In a constituency, 55% of the total number of voters are males and the rest are females. If 40% of the males are illiterate and 40% of the females are literate, then by what percent is the number of literate males more than that of illiterate females?
Let the total number of voters be 100.
Number of male voters = 55
Number of female voters = 100 - 55 = 45
Literate males = 100% - 40% = 60% of the total males
= 55 $$\times \frac{60}{100}$$ = 33
Illiterate females =Â 100% - 40% = 60% of the total females
= 45 $$\times \frac{60}{100}$$ = 27
Required percentage = $$\frac{33 - 27}{27} \times 100 =Â \frac{6}{27} \times 100 = 22 \frac{2}{9}$$
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