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In a company with 48 employees, some part-time and some full

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In a company with 48 employees, some part-time and some full [#permalink] New post 12 May 2012, 05:26
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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
[Reveal] Spoiler: OA
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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 12 May 2012, 10:51
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alexpavlos 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

Found this question; I can do it by picking numbers, however whats the algebraic solution to it?


Say # of part-time employees is p, then # of full-time employees will be 48-p.

We want to maximize \frac{p}{3}+\frac{48-p}{4} --> \frac{p}{3}+\frac{48-p}{4}=\frac{p+3*48}{12}=\frac{p}{12}+12, so we should maximize p, but also we should make sure that \frac{p}{12}+12 remains an integer (as it represent # of people). Max value of p for which p/12 is an integer is for p=36 (p can not be 48 as we are told that there are some # of full-time employees among 48) --> \frac{p}{12}+12=3+12=15.

Answer: D.

Or: since larger share of part-time employees take the subway then we should maximize # of part-time employees, but we should ensure that \frac{p}{3} and \frac{48-p}{4} are integers. So p should be max multiple of 3 for which 48-p is a multiple of 4, which turns out to be for p=36 --> \frac{p}{3}+\frac{48-p}{4}=15.

Similar question to practice:
in-a-certain-class-consisting-of-36-students-some-boys-and-108870.html
in-a-200-member-association-consisting-of-men-and-women-106175.html
one-week-a-certain-truck-rental-lot-had-a-total-of-15505.html

Hope it helps.
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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 12 May 2012, 13:00
Let part time emp=x
Let full time emp=y

then,

48=x+y.........(1)

(1/3)x+(1/4)y=No. of ppl taking the subway

4x+3y/12=No. of ppl taking the subway

using 1

x/12+3*48/12=No. of ppl taking the subway


so, the minimum value of x has to be 12.

Hence to maximize the no. put 36 for x

36/12+12/4=15

Hope that helps!!
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Re: CAN SOMEONE HELP SOLVE THIS PS [#permalink] New post 12 Jun 2012, 10:21
prakash111687 wrote:
Hi can someone let me know how the below problem can be solved efficiently.


A some engineering firm with 48 employees, in that some are part-time and some full-time, exactly 1/3 of the engineers are part-time employees and 1/4 of the engineers are full-time employees take the bus as a mean of transportation to work. What is the greatest possible number of engineers who take the bus as a mean of transportation to work?

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





Step 1) If you know that 1/4 of the employees are full time and ride the bus you can multiply .25*48 = 12
Step 2) find the difference between 1/3 and 1/4 = 1/12 * 48= 3 # of people that take the bus every day regardless of whether they are full or part time


And you can add step one and two together to get 15.

Tricked me up for a bit
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Re: CAN SOMEONE HELP SOLVE THIS PS [#permalink] New post 12 Jun 2012, 22:35
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Step 2) find the difference between 1/3 and 1/4 = 1/12 * 48= [b]3

Hi,

Firstly, 1/12*48 = 4

Secondly, You may find the correct question & solution here:
http://gmatclub.com/forum/in-a-company-with-48-employees-some-part-time-and-some-full-132442.html

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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 13 Jun 2012, 10:39
Hi,just wanted to share my method.

Let the number of part time employees = x
therefore,the number of full time employees = (48 - x)
1/3x + 1/4 (48-x) = number of people who take the subway

Since the number that constitutes "x" needs to be a multiple of both 3 & 4 and needs to be less than 48,
we take the LCM of 3,4 i.e. 12 and consider all common multiples until 48.
Hence we get 12,24,36 which are common multiples and < 48.

The question asks for the max number so plug in :

1/3(12) + 1/4(36) = 4 + 9 = 13
1/3(24) + 1/4(24) = 8 + 6 = 14
1/3(36) + 1/4(12) = 12 + 3 = 15 -----> maximum number and hence the answer :-D
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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 14 Jun 2013, 04:28
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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 16 Jun 2013, 03:20
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An easier approach... We know that the number of part time employees taking the bus must be divisible by 3 and full time by 4 and in order to maximize the number the number of part time employees should be higher. So i just jotted down a table of possible combos from 48 part time 0 full time as below, reducing 3 from 48 as i went down and chose the first option where the number of fulltime employees will be divisible by 4



Part Time Full Time Comment
48 0 Not valid as there is some number of Full time employees as per question
45 3 FT not divisible by 4
42 6 FT not divisible by 4
39 9 FT not divisible by 4
36 12 This is our answer

So number = 36/3 + 12/4 = 15
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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 18 Aug 2014, 19:15
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Re: In a company with 48 employees, some part-time and some full [#permalink] New post 18 Aug 2014, 21:46
alex1233 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


P/3 + F/4 = P/3 + (48-P)/4 = 12 + P/2
P/3 + F/3 = (P+F)/3 = 48/3 = 16
P/4 + F/4 = 12
P/3 + F/3 > P/3 + F/4 > P/4 + F/4
--> 16> 12 + P/12 > 12

GREATEST Possible: 12 + p/12 = 15 --> p = 36 ( integer --> good)
15 or D is the answer
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Re: In a company with 48 employees, some part-time and some full   [#permalink] 18 Aug 2014, 21:46
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