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?

• A. $19\frac{11}{31}$%
• B. $30\frac{5}{9}$%
• C. $40$%
• D. $44$%

Answer 1

Answer: (B)

$30\frac{5}{9}$%

Let the number be a

Number increased twice by $20$%

$\therefore$ Increased number$=a(1+\frac{20}{100})(1+\frac{20}{100})$

$=\frac{36a}{25}$

Let r be percent reduced to get it back to original

$\therefore \frac{36a}{25}(1-\frac{x}{100})=a$

$1-\frac{x}{100}=\frac{25}{36}$

$\therefore x=\frac{11}{36}\times 100$

$x=30\frac{5}{9}$