A alone can do a piece of work in 30 days, while B alone can do it in 15 days, and C alone can do it in 10 days. If in every second day B, and in every third day C helps A in doing the work, how many days will be required to complete the whole work?
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Step-by-step explanation:
Let the total units of work to be done is 30. (LCM of 30, 15 & 10)
A do 1 units of work per day. (30/30)
B do 2 units of work per day. (30/15)
C do 3 units of work per day. (30/10)
Therefore, 1st 6 days (A+B+C) can do= (1×6+2×3+3×1)= 18 units
[1st day A, 2nd day (A+B), 3rd day (A+C), 4th day (A+B), 5th day A, 6th day (A+B+C) work. So, A works 6 days, B works 3 days and C works 2 days.]
Remaining work= (30-18 ) units= 12 units
2nd 3 days (A+B+C) can do= (1×3+2×1+3×1)= 8 units
Remaining work= (12-8 ) units= 4 units
Next 2 days (A+B) can do= (1×2+2×1)= 4 units
Remaining work= (4–4) units = 0 units
Total required days=(6+3+2) days= 11 days
11 days will be required to complete the whole work .
Answer: 11 days.
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