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# At a certain company, the average (arithmetic mean) number

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CEO
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At a certain company, the average (arithmetic mean) number [#permalink]  17 Oct 2007, 18:01
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At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees to the number of the company's female employees?

1) There are 52 male employees at the company.

2) The average number of years of experience for the company's male and female employees combined is 9.3 years.
[Reveal] Spoiler: OA
CEO
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Re: At a certain company, the average (arithmetic mean) number [#permalink]  17 Oct 2007, 18:03
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Is this a correct approach? (I posted OA below, just wanted to see if I got it correctly or just by accident) Thanks.

Let:
X=total yrs for males
Y=total yrs for females
M= males
F= females.

X/M=9.8
Y/F=9.1

(X+Y)/M+F=9.3

The average number of years of experience for the company's male and female employees combined is 9.3 years.

X=9.8M
Y=9.1F

(9.8M+9.1F)/M+F=9.3 ---> 9.8M+9.1F=9.3M+9.3F -->

.5M=.2F---> .5M/F=.2---> M/F=.2/.5 ---> M/F=2/5

From this I get B.
Manager
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Re: At a certain company, the average (arithmetic mean) number [#permalink]  17 Oct 2007, 18:08
Looks like the right approach... I did it without math.

(1) doesn't talk anything about # of years of experience - INSUFFICIENT

(2), the weighted average of 9.1 and 9.8 is 9.3 ... It allows for multiple values of 9.1 and 9.8 to make that 9.3 ... But the ratio will always be constant - SUFFICIENT
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Re: At a certain company, the average (arithmetic mean) number [#permalink]  21 Feb 2008, 07:45
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GMATBLACKBELT wrote:
Big-O wrote:
Looks like the right approach... I did it without math.

(1) doesn't talk anything about # of years of experience - INSUFFICIENT

(2), the weighted average of 9.1 and 9.8 is 9.3 ... It allows for multiple values of 9.1 and 9.8 to make that 9.3 ... But the ratio will always be constant - SUFFICIENT

Can you explain this a bit more?? thx.

it's exactly what you did. its just terminology

weighted avg = (s1 +s2)/(N1+N2)

The combined sums are divided by the combined # of elements to produce a weighted outcome.
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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Re: At a certain company, the average (arithmetic mean) number [#permalink]  21 Feb 2008, 08:39
GMATBLACKBELT wrote:
At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees to the number of the company's female employees?

1) There are 52 male employees at the company.

2) The average number of years of experience for the company's male and female employees combined is 9.3 years.

B is the answer -> weighted average -> ratio of male to female is 2:5
Re: At a certain company, the average (arithmetic mean) number   [#permalink] 21 Feb 2008, 08:39
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