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duahsolo wrote: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?

A. 12
B. 13
C. 14
D. 15
E. 16

This is VERY similar to this Official Guide question: https://www.beatthegmat.com/og-help-problem-t286349.html

The important thing here to recognize here is that the number of part-time employees and the number of full-time employees must be positive INTEGERS. For example, we can't have 5 1/3 part-time employees.

Also recognize that we're told that we have some part-time and some full-time employees
Since "some" means 1 OR MORE, we cannot have zero part-time employees or zero full-time employees.

Okay, now onto the question...

We want to MAXIMIZE the number of employees who take the subway to work.
Since a greater proportion of part-time employees take the subway to work, we want to MAXIMIZE the number of PART-time employees in the class.
The greatest number of part-time employees is 47 (since 48 part-time employees would mean 0 full-time employees, and we must have at least 1 full-time employee)

47 part-time employees
This is no good, because 1/3 of the part-time employees take the subway to work, and 47 is not divisible by 3.

So, let's try ...
46 part-time employees
This is no good, because 1/3 of the part-time employees take the subway to work, and 46 is not divisible by 3.

As you can see, we need only consider values where the number of part-time employees is divisible by 3. So, that's what we'll do from here on...

45 part-time employees
If 1/3 of the part-time employees take the subway to work, then 15 part-time employees take subway. Fine.
HOWEVER, if there are 45 part-time employees, then there must be 3 full-time employees .
If 1/4 of the full-time employees take the subway to work, then there can't be 3 full-time employees, since 3 is not divisible by 4.

42 part-time employees
This means there are 6 full-time employees
If 1/4 of the full-time employees take the subway to work, then there can't be 6 full-time employees, since 6 is not divisible by 4.

39 part-time employees
This means there are 9 full-time employees
If 1/4 of the full-time employees take the subway to work, then there can't be 9 full-time employees, since 9 is not divisible by 4.

36 part-time employees and 12 full-time employees
1/3 of the part-time employees take the subway to work, so 12 part-time employees walk
1/4 of the full-time employees take the subway to work, so 3 full-time employees walk
PERFECT - it works!!
So, a total of 15 employees take the subway to work

Answer: D

Cheers,
Brent

Hi duahsolo,

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

GMAT assassins aren't born, they're made,
Rich