A and B can complete a piece of work together in 5days. If a works at twice his speed and B at half of his Speed, the work can be finished in 4 days. How many days would it take for A alone to complete the Work ?

Work with rates: let A's one-day work be a and B's be b. Together a + b = 1/5. In the changed situation A does 2a and B does b/2, and the job finishes in 4 days, so 2a + b/2 = 1/4.

Substitute b = 1/5 - a: 2a + 1/10 - a/2 = 1/4, so 3a/2 = 3/20, giving a = 1/10. A alone takes 10 days, and B's rate is also 1/10. Check: 2/10 + 1/20 = 1/4, which gives 4 days. Option D is correct.

• Option A (18): A's rate 1/18 gives a changed combined rate of about 0.183, roughly 5.5 days, not 4.
• Option B (15): A's rate 1/15 gives B's rate 2/15; the changed rate becomes 2/15 + 1/15 = 1/5, so 5 days, not 4.
• Option C (12): A's rate 1/12 gives a changed rate of 27/120, about 4.4 days, not 4.

Core concept: work-rate problems with two unknowns, solved as a pair of linear equations. This is routine quantitative reasoning for CSAT, and the same method carries into Prelims numerical questions.

Trap: Aspirants often treat "twice his speed" and "half of his speed" as doubling or halving the days, or they forget to convert the 4-day finish into a rate of 1/4.