When some scores count for more, the ordinary average is the wrong tool.
The problem
Quizzes 85 (30%), exams 78 (50%), project 92 (20%). What is the final grade?
Multiply each score by its weight: 85(0.30) + 78(0.50) + 92(0.20) = 25.5 + 39 + 18.4 = 82.9. B is the plain average of the three, which ignores that the exams alone carry half the grade and pull the total down.
(85 + 78 + 92) ÷ 3 = 85
treats all three equally
to be found
exams count for half
The exam score is the lowest and carries the most weight, so expect the real grade to come out below 85.
A plain average is just a weighted mean where every weight happens to be equal. It is the special case, not the general rule.
| Component | Score | Weight | Score × weight |
|---|---|---|---|
| quizzes | 85 | 0.30 | 25.5 |
| exams | 78 | 0.50 | 39.0 |
| project | 92 | 0.20 | 18.4 |
| total | — | 1.00 | 82.9 |
Because the weights already add to 1, the column total is the answer — no dividing needed.
Divide by the total weight instead.
weighted mean = Σ (value × weight) Σ weight
3 units at 90, 5 units at 82
weights are unit counts
(270 + 410) ÷ 8
total ÷ total weight
= 680 ÷ 8 = 85
the GWA
Weights do not have to be percentages. Units, credits, number of students — any measure of “how much this one counts” works the same way.
Class A: 30 students averaging 80. Class B: 20 students averaging 90.
(80 + 90) ÷ 2 = 85
wrong
(2400 + 1800) ÷ 50
= 84
class A total = 30 × 80 = 2400
all their marks added
class B total = 20 × 90 = 1800
same again
(2400 + 1800) ÷ 50 = 84
50 students in all
You may never average two averages unless the groups are the same size. The bigger class pulls harder, and 85 is printed as a choice precisely to catch that.
“What do I need on the final to pass?” is the same formula rearranged.
quizzes 88 at 40%, final at 60%
target: 85 overall
0.4(88) + 0.6x = 85
set up
35.2 + 0.6x = 85
work the known part
x = 49.8 ÷ 0.6 = 83
the mark needed
If the answer comes out above 100, the target is unreachable — which is itself a legitimate answer to that kind of question.
class total: 50 × 80 = 4000
boys’ total: 30 × 76 = 2280
girls’ total: 4000 − 2280 = 1720
girls’ mean: 1720 ÷ 20 = 86
20 girls
The boys are 4 below 80, but the girls are not 4 above it: the groups are different sizes. 84 would be right only for equal groups.
50a + 90b = 66(a + b)
the totals add up
24b = 16a
collect the terms
a : b = 24 : 16 = 3 : 2
Shortcut: 66 is 16 above 50 and 24 below 90; the sizes are in the reverse ratio of those distances, 24 : 16. Check: 3 people at 50 and 2 at 90 give (150 + 180) ÷ 5 = 66 ✓.
The combined mean sits nearer the bigger group. 66 is closer to 50, so A is the larger group — 2 : 3 points the wrong way.
Expect two groups combined, a grade built from weighted parts, a missing group mean, or the final-exam score needed for a target. Always work with totals (count × mean, or weight × score) and never average the averages. Ten-second check: the combined mean lies between the group means, closer to the bigger group.
The problem
Class standing 90 (40%), midterm 75 (25%), finals 80 (35%). Find the final grade.
90(0.40) + 75(0.25) + 80(0.35) = 36 + 18.75 + 28 = 82.75. C is the unweighted mean (245 ÷ 3). Check the weights total 100% before starting — 40 + 25 + 35 does, so nothing is missing.
