Question

A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes remaining work in 42 days. In how many days could both do it?

Answer 1

Hint: Divide the works in parts and find out how many parts of work A and B do alone in 1 day. Assume that both can do the work in $x$ days. Find out how many parts of the work A will do and how many B will do in these $x$ days. Take their sum and equate it with 80.

Complete step-by-step answer:

Let A do 1 part of work each day. Therefore, in 80 days he will do 80% of the work. Hence, the work contains 80 parts. If he works for 10 days then he will complete only 10 part of the work and therefore the remaining 70 part of the work will be done by B. It is given that B does this work in 42 days. Therefore, in one day B does, $\dfrac{70}{42}$$=\dfrac{5}{3}$ part of the work.
Let us assume that both will complete the work in $x$ days. Therefore, in $x$ days A will do $x$ part of the work and B will do $\dfrac{5x}{3}$ part of the work. The sum of their parts will be equal to the total part of the work, that is 80. Therefore,
$\begin{align}
  & x+\dfrac{5x}{3}=80 \\
 & \dfrac{8x}{3}=80 \\
 & x=30 \\
\end{align}$
Hence, both will complete the work in 30 days.

Note: We have divided the work in parts to make it free from assumption of 2 variables. Further we have used a unitary method to calculate the part of work that B does in one day. The sum of their individual contribution to the work will be equal to the total work.