Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Simple Equations Class 7 Exercise 4.1 Solutions – 7th Class Maths 4.1 Exercise Solutions

Question 1.

Complete the last column of the table.

Solution:

i) x + 3 = 0

∵ x = 3 ⇒ 3 + 3 = 6 ≠ 0 No

ii) x + 3 = 0

∵ x = 0 ⇒ 0 + 3 = 3 ≠ 0 No

iii) x + 3 = 0

∵ x = – 3 ⇒ – 3 + 3 = 0 Yes

iv) x – 7 = 1

∵ x = 7 ⇒ 7 – 7 = 0 ≠ 1 No

v) x – 7 = 1

∵ x = 8 ⇒ 8 – 7 = 1 Yes

vi) 5x = 25

∵ x = 0 ⇒ 5 × 0 = 5 No

vii) 5x = 25

∵ x = 5 ⇒ 5(5) = 25 Yes

viii) 5x = 25

∵ x = – 5 ⇒ 5(-5) = – 25 ≠ 25 No

ix) \(\frac { m }{ 3 }\) = 2

∵ m = – 6 ⇒ \(\frac { -6 }{ 3 }\) = – 2 ≠ 2 No

x) \(\frac { m }{ 3 }\) = 2

∵ m = 0 ⇒ \(\frac { 0 }{ 3 }\) = 0 ≠ 2 No

xi) \(\frac { m }{ 3 }\) = 2

∵ m = 6 ⇒ \(\frac { 6 }{ 3 }\) = 2 Yes

Question 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:

n + 5 = 19

∵ n = 1 ⇒ 1 + 5 = 6 ≠ 19

n = 1 is not the solution.

b) 7n + 5=19(n = – 2)

Solution:

7n + 5 = 19 ∵ n = – 2

7n + 5 = 7(- 2) + 5

= – 14 + 5 = – 9 ≠ 19

∴ n = – 2 is not the solution.

c) 7n + 5 = 19 (n = 2)

Solution:

7n + 5 = 19

∵ n = 2

7(2) + 5 = 14 + 5 = 19

∴ n = 2 is the solution.

d) 4p – 3 = 13 (p = 1)

Solution:

4p – 3 = 13 ∵ p = 1

4p – 3 = 4(1) – 3 = 4 – 3 = 1 ≠ 13

∴ p = 1 is not the solution.

e) 4p – 3 = 13(p = – 4)

Solution:

4p – 3 = 13

4p – 3 = 4(-4) – 3 = – 16 – 3 = – 19 ≠ 13

∴ p = – 4 is not the solution.

f) 4p – 3 = 13(p = 0)

Solution:

4p – 3 = 13 ∵ p = 0

4p – 3 = 4(0) – 3 = – 3 ≠ 13

∴ p = 0 is not the solution.

Question 3.

Solve the following equations by trial and error method:

i) 5p + 2 = 17

Solution:

Trial and error method

Let p = 1

LHS 5p + 2 = 5(1) + 2 = 5 + 2 = 7

RHS = 17

∴ LHS ≠ RHS

p = 1 is not the solution

Let p = 2

LHS 5p + 2 = 5(2) + 2 = 10 + 2 = 12

RHS = 17

LHS ≠ RHS

p = 2 is not the solution

Let p = 3

LHS 5p + 2 = 5(3) + 2 = 15 + 2 = 17

RHS = 17

∴ LHS = RHS

∴ p = 3 is the solution.

ii) 3m – 14 = 4

Solution:

Let m = 1

LHS = 3m – 14 = 3(1) – 14 = 3 – 14 = – 11

RHS = 4

LHS ≠ RHS

∴ m = 1 is not the solution

Let m = 3

LHS = 3m – 14 = 3(3) – 14 = 9 – 14 = – 5

RHS = 4

∴ LHS ≠ RHS

∴ m = 3 is not the solution

Let m = 6

LHS 3m – 14 = 3(6) – 14 = 18 – 14 = 4

RHS = 4

∴ LHS = RHS

∴ m = 6 is the solution

Question 4.

Write equations for the following statements :

i) The sum of numbers x and 4 is 9.

Solution:

x + 4 = 9

ii) 2 subtracted from y is 8.

Solution:

y – 2 = 8

iii) Ten times a is 70.

Solution:

10 a = 70

iv) The number b divided by 5 gives 6 .

Solution:

\(\frac { b }{ 5 }\) = 6

v) Three-fourth of t is 15.

Solution:

\(\frac { 3 }{ 4 }\)(t) = 15

vi) Seven times m plus 7 gets you 77 .

Solution:

7m + 7 = 77

vii) One-fourth of a number x minus 4 gives 4.

Solution:

\(\frac { 1 }{ 4 }\)x – 4 = 4

viii) If you take away 6 from 6 times y, you get 60.

Solution:

6y – 6 = 60

ix) If you add 3 to one-third of z, you get 30 .

Solution:

\(\frac { 1 }{ 3 }\)z + 3 = 30

Question 5.

Write the following equations in statement forms:

i) p + 4 = 15

Solution:

4 is added to p gives 15

ii) m – 7 = 3

Solution:

7 is subtracted from m, the difference is 3 .

iii) 2m = 7

Solution:

Two times of m is 7

iv) \(\frac { m }{ 5 }\) = 3

Solution:

m is divided by 5 gives quotient 3

v) \(\frac { 3m }{ 5 }\) = 6

Solution:

3 times of m divided by 5 gives 6.

vi) 3p + 4 = 25

Solution:

3 times of p is added to 4 to give result 25

vii) 4p – 2 = 18

Solution:

2 is subtracted from 4 times of p gives 18

viii) \(\frac { p }{ 2 }\) + 2 = 8

Solution:

p is divided by 2 and 2 is added the result is 8.

Question 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:

Let m be the number of Parmit marbles

5 times of m = 5m

7 more than 5m = 5m + 7

According to the problem

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:

Let Laxmi’s age to be ‘y’ years 3 times Laxmi’s age = 3y years

4 year older than 3y years = 3y + 4

According to the problem 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:

Let the lowest score = l

Twice of l = 2l

Twice of l plus 7 = 2l + 7

According to the problem 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:

Let the base angle = b°

Twice the base angle = 2b°

We know that the sum of the angles in a triangle = 180°

2b + b + b = 180°

4b = 180°

b = \(\frac{180}{4}\)

b = 45°