NCERT Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.1

NCERT Solutions Class 7 Mathematics 
Chapter – 4 (Simple Equations)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 4 Simple Equations Exercise 4.1 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 4 Simple Equations

Exercise – 4.1 

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 – 7 = 1 x = 7
(v) x – 7 = 1 x = 8
(vi) 5x = 25 x = 0
(vii) 5x = 25 x = 5
(viii) 5x = 25 x = -5
(ix) \frac{m}{3} = 2 m = – 6
(x) \frac{m}{3} = 2 m = 0
(xi) \frac{m}{3} = 2 m = 6

Solution –
(i) x + 3 = 0
LHS = x + 3
By substituting the value of x = 3
LHS = 3 + 3 = 6
By comparing LHS and RHS
6 ≠ 0
LHS ≠ RHS
∴ No, the equation is not satisfied.

(ii) x + 3 = 0
LHS = x + 3
By substituting the value of x = 0
LHS = 0 + 3 = 3
By comparing LHS and RHS
3 ≠ 0
LHS ≠ RHS
∴ No, the equation is not satisfied.

(iii) x + 3 = 0
LHS = x + 3
By substituting the value of x = – 3
LHS = – 3 + 3 = 0
By comparing LHS and RHS
0 = 0
LHS = RHS
∴ Yes, the equation is satisfied

(iv) x – 7 = 1
LHS = x – 7
By substituting the value of x = 7
LHS = 7 – 7 = 0
By comparing LHS and RHS
0 ≠ 1
LHS ≠ RHS
∴ No, the equation is not satisfied

(v) x – 7 = 1
LHS = x – 7
By substituting the value of x = 8
LHS = 8 – 7 = 1
By comparing LHS and RHS
1 = 1
LHS = RHS
∴ Yes, the equation is satisfied.

(vi) 5x = 25
LHS = 5x
By substituting the value of x = 0
LHS = 5 × 0 = 0
By comparing LHS and RHS
0 ≠ 25
LHS ≠ RHS
∴ No, the equation is not satisfied.

(vii) 5x = 25
LHS = 5x
By substituting the value of x = 5
LHS = 5 × 5 = 25
By comparing LHS and RHS
25 = 25 LHS = RHS
∴ Yes, the equation is satisfied.

(viii) 5x = 25
LHS = 5x
By substituting the value of x = -5
LHS = 5 × (-5) = – 25
By comparing LHS and RHS
– 25 ≠ 25
LHS ≠ RHS
∴ No, the equation is not satisfied.

(ix) \mathbf{\frac{m}{3}} = 2
LHS = \frac{m}{3}
By substituting the value of m = – 6
LHS = \frac{-6}{3} = – 2
By comparing LHS and RHS
– 2 ≠ 2
LHS ≠ RHS
∴ No, the equation is not satisfied.

(x) \mathbf{\frac{m}{3}} = 2
LHS = \frac{m}{3}
By substituting the value of m = 0
LHS = 0/3 = 0
By comparing LHS and RHS
0 ≠ 2
LHS ≠ RHS
∴ No, the equation is not satisfied.

(xi) \mathbf{\frac{m}{3}} = 2
LHS = \frac{m}{3}
By substituting the value of m = 6
LHS = \frac{6}{3} = 2
By comparing LHS and RHS
2 = 2
LHS = RHS
∴ 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 – 7 = 1 x = 7 No
(v) x – 7 = 1 x = 8 Yes
(vi) 5x = 25 x = 0 No
(vii) 5x = 25 x = 5 Yes
(viii) 5x = 25 x = -5 No
(ix) \frac{m}{3} = 2 m = – 6 No
(x) \frac{m}{3} = 2 m = 0 No
(xi) \frac{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)
(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)

Solution –
(a) n + 5 = 19 (n = 1)
LHS = n + 5
By substituting the value of n = 1
LHS = n + 5
= 1 + 5
= 6
By comparing LHS and RHS
6 ≠ 19
LHS ≠ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)
LHS = 7n + 5
By substituting the value of n = -2
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing LHS and RHS
– 9 ≠ 19
LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
LHS = 7n + 5
By substituting the value of n = 2
LHS = 7n + 5
= (7 × (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 – 3 = 13 (p = 1)
LHS = 4p – 3
By substituting the value of p = 1
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing LHS and RHS
1 ≠ 13
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
LHS = 4p – 3
By substituting the value of p = – 4
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing LHS and RHS
– 19 ≠ 13
LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
LHS = 4p – 3
By substituting the value of p = 0
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.

