BTGModeratorVI wrote: ↑

Wed Dec 16, 2020 12:38 pm

A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80
(B) 82
(C) 84
(D) 86
(E) 88

Answer: E
Source: official guide

One option is to use weighted averages formula (for kicks):

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

We want the TOTAL average to equal 80
4/5 of the 5 scores have an average of 78
Let x = average of the 1 remaining test, which is worth 1/5 of the total average.

We get: 80 = (4/5)(78) + (1/5)(x)
Multiply both sides by 5 to get: 400 = 312 + x
Solve: x = 88

Answer: E

Cheers,
Brent

BTGModeratorVI wrote: ↑

Wed Dec 16, 2020 12:38 pm

A student's average (arithmetic mean) test score on 4 tests is 78. What must be the student's score on a 5th test for the student's average score on the 5 tests to be 80?

(A) 80
(B) 82
(C) 84
(D) 86
(E) 88

Answer: E
Source: official guide

Solution:

We use the formula: average = sum/quantity, or, equivalently, average x quantity = sum.

Even though we don't know the individual scores for each of the first 4 tests, we know the sum of the 4 scores must be 78 x 4 = 312.

Let's let n = the score on the 5th test. Thus, the sum of the 5 tests is 312 + 5th test score, or 312 + n. Also, because the new average will be 80, we substitute 80 for the new average. We can now solve for n.

Average = sum/quantity

80 = (312 + n)/5

400 = 312 + n

88 = n

Answer: E