Question

A and B can do a piece of work in 30 days. While B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days before B and C leave. How many days more will A take to finish the work?
A.18 days
B.24 days
C.30 days
D.36 days

Answer 1

Hint: In this question, we need to determine the additional number of days that A will require to complete the remaining work as B and C left the work after 10 days. For this, we will use the unitary method and the concept which states that the total work completed in a day is the summation of the work done by the individual workers in a day.

Complete step-by-step answer:
Let A, B and C complete the work while working alone be A, B and C respectively.
According to the question,
A and B completes a work in 30 days working together. So, the amount of work completed by A and B in a day is given as:
$\dfrac{1}{A} + \dfrac{1}{B} = \dfrac{1}{{30}} - - - - (i)$
Similarly, B and C complete a work in 24 days working together. So, the amount of work completed by B and C in a day is given as:
$\dfrac{1}{B} + \dfrac{1}{C} = \dfrac{1}{{24}} - - - - (ii)$
Again, A and C complete work in 20 days working together. So, the amount of work completed by A and C in a day is given as:
$\dfrac{1}{A} + \dfrac{1}{C} = \dfrac{1}{{20}} - - - - (iii)$
Summation of all the one-day work by A, B and C gives the total of one-day work completed. So, adding equation (i) , (ii) and (iii) we get
$
  \dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{B} + \dfrac{1}{C} + \dfrac{1}{A} + \dfrac{1}{C} = \dfrac{1}{{30}} + \dfrac{1}{{24}} + \dfrac{1}{{20}} \\
  2\left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right) = \dfrac{1}{{30}} + \dfrac{1}{{24}} + \dfrac{1}{{20}} \\
  2\left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right) = \dfrac{{20 + 25 + 30}}{{600}} \\
  2\left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right) = \dfrac{{75}}{{600}} \\
  \left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right) = \dfrac{{75}}{{2 \times 600}} \\
   = \dfrac{1}{{16}} - - - - (iv) \\
 $
Hence, the total piece of work done by A, B and C together in a day is $\dfrac{1}{{16}}$.
Now, it is also given in the question that A, B and C work together only for 10 days and then, B and C leave. So, total work done in 10 days by A, B and C is given as:
$
  {\left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right)_{one{\text{ day}}}} = \dfrac{1}{{16}} \\
  {\left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right)_{{\text{10 days}}}} = \dfrac{{10}}{{16}} = \dfrac{5}{8} \\
 $
As, the total work is dented by 1 so, the remaining work that has to be done by A alone is given as$1 - \dfrac{5}{8} = \dfrac{{8 - 5}}{8} = \dfrac{3}{8}$
Now, substituting equation (ii) in the equation (iv) to determine the one-day work of A alone.
$
  \left( {\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}} \right) = \dfrac{1}{{16}} \\
  \dfrac{1}{A} + \left( {\dfrac{1}{B} + \dfrac{1}{C}} \right) = \dfrac{1}{{16}} \\
  \dfrac{1}{A} + \dfrac{1}{{24}} = \dfrac{1}{{16}} \\
  \dfrac{1}{A} = \dfrac{1}{{16}} - \dfrac{1}{{24}} \\
   = \dfrac{{3 - 2}}{{48}} \\
   = \dfrac{1}{{48}} \\
 $

Hence, the time taken by A to complete the work alone is 48 days.
As calculated earlier, that $\dfrac{3}{8}$ of the work needs to be done by A alone. So, by following the unitary method,
Total number of days in which A will complete the remaining work is $\dfrac{3}{8} \times 48 = 3 \times 6 = 18{\text{ days}}$
Hence, A will complete the remaining work in 18 days.
So, the correct answer is “Option A”.

Note: It is worth noting down here that, as B and C left the work so, the time of completion of the work will be more as compared to the time if they would work together all the time to complete the work.