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by khizarj » Tue Jun 14, 2011 2:38 pm
Machine A and B can finish a task in 6/5 of an hour. A and C and finish in 2/3 of an hour and B and C can finish in 2 hours. How long does it take for all three together to finish?

oops..posted in DS by mistake..how do i delete this?
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by cans » Tue Jun 14, 2011 9:14 pm
a&b = 6/5 hours
a&c = 2/3 hours
b&c = 2 hours
(A&b)&(a&c)&(b&c) can finish in 1/ ( 1/(6/5) + (1/(2/3) + 1/2 ) = 1/ ( 5/6 + 3/2 + 1/2)) = 6/17
thus a&b&c can do in 3/17 hours
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by sourabh33 » Wed Jun 15, 2011 4:13 am
IMO 12/17

Let the total work be 30 units

1. A+B -> Time taken = 6/5 -> Rate of work = 30/6/5 = 25 units/hr
2. A+C -> Time taken = 2/3 -> Rate of work = 30/2/3 = 45 units/hr
3. B+C -> Time taken = 2 -> Rate of work = 30/2 = 15 units/hr

Adding rates of all three we get 2(A+B+C)= 85
Therefore Rate f a+b+c = 85/2

Now to find the time taken by a+b+c -> work/rate -> 30/85/2 -> 60/85 -> 12/17
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