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

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At a certain company, average (arithmetic mean)number of [#permalink]

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25 Aug 2012, 15:38
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Question Stats:

62% (01:48) correct 38% (01:10) wrong based on 146 sessions

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At a certain company, average (arithmetic mean) number of years of experience is 9.8 years for male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees and 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

Last edited by Bunuel on 24 Nov 2013, 11:03, edited 2 times in total.
Renamed the topic.
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Re: At a certain company, average (arithmetic mean)number of [#permalink]

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26 Aug 2012, 00:30
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prashi82 wrote:
At a certain company, average (arithmetic mean)number of years of experience is 9.8 years for male employees and 9.1 years for the female employees. What is the ratio of the number of the company's male employees and 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.

(1) Definitely not sufficient, we don't know neither the number of women, nor the combined average for the company.

(2) Sufficient. The combined average being 9.3, the differences form this average are 0.5 and 0.2, therefore the ratio between the male and female employees is 0.2:0.5=2:5.

This is a question involving weighted average.
Having two quantities $$Q_1$$ and $$Q_2$$ with averages $$a_1$$ and $$a_2$$ respectively, if the combined average is $$a$$, and let assume that $$a_1>a>a_2,$$ then we can write:

$$\frac{a_1Q_1+a_2Q_2}{Q_1+Q_2}=a$$ from which $$a_1Q_1+a_2Q_2=aQ_1+aQ_2$$ or $$(a_1-a)Q_1=(a-a_2)Q_2,$$ which means that the distances from the combined average are inversely proportional to the quantities.
This equality we can also be written as $$\frac{a_1-a}{a-a_2}=\frac{Q_2}{Q_1}.$$

In the above question, the differences were 0.5 and 0.2 (men's average to women's average differences), which give a ratio of 5:2, then the quantities (numbers of employees) are in a ratio of 2:5 (male to women).
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Re: At a certain company, average (arithmetic mean)number of [#permalink]

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14 Feb 2015, 15:27
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Re: At a certain company, average (arithmetic mean)number of [#permalink]

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14 Feb 2015, 21:08
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Hi All,

This DS question is ultimately about Weighted Averages.

We're told that the average work experience (at a particular company) for men is 9.8 years and the average work experience for women is 9.1 years. We're asked for the RATIO of Men to Women at this company.

In these types of question, we don't necessarily need the number of men and the number of women to find the ratio of men to women...

Fact 1: 52 males work at the company.

This tells us nothing about the number of women working at the company.
Fact 1 is INSUFFICIENT.

Fact 2: The average work experience for ALL employees is 9.3 years.

Using this Fact, and the information in the prompt, I'm going to set up a Weighted Average calculation:

M = # of male employees
F = # of female employees

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

9.8M + 9.1F = 9.3M + 9.3F
0.5M = 0.2F

5M = 2F

At this point, if you recognize that we have a ratio, then you can stop working. If you want to do the next "math step", then that's fine though...

M/F = 2/5

We now have the ratio that we're asked for.
Fact 2 is SUFFICIENT.

[Reveal] Spoiler:
B

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Re: At a certain company, average (arithmetic mean)number of   [#permalink] 14 Feb 2015, 21:08
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