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by khizarj » Tue Jun 14, 2011 2:40 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?
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by vineeshp » Tue Jun 14, 2011 6:26 pm
1/A + 1/B = 5/6
1/A + 1/C = 3/2
1/B + 1/C = 1/2.
Add the three equations and you will get

2 (1/A + 1/B + 1/C) = 5/6 + 3/2 + 1/2 = 17/6

(1/A + 1/B + 1/C) = 5/6 + 3/2 + 1/2 = 17/12

So Together they can finish in 12/17.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by Mom4MBA » Wed Jun 15, 2011 3:20 am
same way, same answer 12/17 of hour
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by GMATGuruNY » Wed Jun 15, 2011 6:17 am
khizarj wrote: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?
Let job = 12 units.

Rate for A and B = w/t = 12/(6/5) = 10 units per hour.
Rate for A and C = w/t = 12/(2/3) = 18 units per hour.
Rate for B and C = w/t = 12/2 = 6 units per hour.

Combining the rates:
(A+B) + (A+C) + (B+C) = 10+18+6
2(A+B+C) = 34
A+B+C = 17 units per hour.

Time for A+B+C = w/r = 12/17 hours.
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