Question

A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in

Answer 1

Hint: We will be using a unitary method to solve this question. From the question we can see that B is twice efficient than A and C is thrice efficient than A. If A does 1 unit of work per day, B will do 2 units and similarly C will be doing 3 units.

Complete step-by-step answer:
Let us consider that B takes \[x\] days to complete the work alone, so A takes \[2x\] days to complete the work.
As it is given in the question that C is thrice as efficient as A, so we get that C takes \[\dfrac{2x}{3}\] days to complete the work.
Work done by A, B and C in one day \[=\dfrac{1}{2x}+\dfrac{1}{x}+\dfrac{3}{2x}......(1)\]
Simplifying equation (1) and equating it to \[\dfrac{1}{2}\] because the work is completed in 2 days so this is half the work. So we get,
 \[\,\Rightarrow \dfrac{6}{2x}=\dfrac{1}{2}......(2)\]
Now cancelling similar terms in equation (2) and then solving for x we get,
 \[\,\Rightarrow x=6\]
So B can complete the work in 6 days.

Note: Whatever the question is asking us to find we will take it to be x, this way it consumes less time. There can be an alternate and less time consuming solution. If A does 1 unit of work per day, B will do 2 units and similarly C will be doing 3 units of work.
Now, A+B+C one-day work \[=6\] units.
Since they complete the work in 2 days, so total work will be \[2\times 6=12\] units.
B alone will take \[\dfrac{12}{2}\] i.e. 6 days.