Question

$B$ takes $16$ less days than $A$ to do a piece of work. If both work together they finish the work in $15$ days. How much time will $B$ alone take to do the work?

Answer 1

Hint: For this question we will first assume the total work done and then assume the time taken by A to do the total work and then subtract $16$ from it to find out the time taken by B to do the complete work. Finally, we will apply the unitary method and find the time taken by both of them to complete the work and equate it to $15$ as given in the question.

Complete step-by-step answer:
Let’s say that the total work done is $W$ and suppose $A$ does it in $x$ days. Therefore according to the condition given in the question $B$ will do it in $x-16$ days.
Now, as we assumed that $A$ completes the work $W$ in $x$ days,
Therefore, $1$ unit of work will be done in $=\dfrac{W}{x}$ days.
Now, as we assumed that $B$ will take $\left( x-16 \right)$ days and hence, $1$ unit of work will be done in $=\dfrac{W}{\left( x-16 \right)}$ days.
Now together they will do $\left( \dfrac{W}{x}+\dfrac{W}{\left( x-16 \right)} \right)$ units of work in one day that is:
$\left( \dfrac{W}{x}+\dfrac{W}{\left( x-16 \right)} \right)=W\left( \dfrac{1}{x}+\dfrac{1}{\left( x-16 \right)} \right)=\dfrac{W\left( x-16+x \right)}{x\left( x-16 \right)}=\dfrac{W\left( 2x-16 \right)}{x\left( x-16 \right)}$
If $A$ and $B$ do $\dfrac{W\left( 2x-16 \right)}{x\left( x-16 \right)}$ units of work in one day and they need to do a total of $W$ units.
Therefore for $W$ units, they will take: \[\dfrac{W}{\dfrac{W\left( 2x-16 \right)}{x\left( x-16 \right)}}=\dfrac{W}{W}\times \dfrac{x\left( x-16 \right)}{\left( 2x-16 \right)}=\dfrac{x\left( x-16 \right)}{\left( 2x-16 \right)}\] days
It is given in the question that they both do the work $15$ days when they work together, therefore we will equate the above expression to $15$ that is \[\dfrac{x\left( x-16 \right)}{\left( 2x-16 \right)}=15\],

We will now multiply x into the bracket: $\dfrac{{{x}^{2}}-16x}{2x-16}=15$ ,
Now we will take the denominator to the RHS: ${{x}^{2}}-16x=15\left( 2x-16 \right)$, after this again multiply $15$ into the bracket and take the term to left hand side of the equation:
$\begin{align}
  & {{x}^{2}}-16x=30x-240\Rightarrow {{x}^{2}}-16x-30x+240=0 \\
 & \Rightarrow {{x}^{2}}-46x+240=0 \\
\end{align}$
Now, we will split the middle term such that it’s product is $240$ and the sum should be $46$ :
Therefore:
 $\begin{align}
  & {{x}^{2}}-46x+240=0 \\
 & {{x}^{2}}-40x-6x+240=0 \\
 & x\left( x-40 \right)-6\left( x-240 \right)=0 \\
 & \left( x-40 \right)\left( x-6 \right)=0 \\
\end{align}$
Hence the value of $x$ will be $40,6$ .
If $x=6$ then $\left( x-16 \right)=\left( 6-16 \right)=-10$ , since days cannot be negative therefore we will reject$x=6$.
 Now, If $x=40$ then $\left( x-16 \right)=\left( 40-16 \right)=24$ .
As we assumed that $B$ will do the work in $x-16$ days, therefore the time taken by $B$ is $24$ days.

Note: As we see that unitary method is basically an algorithm we generally do not describe it but it should be explained so that the logic is clear to the examiner. Also, the obtained quadratic equation must be solved with care as the last term is bigger therefore splitting the middle term can be difficult while solving the equation.