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Mary purchased 3 theater tickets with an average (arithmetic mean) price of $8. If Mary also purchases...

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Mary purchased 3 theater tickets with an average (arithmetic mean) price of $8. If Mary also purchases...

by AAPL » Fri Sep 11, 2020 5:27 pm

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Mary purchased 3 theater tickets with an average (arithmetic mean) price of $8. If Mary also purchases a fourth theater ticket with a price of $16, what is the average price of all 4 theater tickets?

A. $4
B. $6
C. $8
D. $10
E. $12

OA D
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Re: Mary purchased 3 theater tickets with an average (arithmetic mean) price of $8. If Mary also purchases...

by Nirav Umaretiya » Sat Sep 12, 2020 8:31 pm
In this PS, they are testing the basics of Statistics.

Mary purchased the 3 tickets whose average is 8. They have not mentioned the price of these tickets.

\(\frac{x+y+z}{3}\) = 8

We don't need that either. We just need to know that
---> x+y+z = 8(3) = 24

Now she purchased a 4th ticket of $16, so the average is...

\(\frac{x+y+z+16}{4}\)
= \(\frac{24+16}{4}\)
= \(\frac{40}{4}\)
= 10

Correct answer = D
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