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Que: An instructor gave the same test to three groups: A, B, and C. The average (arithmetic mean).......

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by Max@Math Revolution » Sat Jul 17, 2021 9:24 pm

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Que: An instructor gave the same test to three groups: A, B, and C. The average (arithmetic mean) scores for the three groups were 40, 50, and 60, respectively. The ratio of the numbers of candidates in the A, B, and C groups were 1 : 3: 5, respectively. What was the average score for the three groups combined?

(A) 21.11
(B) 36.66
(C) 50
(D) 54.44
(E) 80
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Source: — Problem Solving |

Solution: Let the number of candidates in groups A, B, and C be 1k, 3k, and 5k, respectively, where ‘k's is a constant of proportionality.

The average (arithmetic mean) scores for the three groups were 40, 50, and 60, respectively. The combined average for the three groups is

=> \(\frac{\left(40\cdot k\right)+\left(50\cdot3k\right)+\left(60\cdot5k\right)}{\left(k+3k+5k\right)}\)

=> \(\frac{\left[\left(40k+150k+300k\ \right)\right]}{\left(9k\right)}\)

=> \(\frac{490k}{9k}=54.44\)

Therefore, D is the correct answer.

Answer D
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