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Question

A and B can do a piece of work in 30 days. B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days then B and C leave. How many days more will A take to finish the remaining work?


Solution

Answer

Let total work be 120 units  (LCM of 30, 24 and 20).

(A+B)'s one day work = 120/30=4 units

(B+C)'s one day work = 120/24=5 units

(C+A)'s one day work = 120/20=6 units

Adding all three equations,
2A+2B+2C=15
2(A+B+C)=15
A+B+C=7.5

SO,
(A+B+C)'s one day work = 7.5 units

A's one day work = (A+B+C)'s one day work-(B+C)'one day work=7.5 - 5 = 2.5 units

In 10 days, all three complete 75 (10×7.5) units of work. 
Remaining 45 units can be completed by A in 18 days (at rate of 2.5 units per day).

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