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marcodonzelli
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 25 Jan 2008, 09:34
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 people. The ratio of men to women is 7/5 in Finance, 3/4 in Marketing and 8/7 in R&D. If every Hardsoft employee works for exactly one of the 3 divisions, are there fewer than 120 women working in Hardsoft?

1. There are 100 men working in Hardsoft
2. More women work in Finance than in R&D



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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 25 Jan 2008, 10:28
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marcodonzelli
Hardsoft Ltd. has 3 divisions, each staffed by at least 10 people. The ratio of men to women is 7/5 in Finance, 3/4 in Marketing and 8/7 in R&D. If every Hardsoft employee works for exactly one of the 3 divisions, are there fewer than 120 women working in Hardsoft?

1. There are 100 men working in Hardsoft
2. More women work in Finance than in R&D
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E - insufficient, x - number of people in Finance, y - marketing

(24, 182, 15) yield more than 120
(72, 98, 30) yield fewer than 120

when y is high, the the number of females will greater than 120, when y is low the number of females is going to be fewer than 120. Then I did some number manipulations ...
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 25 Jan 2008, 10:46
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C

1.
minimum of women=82: (77:55, 15:20, 8:7)
maximum of women=121: (14:10, 78:104, 8:7)
insufficient.

2. insufficient

1&2. maximum of women=121: (14:10, 78:104, 8:7) is not work. Therefore, <120
sufficient.

I think It is tough problem to be solve for 2 min. Because we have to care of integers (there is no "half man or 2/3 woman"). +1
eschn3am, what do you think?
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 02 Feb 2008, 16:28
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walker
C

1.
minimum of women=82: (77:55, 15:20, 8:7)
maximum of women=121: (14:10, 78:104, 8:7)
insufficient.

2. insufficient

1&2. maximum of women=121: (14:10, 78:104, 8:7) is not work. Therefore, <120
sufficient.

I think It is tough problem to be solve for 2 min. Because we have to care of integers (there is no "half man or 2/3 woman"). +1
eschn3am, what do you think?
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Walker, I am not questioning your approach or the answer you gave. But can you explain how did you go about estimating the max and min number of women. For max, did you consider, keeping the factor of 3 maximum because 4 is smallest of 5,4 and 7. ?
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 03 Feb 2008, 01:30
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rgajare14
But can you explain how did you go about estimating the max and min number of women. For max, did you consider, keeping the factor of 3 maximum because 4 is smallest of 5,4 and 7. ?
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You are right.

The ratio of men to women is 7/5 in Finance, 3/4 in Marketing and 8/7 in R&D

3/4 is the most appropriate to yield maximum number of women.
we try to find max possible configuration:

7/5 -7 men; 3/4 - 84 men; 8/7 - 8 men: 7+84+8=99<100 - is not satisfy the condition.
7/5 (+7) men; 3/4 (-6) men; 8/7 - (0) men: add 7 men in Finance and subtract 6 men from Marketing
7/5 -14 men; 3/4 - 78 men; 8/7 - 8 men: 14+78+8=100 - satisfies the condition.
7/5 -10 women; 3/4 - 104 women; 8/7 - 7 women: 10+104+7=121 - satisfies the condition.
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 03 Feb 2008, 06:03
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marcodonzelli
Hardsoft Ltd. has 3 divisions, each staffed by at least 10 people. The ratio of men to women is 7/5 in Finance, 3/4 in Marketing and 8/7 in R&D. If every Hardsoft employee works for exactly one of the 3 divisions, are there fewer than 120 women working in Hardsoft?

1. There are 100 men working in Hardsoft
2. More women work in Finance than in R&D
Show more

70/50 + 6/8 + 24/21 = 100 men and 79 women <-- minimum women
14/10 + 78/104 + 8/7 = 100 men and 121 women <-- maximum women

In each of the above examples more women work in finance than in R&D. Since each of the examples work while using both statements and we get different results as to whether fewer than 120 women work at Hardsoft...

Answer E
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Hardsoft Ltd. has 3 divisions, each staffed by at least 10 03 Feb 2008, 06:50
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Yeah, You are right...



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