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# A software company named Hardsoft has three divisions, each staffed by

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Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
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Kudos [?]: 866 [0], given: 28

A software company named Hardsoft has three divisions, each staffed by [#permalink]

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26 May 2010, 22:57
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Difficulty:

45% (medium)

Question Stats:

24% (00:03) correct 76% (02:59) wrong based on 15 sessions

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A software company named Hardsoft has three divisions, each staffed by at least 10 people. The ratio of men to women is 7:5 in Finance and Administration, 3:4 in Marketing and Sales and 8:7 in Research and Development. If there are 100 men working in Hardsoft and every Hardsoft employee works for one of the three divisions, then the difference between the maximum and minimum possible number of women working in Hardsoft is

(A) 39
(B) 40
(C) 41
(D) 42
(E) more than 42
[Reveal] Spoiler: OA

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press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Kudos [?]: 866 [0], given: 28

Manager
Joined: 04 May 2010
Posts: 86

Kudos [?]: 129 [2], given: 7

WE 1: 2 yrs - Oilfield Service
Re: A software company named Hardsoft has three divisions, each staffed by [#permalink]

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27 May 2010, 04:29
2
KUDOS
This is a tough question, and involves a bit of number crunching.

We know the total number of men, and using this we need to calculate the maximum and minimum possible number of women, provided the total number of men is 100.

The primary criteria is therefore: $$7x + 3y + 8z = 100$$ where $$x,y & z$$ are positive integers. Note that $$y>=2$$ to maintain the condition that each division has AT LEAST 10 members.

Now to maximize the number of women, we must invest the highest proportion of men into the division with the higher ratio of women to men. This happens to be the Marketing & Sales division of 3:4

What we are trying to do therefore is maximize $$y$$ and minimize $$x & z$$

From the equation we set up: $$y = \frac{(100 - 7x - 8z)}{3}$$
Let's use low values of x and z to get an integer value of y.

A bit of trial and error yields x = 2 and z = 1. which gives y = 26

For this value, the number of women, which is given by $$5x + 4y + 7z$$ comes out to be 121.

Now let's minimize the number of women. For this we invest the highest proportion of men into the division with the lowest ratio of women to men. This happens to be the Finance division..

What we are trying to do here is maximize $$x$$ and minimize $$z$$, and more importantly, $$y$$.

$$x = \frac{(100 - 3y - 8z)}{7}$$

A bit of trial and error yields y = 5 and z = 1 and x = 11; OR y = 2 and z = 3 and x = 10
We would prefer to reduce y, since it has a higher proportion of women, so choose the second set of values.

This yields a minimum number of women of 79.

Difference = 42. PICK D.

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Manager
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Re: A software company named Hardsoft has three divisions, each staffed by [#permalink]

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08 Jun 2010, 04:49
Hey
Is this the only way to solve this problem?? I find this approach difficult and lengthy.

Kudos [?]: 233 [0], given: 1

Manager
Joined: 04 May 2010
Posts: 86

Kudos [?]: 129 [0], given: 7

WE 1: 2 yrs - Oilfield Service
Re: A software company named Hardsoft has three divisions, each staffed by [#permalink]

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08 Jun 2010, 05:15
bibha wrote:
Hey
Is this the only way to solve this problem?? I find this approach difficult and lengthy.

I really can't think of an easier way - if it's any consolation, I really doubt this problem will ever show up on a GMAT test. There's too much trial and error required because of the different ratios that contribute to minimizing or maximizing the answer and the very rigid constraint on the total number of people that makes it even tougher. Coming to an exact answer ain't gonna be easy.

Do post any other solution you find though; it would be a great help to me too

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Intern
Affiliations: NYSSA
Joined: 07 Jun 2010
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Location: New York City
Schools: Wharton, Stanford, MIT, NYU, Columbia, LBS, Berkeley (MFE program)
WE 1: Senior Associate - Thomson Reuters
WE 2: Analyst - TIAA CREF
Re: A software company named Hardsoft has three divisions, each staffed by [#permalink]

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18 Jun 2010, 12:07
dimitri92 wrote:
A software company named Hardsoft has three divisions, each staffed by at least 10 people. The ratio of men to women is 7:5 in Finance and Administration, 3:4 in Marketing and Sales and 8:7 in Research and Development. If there are 100 men working in Hardsoft and every Hardsoft employee works for one of the three divisions, then the difference between the maximum and minimum possible number of women working in Hardsoft is
(A) 39 (B) 40 (C) 41 (D) 42 (E) more than 42

This crap took forever!!!

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Re: A software company named Hardsoft has three divisions, each staffed by [#permalink]

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23 Sep 2017, 20:56
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Re: A software company named Hardsoft has three divisions, each staffed by   [#permalink] 23 Sep 2017, 20:56
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