Question

A can do a piece of work in ‘x’ days and B can do the same work in ‘x+16’ days. If both working together can do it in 15 days. Calculate x.
a. 24
b. 25
c. 27
d. None of the above

Answer 1

Hint: We will find the work done by A in 1 day, then we will find the work done by B in 1 day using the unitary method. In the last step, we will find the work done by both of them in 1 day and the time required by both of them to complete the work. Finally, we will equate the obtained equation to 15 and solve the quadratic equation to get the value of x.

Complete step-by-step answer:
It is given in the question that A can do a piece of work in ‘x’ days and B can do the same work in ‘x+16’ days. If both working together can do it in 15 days. And we have been asked to find the value of x.
We will use the unitary method to find the work done by A and B in 1 day. Basically, the unitary method is a mathematical approach in which we will first find the value of 1 unit and then multiply it with the required value to get the desired values.
As A can do all the work in x days. So, work done by A in 1 day, is $\dfrac{1}{x}$.
Also, B can do all the work in (x+16) days, so the work done by B in 1 day is $\dfrac{1}{\left( x+16 \right)}$.
If both A and B work together, then work done in 1 day will be,
\[\dfrac{1}{x}+\dfrac{1}{\left( x+16 \right)}\]
On taking the LCM, we get,
\[\begin{align}
  & =\dfrac{x+16+x}{x\left( x+16 \right)} \\
 & =\dfrac{2x+16}{{{x}^{2}}+16x} \\
\end{align}\]
As \[\dfrac{2x+16}{{{x}^{2}}+16x}\] gives us the work done by A and B in 1 day, the total time taken by A and B to complete the work is given by the reverse of \[\dfrac{2x+16}{{{x}^{2}}+16x}\]. It is because the reverse of $\dfrac{1}{x}$ is x and the reverse of $\dfrac{1}{\left( x+16 \right)}$ is (x+16), which gives us the data regarding the time taken by A and B individually to complete the given work.
So, the time taken by both A and B to complete the work is, \[\dfrac{{{x}^{2}}+16x}{2x+16}\].
We also know that both A and B require 15 days to complete the work.
So, \[\dfrac{{{x}^{2}}+16x}{2x+16}\] must be equal to 15 days. So, we can write,
\[\dfrac{{{x}^{2}}+16x}{2x+16}=15\]
On cross multiplying, we get,
\[\begin{align}
  & {{x}^{2}}+16x=15\left( 2x+16 \right) \\
 & {{x}^{2}}+16x=30x+240 \\
\end{align}\]
On transposing 30x + 240 from RHS to LHS, we get,
\[\begin{align}
  & {{x}^{2}}+16x-30x-240=0 \\
 & {{x}^{2}}-14x-240=0 \\
\end{align}\]
We can also write -14x as -24x + 10x, so we get,
\[\begin{align}
  & {{x}^{2}}-24x+10x-240=0 \\
 & x\left( x-24 \right)+10\left( x-24 \right) \\
 & \left( x+10 \right)\left( x-24 \right)=0 \\
 & x=-10,24 \\
\end{align}\]
We know that work done cannot be negative, so we will neglect -10.
Thus, the value of x = 24.
So, A can do the work in 24 days.
Hence, option (a) is the correct answer.

Note: Many students make mistake by equating \[\dfrac{2x+16}{{{x}^{2}}+16x}\] with 15, this is wrong, as \[\dfrac{2x+16}{{{x}^{2}}+16x}\] gives us the value of the work done by A and B in one day and 15 days is the total time taken to complete the work. So, this mistake must be avoided. Also, after finding the value of x = 24 days, that is the time taken by A to complete the work, we can also find the time taken by B to do the work as (x+16) = (24+16) = 40 days.