 # RS Aggarwal Solutions for Class 6 Maths Chapter 9 Linear Equations in One Variable

## Class 6 RS Aggarwal Chapter 9 – Linear Equations in One Variable

RS Aggarwal Solutions for Class 6 Maths Chapter 9 are provided here. Students are advised to go through the RS Aggarwal Solutions for Class 6 Maths to prepare well for the exam and to gain high marks. Students need to practice RS Aggarwal Solutions diligently to score high marks. By going through RS Aggarwal Solutions, students will clearly understand the chapter. Practising textbook questions will help the students in boosting their self- confidence and to understand the topics discussed in this chapter in detail.

Students are advised to go through the RS Aggarwal Solutions for Class 6 Chapter 9 which have been solved by the BYJU’S experts in pdf format. Download pdf of Class 6 Chapter 9 in their respective links.

## Download PDF of RS Aggarwal Solutions for Class 6 Chapter 9 Linear Equations in One Variable           ## Exercise 9A PaGE no: 138

1. Write each of the following statements as an equation:

(i) 5 times a number equals 40.

(ii) A number increased by 8 equals 15.

(iii) 25 exceeds a number by 7.

(iv) A number exceeds 5 by 3.

(v) 5 subtracted from thrice a number is 16.

(vi) If 12 is subtracted from a number, the result is 24.

(vii) Twice a number subtracted from 19 is 11.

(viii) A number divided by 8 gives 7.

(ix) 3 less than 4 times a number is 17.

(x) 6 times a number is 5 more than the number.

Solution

(i) Let required number be x

5 times a number = 5x

∴ 5 times a number equals 40 can be written as 5x = 40

(ii) Let the number be x

A number increased by 8 = x + 8

∴ A number increased by 8 equals 15 can be written as x + 8 = 15

(iii) Let the number be x

25 exceeds a number = 25 – x

∴ 25 exceeds a number by 7 can be written as 25 – x = 7

(iv) Let the required number be x

A number exceeds 5 = x – 5

∴ A number exceeds 5 by 3 can be written as x – 5 = 3

(v) Let the required number be x

Thrice a number = 3x

5 subtracted from thrice a number = 3x – 5

∴ 5 subtracted from thrice a number is 16 can be written as 3x – 5 = 16

(vi) Let the number be x

12 subtracted from a number = x – 12

∴ If 12 is subtracted from a number, the result is 24 can be written as x – 12 = 24

(vii) Let the number be x

Twice a number = 2x

Twice a number subtracted from 19 = 19 – 2x

∴ Twice a number subtracted from 19 is 11 can be written as 19 – 2x = 11

(viii) Let the number be x

A number divided by 8 = x / 8

∴ A number divided by 8 gives can be written as x / 8 = 7

(ix) Let he number be x

4 times a number = 4x

3 less than 4 times a number = 4x – 3

∴ 3 less than 4 times a number is 17 can be written as 4x – 3 = 17

(x) Let the number be x

6 times a number = 6x

5 more than the number = x + 5

∴ 6 times a number is 5 more than the number can be written as 6x = x + 5

2. Write a statement for each of the equations, give below:

(i) x – 7 = 14

(ii) 2y = 18

(iii) 11 + 3x = 17

(iv) 2x – 3 = 13

(v) 12y – 30 = 6

(vi) 2z / 3 = 8

Solutions

(i)The statement of equation x – 7 = 14 can be written as 7 less from the number x is 14

(ii) The statement of equation 2y = 18 can be written as twice a number y is 18

(iii) The statement of equation 11 + 3x = 17 can be written as 11 increased by thrice a number x is 17

(iv) The statement of equation 2x – 3 = 13 can be written as 3 less from twice the number x is 13

(v) The statement of equation 12y – 30 = 6 can be written as 12 times the number y decreased by 30 is 6

(vi) The statement of equation 2z / 3 = 8 can be written as twice the number z divided by 3 is 8

## Exercise 9B Page no: 143

Solve each of the following equations and verify the answer in each case:

