Question

 A started work and left after working for 2 days. Then B was called and he finished the work in 9 days. Had A left the work after working for 3 days, B would have finished the remaining work in 6 days. In how many days can each of them working alone to finish the whole work?
A. 5 days, 8.5 days
B. 2.5 days, 7.5 days
C. 5 days. 15 days
D. 3 days, 15 days

Answer 1

Hint: In the above question, first of all, we will suppose A and B can do the work in x and y days. Then, we will find the work done by A or B in the given days by using the simply unitary method and write the sum of work that is equal to 1.

Complete step-by-step answer:

We have been given that, A started a work and left after working for 2 days and the remaining work is completed by B in 9 days. Also, if A left the work after working for 3 days, B would have finished the remaining work in 6 days.
Let us suppose, A and B can do the work in x and y days respectively.
Now, we know that the sum of work done by A and B is equal to 1.
In x days work done by \[\text{A }=\text{ 1}\]
In 1 day work done by \[\text{A }=\dfrac{1}{x}\]
So, in 2 day work done by \[\text{A }=\dfrac{2}{x}\]
Similarly, in 3 day work done by \[\text{A }=\dfrac{3}{x}\]
Again, In y days work done by \[\text{B }=\text{ 1}\]
In 1 day work done by \[\text{B }=\dfrac{1}{y}\]
So, in 9 days work done by \[\text{B }=\dfrac{9}{y}\]
Similarly, in 6 days work done by \[\text{B }=\dfrac{6}{y}\]
According to question, we have,
\[\begin{align}
  & \dfrac{2}{x}+\dfrac{9}{y}=1\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
 & \dfrac{3}{x}+\dfrac{6}{y}=1\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
Let us suppose, \[\dfrac{1}{x}=a\text{ and }\dfrac{1}{y}=b,\] then, the equations become,
\[\begin{align}
  & 2a+9b=1\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)} \\
 & 3a+6b=1\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)} \\
\end{align}\]
On performing \[(iv)\times 2-(iii)\times 3,\] we get,
\[\begin{align}
  & \left( 6a+12b \right)-\left( 6a+27b \right)=2-3 \\
 & 6a+12b-6a-27b=-1 \\
\end{align}\]
On cancelling similar terms, we get,
\[\begin{align}
  & -15b=-1 \\
 & b=\dfrac{1}{15} \\
\end{align}\]
Since, \[b=\dfrac{1}{y}\]
\[\begin{align}
  & \dfrac{1}{y}=\dfrac{1}{15} \\
 & y=15\text{ }days \\
\end{align}\]
On substituting \[b=\dfrac{1}{15}\] in equation (iii), we get,
\[\begin{align}
  & \Rightarrow 2a+9\times \dfrac{1}{15}=1 \\
 & \Rightarrow 2a+\dfrac{3}{5}=1 \\
 & \Rightarrow 2a=1-\dfrac{3}{5} \\
 & \Rightarrow 2a=\dfrac{5-3}{5} \\
 & \Rightarrow 2a=\dfrac{2}{5} \\
 & \Rightarrow a=\dfrac{1}{5} \\
\end{align}\]
Since, \[a=\dfrac{1}{x}\]
\[\begin{align}
  & \dfrac{1}{x}=\dfrac{1}{3} \\
 & x=5\text{ days} \\
\end{align}\]
Therefore, A and B finish the work in 5 days and 15 days respectively. So, the correct option is C.

Note: While solving the equation \[\dfrac{1}{x}\text{ and }\dfrac{1}{y}\] it is better to take \[\dfrac{1}{x}\text{=a and }\dfrac{1}{y}=b\]. Otherwise, it is very difficult to solve the equation also, it will take you a lot of time. Also, remember that after solving the equation in a and b you have to further substitute the value of a and b in terms of x and y, which will give us the final answer.