Three pipes A B and C can fill the cistern in 10 12 and 15 hours respectively In how much time the c...........

Quantitative Aptitude ( CCS) Pipes&cistern

Three pipes A, B and C can fill the cistern in 10, 12, and 15 hours respectively. In how much time the cistern will be full if A is operated for the whole time and B and C are operated alternately which B being first?

10 hours 32 minutes

6 hours

5 1/2 hours

5 7/10 hours

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Explanation:

Answer & Explanation
5 7/10 hours
Explanation:
In first hour, part of cistern filled is (1/10 + 1/12) = 11/60
In second hour, part of cistern filled is (1/10 + 1/15) = 1/6
So in 2 hours, part of cistern filled is 11/60 + 10/60 = 21/60 = 7/20
now in 2*2 (4) hours, part of cistern filled is (7/20)*2 = 14/20 = 7/10
now in the 5th hour, A+B’s turn which fill 11/60 in that hour, but the cistern remaining to be filled is (1 – 7/10) = 3/10, since 3/10 is more than 11/60, so after 5th hour remaining part to be filled is 3/10 – 11/60 = 7/60
now in 6th hour, (A+C)’s turn, it will fill remaining 7/60 in (7/60)*(6/1) = 7/10
so total 5 7/10 hours