1. x+ 5 = 12

Solution

Given x + 5 = 12

Subtracting -5 from both sides

x + 5 – 5 = 12 – 5

x = 7

Check

Substituting x = 7 in equation x + 5 = 12

We get

7 + 5 = 12

12 = 12

LHS = RHS

∴ LHS = RHS, when x = 7

2. x + 3 = -2

Solution

Given

x + 3 = – 2

Subtracting -3 from both sides

x + 3 – 3 = -2 – 3

x = -5

Check

Substituting x = -5 in equation x + 3 = – 2

We get,

x + 3 = -2

-5 + 3 = -2

-2 = -2

LHS = RHS

∴ LHS = RHS, when x = -5

3. x – 7 = 6

Solution

Given

x – 7 = 6

x – 7 + 7 = 6 + 7

x = 13

Check

Substituting x = 13 in equation x -7 = 6

We get,

x – 7 = 6

13 – 7 = 6

6 = 6

LHS = RHS

∴ LHS = RHS, when x = 13

4. x – 2 = -5

Solution

Given

x – 2 = -5

x – 2 + 2 = -5 + 2

x = -3

Check

Substituting x = -3 in equation x – 2 = -5

We get,

x – 2 = -5

-3 – 2 = -5

-5 = -5

LHS = RHS

∴ LHS = RHS, when x = -3

5. 3x – 5 = 13

Solution

Given

3x – 5 = 13

3x – 5 + 5 = 13 + 5

3x = 18

x = 18 / 3

x = 6

Check

Substituting x = 6 in equation 3x – 5 = 13

We get,

3x – 5 = 13

3 (6) – 5 = 13

3 × 6 – 5 = 13

18 – 5 = 13

13 = 13

LHS = RHS

∴ LHS = RHS, when x = 6

6. 4x + 7 = 15

Solution

Given

4x + 7 = 15

Subtracting 7 from both sides

4x + 7 – 7 = 15 – 7

4x = 8

x = 8 / 4

x = 2

Check

Substituting x = 2 in equation 4x + 7 = 15

We get,

4x + 7 = 15

4 (2) + 7 = 15

4 × 2 + 7 = 15

8 + 7 = 15

15 = 15

LHS = RHS

∴ LHS = RHS, when x = 2

7. x / 5 = 12

Solution

Given

x / 5 = 12

Multiplying both sides by 5

x / 5 × 5 = 12 × 5

x = 60

Check

Substitute x = 60 in equation x / 5 = 12

60 / 5 = 12

12 = 12

LHS = RHS

∴ LHS = RHS, when x = 60

8. 3x / 5 = 15

Solution

Given

3x / 5 = 15

Multiplying both sides by 5

3x / 5 × 5 = 15 × 5

3x = 75

x = 75 / 3

x = 25

Check

Substitute x = 25 in equation 3x / 5 = 15

3x / 5 = 15

3 × 25 / 5 = 15

3 × 5 = 15

15 = 15

LHS = RHS

∴ LHS = RHS, when x = 25

9. 5x – 3 = x + 17

Solution

Given

5x – 3 = x + 17

Transposing x to LHS and -3 to RHS

5x – x = 17 + 3

4x = 20

x = 20 / 4

x = 5

Check

Substituting x = 5 in equation 5x – 3 = x + 17

5x -3 = x + 17

5 (5) – 3 = 5 + 17

5 × 5 – 3 = 22

25 – 3 = 22

22 = 22

LHS = RHS

∴ LHS = RHS, when x = 5

10. 2x – 1 / 2 = 3

Solution

Given

2x – 1 / 2 = 3

Adding 1 / 2 to both sides

2x – 1 / 2 + 1 / 2 = 3 + 1 / 2

2x – 0 = (6 +1) / 2 [By taking LCM]

