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 = [latex]\frac{1}{4}[/latex] ∙ z = [latex]\frac{z}{4}[/latex]
One fourth of z is added to 3 = [latex]\frac{z}{4}[/latex] + 3
Result is 7
∴ [latex]\frac{z}{4}[/latex] + 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) [latex]\frac{\mathbf{a}}{\mathbf{3}}[/latex] = 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) [latex]\frac{p}{4}[/latex] – 7 = 5 (p = 8)
Answer:
Given [latex]\frac{p}{4}[/latex] – 7 = 5; When p = 8
LHS = [latex]\frac{p}{4}[/latex] – 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.
∴ 0.45 ÷ 0.9 = 0.5
