Bunuel wrote:

The average (arithmetic mean) score of b quizzes for a group of students is 6.4. When one quiz score is discarded, the average of the remaining quiz scores becomes 4.8. What is the discarded quiz score?

A. 1.6b + 24

B. 1.6b – 4.8

C. 1.6b + 4.8

D. 6.4b – 24

E. 6.4b +4.8

A*n = S

First set of scores

A = 6.4, n = b

Sum, S = 6.4b

Second set, one quiz discarded

A = 4.8, n = (b - 1)

Sum, S = 4.8(b - 1) = (4.8b - 4.8)

The difference in sums is the discarded score:

6.4b - (4.8b - 4.8) =

6.4b - 4.8b + 4.8 =

1.6b + 4.8

Answer C

Algebra is faster, but if you're really stuck, assign values.

(Average, A) * (number of terms,n) = Sum, S

Let b = 2

Then the sum for the first set of scores is A*n = S: (6.4 * 2) = 12.8

One score is removed. Number, n, of scores is now (2-1)= 1

Sum for second set of scores:(4.8 * 1) = 4.8

The score that was removed is

(Sum1 - Sum 2) = (12.8 - 4.8) = 8

With b = 2, find the answer that yields 8

Discard A and E (they are huge)

Discard B and D (with a little mental math, it's clear they're negative)

Answer C by POE. CHECK

C. 1.6b + 4.8

(1.6)(2) + 4.8 = (3.2 + 4.8) = 8 MATCH

Answer C

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