Question

A alone can do a piece of work in $ 8 $ days. A and B together can finish the same work in $ 6 $ days. If B alone is doing the work, how long will it take?

Answer 1

Hint: The given problem is a work and time problem. We will use the unitary method to solve this problem. We are given the time consumed by A to finish a work and the time consumed by A and B together to finish the work. Using these given things we are asked to find the time consumed by B alone to finish the work. We will also introduce algebraic expressions to solve this problem.

Complete step-by-step answer:
A can do a piece of work in $ 8 $ days.
Let the work done by A in one day is $ \dfrac{1}{8} $ th of the work done.
A and B together can do the same work in 6 days.
So, the work done by A alone is $ 6 \times \dfrac{1}{8} $ of the total work.
 $ 6 \times \dfrac{1}{8} = \dfrac{6}{8} $
Now the rest work be done by B let it be $ x $ .
So, $ \dfrac{6}{8} + x = 1 $
 $ \Rightarrow x = 1 - \dfrac{6}{8} = \dfrac{2}{8} $
The work is for $ 6 $ days.
So, the actual work done by B for 1 day is $ \dfrac{1}{6} \times \dfrac{2}{8} = \dfrac{1}{{24}} $
Since, B takes $ 1 $ a day to complete $ \dfrac{1}{{24}} $ part of the work.
This means B takes $ 24 $ days to finish the work alone.
So, the correct answer is “$ 24 $”.

Note: The given problem is based upon a statement containing some information which we have used to find an algebraic expression. So, formulate the question properly to calculate the result. The result we are obtaining is for $ 6 $ days which we still have to calculate for one day and this can be done by multiplying the reciprocal of $ 6 $ in it.