/400

x = 1000

Number of ₹100 notes = 2x = 2000

Number of ₹50 notes = 3x = 3000

Number of ₹10 notes = 5x = 5000

**15. I have a total of ₹ 300 in coins of ₹ 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 do I have?**

**Solutions to:**

Let x be the number of ₹5 coins.

Then,

No, ₹2 coins = 3x

And, number of ₹1 coins = (160 – 4x) Now,

Cost of ₹5 coins = x × 5 = 5x

Cost of ₹2 coins = 3x × 2 = 6x

Cost of ₹1 coins = (160 – 4x) × 1 = (160 – 4x)

according to the question

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 organizers of an essay competition decide that the winner of the competition gets a prize of ₹ 100 and the 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**.

**Solutions to**:

Let the number of winners be x.

Then, number of winning participants = 63 – x

Total amount given to the winner = x × 100 = 100x

Total amount given to the winning participant = 25 × (63-x)

according to the question

100x + 25 × (63-x) = 3,000

100x + 1575 – 25x = 3,000

75x = 3,000 – 1575

75x = 1425

⇒ X = 1425/75

x = 19

Hence the number of winners is 19.

**Exercise 2.3 Page: 30**

**Solve the following equations and check your results.**

**1. 3x = 2x + 18**

**Solutions to:**

3x = 2x + 18

⇒ 3x – 2x = 18

x = 18

Substituting the value of x in the RHS and LHS, we get, 3 × 18 = (2 × 18) + 18

54 = 54

LHS = RHS

**2. 5t – 3 = 3t – 5**

**Solutions to:**

5t – 3 = 3t – 5

5t – 3t = -5 + 3

2t = -2

t = -1

Substituting 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**

**Solutions to:**

5x + 9 = 5 + 3x

5x – 3x = 5 – 9

⇒ 2x = -4

x = -2

Substituting 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**

**Solutions to:**

4z + 3 = 6 + 2z

4z – 2z = 6 – 3

⇒ 2z = 3

⇒ Z = 3/2

Substituting the value of z in the 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**

**Solutions to:**

2x – 1 = 14 – x

⇒ 2x + x = 14 + 1

3x = 15

x = 5

Substituting the value of x in the RHS and LHS, we get (2×5) – 1 = 14 – 5

10 – 1 = 9

9 = 9

LHS = RHS

**6. 8x + 4 = 3(x – 1) + 7**

**Solutions to:**

8x + 4 = 3(x – 1) + 7

⇒ 8x + 4 = 3x – 3 + 7

⇒ 8x + 4 = 3x + 4

⇒ 8x – 3x = 4 – 4

5x = 0

x = 0

Substituting the value of x in the 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)**

**Solutions to:**

x = 4/5 (x + 10)

x = 4x/5 + 40/5

⇒ x – (4x/5) = 8

(5x – 4x)/5 = 8

x = 8 × 5

x = 40

Substituting the value of x in the 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**

**Solutions to:**

2x/3 + 1 = 7x/15 + 3

⇒ 2x/3 – 7x/15 = 3 – 1

(10x – 7x)/15 = 2

3x = 2 × 15

3x = 30

⇒ X = 30/3

x = 10

Substituting the value of x in the RHS and LHS, we get,

**9. 2y + 5/3 = 26/3 – y**

**Solutions to:**

2y + 5/3 = 26/3 – y

2y + y = 26/3 – 5/3

3y = (26 – 5)/3

3y = 21/3

3y = 7

y = 7/3

Substituting the value of y in the RHS and LHS, we get,

(2 × 7/3) + 5/3 = 26/3 – 7/3

⇒ 14/3 + 5/3 = 26/3 – 7/3

(14 + 5)/3 = (26 – 7)/3

⇒ 19/3 = 19/3

LHS = RHS

**10. 3m = 5m – 8/5**

**Solutions to:**

3m = 5m – 8/5

⇒ 5m – 3m = 8/5

2m = 8/5

⇒ 2m × 5 = 8

10 meters = 8

M = 8/10

M = 4/5

Substituting 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

**Exercise 2.4 Page: 31**

**1. Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. Now the result obtained is 3 times more than what he thought. what is the number?**

**Solutions to:**

Let the number be x,

according to the question

(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 new number becomes double of the other new number. What are the numbers?**

**Solutions to:**

Let x be a positive number, then the other number will be 5x. according to the question

5x + 21 = 2(x + 21)

⇒ 5x + 21 = 2x + 42

5x – 2x = 42 – 21

⇒ 3x = 21

x = 7

a number = x = 7

Second number = 5x = 5×7 = 35. There are two numbers 7 and 35.

**3. The 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 27 more than the original number. What is a two digit number?**

**Solution:**

Let the tens place digit be x, then the units place digit will be (9-x).

Original two digit number = 10x + (9-x)

After interchanging the digits, the new number = 10(9-x) + x

As per the question,

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

So the number is 36.

**4. One of the two digits of a two digit number is thrice the other digit. If you add the number obtained by replacing the digits of this two digit number to the original number, you get 88. What is the original number?**

**Solution:**

Let the digit in tens place place be x, then the digit in units place will be 3x.

