NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

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.

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NCERT Solutions Class 7 Maths Chapter 4 Simple Equations

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Class	7
Subject	Maths
Book	Mathematics
Chapter Number	4
Chapter Name	Simple Equations

NCERT Solutions Class 7 Maths chapter 4 Simple Equations

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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.

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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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