SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.1 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.1

Question 1.

Write the equations of the following mathematical statements.

(i) A number x decreased by 5 is 14.

Answer:

Given number = x

Number decreased by 5 = x – 5

∴ x – 15 = 14.

(ii) Eight times of y plus 3 is -5.

Answer:

Given number = y

Eight times of y = 8 ∙ y

Eight times of y plus 3 = 8y + 3

∴ 8y + 3 = – 5

(iii) If you add one fourth of z to 3 you get 7.

Answer:

Given number = z

One fourth of z = \(\frac{1}{4}\) ∙ z = \(\frac{z}{4}\)

One fourth of z is added to 3 = \(\frac{z}{4}\) + 3

Result is 7

∴ \(\frac{z}{4}\) + 3 = 7.

(iv) If you take away 5 from 3 times of m, you get 11.

Answer:

Given number = m

3 times of m = 3m

Take away 5 from 3 times of m = 3m – 5

Result is 11

∴ 3m – 5 = 11

(v) Sum of angles 2x, (x – 30) is a right angle.

Answer:

Given angles 2x, (x – 30)

Sum of angles 2x, (x – 30)

= 2x + x – 30 = 3x – 30

Sum of angles is right angle (90°).

∴ 3x – 30 = 90°

(vi) The perimeter of a square of side ‘a’ is 14 m.

Answer:

Given side of a square = a

Perimeter = 4 ∙ side = 4 ∙ a

Given perimeter = 14 m

∴ 4a = 14 m

Question 2.

Write the following equations in statement form.

(i) m – 5 = 12

Answer:

A number m is decreased by 5 is 12.

(ii) \(\frac{\mathbf{a}}{\mathbf{3}}\) = 4

Answer:

One third of a is 4.

(iii) 4x + 7 = 15

Answer:

Sum of 4 times of x and 7 is 15.

(or)

7 is added to 4 times of x is 15.

(iv) 2 – 3y = 11

Answer:

2 is decreased by 3 times of y is 11.

(or)

3 times of y is subtracted from 2 is 11.

Question 3.

Check whether the value given in the brackets is a solution to the given equation or not.

(i) 5n – 7 = 23 (n = 6)

Answer:

Given 5n – 7 = 23

When n = 6

L.H.S = 5n – 7

= 5(6) – 7

= 30 – 7

= 23

R.H.S =23

Here, L.H.S = R.H.S

So, n = 6 is a solution of the given equation.

(ii) \(\frac{p}{4}\) – 7 = 5 (p = 8)

Answer:

Given \(\frac{p}{4}\) – 7 = 5; When p = 8

LHS = \(\frac{p}{4}\) – 7

RHS = 5

– 5 ≠ 5

Here, LHS ≠ RHS

So, p = 8 is not a solution of the given equation.

(iii) 5 – 2x = 19 -7

Answer:

Given 5 – 2x = 19

When x = – 7

LHS = 5 – 2x

= 5 – 2(- 7) = 5 + 14 = 19

RHS = 19

19 = 19

Here, LHS = RHS

So, x = – 7 is a solution of the given equation.

(iv) 2 + 3(m – 1) = 5 (m = -2)

Answer:

Given 2 + 3(m – 1) = 5

When m = – 2

LHS = 2 + 3(m – 1)

= 2 + 3(- 2 – 1)

= 2 + 3(- 3) = 2 – 9 = – 7

RHS = 5

– 7 ≠ 5

Here, LHS ≠ RHS

So, m = – 2 is not a solution of given equation.

Question 4.

Solve the following equations using trial and error method,

(i) 3x – 7 = 5

Answer:

For x = 4, LHS = RHS

So, x = 4 is the solution of given equation.

(ii) 5 – y = – 1

Answer:

For y = 6, LHS = RHS.

So, y = 6 is the solution of given equation.