Question

A number is increased by 20% and then again by 20%. By what percent should the increased number be reduced so as to get back the original number?

Answer 1

$\mathcal{\underline{\purple{Steps :}}}$

• Find the Net %Change when successive percentage increase are applied

• This net % change is the difference of Original No. to the last No obtained after a final percentage increase.

•°• To get back the original No., the Increased No. must be reduced by net % increase.

$\mathcal{\underline{\orange{Points To Remember:}}}$

¶ Effect of %change Increase on a No
When a No. is increased by S%
$New No.= {N(1 +\frac{S}{100}})$

OR

¶ When a No. is increased successively by a% & b%

$Net percentage increase = a + b +\frac{ab}{100}$

$\mathcal{\red{\underline{SOLUTION:}}}$

$\mathcal{\blue{\underline{Method -1:}}}$

Let the original No. = 100

By Straight Line Method Of solving :
[100] ➨ 20%↑ ➨[120] ➨20%↑ ➨ [144]

144 - 100 = 44
100 = 100% => 1 = 1%
•°• 44 = 44%
Net Increase in percent= 44%

$\mathcal{\blue{\underline{Method -2:}}}$
Using Formula,
Here, a= b

$Net_{Percent Increase} = (2a + \frac{a^{2}}{100})$

$= [(2×20) +\frac{20^{2}}{100}]$

$=[(40)+\frac{400}{100}]$

= 40 + 4

= 44

•°• The Increased No. should be reduced by 44% to get back the original No.

$\mathcal{\huge{\purple{Hope it helps}}}$