A is twice as fast as B and B is thrice as fast as C. If C alone can complete the work in 24 days, in how many days can they together finish the work? MCQ with Answer and Explanation
A is twice as fast as B and B is thrice as fast as C. If C alone can complete the work in 24 days, in how many days can they together finish the work?
A. 3 days
B. 2.4 days
C. 4 days
D. 2.6 days
Answer: Option B
Solution (By JKSSB Mock Tests)
Let efficiency of C = 1. Then efficiency of B = 3 * 1 = 3. Efficiency of A = 2 * 3 = 6. Total work = Efficiency of C * Time of C = 1 * 24 = 24 units. Combined efficiency of A, B and C = 6 + 3 + 1 = 10. Time taken together = 24 / 10 = 2.4 days.
A can complete a piece of work in 10 days, B in 15 days, and C in 20 days. A and C worked together for 2 days and then A was replaced by B. In how many days altogether was the work completed?
Explanation:
Total work = LCM(10, 15, 20) = 60 units. Efficiency of A = 6, B = 4, C = 3. A and C work for 2 days: work done = 2 * (6 + 3) = 18 units. Remaining work = 60 - 18 = 42 units. B and C complete remaining work together with efficiency = 4 + 3 = 7 units/day. Time taken for remaining work = 42 / 7 = 6 days. Total time = 2 + 6 = 8 days.
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