**NCERT Solutions Class 7 Maths Chapter 4 Simple Equations** – Here are all the NCERT solutions for Class 7 Maths Chapter 4. This solution contains questions, answers, images, explanations of the complete chapter 4 titled Simple Equations of Maths taught in class 7. If you are a student of class 7 who is using NCERT Textbook to study Maths, then you must come across chapter 4 Simple Equations. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations in one place.

## NCERT Solutions Class 7 Maths Chapter 4 Simple Equations

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 7 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 4 Simple Equations , Maths, Class 7.

Class | 7 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 4 |

Chapter Name |
Simple Equations |

### NCERT Solutions Class 7 Maths chapter 4 Simple Equations

Class 7, Maths chapter 4, Simple Equations solutions are given below in PDF format. You can view them online or download PDF file for future use.

Did you find NCERT Solutions Class 7 Maths chapter 4 Simple Equations helpful? If yes, please comment below. Also please like, and share it with your friends!

### NCERT Solutions Class 7 Maths chapter 4 Simple Equations- Video

You can also watch the video solutions of NCERT Class7 Maths chapter 4 Simple Equations here.

If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.

### Download NCERT Solutions Class 7 Maths chapter 4 Simple Equations In PDF Format

You can also download here the **NCERT Solutions Class 7 Maths chapter 4 Simple Equations** in PDF format.

Click Here to download NCERT Solutions for Class 7 Maths chapter 4 Simple Equations

### Question & Answer

Q.1:The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10 y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.

Ans :Missing

Q.2:Complete the last column of the table

Ans :(i) L.H.S. = x + 3 By putting x = 3, L.H.S. \(= 3 + 3 = 6 \neq\) R.H.S No, the equation is not satisfied. (ii) x + 3 = 0 L.H.S. = x +3 By putting x = 0, L.H.S. \(= 0 + 3 = 3 \neq\) R.H.S. No, the equation is not satisfied. (iii) x + 3 = 0 L.H.S. = x +3 By putting x = -3, L.H.S. = - 3 + 3 = 0 R.H.S. Yes, the equation is satisfied. (iv) x - 7 = 0 L.H.S. = x - 7 By putting x = 7, L.H.S. \(= 7 - 7 = 0 \neq\) R.H.S. No, the equation is not satisfied. (v) x - 7 = 1 L.H.S. = x - 7 By putting x = 8, L.H.S. = 8 - 7 = 1 R.H.S. Yes, the equation is satisfied. (vi) 5x = 25 L.H.S. = 5x By putting x = 0, L.H.S. \(= 5 \times 0 = 0 \neq\) R.H.S. No, the equation is not satisfied. (vii) 5x = 25 L.H.S. = 5x By putting x = 5, L.H.S. = 5 x 5 = 25 R.H.S. Yes, the equation is satisfied. (viii) 5x = 25 L.H.S. = 5x By putting x = 5, L.H.S. \(= 5 \times (-5) = 25 \neq\) R.H.S. No, the equation is not satisfied. (ix) m/3 = 2 L.H.S = m/3 By putting m = - 6, L.H.S \(= \frac { - 6 } { 3 } = - 2\) R.H.S No, the equation is not satisfied. (x) m/3 = 2 L.H.S = m/3 By putting m = 0, L.H.S \(= \frac { - 0 } { 3 } = 0\) R.H.S No, the equation is not satisfied. (x) m/3 = 2 L.H.S = m/3 By putting m = 6, L.H.S \(= \frac { 6 } { 3 } = 2\) R.H.S Yes, the equation is satisfied.

Q.3:Check whether the value given in the brackets is a solution to the given equation or not: (a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = – 2) (c) 7n + 5 = 19 (n = 2) (d) 4p – 3 = 13 (p = 1) (e) 4p – 3 = 13 (p = – 4) (f) 4p – 3 = 13 (p = 0)

