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
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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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