Question : A and B can do a job in 10 days and 5 days, respectively. They worked together for two days, after which B was replaced by C and the work was finished in the next three days. How long will C alone take to finish 60% of the job?
Option 1: 18 days
Option 2: 30 days
Option 3: 25 days
Option 4: 24 days
Correct Answer: 18 days
Solution :
Time taken by A to complete the work = 10 days
Part of work done by A in a day = $\frac{1}{10}$
Time taken by B to complete the work = 5 days
Part of work done by B in a day = $\frac{1}{5}$
Part of work done by A and B in a day = $\frac{1}{10}+\frac{1}{5} = \frac{3}{10}$
Part of work done by A and B in 2 days = $\frac{6}{10}$
Remaining work = $\frac{10-6}{10} = \frac{4}{10} = \frac{2}{5}$
Let the time taken by C to complete the work be $x$ days.
Part of work done by C in a day = $\frac{1}{x}$
Part of work done by A and C in a day = $\frac{1}{10}+\frac{1}{x}$
Part of work done by A and C in 3 days = $\frac{3}{10}+\frac{3}{x}$
$\frac{2}{5} = \frac{3}{10}+\frac{3}{x}$
$⇒\frac{3}{x} = \frac{2}{5} - \frac{3}{10} = \frac{4-3}{10} = \frac{1}{10}$
$\therefore x = 3 × 10 = 30$ days
Time taken by C to complete 60% of work = 0.6 × 30 = 18 days
Hence, the correct answer is 18 days.
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