Question

A takes 10 days less than the time taken by B to finish a piece of work if both A and B together can finish the work in 12 days the time taken by B to finish the work is
$
  (a){\text{ 20 days}} \\
  (b){\text{ 30 days}} \\
  (c){\text{ 40 days}} \\
  (d){\text{ 50 days}} \\
 $

Answer 1

Hint: In this question there is a relation given between the time taken by A and B to finish a work, so let the days taken by B to finish the work alone be some variable. Use the information provided in the question to formulate an equation in order to solve this variable.

Complete step-by-step answer:

Let B alone finish the work in x days.
And it is given that A takes 10 days less to finish the work from B.

Therefore A can finish the work in (x - 10) days.
So A’s one day work is $ = \dfrac{1}{{x - 10}}$………………… (1)
And B’s one day work is $ = \dfrac{1}{x}$……………………… (2)

Now it is given that both A and B together can finish the work in 12 days.
Therefore together A and B one day work is =$\dfrac{1}{{12}}$.

So A’s one day work + B’s one day work = together A and B one day work.
$ \Rightarrow \dfrac{1}{{x - 10}} + \dfrac{1}{x} = \dfrac{1}{{12}}$

Now simplify the above equation we have,
$ \Rightarrow x + x - 10 = \dfrac{{x\left( {x - 10} \right)}}{{12}}$
$ \Rightarrow 24x - 120 = {x^2} - 10x$
$ \Rightarrow {x^2} - 34x + 120 = 0$

Now factorize the above equation we have,
$ \Rightarrow {x^2} - 4x - 30x + 120 = 0$
$ \Rightarrow x\left( {x - 4} \right) - 30\left( {x - 4} \right) = 0$
$ \Rightarrow \left( {x - 4} \right)\left( {x - 30} \right) = 0$
$ \Rightarrow x = 4,30$

As x = 4 cannot possible because they together finish the work in 12 days,
Therefore the possible value of x is 30 days.
So, B can finish the work in 30 days.

Hence option (B) is correct.

Note: Whenever we come across such basic day and time questions the key concept to understand is that the work done by a person in one day in addition to the work done by another person in one day will surely be equal to the overall work done on that particular day. This concept will help getting on the right track while solving problems of such type.