Question

A, B, C can do a piece of work individually in 8, 10 and 15 days respectively. If A and B start working together but A quits after 2 days. After this C joins B till it has completion of work. In how many days will the work be completed?
A. \[\dfrac{{53}}{9}\]
B. \[\dfrac{{34}}{7}\]
C. \[\dfrac{{85}}{3}\]
D. \[\dfrac{{53}}{{10}}\]

Answer 1

Hint: Here we use unitary method to find the amount of work completed by each A, B and C separately. Add the amount of work done by A and B together in one day and multiply it by the number of days till A and B worked together. Subtract the value of work done by A and B from 1 to find the amount of remaining work. In the same way, find the amount of work done by B and C together in one day. Assume the number of days to complete the work as a variable we equate the amount of remaining work to the amount of work obtained from B and C in the number of days.
* Unitary method helps us to find the value of a single unit when we are given the value of multiple units by dividing the value of multiple units by the number of units.

Complete step-by-step answer:
We are given A, B, C can do a piece of work individually in 8, 10 and 15 days respectively.
Let us assume work done by A in 8 days is 1 piece
Then using unitary method, we can write
\[ \Rightarrow \]Work done by A in 1 day is \[\dfrac{1}{8}\]
Work done by B in 10 days is 1 piece.
Then using unitary method, we can write
\[ \Rightarrow \]Work done by B in 1 day is \[\dfrac{1}{{10}}\]
Work done by C in 15 days is 1 piece.
Then using unitary method, we can write
\[ \Rightarrow \]Work done by C in 1 day is \[\dfrac{1}{{15}}\]
Now we are given A and B work together.
Work done by A and B together in one day will be the sum of work done by A and work done by B individually in one day i.e. \[\dfrac{1}{8} + \dfrac{1}{{10}}\]
Take LCM to calculate the value
\[ \Rightarrow \dfrac{1}{8} + \dfrac{1}{{10}} = \dfrac{{10 + 8}}{{8 \times 10}}\]
\[ \Rightarrow \dfrac{1}{8} + \dfrac{1}{{10}} = \dfrac{{18}}{{80}}\]
Cancel the same factors from numerator and denominator.
\[ \Rightarrow \dfrac{1}{8} + \dfrac{1}{{10}} = \dfrac{9}{{40}}\]
Therefore work done by A and B two days will be given by \[2 \times \]work done by A and B in one day
\[ = 2 \times \dfrac{9}{{40}}\]
\[ = \dfrac{9}{{20}}\]
Then the amount of work remaining can be calculated by 1 minus the amount of work completed.
Amount of work remaining \[ = 1 - \dfrac{9}{{20}}\]
Take LCM to calculate the value
\[ = \dfrac{{20 - 9}}{{20}}\]
Therefore, amount of work remaining\[ = \dfrac{{11}}{{20}}\] … (1)
Now we know B and C work together.
Work done by B and C together in one day will be the sum of work done by B and work done by C individually in one day i.e. \[\dfrac{1}{{10}} + \dfrac{1}{{15}}\]
Take LCM to calculate the value
\[ \Rightarrow \dfrac{1}{{10}} + \dfrac{1}{{15}} = \dfrac{{15 + 10}}{{10 \times 15}}\]
\[ \Rightarrow \dfrac{1}{{10}} + \dfrac{1}{{15}} = \dfrac{{25}}{{150}}\]
Cancel the same factors from numerator and denominator.
\[ \Rightarrow \dfrac{1}{{10}} + \dfrac{1}{{15}} = \dfrac{1}{6}\]
Let B and C work together for n days.
Then the amount of work done by B and C in n days is given by \[\dfrac{1}{6}n\].
Since we know B and C work together till the work is complete, so they remaining work is completed by B and C
Equate the value of work done by B and C to the remaining work in equation (1)
\[ \Rightarrow \dfrac{1}{6}n = \dfrac{{11}}{{20}}\]
Cross multiply the value of 6 from LHS to RHS
\[ \Rightarrow n = \dfrac{{11}}{{20}} \times 6\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow n = \dfrac{{33}}{{10}}\]
Now we know A and B worked for 2 days and B and C worked for \[\dfrac{{33}}{{10}}\]days.
So, total number of days to complete the work \[ = 2 + \dfrac{{33}}{{10}}\]
Take LCM to calculate the value
\[ = \dfrac{{20 + 33}}{{10}}\]
\[ = \dfrac{{53}}{{10}}\]

So, the correct answer is “Option D”.

Note: Students can get confused in the part where we are finding the value of remaining work, we are subtracting the amount of work done from 1 because we consider the complete work as one unit and the fractions of work done denote the part of that one unit that is completed.