nalinnair wrote:
Five machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 30 hours to fill a certain production order, how many fewer hours does it take all five machines, operating simultaneously, to fill the same production order?
(A) 3
(B) 5
(C) 6
(D) 16
(E) 24
chetan2u BunuelMy question is nothing new but 'Old wine new bottle'. Can we say this?
Since all 5 machines work at a same constant rate, it can be said that either 5 do the work of 4 or 4 do the same work more(+25%) efficiently. If 4 machines do the work they perform the work with an individual efficiency that is 25% better than each machine's previous efficiency -
As the word 'simultaneously' is used each of the 4 machines do 30 hours work, totalling 120 hours of work. So, 120 hours of work when completed by 5 machines each work 24 hours. Thus, they each took 6 hours less or in other words they performed the work more efficiently - together efficiency improved by \(\frac{6}{30}*100 = 20%\) for 5 machines. We can also say that individually each machine performed with 25% more(1 machine's work divided among 4 machines) efficiency. Thus, each take \(\frac{30}{1.25} = 24\) hours. Answer is 30 - 24 = 6 hours.
OR
If we ignore(only when all the machines work at same constant rate) the the word 'simultaneously' and take machines to be working consecutively one after the other, each of the 4 machines take \(\frac{30}{4} = 7.5\) hours. Now that 5 machines are going to do the same work, it can be said that 4 can do the same work if each of them work at 1.25times rate(since 1 machine's work is distributed among 4). So the 4 machines now take \(\frac{7.5}{1.25} = 6\) hours and total time is 24 hours. Hence answer is 30 - 24 = 6 hours.
Second scenario is just to understand the question better. Can we say that?
Please comment wherever I'm wrong.