NCERT Solutions for Class 7 Maths Exercise 4.1 Chapter 4 Simple Equations in simple PDF are given here. In this exercise of NCERT Solutions for Class 7 Chapter 4, students learn about what actually equation means. In an equation there is always an equality sign. The equality sign shows that the value of the expression to the left of the sign is equal to the value of the expression to the right of the sign. Learn more about these topics by solving the questions of NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations with the help of solutions provided here.
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1. Complete the last column of the table.
S. No. 
Equation 
Value 
Say, whether the equation is satisfied. (Yes/No) 
(i) 
x + 3 = 0 
x = 3 

(ii) 
x + 3 = 0 
x = 0 

(iii) 
x + 3 = 0 
x = 3 

(iv) 
x â€“ 7 = 1 
x = 7 

(v) 
x â€“ 7 = 1 
x = 8 

(vi) 
5x = 25 
x = 0 

(vii) 
5x = 25 
x = 5 

(viii) 
5x = 25 
x = 5 

(ix) 
(m/3) = 2 
m = â€“ 6 

(x) 
(m/3) = 2 
m = 0 

(xi) 
(m/3) = 2 
m = 6 
Solution:
(i) x + 3 = 0
LHS = x + 3
By substituting the value of x = 3
Then,
LHS = 3 + 3 = 6
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied.
(ii) x + 3 = 0
LHS = x + 3
By substituting the value of x = 0
Then,
LHS = 0 + 3 = 3
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied.
(iii) x + 3 = 0
LHS = x + 3
By substituting the value of x = â€“ 3
Then,
LHS = â€“ 3 + 3 = 0
By comparing LHS and RHS
LHS = RHS
âˆ´Yes, the equation is satisfied
(iv) x â€“ 7 = 1
LHS = x â€“ 7
By substituting the value of x = 7
Then,
LHS = 7 â€“ 7 = 0
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied
(v) x â€“ 7 = 1
LHS = x â€“ 7
By substituting the value of x = 8
Then,
LHS = 8 â€“ 7 = 1
By comparing LHS and RHS
LHS = RHS
âˆ´Yes, the equation is satisfied.
(vi) 5x = 25
LHS = 5x
By substituting the value of x = 0
Then,
LHS = 5 Ã— 0 = 0
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied.
(vii) 5x = 25
LHS = 5x
By substituting the value of x = 5
Then,
LHS = 5 Ã— 5 = 25
By comparing LHS and RHS
LHS = RHS
âˆ´Yes, the equation is satisfied.
(viii) 5x = 25
LHS = 5x
By substituting the value of x = 5
Then,
LHS = 5 Ã— (5) = â€“ 25
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied.
(ix) m/3 = 2
LHS = m/3
By substituting the value of m = â€“ 6
Then,
LHS = 6/3 = â€“ 2
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied.
(x) m/3 = 2
LHS = m/3
By substituting the value of m = 0
Then,
LHS = 0/3 = 0
By comparing LHS and RHS
LHS â‰ RHS
âˆ´No, the equation is not satisfied.
(xi) m/3 = 2
LHS = m/3
By substituting the value of m = 6
Then,
LHS = 6/3 = 2
By comparing LHS and RHS
LHS = RHS
âˆ´Yes, the equation is satisfied.
S. No. 
Equation 
Value 
Say, whether the equation is satisfied. (Yes/No) 
(i) 
x + 3 = 0 
x = 3 
No 
(ii) 
x + 3 = 0 
x = 0 
No 
(iii) 
x + 3 = 0 
x = 3 
Yes 
(iv) 
x â€“ 7 = 1 
x = 7 
No 
(v) 
x â€“ 7 = 1 
x = 8 
Yes 
(vi) 
5x = 25 
x = 0 
No 
(vii) 
5x = 25 
x = 5 
Yes 
(viii) 
5x = 25 
x = 5 
No 
(ix) 
(m/3) = 2 
m = â€“ 6 
No 
(x) 
(m/3) = 2 
m = 0 
No 
(xi) 
(m/3) = 2 
m = 6 
Yes 
2. Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
Solution:
LHS = n + 5
By substituting the value of n = 1
Then,
LHS = n + 5
= 1 + 5
= 6
By comparing LHS and RHS
6 â‰ 19
LHS â‰ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b) 7n + 5 = 19 (n = â€“ 2)
Solution:
LHS = 7n + 5
By substituting the value of n = 2
Then,
LHS = 7n + 5
= (7 Ã— (2)) + 5
= â€“ 14 + 5
= â€“ 9
By comparing LHS and RHS
9 â‰ 19
LHS â‰ RHS
Hence, the value of n = 2 is not a solution to the given equation 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
Solution:
LHS = 7n + 5
By substituting the value of n = 2
Then,
LHS = 7n + 5
= (7 Ã— (2)) + 5
= 14 + 5
= 19
By comparing LHS and RHS
19 = 19
LHS = RHS
Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.
(d) 4p â€“ 3 = 13 (p = 1)
Solution:
LHS = 4p â€“ 3
By substituting the value of p = 1
Then,
LHS = 4p â€“ 3
= (4 Ã— 1) â€“ 3
= 4 â€“ 3
= 1
By comparing LHS and RHS
1 â‰ 13
LHS â‰ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p â€“ 3 = 13.
(e) 4p â€“ 3 = 13 (p = â€“ 4)
Solution:
LHS = 4p â€“ 3
By substituting the value of p = â€“ 4
Then,
LHS = 4p â€“ 3
= (4 Ã— (4)) â€“ 3
= 16 â€“ 3
= 19
By comparing LHS and RHS
19 â‰ 13
LHS â‰ RHS
Hence, the value of p = 4 is not a solution to the given equation 4p â€“ 3 = 13.
(f) 4p â€“ 3 = 13 (p = 0)
Solution:
LHS = 4p â€“ 3
By substituting the value of p = 0
Then,
LHS = 4p â€“ 3
= (4 Ã— 0) â€“ 3
= 0 â€“ 3
= 3
By comparing LHS and RHS
