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# A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease per cent

**Solution:**

Let the number be x.

If x is increased by 20% we get:

= x + (20/100) × (x)

= x + (1/5)x

= (6/5)x

The above number is decreased by 20% and we get

= (6/5)x - 20% of 6x/5

= (6x/5) - (1/5)(6x/5)

= (6x/5) - (6x/25)

= 30x - 6x / 25

= 24x/25

The net decrease is calculated as:

Decrease = x - 24x/25

= (25x - 24x)/25

= x/25

The % decrease is therefore calculated as:

Decrease % = [(x/25)/x] × 100

= [1/25] × 100

= 4%

**✦ ****Try this:** A number is increased by 10% and then it is decreased by 10%. Find the net increase or decrease per cent

Let the number be 100

By increasing it by 10% we get:

Increasing 100 by 10% = 1.10 × 100

= 110

By decreasing it by 10% we get

Decreasing 110 by 10% we get = 0.9 × 110

= 99

The net change is a decrease of 1%as shown below:

(99 - 100)/100

= 1/100

= 1%

**☛ Also Check: **NCERT Solutions for Class 8 Maths Chapter 8

**NCERT Exemplar Class 8 Maths Chapter 9 Sample Problem 12**

## A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease per cent

**Summary:**

If A number is increased by 20% and then it is decreased by 20%. Then there is a net decrease of 4% in the original number

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