Question

A and B can do a piece of work in 25 days and 30 days, respectively. They worked together for 6 days, after while B was replaced by P and the work was finished in the next 8 days. How long will P alone take to complete the same work?

• A. 25 3/4 days
• B. 15 1/2 days
• C. 33 1/3 days
• D. 30 2/5 days

Answer 1

Answer:

C. 33 1/3 days

Let total work = 1 unit.

• A’s 1-day work = ( \frac{1}{25} )

• B’s 1-day work = ( \frac{1}{30} )

  Work done by A + B in 6 days:

$\left(\frac{1}{25} + \frac{1}{30}\right) \times 6$
$= \left(\frac{6+5}{150}\right) \times 6$
$= \frac{11}{150} \times 6$
$= \frac{66}{150}$
$= \frac{11}{25}$

Remaining work:
$1 - \frac{11}{25} = \frac{14}{25}$

Now, A + P complete this in 8 days:

$\text{(A + P one day work)} = \frac{14}{25} \div 8 = \frac{14}{200} = \frac{7}{100}$

So,

$\frac{1}{25} + \frac{1}{P} = \frac{7}{100}$
$\frac{1}{P} = \frac{7}{100} - \frac{1}{25}$
$= \frac{7}{100} - \frac{4}{100}$
$= \frac{3}{100}$

$P = \frac{100}{3} = 33 \tfrac{1}{3} \text{ days}$

P alone will take$(33 \tfrac{1}{3})$ days