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Bunuel
A company composed of two divisions is reviewing managerial salaries. The two managers in Division A of the company earn an average annual salary of $75,000. The three managers in Division B of the company earn an average annual salary of $100,000. What is the average salary of all the managers in the company’s two divisions?

A. $75,000
B. $85,000
C. $87,500
D. $90,000
E. $100,000

One thing to look at is to understand the average will not be 75,000 or 100,000. It is a weighted average so it will be closer to 100 than 75, this is because we have 3*100 and 2*75

Since it is closer to 100,000 than 75,000 it cannot be 85,000 because 85,000 - 75,000 = 10,000 and 100,000 - 85,000 = 15,000 so not possible. Also it cannot be 87,500 because 87,500 - 75,000 = 12,500 and 100,000 - 87,500 = 12,500 which makes the weighted average equal. (not true again).

2 * 75 = 150
3 * 100 = 300

450/5 = 90

Answer choice D
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Bunuel
A company composed of two divisions is reviewing managerial salaries. The two managers in Division A of the company earn an average annual salary of $75,000. The three managers in Division B of the company earn an average annual salary of $100,000. What is the average salary of all the managers in the company’s two divisions?

A. $75,000
B. $85,000
C. $87,500
D. $90,000
E. $100,000


This question can be solved through Weighted Average Method.

\(\frac{2*75000 + 3*100000}{5}\)

As max portion is 3 , there is a great possibility that answer will centered to 100000. It must be 90000.

The best answer is D.
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