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.

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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PhD in Applied Mathematics

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