Original two digit number = 10x + 3x

After changing the digits, the new number = 30x + x

As per the question,

(30x + x) + (10x + 3x) = 88

31x + 13x = 88

44x = 88

x = 2

Original number = 10x + 3x = 13x = 13×2 = 26

**5. The present age of Shobo’s mother is six times Shobo’s present age. Five years from now Shobo’s age will be one-third of his mother’s present age. What is his present age?**

**Solution:**

Let Shobo’s present age be x, then his mother’s age will be 6x.

Shobo’s age after 5 years = x + 5

As per the question,

(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. A narrow rectangular plot of land is reserved for a school in Mahuli village. The ratio of the length and breadth of the plot is 11:4. For fencing the plot, the Gram Panchayat will have to spend ₹ 75000 at the rate of ₹ 100 per meter. What are the dimensions of the plot?**

**Solution:**

Let the length of the rectangular plot be 11x and breadth 4x.

Fencing rate per meter = ₹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)

As per the question,

(30x × 100) = 75000

3000x = 75000

x = 75000/3000

x = 25

Length of plot = 11x = 11 × 25 = 275 m

Width of plot = 4 × 25 = 100 m.

**7. Hassan buys two types of textile materials for school uniforms; Shirt cloth costing ₹ 50 per meter and trouser cloth costing ₹ 90 per metre. For every 3 meters of shirt fabric, he buys 2 meters of trousers. He sells the material at 12% and 10% profit respectively. Their total sales are ₹36,600. How much trouser material did he buy?**

**Solution:**

Suppose he bought 2x meter trouser cloth and 3x meter shirt cloth

Selling price per meter of shirt cloth = ₹ 50 + 50 × (12/100) = ₹ 56

Selling price per meter of trouser cloth = ₹ 90 + 90 × (10/100) = ₹ 99

Total amount of sale = ₹36,600

As per the question,

(2x × 99) + (3x × 56) = 36600

198x + 168x = 36600

366x = 36600

x = 36600/366

x = 100

Total material of pants purchased by him = 2x = 2 × 100 = 200 m.

**8. Half of the herd of deer are grazing in the field, and three-fourths of the remaining are playing nearby. Rest 9 are drinking pond water. Find the number of deer in the herd.**

**Solution:**

Let the total number of deer be x.

Deer grazing in the field = x/2

Deer playing nearby = x/2 × = 3x/8

deer drinking water = 9

As per the question,

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 as old as his granddaughter. He is also 54 years older than her. Find their present age.**

**Solution:**

Let the age of granddaughter be x and age of grandfather be 10x.

Also he is 54 years older than her.

According to the question 10x = x + 54

10x – x = 54

9x = 54

x = 6

Grandfather’s age = 10x = 10×6 = 60 years.

Granddaughter’s age = x = 6 years.

**10. Aman’s age is three times that of his son. Ten years ago, he was five times the age of his son. Find their present age.**

**Solution:**

Let the age of Aman’s son be x, then the age of Aman will be 3x.

As per the question,

5(x – 10) = 3x – 10

5x – 50 = 3x – 10

5x – 3x = -10 + 50

2x = 40

x = 20

Age of Aman’s son = x = 20 years

Age of Aman = 3x = 3×20 = 60 years

Exercise 2.5 Page: 33

Solve the following linear equations.

**1. x/2 – 1/5 = x/3 +**

**Solution:**

x/2 – 1/5 = x/3 +

x/2 – x/3 = + 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**

**Solution:**

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**

**Solution:**

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**

**Solution:**

(X – 5)/3 = (X – 3)/5

5(x-5) = 3(x-3)

5x-25 = 3x-9

5x – 3x = -9+25

2x = 16

x = 8

**5. (3t – 2)/4 – (2t + 3)/3 = 2/3 – t**

**Solution:**

(3t – 2)/4 – (2t + 3)/3 = 2/3 – t

((3t – 2)/4) × 12 – ((2t + 3)/3) × 12

(3t – 2) × 3 – (2t + 3) × 4 = 2 × 4 – 12t

9t – 6 – 8t – 12 = 8 – 12t

9t – 6 – 8t – 12 = 8 – 12t

t – 18 = 8 – 12t

t + 12t = 8 + 18

13t = 26

t = 2

**6. m – (m – 1)/2 = 1 – (m – 2)/3**

**Solution:**

m – (m – 1)/2 = 1 – (m – 2)/3

m – m / 2 – 1/2 = 1 – (m / 3 – 2/3)

m – m/2 + = 1 – m/3 + 2/3

m – m/2 + m/3 = 1 + 2/3 –

m/2 + m/3 = + 2/3

(3m + 2m)/6 = (3 + 4)/6

5m/6 = 7/6

m = 7/6 × 6/5

m = 7/5

Simplify and solve the following linear equations.

**7. 3 (t – 3) = 5 (2t + 1)**

**Solution:**

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**

**Solution:**

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**

**Solution:**

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)**

**Solution:**

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 = 3/5

f = 0.6

**Exercise 2.6 page: 35**

**Solve the following equations.**

**1. (8x – 3)/3x = 2**

**Solution:**

**2. 9x/(7 – 6x) = 15**

**Solution:**

**3. z/(z + 15) = 4/9**

**Solution:**

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**

**Solution:**

**5. (7y + 4)/(y + 2) = -4/3**

**Solution:**

**6. The ratio of the ages of Hari and Harry is 5:7. Four years from now, the ratio of their ages will be 3:4 Find their present age.**

**Solution:**

**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.**

**Solution:**

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