Question

Two painters were painting a house. X finished his work in \[10{\rm{ days}}\]. The same amount of work was done by painter Y in \[12{\rm{ days}}\]. How much more work does painter X do in a day than painter Y.

Answer 1

Hint: : In the solution, the two painters X and Y are worked on the same and completed in 10 and 12 days, it means the painter X works faster than Y because he took only 10 days to complete the work. So we have to subtract the work of Y from the X.

Complete step-by-step answer:
The number of days that painter X finishes a particular work is \[X = 10\,{\rm{days}}\].
The number of days that painter Y finishes a particular work is \[Y = 12\,{\rm{days}}\].
So the X can complete certain work in 1 day is \[{X_1} = \dfrac{1}{{10}}\]
So the Y can complete the same work in 1 day is \[{Y_1} = \dfrac{1}{{12}}\]
The equation to find the work does painter X do in a day than painter Y is, (let us assume Z be the one day work is that X do more than Y)
\[Z = {X_1} - {Y_1}\]
Substituting the values in the above equation, we get
\[Z = \dfrac{1}{{10}} - \dfrac{1}{{12}}\]
Taking the LCM of 10 and 12, we get 60, then
\[\begin{array}{c}
Z = \dfrac{{6 - 5}}{{60}}\\
 = \dfrac{1}{{60}}
\end{array}\]
Therefore, the work that painter X does more than the Y is\[\dfrac{1}{{60}}\].

Note: Here, in the question, it is mentioned only the working day so to find the one-day work of both painters, we have to reverse it (it means changing 10 as \[\dfrac{1}{{10}}\]). Be sure about the LCM process, we have to take the least numbers firstly to find the LCM. It should be noted that the value of X is kept first and the value of Y should be done from it.