3 men or 7 women finish a work in 32 days. 7 men and 5 women together will finish the twice of this work in

Let one man do m units of work per day and one woman do w units per day. Since 3 men and 7 women finish the same work in 32 days, 3m = 7w, so w = 3m/7.

Original work = 3m × 32 = 96m units. The question asks for twice this, so the required work is 192m units.

Daily work of 7 men and 5 women = 7m + 5(3m/7) = 49m/7 + 15m/7 = 64m/7.

Days needed = 192m ÷ (64m/7) = 192 × 7 / 64 = 21 days. So B is correct.

The core concept is man-day and woman-day equivalence (work-equivalence ratios), which is a routine Quantitative Aptitude and Mental Ability topic in Prelims and also useful for simple resource-allocation reasoning in Mains. The other options usually come from mishandling the ratio or skipping the doubling of work.

Trap: forgetting to double the work, or inverting the men-women ratio (writing 3w = 7m instead of 3m = 7w). Both lead to a wrong day count even though the setup looks right.