2x = 7 / 2

Dividing both sides by 2

2x / 2 = 7 / 2 × 2

x = 7 / 4

Check

Substituting x = 7 / 4 in equation 2x – 1 / 2 = 3

2x – 1 / 2 = 3

2 (7 / 4) – 1 / 2 = 3

2 × 7 / 4 – 1 / 2 = 3

7 / 2 – 1 / 2 = 3

(7 – 1) / 2 = 3

6 / 2 = 3

3 = 3

LHS = RHS

∴ LHS = RHS, when x = 7 / 4

11. 3(x + 6) = 24

Solution

Given

3(x + 6) = 24

3x + 18 = 24 [removing parentheses]

Subtracting 18 from both sides

3x + 18 – 18 = 24 – 18

3x = 6

x = 6 / 3

x = 2

Check

Substituting x = 2 in equation 3(x + 6) = 24

3(x + 6) = 24

3(2 + 6) = 24

3 (8) = 24

3 × 8 = 24

24 = 24

LHS = RHS

LHS = RHS

∴ LHS = RHS, when x = 2

12. 6x + 5 = 2x + 17

Solution

Given

6x + 5 = 2x + 17

Transposing 2x to LHS and 5 to RHS

6x – 2x = 17 – 5

4x = 12

x = 12 / 4

x = 3

Check

Substituting x = 3 in equation 6x + 5 = 2x + 17

LHS = 6x + 5

= 6 (3) + 5

= 6 × 3 + 5

= 18 + 5

= 23

RHS = 2x + 17

= 2 (3) + 17

= 2 × 3 + 17

= 6 + 17

= 23

LHS = RHS

∴ LHS = RHS, when x = 3

13. x / 4 – 8 = 1

Solution

Given

x / 4 – 8 = 1

x / 4 – 8 + 8 = 1 + 8

x / 4 = 9

Multiplying both sides by 4

x / 4 × 4 = 9 × 4

x = 36

Check

Substituting x = 36 in equation x / 4 – 8 = 1

x / 4 – 8 = 1

36 / 4 – 8 = 1

9 – 8 = 1

1 = 1

LHS = RHS

∴ LHS = RHS, when x = 36

## Exercise 9C PAGE no: 144

1. If 9 is added to certain number, the result is 36. Find the number.

Solution

Let the number be x

9 added to a number = x + 9

Given

x + 9 = 36

x = 36 – 9

x = 27

∴ The number when added to 9 gives 36 is 27

2. If 11 is subtracted from 4 times a number, the result is 89. Find the number.

Solution

Let the number be x

4 times a number = 4x

Given

4x – 11 = 89

4x = 89 + 11

4x = 100

x = 100 / 4

x = 25

3. Find a number which when multiplied by 5 is increased by 80.

Solution

Let the number be x

Multiplied by 5 = 5x

According to the question

5x = x + 80

5x – x = 80

4x = 80

x = 80 / 4

x = 20

∴ A number which when multiplied by 5 is increased by 80 is 20

4. The sum of three consecutive natural numbers is 114. Find the numbers.

Solution

Let the three consecutive natural numbers be x, (x + 1), and (x + 2)

Given

x + (x + 1) + (x + 2) = 114

x + x + 1 + x + 2 = 114

3x + 3 = 114 [subtracting 3 from both sides]

3x + 3 – 3 = 114 – 3

3x = 111

Dividing both sides by 3

3x / 3 = 111

x = 111 / 3

x = 37

x + 1 = 37 + 1

= 38

x + 2 = 37 + 2

= 39

The three consecutive natural numbers are 37, 38 and 39

5. When Raju multiplies certain number by 17 and adds 4 to the product, he gets 225. Find the number.

Solution

Let the number be x

When multiplied by 17 becomes 17x

Given

17x + 4 = 225

Subtracting 4 from both sides

17x + 4 – 4 = 225 – 4

17x = 221

Divide both sides by 17

17x / 17 = 221 / 17

x = 221 / 17

x =13

∴ The number is 13 when Raju multiplies by 17 and adds to the product, he gets 225

