The average (arithmetic mean) score of b quizzes for a group of studen

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

\((6.4b - x)/ (b-1) = 4.8\)
x = 1.6b + 4.8
C

\(\frac{6.4b - d}{( b -1 )} = 4.8\)

\(6.4b - d = 4.8b - 4.8\)

Or, \(1.6b + 4.8 = d\)

So, the discarded quiz score is 1.6b + 4.8 , answer will be (C)

average = total sum / total q

1) 6.4 = x/b = 6.4b-x
2) 4.8 = x/b-1 = 4.8b - 4.8 - x

6.4b-x=4.8b-4.8-x
1.6b+4.8

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