A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?

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Solution

$Time taken by A to do the work = 20 hours Time taken by B to do the work = 24 hours Time taken by (A + B + C) to do the work = 8 hours Now, Work done by A = \frac{1}{20} Work done by B = \frac{1}{24} Work done by (A + B + C) = \frac{1}{8} ∴ Work done by C = \frac{1}{8} - (\frac{1}{20} + \frac{1}{24}) = \frac{1}{8} - (\frac{6}{120} + \frac{5}{120}) = \frac{1}{8} - (\frac{11}{120}) = \frac{15 - 11}{120} = \frac{4}{120} = \frac{1}{30} Thus, C can do the work in 30 hours .$

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can do a piece of work in

4

hours;

B

and

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3

hours, while

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C

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2

hours. How long will

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A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?

Time taken by A to do the work=20 hoursTime taken by B to do the work=24 hoursTime taken by A+B+C to do the work=8 hoursNow,Work done by A=120Work done by B=124Work done by A+B+C=18∴ Work done by C =18-120+124=18-6120+5120=18-11120=15-11120=4120=130Thus, C can do the work in 30 hours.