A can do a job in the same time as (B + C) together. A + B finish it in 10 days, and C alone takes 50 days. In how many days can B alone finish the job?

Problem restatementRelate the individual daily work rates of A, B, and C from the given joint-time facts, then isolate B's rate.

Given data

• A alone takes the same time as (B + C) together → rates satisfy a = b + c.
• A + B complete the work in 10 days → a + b = 1/10 (work/day).
• C alone takes 50 days → c = 1/50 (work/day).

Concept/ApproachUse simple rate algebra. From a = b + c, substitute into a + b = 1/10 to solve for b.

Step-by-step calculation a = b + c a + b = 1/10 (b + c) + b = 1/10 → 2b + c = 1/10 c = 1/50 → 2b + 1/50 = 1/10 2b = 1/10 − 1/50 = (5 − 1)/50 = 4/50 = 2/25 b = 1/25

Hence, B alone needs 25 days.

VerificationThen a = b + c = 1/25 + 1/50 = 3/50. A alone needs 50/3 ≈ 16.67 days. Check A + B rate: 3/50 + 1/25 = 3/50 + 2/50 = 1/10 (matches).

Common pitfalls

• Confusing “same time” with “same rate”. Here “same time” means a = b + c for rates, not a = b or c.

Final Answer25 days