# SCERT Class 7 Mathematics/ Unchanging Relations – Chapter 3

**SCERT Class 7 Mathematics/ Unchanging Relations is designed for Class 7 Students. Here you can find out practice problems for Class 7 students.**

**SCERT Class 7 Mathematics /Unchanging Relations – Chapter 3Textbook Solutions/Model Questions/Worksheet**

**Using the rule (x + y) + (x – y) = 2x, solve the following:**

- The sum of two numbers is 10 and difference is 2. Find the numbers?

Solution:

Let’s take the numbers as x and y.

Then x + y = 10 and x – y = 2

But we know that (x + y) + (x – y) = 2x, so 10 + 2 = 2x which gives 12 = 2x, x = 6.

So largest number is 6 and smallest number is 10 – 6 = 4. - The sum of two numbers is 16 and difference is 5. Find the numbers?

Solution:

Given x + y = 16 and x – y = 5

(x + y) + (x – y) = 16 + 5 = 21 = 2x

x = 21/2 = 10 ½

So largest number is 10 ½ and smallest number is 16 – 10 ½ = 5 ½ - The sum and difference of some pairs of numbers are given below:

a) Sum 12, Difference 8

b) Sum 140, Difference 80

c) Sum 23, Difference 11

d) Sum 20, Difference 5

Solution:

a) Given x + y = 12 and x – y = 8

(x + y) + (x – y) = 12 + 8 = 20 = 2x

x = 20/2 = 10

So largest number is 10 and the smallest number is 12 – 10 = 2.

b) Given x + y = 140, Difference = 80

(x + y) + (x – y) = 140 + 80 = 220 = 2x

x = 220/2 = 110

So largest number is 110 and the smallest number is 140 – 110 = 30

c) Given x + y = 23 and x – y = 11

(x + y) + (x – y) = 23 + 11 = 34 = 2x

x = 34/2 = 17

So largest number is 17 and the smallest number is 23 – 17 = 6

d) Given x + y = 20 and x – y = 5

(x + y) + (x – y) = 20 + 5 = 25 = 2x

x = 25/2

So largest number is 25/2 and the smallest number is 20 – (25/2) = 15/2.

Multiplying two numbers by a number separately and adding give the same result as multiplying their sum by the number.

i.e., xz + yz = (x + y) z, for all numbers x, y, z.

Multiplying two numbers by a number separately and subtracting give the same result as multiplying their difference by the number.

i.e. xz – yz = (x – y) z, for all numbers x, y, z.**Using the above rules, find:**

- (63 x 12) + (37 x 12)
- (1/3 x 20) + (2/3 x 20)
- (2 ½ x 23) – (1 ½ x 23)
- (15 x ¾) + (5 x ¾)
- (65 x 11) – (55 x 11)
- (13.5 x 40) – (3.5 x 40)

ANSWERS:

- (63 x 12) + (37 x 12) = (63 + 37) 12 = 100 x 12 = 1200
- (1/3 x 20) + (2/3 x 20) = (1/3 + 2/3) 20 = 3/3 x 20 = 1 x 20 = 20
- (2 ½ x 23) – (1 ½ x 23) = (2 ½ – 1 ½) 23 = 1 x 23 = 23
- (15 x ¾) + (5 x ¾) = (15 + 5) ¾ = 20 x ¾ = 60/4 = 15
- (65 x 11) – (55 x 11) = (65 – 55) 11 = 10 x 11 = 110
- (13.5 x 40) – (3.5 x 40) = (13.5 – 3.5) 40 = 10 x 40 = 400