heyholetsgo wrote:
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets.
A. 4
B. 6
C. 8
D. 10
E. 12
An alternative approach—For some reason, many test-takers seem hesitant to alter information given in PS problems, but they have no trouble doing so in DS problems. In this question, we can reduce the confusing rate in the middle of the problem to one that is a much more relatable
production of w widgets.
Take the original rate—
\(\frac{5w}{4}\) in 3 days
Divide both parts by 5/4—
w in
\(\frac{12}{5}\) (or 2.4) days
Since the question asks us about
the slower machine producing
twice the number of widgets, we can deduce that the answer must be greater than 4 (since 2 * 2.4 is already nearly 5). Get rid of (A). At this point, we can test the answers in the following manner. I like to start in the middle of the original five options with either (B) or (D), since, if I get lucky, I will either choose the answer directly or know which option, greater than (D) or less than (B),
must be the answer without doing additional work. I would start with (D), given that, again, the question is asking about the slower machine at twice the production.
a) If I assume that Machine X takes 10 days to produce 2
w widgets, then it produces
w widgets in 5 days.
b) If Machine X takes 5 days to produce
w widgets, then Machine Y takes 5 - 2, or 3 days to produce the same
w widgets.
Putting everything together, the daily production of Machine X + Machine Y should equal the known production rate of
w widgets. Just invert the rates we have worked out so far to get the daily production.
\(\frac{1}{5}w+\frac{1}{3}w=\frac{5}{12}w\)
Divide out the
w—
\(\frac{1}{5}+\frac{1}{3}=\frac{5}{12}\)
\(\frac{3}{15}+\frac{5}{15}=\frac{5}{12}\)
\(\frac{8}{15}≠\frac{5}{12}\)
In fact,
\(\frac{8}{15}>\frac{5}{12}\)
One fraction is greater than half, while the other is less than half, so we know
the calculated production of Machine X is too high. It must be slower, so the answer has to be 12. It would not take long to run the numbers:
c) X: 6 days, Y: 4 days
\(\frac{1}{6}w+\frac{1}{4}w=\frac{5}{12}w\)
\(\frac{1}{6}+\frac{1}{4}=\frac{5}{12}\)
\(\frac{2}{12}+\frac{3}{12}=\frac{5}{12}\)
\(\frac{5}{12}=\frac{5}{12}\)Yes, a little knowledge of work rates helps. But algebra and logic can help you reach the answer
with certainty without too much effort or time, and keeping calm under pressure separates a 50-51 test-taker from someone with the same mathematical skill who balks at seeing a familiar concept presented in a less familiar way.
Good luck with your studies.
- Andrew