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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.

Simple Equations Download

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Download

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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 \(\frac { z } { 3 } + 3 = 30\)

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