A can do a piece of work in 24 days, B in 32 days, and C in 64 days. They start together but A leaves after 6 days and B leaves 6 days before the completion of the work. Find the total number of days taken to complete the work. MCQ with Answer and Explanation
A can do a piece of work in 24 days, B in 32 days, and C in 64 days. They start together but A leaves after 6 days and B leaves 6 days before the completion of the work. Find the total number of days taken to complete the work.
A. 16 days
B. 18 days
C. 20 days
D. 24 days
Answer: Option C
Solution (By JKSSB Mock Tests)
Total work = LCM(24, 32, 64) = 192 units. Efficiency of A = 8, B = 6, C = 3. Let total days be x. A works for 6 days. B works for (x - 6) days. C works for x days. Total work = 6*8 + (x - 6)*6 + x*3 = 192 => 48 + 6x - 36 + 3x = 192 => 9x + 12 = 192 => 9x = 180 => x = 20 days.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. If A works everyday and is assisted by B and C together on every third day, then in how many days will the work be completed?
Explanation:
Total work = LCM(20, 30, 60) = 60 units. Efficiency of A = 3, B = 2, C = 1. Day 1: A works = 3 units. Day 2: A works = 3 units. Day 3: A+B+C work = 3 + 2 + 1 = 6 units. Work done in 3 days = 3 + 3 + 6 = 12 units. Number of 3-day cycles to complete 60 units = 60 / 12 = 5 cycles. Total days = 5 * 3 = 15 days.
A can do a piece of work in 40 days. He works at it for 8 days and then B finishes it in 16 days. How long will A and B together take to complete the work?
Explanation:
Let total work be 40 units. Efficiency of A = 1 unit/day. In 8 days, A completes 8 units. Remaining work = 40 - 8 = 32 units. B finishes 32 units in 16 days, so B's efficiency = 32 / 16 = 2 units/day. Combined efficiency of A and B = 1 + 2 = 3 units/day. Time taken together = 40 / 3 = 13(1/3) days.
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