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

## AP 7th Class Maths 4th Chapter Simple Equations InText Questions

Try These (Page No. 120)

Question 1.

The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of ( 10 y 20). From the different values of (10y -20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.

Solution:

Let us find the value of the expression (10y – 20) for different values of y.

Let y = 2.

We have : 10y – 20 = 10(2) – 20 = 0

Let y = 3.

We have : 10y – 20 = 10(3) – 20 = 10

Let y = 4.

We have : 10y – 20 = 10(4) – 20 = 20

Let y = 5.

We have: 10y – 20 = 10(5) – 20 = 30

Let y = 6.

We have : 10y – 20 = 10(6) – 20 = 40

Hence 10y – 20 depends upon y

Now, consider 10y – 20 = 50

Transpose – 20 to the RHS :

10y = 50 + 20 = 70

or y =7

Try These (Page No. 124)

Question 1.

Write atleast one other form for each equation (ii), (iii) and (iv).

Solution:

Other forms are as unde r:

i) p multiplied by 5 gives 20

ii) 7 added to thrice of n gives 1

iii) one-fifth of m less 2 gives 6 .

Try These (Page No: 140)

Question 1.

i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is ?

Solution:

Let the number required be x

x is multiplied by 6 , the product = 6x.

5 is subtracted from 6x, then difference = 6x – 5

Result = 7

6x – 5 = 7 ⇒ 6x = 7 + 5

6x = 12 ∴ x = \(\frac{12}{6}\) = 2

∴ Required number = 2

ii) What is that number one third of which added to 5 gives 8 ?

Solution:

Let the number required be x

One third of number = \(\frac{x}{3}\)

According to the problem

\(\frac{x}{3}\) + 5 = 8 ⇒ \(\frac{x}{3}\) = 8 – 5

\(\frac{x}{3}\) = 3 ⇒ x= 3 × 3 ⇒ x = 9

∴ The required number = 9

Try These (Page No : 142)

Question 1.

There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?

Solution:

Let the number of mangoes in the smaller box be x

According to the question

8x + 4 = 100 ⇒ 8x = 100 – 4

8x = 96 ⇒ x = \(\frac{96}{8}\) ∴ x = 12

∴ The number of mangoes in the smaller box = 12