NCERT Solutions Class 7th Math Chapter – 4 Simple Equations Exercise – 4.1

1. Complete the last column of the table.

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

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

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

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

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

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

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

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

(ix) m/3 = 2
Solution: LHS = m/3
By substituting the value of m = – 6
Then,
LHS = -6/3 = – 2
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.

(x) m/3 = 2
Solution: LHS = m/3
By substituting the value of m = 0
Then,
LHS = 0/3 = 0
By comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.

(xi) m/3 = 2
Solution: LHS = m/3
By substituting the value of m = 6
Then,
LHS = 6/3 = 2
By comparing LHS and RHS
LHS = RHS
∴Yes, the equation is satisfied.

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 × 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
Solution:
LHS = 3m – 14
By substituting the value of m = 3
Then,
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.
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 – 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) Three-fourth of t is 15.
Solution: The above statement can be written in the equation form as,
= ¾t = 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 + 7 = 77

(vii) One-fourth of a number x minus 4 gives 4.
Solution: The above statement can be written in the equation form as,
One-fourth of a number x is x/4 – 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 – 6 = 60

(ix) If you add 3 to one-third of z, you get 30.
Solution: The above statement can be written in the equation form as,
One-third of z is 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 – 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: Three-fifth of m is 6.

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

(vii) 4p – 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 Parmit’s marbles.)
Solution:
From the question it is given that,
Number of Parmit’s marbles = m
Then,
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.)
Solution:
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.)
Solution:
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).
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 × base angle = 2b
= b + b + 2b = 180^o
= 4b = 180^o