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

Solution –
(i) 5p + 2 = 17
LHS = 5p + 2
By substituting the value of p = 0
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
By comparing LHS and RHS
2 ≠ 17
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation.

Let, p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing LHS and RHS
7 ≠ 17
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation.

Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing LHS and RHS
12 ≠ 17
LHS ≠ RHS
Hence, the value of p = 2 is not a solution to the given equation.

Let, p = 3
LHS = 5p + 2
= (5 × 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 – 14 = 4
LHS = 3m – 14
By substituting the value of m = 3
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS and RHS
-5 ≠ 4
LHS ≠ RHS
Hence, the value of m = 3 is not a solution to the given equation.

Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing LHS and RHS
– 2 ≠ 4
LHS ≠ RHS
Hence, the value of m = 4 is not a solution to the given equation.

Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS and RHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m = 5 is not a solution to the given equation.

Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 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.
(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.

Solution – 

(i) The sum of numbers x and 4 is 9.
The above statement can be written in the equation form as,
⇒ x + 4 = 9

(ii) 2 subtracted from y is 8.
The above statement can be written in the equation form as,
⇒ y – 2 = 8

(iii) Ten times a is 70.
The above statement can be written in the equation form as,
⇒ 10a = 70

(iv) The number b divided by 5 gives 6.
The above statement can be written in the equation form as,
⇒ \frac{b}{5} = 6

(v) Three-fourth of t is 15.
The above statement can be written in the equation form as,
⇒ \frac{3}{4}\times t = 15

(vi) Seven times m plus 7 gets you 77.
The above statement can be written in the equation form as,
Seven times m is 7m
⇒ 7m + 7 = 77

(vii) One-fourth of a number x minus 4 gives 4.
The above statement can be written in the equation form as,
One-fourth of a number x is \frac{x}{4}
⇒ \frac{x}{4} – 4 = 4

(viii) If you take away 6 from 6 times y, you get 60.
The above statement can be written in the equation form as,
6 times of y is 6y
⇒ 6y – 6 = 60

(ix) If you add 3 to one-third of z, you get 30.
The above statement can be written in the equation form as,
One-third of z is \frac{z}{3}
⇒ 3 + \frac{z}{3} = 30

5. Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) \mathbf{\frac{m}{5}} = 3
(v) \mathbf{\frac{3m}{5}} = 6
(vi) 3p + 4 = 25
(vi) 3p + 4 = 25
(viii) \mathbf{\frac{p}{2}} + 2 = 8

Solution –
(i) p + 4 = 15
The sum of numbers p and 4 is 15.

(ii) m – 7 = 3
7 subtracted from m is 3.

(iii) 2m = 7
Twice of number m is 7.

(iv) m/5 = 3
The number m divided by 5 gives 3.

(v) \mathbf{\frac{3m}{5}} = 6
Three-fifth of m is 6.

(vi) 3p + 4 = 25
Three times p plus 4 gives you 25.

(vii) 4p – 2 = 18
Four times p minus 2 gives you 18.

(viii) \mathbf{\frac{p}{2}} + 2 = 8
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 Parmit’s marbles.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
(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.)
(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 –
(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 Parmit’s marbles.)
From the question it is given that,
Number of Parmit’s marbles = m
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having
= (5 × m) + 7 = 37
= 5m + 7 = 37

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
From the question it is given that,
Let Laxmi’s age to be = y years old
Then,
Lakshmi’s father is 4 years older than three times of her age
= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father
= (3 × 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.)
From the question it is given that,
Highest score in the class = 87
Let lowest score be l
= 2 × Lowest score + 7 = Highest score in the class
= (2 × 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).
From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Vertex angle = 2 × base angle = 2b
= b + b + 2b = 180o
= 4b = 180o

 

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