Question

If Logan does a job in 63 hours and Rose does a job in 18 hours, how long will it take them to do the job together?

Answer 1

Hint: The amount of work done by a person varies directly with the time taken by him/her to complete it. : If two quantities a and b vary with each other in such a manner that the ratio a : b remains constant and positive, then we can say that a and b vary directly with each other or a and b are in direct variation.
Thus, if a man can complete a piece of work in $10days$, then by unitary method, we can say that in a day he will do only $\dfrac{1}{{10}}th$ part of the total work.
On the other hand, if a man completes $\dfrac{1}{{10}}th$ of the work in one day, then he will take $10days$ to complete the work.

Complete step-by-step answer:
We have,
Time taken by Logan to complete the work= $63hours$.
Time taken by Rose to complete the work= $18hours$.
Therefore, Work done by Logan in$1hour = \dfrac{1}{{63}}$.
Work done by Rose in $1hour = \dfrac{1}{{18}}$.
So, work done by Logan and Rose in$1hour = \dfrac{1}{{63}} + \dfrac{1}{{18}}$.
Now, L.C.M of 63 and 18 is 126.
$\therefore \dfrac{1}{{63}} + \dfrac{1}{{18}} = \dfrac{{2 + 7}}{{126}}$
By adding the numerator we get,
\[ \Rightarrow \dfrac{1}{{63}} + \dfrac{1}{{18}} = \dfrac{9}{{126}}\]
Converting the fraction in its lowest form we get,
\[\dfrac{1}{{63}} + \dfrac{1}{{18}} = \dfrac{1}{{14}}\]
Logan and Rose will take $\dfrac{1}{{\dfrac{1}{{14}}}}hours$ or reciprocal of $\dfrac{1}{{14}}hours$to complete the job together.$14hours$

Hence, Logan and Rose will take $14hours$ to complete the job together.$14hours$

Note: Rule-1: If a person $X$ completes a piece of work in $n$ days, then work done by person $X$in one day is $\dfrac{1}{n}th$part of the work.
Rule-2: If a person $X$ completes $\dfrac{1}{n}th$ part of the work in one day, then person $X$ will take $n$ days to complete the work.