In a constituency, 45% of the total number of voters are males and the rest are females. If 60% of the males are illiterate and 60% of the females are literate, then by what percent is the number of illiterate females more than that of literate males? (Rounded off to 2 decimal places)

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Accepted Answer

Step 1: Let the total number of voters be 100.
Step 2: Number of males = 45% of 100 = 45
Step 3: Number of females = 100 - 45 = 55
Step 4: Illiterate males = 60% of 45 = 0.6 × 45 = 27
Step 5: Literate females = 60% of 55 = 0.6 × 55 = 33
Step 6: Illiterate females = Total females - Literate females = 55 - 33 = 22
Step 7: Literate males = Total males - Illiterate males = 45 - 27 = 18
Step 8: We need to find by what percent illiterate females are more than literate males.
Percentage increase = ((Illiterate females - Literate males) / Literate males) × 100
= ((22 - 18) / 18) × 100 = (4 / 18) × 100 ≈ 22.22%

Therefore, the number of illiterate females is approximately 22.22% more than the number of literate males.

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