LIC AAO 06 Mar 2016 Paper

Let the population of town be x
Males = $\frac{55\mathit{x}}{100}$ = $\frac{11\mathit{x}}{20}$
Females = $\frac{45\mathit{x}}{100}$ = $\frac{9\mathit{x}}{20}$
Illeterate females = $\frac{40}{100}$ × $\frac{9\mathit{x}}{20}$ = $\frac{9\mathit{x}}{50}$
According to the question,
$\frac{9\mathit{x}}{50}$ = 360 ⇒ 9x = 360 × 50
⇒ x = $\frac{36\times 50}{9}$ = 2000
Number of males = $\frac{11\mathit{x}}{20}$ = $\frac{11}{20}$ × 2000 = 1100
∴ Illeterate males = 30% of 1100
= $\frac{1100\times 30}{100}$ = 300