6. If the number is tripled and the result is increased by 5, we get 50. Find the number.

Solution

Let x be the number

According to the question, the number is tripled and increased by 5 we get 50

3x + 5 = 50

Subtracting -5 from both sides

3x + 5 – 5 = 50 – 5

3x = 45

Divide 3 from both sides

3x / 3 = 45 / 3

x = 15

∴ 15 is the number when it is tripled and increased by 5 results in 50

7. Find two numbers such that one of them exceeds the other by 18 and their sum is 92.

Solution

Let one of the number be x

Exceeds the other by 18 = x + 18

According to the question

x + (x + 18) = 92

2x + 18 = 92

Subtracting -18 from both sides

2x + 18 – 18 = 92 – 18

2x = 74

Dividing both sides by 2

x = 74 / 2

x = 37

x = 37

(x + 18) = 37 + 18

= 55

∴ The two numbers are 37 and 55

8. One out of two numbers is thrice the other. If their sum is 124, find the numbers.

Solution

Let one number be x

According to the question

x + 3x = 124

4x = 124

Dividing both sides by 4

4x / 4 = 124 / 4

x = 31

x = 31 and 3x = 3 × 31

= 93

∴ Required numbers are 31 and 93

9. Find two numbers such that one of them is five times the other and their difference is 132.

Solution

Let one number be x

The other number is 5x

According to the question

5x – x = 132

4x = 132

Dividing both sides by 4

4x / 4 = 132 / 4

x = 33

x = 33 and 5x = 5 (33)

= 5 × 33

= 165

∴ required two numbers are 33 and 165

10. The sum two consecutive even numbers is 74. Find the numbers.

Solution

Let one of the even number be x

The other consecutive even number (be x + 2)

As per the question

x + (x + 2) = 74

2x + 2 = 74

Subtracting -2 from both sides

2x + 2 – 2 = 74 – 2

2x = 72

Dividing 2 from both sides

2x / 2 = 72 / 2

x = 36

x = 36 and (x + 2) = 36 + 2

= 38

∴ 36 and 38 are the two consecutive even number

11. The sum of three consecutive odd numbers is 21. Find the numbers.

Solution

Let the one of the required odd number be x

The other two consecutive odd numbers be (x + 2) and (x + 4)

As per the question

x + (x + 2) + (x + 4) = 21

2x + 2 + x + 4 = 21

2x + x + 2 + 4 = 21

3x + 6 = 21

Subtracting both sides by -6

3x + 6 – 6 = 21 – 6

3x = 15

Dividing both sides by 3

3x / 3 = 15 / 3

x = 5

x = 5 x + 2 = 5 + 2 = 7 x + 4 = 5 + 4 = 9

∴ 5, 7 and 9 are the three consecutive odd numbers

12. Reena is six year older than her brother Ajay. If sum of their ages is 28 years, what are their present ages.

Solution

Let x years be the present age of Ajay

Reena is 6 years older than Ajay shows (x + 6) years

According to the question

x + (x + 6) = 28

2x + 6 = 28

Subtracting -6 from both sides

2x + 6 – 6 = 28 – 6

2x = 22

Dividing both sides by 2

2x / 2 = 22 / 2

x = 11

x = 11 years and (x + 6) = 11 + 6 = 17 years

∴ Present age of Ajay is 11 years and Reena’s age is 17 years

### RS Aggarwal Solutions for Class 6 Maths Chapter 9 Linear Equations in One Variable

Chapter 9 – Linear Equations in One Variable consists of 3 exercises. RS Aggarwal Solutions are solved in detail for each question in every exercise. Let’s have a look at the topics which are included in this chapter

• Systematic method for solving an equation
• Applications of equations

### Chapter Brief of RS Aggarwal Solutions for Class 6 Maths Chapter 9 – Linear Equations in One Variable

A linear equation is an equation in which the highest power of the variables is 1. Variables mentioned in terms of any alphabets such as a, b, c, x, y. There are four rules to solve an equation

(i) We can add the same number to both sides of an equation

(ii) We can subtract the same number from both sides of an equation

(iii) We can multiply both sides of an equation by the same non zero number

(iv) We can divide both sides of an equation by the same non zero number

We can solve linear equations easily by using these rules. They are used for comparing rates of pay, budgeting, making predictions.