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Question: \(B\) takes \(16\) less days than \(A\) to do a piece of work. If both work together they finish the...

BB takes 1616 less days than AA to do a piece of work. If both work together they finish the work in 1515 days. How much time will BB alone take to do the work?

Explanation

Solution

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 1616 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 1515 as given in the question.

Complete step-by-step answer :
Let’s say that the total work done is WW and suppose AA does it in xx days. Therefore according to the condition given in the question BB will do it in x−16x-16 days.
Now, as we assumed that AA completes the work WW in xx days,
Therefore, 11 unit of work will be done in =Wx=\dfrac{W}{x} days.
Now, as we assumed that BB will take (x−16)\left( x-16 \right) days and hence, 11 unit of work will be done in =W(x−16)=\dfrac{W}{\left( x-16 \right)} days.
Now together they will do (Wx+W(x−16))\left( \dfrac{W}{x}+\dfrac{W}{\left( x-16 \right)} \right) units of work in one day that is:
(Wx+W(x−16))=W(1x+1(x−16))=W(x−16+x)x(x−16)=W(2x−16)x(x−16)\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 AA and BB do W(2x−16)x(x−16)\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 WW units.
Therefore for WW units, they will take: WW(2x−16)x(x−16)=WW×x(x−16)(2x−16)=x(x−16)(2x−16)\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 1515 days when they work together, therefore we will equate the above expression to 1515 that is x(x−16)(2x−16)=15\dfrac{x\left( x-16 \right)}{\left( 2x-16 \right)}=15,

We will now multiply x into the bracket: x2−16x2x−16=15\dfrac{{{x}^{2}}-16x}{2x-16}=15 ,
Now we will take the denominator to the RHS: x2−16x=15(2x−16){{x}^{2}}-16x=15\left( 2x-16 \right), after this again multiply 1515 into the bracket and take the term to left hand side of the equation:
x2−16x=30x−240⇒x2−16x−30x+240=0 ⇒x2−46x+240=0 \begin{aligned} & {{x}^{2}}-16x=30x-240\Rightarrow {{x}^{2}}-16x-30x+240=0 \\\ & \Rightarrow {{x}^{2}}-46x+240=0 \\\ \end{aligned}
Now, we will split the middle term such that it’s product is 240240 and the sum should be 4646 :
Therefore:
x2−46x+240=0 x2−40x−6x+240=0 x(x−40)−6(x−240)=0 (x−40)(x−6)=0 \begin{aligned} & {{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{aligned}
Hence the value of xx will be 40,640,6 .
If x=6x=6 then (x−16)=(6−16)=−10\left( x-16 \right)=\left( 6-16 \right)=-10 , since days cannot be negative therefore we will rejectx=6x=6.
Now, If x=40x=40 then (x−16)=(40−16)=24\left( x-16 \right)=\left( 40-16 \right)=24 .
As we assumed that BB will do the work in x−16x-16 days, therefore the time taken by BB is 2424 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.