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# 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?

Last updated date: 20th Jun 2024
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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.

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.