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**Number Line Prime Class 7 Solutions Chapter 13 Linear Equations**

Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 13, Linear Equations. Here students can easily find step by step solutions of all the problems for Linear Equations, Exercise 13A, 13B and 13C Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 13 solutions. Here in this post all the solutions are based on latest Syllabus.

**Linear Equations Exercise 13A Solution :**

**Question no – (1)**

**Solution :**

(a) ∴ 5x = 20

∴ x = 20/5

∴ x = 4

(b) ∴ 2y – 3 = 7

∴ 2y = 7 + 3

∴ y = 8/2

(c) ∴ 2/5x + 7 = 30

∴ 2x/5 = 30 – 7

∴ x = 73 × 5/2

(d) ∴ x + 1/3x = 32

∴ 4x/3 = 32

∴ x = 24

**Question no – (2)**

**Solution :**

(a) Given, p – 10 = 50

∴ 10 taken away from a number is 50

(b) Given, m/3 + 2 = 14

∴ One-third of a number added to 2 gives 14

(c) As per the question, 6x + 3 = 21

∴ 3 added to six times a number give 21.

(d) Given, t/5 = 6

∴ One-fifth of a number is 6

**Question no – (3)**

**Solution :**

∴ Let the smaller number = p

∴ Greater number = 15 + p

And 15 + p + p = 65

∴ 2p = 65 – 15

∴ P = 50/2

∴ p = 25

∴ The smaller number = 25

**Question no – (4)**

**Solution :**

(a) y + 3 = 10

∴ y + 3 – 3 = 10 – __3__

(b) x – 2 = 5

∴ x – 2 + 1 = 5 + __1__

(c) a/2 = 9

∴ a/2 × 2 = 9 × __2__

(d) 3k = 27

∴ 3k ÷ 3 = 27 ÷ __3__

**Linear Equations Exercise 13B Solution :**

**Question no – (1) **

**Solution :**

(a) ∴ P = – 2

∴ LHS = 7p + 5

= 7 (- 2) + 5

= – 14 + 5

= – 9 ≠ RHS

∴ – 2 is not a roof of the given equation.

(b)∴ m = – 3

∴ LHS = 4m -3

= 4 (- 3) – 3

= -12 – 3

= – 15 ≠ – 5

∴ – 3 is not a roof of the given equation.

(c)∴ x = – 2

∴ RHS = 3 – 4 × (- 2)

= 3 + 8

= 11 ≠ LHS

∴ – 2 is not a roof of the given equation.

(d) ∴ T = 18

∴ LHS = T/3

= 18/3

= 6 = RHS

∴ 18 is a roof of the given equation.

**Question no – (2)**

**Solution :**

(a)