â€“ 3 â‰ 13
LHS â‰ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p â€“ 3 = 13.
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
Solution:
LHS = 5p + 2
By substituting the value of p = 0
Then,
LHS = 5p + 2
= (5 Ã— 0) + 2
= 0 + 2
= 2
By comparing LHS and RHS
2 â‰ 17
LHS â‰ RHS
Hence, the value of p = 0 is not a solution to the given equation.
Let, p = 1
LHS = 5p + 2
= (5 Ã— 1) + 2
= 5 + 2
= 7
By comparing LHS and RHS
7 â‰ 17
LHS â‰ RHS
Hence, the value of p = 1 is not a solution to the given equation.
Let, p = 2
LHS = 5p + 2
= (5 Ã— 2) + 2
= 10 + 2
= 12
By comparing LHS and RHS
12 â‰ 17
LHS â‰ RHS
Hence, the value of p = 2 is not a solution to the given equation.
Let, p = 3
LHS = 5p + 2
= (5 Ã— 3) + 2
= 15 + 2
= 17
By comparing LHS and RHS
17 = 17
LHS = RHS
Hence, the value of p = 3 is a solution to the given equation.
(ii) 3m â€“ 14 = 4
Solution:
LHS = 3m â€“ 14
By substituting the value of m = 3
Then,
LHS = 3m â€“ 14
= (3 Ã— 3) â€“ 14
= 9 â€“ 14
= â€“ 5
By comparing LHS and RHS
5 â‰ 4
LHS â‰ RHS
Hence, the value of m = 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m â€“ 14
= (3 Ã— 4) â€“ 14
= 12 â€“ 14
= â€“ 2
By comparing LHS and RHS
2 â‰ 4
LHS â‰ RHS
Hence, the value of m = 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m â€“ 14
= (3 Ã— 5) â€“ 14
= 15 â€“ 14
= 1
By comparing LHS and RHS
1 â‰ 4
LHS â‰ RHS
Hence, the value of m = 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m â€“ 14
= (3 Ã— 6) â€“ 14
= 18 â€“ 14
= 4
By comparing LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is a solution to the given equation.
4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
Solution:
The above statement can be written in the equation form as,
= x + 4 = 9
(ii) 2 subtracted from y is 8.
Solution:
The above statement can be written in the equation form as,
= y â€“ 2 = 8
(iii) Ten times a is 70.
Solution:
The above statement can be written in the equation form as,
= 10a = 70
(iv) The number b divided by 5 gives 6.
Solution:
The above statement can be written in the equation form as,
= (b/5) = 6
(v) Threefourth of t is 15.
Solution:
The above statement can be written in the equation form as,
= Â¾t = 15
(vi) Seven times m plus 7 gets you 77.
Solution:
The above statement can be written in the equation form as,
Seven times m is 7m
= 7m + 7 = 77
(vii) Onefourth of a number x minus 4 gives 4.
Solution:
The above statement can be written in the equation form as,
Onefourth of a number x is x/4
= x/4 â€“ 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
Solution:
The above statement can be written in the equation form as,
6 times of y is 6y
= 6y â€“ 6 = 60
(ix) If you add 3 to onethird of z, you get 30.
Solution:
The above statement can be written in the equation form as,
Onethird of z is z/3
= 3 + z/3 = 30
5. Write the following equations in statement forms:
(i) p + 4 = 15
Solution:
The sum of numbers p and 4 is 15.
(ii) m â€“ 7 = 3
Solution:
7 subtracted from m is 3.
(iii) 2m = 7
Solution:
Twice of number m is 7.
(iv) m/5 = 3
Solution:
The number m divided by 5 gives 3.
(v) (3m)/5 = 6
Solution:
Threefifth of m is 6.
(vi) 3p + 4 = 25
Solution:
Three times p plus 4 gives you 25.
(vii) 4p â€“ 2 = 18
Solution:
Four times p minus 2 gives you 18.
(viii) p/2 + 2 = 8
Solution
If you add half of a number p to 2, you get 8.
6. Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmitâ€™s marbles.)
Solution:
From the question it is given that,
Number of Parmitâ€™s marbles = m
Then,
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 Ã— Number of Parmitâ€™s marbles + 7 = Total number of marbles Irfan having
= (5 Ã— m) + 7 = 37
= 5m + 7 = 37
(ii) Laxmiâ€™s father is 49 years old. He is 4 years older than three times Laxmiâ€™s age. (Take Laxmiâ€™s age to be y years.)
Solution:
From the question it is given that,
Let Laxmiâ€™s age to be = y years old
Then,
Lakshmiâ€™s father is 4 years older than three times of her age
= 3 Ã— Laxmiâ€™s age + 4 = Age of Lakshmiâ€™s father
= (3 Ã— y) + 4 = 49
= 3y + 4 = 49
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
Solution:
From the question it is given that,
Highest score in the class = 87
Let lowest score be l
= 2 Ã— Lowest score + 7 = Highest score in the class
= (2 Ã— l) + 7 = 87
= 2l + 7 = 87
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Solution:
From the question it is given that,
We know that, the sum of angles of a triangle is 180^{o}
Let base angle be b
Then,
Vertex angle = 2 Ã— base angle = 2b
= b + b + 2b = 180^{o}
= 4b = 180^{o}