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

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

25 Aug 2012, 15:38

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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 25 Aug 2012, 15:43, edited 1 time in total.

Renamed the topic.

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

26 Aug 2012, 00:30

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prashi82 wrote:

(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.

Answer B

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] 26 Aug 2012, 00:30

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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

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