Value of x |
LHS |
Conclusion |

1 | 2.1 + 10 = 12 | 12 ≠ 18 |

2 | 4 + 10 = 14 | 14 ≠ 18 |

3 | 6 + 10 = 16 | 16 ≠ 18 |

4 | 8 + 10 = 18 | 18 = 18 |

∴ 4 is the solution

(b) Given, 3(x – 1) = 6

3x – 3 = 6

Value of x |
LHS 3x – 3 |
conclusion |

1 | 3 – 3 = 0 | 0 ≠ 6 |

2 | 6 – 3 = 3 | 3 ≠ 6 |

3 | 9 – 3 = 6 | 6 = 6 |

∴ 3 is the solution

(c) Given, 5a – 12 = 23

Value of a |
LHS 5a – 12 |
Conclusion |

1 | 5 – 12 = – 7 | – 7 ≠ 23 |

2 | 10 – 12 = – 2 | – 2 ≠ 23 |

3 | 15 – 12 = 3 | 3 ≠ 23 |

4 | 20 – 12 = 8 | 8 ≠ 23 |

5 | 25 – 12 = 13 | 13≠ 23 |

6 | 30 – 12 = 18 | 18 ≠ 23 |

7 | 35 – 12 = 23 | 23 = 23 |

∴ 7 is the solution

(d) Given, 5x = – 75

Value of x |
LHS 5x |
Conclusion |

– 1 | – 5 | – 5 ≠ – 75 |

– 5 | – 25 | – 25 ≠ – 75 |

– 10 | – 50 | – 50 ≠ – 75 |

– 13 | – 65 | – 65 ≠ – 75 |

– 15 | – 75 | – 75 = – 75 |

∴ – 15 is the conclusion

**Question no – (3) **

**Solution :**

(a) Given, x + 7 = 10

∴ Subtracting 7

∴ x + 7 – 7 = 10 – 7

∴ x = 3

(b) As per the question, 3x + 2 = 8

∴ Subtracting 2

∴ 3x + 2 – 2 = 8 – 2

∴ 3x = 6

∴ Dividing by 3

∴ x = 2

(c) Given, 5x – 9 = 1

∴ Adding 9

∴ 5x = 1 + 9

∴ Dividing by 5

∴ x = 10/5

∴ x = 5

(d) As per the question, 7(x – 1) = 20

∴ Dividing by 7

∴ x – 12 = 20/7

∴ Adding 1

∴ x = 20/7 + 1

∴ x = 27/7

(e) As per the question, 6x = 54

∴ Dividing by 6

∴ x = 9

(f) As per the question, x- 4/2 = 1

∴ Multiplying by 2

∴ x – 4 = 2 adding by 4

∴ x = 2 + 4

∴ x = 6

**Question no – (4)**

**Solution :**

(a) Given, 6(t – 2) = 18

∴ 6t – 12 = 18

∴ 6t = 12 + 18

∴ t = 30/6

∴ t = 5

(b) Given, 3/2 p – 4 = 20

∴ 3p/2 = 20 + 4

∴ P = 24 × 2/3

∴ P = 16

(c) Given, 6(x + 1) = 7(x – 2)

∴ 6x + 6 = 7x -14

∴ 7x – 6x = 6 + 14

∴ x = 20

(d) As per the question, 2(x + 4) – 10 = 44

∴ 2x + 8 = 44 + 10

∴ 2x = 54 – 8

∴ x = 46/2

∴ x = 23

(e) As per the question, 4 – x/6 = 7

∴ 4 – x = 42

∴ x = 4 – 42

∴ x = – 38

(f) As per the question, x/2 = – x – 2

∴ x = – 2x – 4

∴ 3x = – 4

∴ x = – 4/3

**Question no – (5) **

**Solution :**

(a) As per the question, x + 4 = – 5

∴ x = – 5 – 4

∴ x = – 9

(b) As per the question, 3/2 x = 15

∴ x = 15 × 2/3

∴ x = 10

(c) As per the question, 4/7 = 20/- x

∴ x = 20/- 4 × 7

∴ x = – 35

(d) As per the question, 2x + 4/8 = 18/8

∴ 2x = 9/4 – 2/4

∴ 2x = 7/4

∴ x = 7/8

(e) Given, 4x/9 + 3/2 = 11

∴ 4x/3 = 11 – 3/2

∴ 4x/3 = 19/2

∴ x = 19 × 3/2 × 4

∴ x = 57/8

(f) Given, 5x + 11 = – 6

∴ 5x= – 15

∴ x = – 3

**Linear Equations Exercise 13C Solution :**

**Question no – (1)**

**Solution :**

∴ m = – 5

Adding by 4

∴ m + 4 = – 1

Multiplying by 2

∴ 2m + 8 = – 2

Dividing by 4

∴ m/2 + 2 = – 1/2

**Question no – (2)**

**Solution :**

∴ t = 1/2

Multiplying by 2

∴ 2t = 1

Adding with 4

∴ 2t + 4 = 5

Subtract by 3

∴ 2t + 1 = 2

**Question no – (3)**

**Solution :**

Let Soni’s age = x

∴ 4/5x + 8 32

∴ 4x/5 = 32 – 8

∴ x = 24 × 5/4

∴ x = 30 years

Therefore, Soni’s age 30 years.

**Question no – (4)**

**Solution :**

∴ This is a right – angled is scales triangle

∴ One angle will be – 90°

∴ The another age will = 180 – (90 + 45)

= 180 – 135

= 45°

**Question no – (5)**

**Solution :**

∴ Let Neena has x toffees.

∴ 4x – 2 = 14

∴ 4x = 16

∴ x = 4

Therefore, Neena has 4 toffees.

**Question no – (6)**

**Solution :**

∴ Together score = (100 + 11) = 11

∴ Let, Rajesh’s score – x

∴ Raj’s Score – 2x

And,

∴ 2x + x = 111

∴ 3x = 111

∴ x = 111/3

∴ x = 37

∴ Rajesh’s score = 37 ,

∴ Raj’s score = 2 × 37

= 74

**Question no – (7) **

**Solution :**

∴ Let, breadth is x m.

∴ Length is = (3x + 6) m

∴ Perimeter = 2 × (x + 3x + 6) m

= 2 (4x + 6)

= 4(2x + 3)

∴ 4(2x + 3) = 100

∴ 2x + 3 = 100/4

∴ 2x = 25 – 3

∴ x = 22/2

∴ x = 11

∴ Length = 33 + 6 = 39 m,

∴ Breath is = 11 m

**Question no – (8)**

**Solution :**

∴ Let, Shreya’s present age = x

∴ 2x + 10 = 40

∴ 2x = 30

∴ x = 15

Therefore, Shreya’s present age will be 15 years.

**Next Chapter Solution : **