Ans :(a) n + 5 -19 (n = 1) Putting n = 1 in L.H.S., \(n + 5 = 1 + 5 = 6 \neq 19\) As L.H.S \(\neq \mathrm { R.H.S. }\) Therefore, n = 1 is not a solution of the given equation, n + 5 =19. (b) 7n + 5 = 19 (n = -2) Putting n = -2 in L.H.S., \(7 n + 5 = 7 \times ( - 2 ) + 5 = - 14 + 5 = - 9 \neq 19\) As L.H.S \(\neq \mathrm { R.H.S. }\) Therefore, n = -2 is not a solution of the given equation, 7n + 5 =19. (c) 7n + 5 = 19 (n = 2) Putting n = 2 in L.H.S., 7 n + 5 = 7 x ( 2 ) + 5 = 14 + 5 = 19 = R.H.S As L.H.S = R.H.S. Therefore, n = 2 is a solution of the given equation, 7n + 5 =19. (d) 4p - 3 = 13 (p = 1) Putting p = 1 in L.H.S., \(4 p - 3 = ( 4 \times 1 ) - 3 = 1 \neq 13\) As L.H.S \(\neq \mathrm { R.H.S. }\) Therefore, p = 1 is not a solution of the given equation, 4p - 3 = 13. (e) 4p - 3 = 13 (p = -4) Putting p = -4 in L.H.S., \(4 p - 3 = 4 \times ( - 4 ) - 3 = - 16 - 3 = - 19 \neq 13\) As L.H.S \(\neq \mathrm { R.H.S. }\) Therefore, p = -4 is not a solution of the given equation, 4p - 3 = 13. (f) 4p - 3 = 13 (p = 0) Putting p = 0 in L.H.S., \(4 p - 3 = ( 4 \times 0 ) - 3 = - 3 \neq 13\) As L.H.S \(\neq \mathrm { R.H.S. }\) Therefore, p = 0 is not a solution of the given equation, 4p - 3 = 13.

Q.4:Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4

Ans :(i) 5p + 2 = 17 Putting p = 1 in L.H.S., \(( 5 \times 1 ) + 2 = 7 \neq \mathrm { R.H.S. }\) Putting p = 2 in L H.S., \(( 5 \times 2 ) + 2 = 10 + 2 = 12 \neq \mathrm { R } . \mathrm { H.S. }\) Putting p = 3 in L H.S., \(( 5 \times 3 ) + 2 = 17 = R . H . S\) Hence, p = 3 is a solution of the given equation. (ii) 3m – 14 = 4 Putting m = 4, \(( 3 \times 4 ) - 14 = - 2 \neq R . \mathrm { H.S. }\) Putting m =5, \(( 3 \times 5 ) - 14 = 1 \neq \mathrm { R } . \mathrm { H.S. }\) Putting m = 6, \(( 3 \times 6 ) - 14 = 18 - 14 = 4 = R . H . S\) Hence, m = 6 is a solution of the given equation.

Q.5:Write equations for the following statements: (i) The sum of numbers x and 4 is 9. (ii) 2 subtracted from y is 8. (iii) Ten times a is 70. (iv) The number b divided by 5 gives 6. (v) Three-fourth of t is 15. (vi) Seven times m plus 7 gets you 77. (vii) One-fourth of a number x minus 4 gives 4. (viii) If you take away 6 from 6 times y, you get 60. (ix) If you add 3 to one-third of z, you get 30.

Ans :(i) x + 4 = 9 (ii) y - 2 =8 (iii) 10a = 70 (iv)\(\frac { b } { 5 } = 6\) (v) \(\frac { 3 } { 4 } t = 15\) (vi) Seven times of m is 7m. 7m +7 = 77 (vii) One-fourth of a number x is x/4. \(\frac { x } { 4 } - 4 = 4\) (viii) Six times of y is 6y. 6Y -6 = 60 (ix) One-third of z is z/3 \(\frac { z } { 3 } + 3 = 30\)

## NCERT / CBSE Book for Class 7 Maths

You can download the NCERT Book for Class 7 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 7 Maths
- Click here to buy NCERT Book for Class 7 Maths

### All NCERT Solutions Class 7

- NCERT Solutions for Class 7 English
- NCERT Solutions for Class 7 Hindi
- NCERT Solutions for Class 7 Maths
- NCERT Solutions for Class 7 Science
- NCERT Solutions for Class 7 Social Science
- NCERT Solutions for Class 7 Sanskrit

### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 | Class 2 | Class 3 |

Class 4 | Class 5 | Class 6 |

Class 7 | Class 8 | Class 9 |

Class 10 | Class 11 | Class 12 |

Download the NCERT Solutions app for quick access to NCERT Solutions Class 7 Maths Chapter 4 Simple Equations. It will help you stay updated with relevant study material to help you top your class!

Previous Next

To get fastest exam alerts and government job alerts in India, join our Telegram channel.

## Discussion about this post