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1. A can complete a piece of work in 18 days, B in 20 days and C in 30 days, B and C together start the work and forced to leave after 2 days. The time taken by A alone to complete the remaining work is:

A. 10 days

B. 12 days

C. 15 days

D. 16 days

Solution:

1^st Method:

\begin{aligned} (B + C) 2 days work \\ = 2 \times (\frac{1}{20} + \frac{1}{30}) \\ = 2 \times \frac{3 + 2}{60} \\ = \frac{1}{6} part \\ Remaining work \\ = 1 - \frac{1}{6} \\ = \frac{5}{6} part \\ A's one day's work \\ = \frac{1}{18} part \\ Time taken to complete the work \\ = \frac{\frac{5}{6}}{\frac{1}{18}} days \\ Hence, \\ Time taken to complete the work \\ = \frac{5}{6} \times 18 \\ = 15 days \end{aligned}

2^nd Method:
% of work B completes in one day =

\frac{100}{20}

= 5%;
% of work C completes in one day =

\frac{100}{30}

= 3.33%;
% of work (A + B) completes together in one day = 5 + 3.33 = 8.33%;
% work (A + B) completes together in 2 days = 8.66 × 2 = 16.66%;
Remaining work = 100 - 16.66 = 83.34%;
% of work A completes in 1 day =

\frac{100}{18}

= 5.55%
Time taken to complete the remaining work by A
=

\frac{83.34}{5.55}

= 15 days

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