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In a company with 48 employees, some parttime and some full [#permalink]
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12 May 2012, 05:26
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In a company with 48 employees, some parttime and some fulltime, exactly (1/3) of the parttime employees and (1/4) of the fulltime 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
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Re: In a company with 48 employees, some parttime and some full [#permalink]
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12 May 2012, 10:51
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alexpavlos wrote: In a company with 48 employees, some parttime and some fulltime, exactly (1/3) of the parttime employees and (1/4) of the fulltime 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 parttime employees is \(p\), then # of fulltime employees will be \(48p\). We want to maximize \(\frac{p}{3}+\frac{48p}{4}\) > \(\frac{p}{3}+\frac{48p}{4}=\frac{p+3*48}{12}=\frac{p}{12}+12\), 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 fulltime employees among 48) > \(\frac{p}{12}+12=3+12=15\). Answer: D. Or: since larger share of parttime employees take the subway then we should maximize # of parttime employees, but we should ensure that \(\frac{p}{3}\) and \(\frac{48p}{4}\) are integers. So \(p\) should be max multiple of 3 for which \(48p\) is a multiple of 4, which turns out to be for \(p=36\) > \(\frac{p}{3}+\frac{48p}{4}=15\). Similar question to practice: inacertainclassconsistingof36studentssomeboysand108870.htmlina200memberassociationconsistingofmenandwomen106175.htmloneweekacertaintruckrentallothadatotalof15505.htmlHope it helps.
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Re: In a company with 48 employees, some parttime and some full [#permalink]
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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]
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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 parttime and some fulltime, exactly 1/3 of the engineers are parttime employees and 1/4 of the engineers are fulltime 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 timeAnd 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]
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12 Jun 2012, 22:35
Quote: 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/inacompanywith48employeessomeparttimeandsomefull132442.htmlRegards,



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Re: In a company with 48 employees, some parttime and some full [#permalink]
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13 Jun 2012, 10:39
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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 (48x) = 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



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Re: In a company with 48 employees, some parttime and some full [#permalink]
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14 Jun 2013, 04:28



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Re: In a company with 48 employees, some parttime and some full [#permalink]
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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 parttime and some full [#permalink]
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18 Aug 2014, 21:46
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alex1233 wrote: In a company with 48 employees, some parttime and some fulltime, exactly (1/3) of the parttime employees and (1/4) of the fulltime 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 + (48P)/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 parttime and some full [#permalink]
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13 Jan 2018, 16:27
Hi All, 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 parttime and some fulltime. Since 1/3 of the parttimers take the subway to work, we know that the number of parttimers MUST be a MULTIPLE OF 3. Since 1/4 of the fulltimers take the subway to work, we know that the number of fulltimers 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 parttimers (since a greater fraction of that group (than the fraction of fulltimers) 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 fulltimers + 1/3 of 36 parttimers = 3 + 12 = 15 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: In a company with 48 employees, some parttime and some full [#permalink]
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13 Jan 2018, 16:49
Gonnaflynow wrote: 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!! In the last step, it should be 36/12 + 48/4 = 15. Cheers, Kabir




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