Question

Sunil takes 5 days more than Anil to complete a certain work. 4 days after starting the work, Anil left the work. The remaining work was completed by Sunil in 5 days. Find how many days each will take to complete the work.

Answer 1

Hint: The work done and the time taken are always reciprocals. So we first find the work done by each one of them in one day and then equate it to the amount of work expected to be done in one day.

Complete step-by-step answer:
Let the number of days taken by Anil to complete the work be x.
Thus the number of days taken by Sunil to complete the work will be x+5.
Now the work is completed in 9 days. So every day the work done would be \[\dfrac{1}{9}\] .
Also the work done by Anil in one day is \[\dfrac{1}{x}\] and the work done by Sunil in one day will be \[\dfrac{1}{x+5}\] .
Now they work together for 4 days and then only Sunil works for 5 days.
Thus the work done is \[4\left( \dfrac{1}{x}+\dfrac{1}{x+5} \right)+\dfrac{5}{x+5}\] .
The work expected to complete in 9 days will therefore be \[9\times \dfrac{1}{9}\] which is 1.
Equating these to we get,
\[4\left( \dfrac{1}{x}+\dfrac{1}{x+5} \right)+\dfrac{5}{x+5}=9\times \dfrac{1}{9}\]
Taking LCM and cancelling the nines on the RHS, we get,
\[\begin{align}
  & 4\left( \dfrac{2x+5}{x\left( x+5 \right)} \right)+\dfrac{5x}{x(x+5)}=1 \\
 & \left( \dfrac{8x+20+5x}{x\left( x+5 \right)} \right)=1 \\
\end{align}\]

Cross multiplying we get,
\[\begin{align}
  & 13x+20=x\left( x+5 \right) \\
 & 13x+20={{x}^{2}}+5x \\
\end{align}\]
Taking all terms to one side we get,
\[{{x}^{2}}-8x-20=0\]
\[\begin{align}
  & {{x}^{2}}-10x+2x-20=0 \\
 & x\left( x-10 \right)+2\left( x-10 \right)=0 \\
 & \left( x-10 \right)\left( x+2 \right)=0 \\
\end{align}\]
Thus x=10 or x=-2. But since the work done and time are both positive we take x=10.
Thus Anil takes 10 days to complete that work and Sunil takes 10+5, which is 15 days to complete the same work individually.

Note: Be careful while applying the work time concept. This concept is very difficult to apply and thus can cause a lot of confusion. You can commit mistakes by not multiplying 9 on the RHS of the equation. This multiplication is must as it deals with the total work done in 9 days and not just in one day. If you do not multiply by 9 it means the work done in one day is equated to work done in 9 days as the LHS of the equation deals with the work done in 9 days.