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Problem Solving

Expert replies
by RiyaR » Sat Oct 18, 2014 9:05 am
In a company with 48 employees, some part-time and some full-time, exactly 1/3 of the part-time employees and 1/4 of the full-time employees take the subway to work. What is the greatest possible number of employees who take the subway to work?

12

13

14

15

16
Is the answer 12?
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Source: — Problem Solving |

by [email protected] » Sat Oct 18, 2014 10:51 am
Hi RiyaR,

This type of question is rare on Test Day (you might see 1) and the shortcuts that are built into it are more about logic than anything else. If you're not sure how to start off this question, then you might have to do a bit of "brute force" (throw some numbers at it and see if a pattern emerges.

We know that there are 48 employees, some part-time and some full-time.

Since 1/3 of the part-timers take the subway to work, we know that the number of part-timers MUST be a MULTIPLE OF 3.
Since 1/4 of the full-timers take the subway to work, we know that the number of full-timers MUST be a MULTIPLE OF 4.

So we need a multiple of 4 added to a multiple of 3 that totals 48. We also want to MAXIMIZE the number of workers that take the subway, which means that we want to maximize the number of part-timers (since a greater fraction of that group (than the fraction of full-timers) takes the subway).

To find that perfect set of numbers, I'm going to start with multiples of 4 and see what happens....

4 --> 44 left (not a multiple of 3)
8 --> 40 left (not a multiple of 3)
12 -> 36 left (this IS a multiple of 3)

So 1/4 of 12 full-timers + 1/3 of 36 part-timers =

3 + 12 = 15

Final Answer: D

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