A can do a work in 4 days, B in 5 days and C in 10 days. Find the time taken by A, B and C together to do the work.
A. 1(9/11) days
B. 2(1/11) days
C. 1(7/11) days
D. 2 days
Answer: Option A
Solution (By JKSSB Mock Tests)
Total work = LCM(4, 5, 10) = 20 units. Efficiency of A = 5, B = 4, C = 2. Combined efficiency = 5 + 4 + 2 = 11 units/day. Time taken together = 20 / 11 = 1(9/11) days.
Explanation:
Total work = LCM(20, 30) = 60 units. Efficiency of A = 3, B = 2. In a 2-day cycle (A then B), work completed = 3 + 2 = 5 units. Number of cycles required = 60 / 5 = 12 cycles. Total time = 12 * 2 = 24 days.
A is twice as efficient as B, and C is thrice as efficient as B. If they working together can finish a work in 5 days, in how many days can A alone finish it?
Explanation:
Let efficiency of B = 1. Then efficiency of A = 2, and efficiency of C = 3. Combined efficiency = 1 + 2 + 3 = 6. Total work = 5 * 6 = 30 units. Time taken by A alone = 30 / 2 = 15 days.
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