ACT Math study guide · Chapter 40
Averages in Context
A missing-value average problem gives the average and all but one value; multiplying the average by the count recovers the total, and subtracting the known values isolates the missing one. This is the same technique used in Part V, applied here inside word problems about grades, tests, and classes.
A weighted average is not the same as a simple average of two numbers. When two groups contribute unequally, whether by credit hours, class size, or exam weight, each value must be multiplied by its weight before dividing by the total weight. Averaging the two raw numbers directly, ignoring the weights, is one of the most common errors on this topic.
Average speed for a full trip is total distance divided by total time, not the average of the individual speeds. A trip that covers equal distances at two different speeds spends more time at the slower speed, which pulls the true average speed below the simple average of the two speeds.
Where this appears as a model: a trip described in stages. A trip covering 240 miles in 4 hours followed by 180 miles in 4 hours has an average speed of 52.5 miles per hour, found from total distance divided by total time instead of by averaging 60 and 45 directly.
▸ Weighted averages
Worked Example A final grade is weighted 30% on a midterm and 70% on a final exam. A student scores 80 on the midterm and 95 on the final. Find the final grade.
Multiply each score by its weight, then add the results.
0.30(80) + 0.70(95) = 24 + 66.5 = 90.5
▸ Average speed for a full trip
Worked Example A drive covers 120 miles at 40 miles per hour, then 120 miles at 60 miles per hour. Find the average speed for the entire trip.
Average speed is total distance divided by total time, not the average of the two speeds.
Time: 120/40 = 3 hr and 120/60 = 2 hr
Average speed = 240/5 = 48
▸ Practice
Question 1
- AA. 15
- BB. 19
- CC. 13
- DD. 17
Question 2
- FF. 77.5
- GG. 80
- HH. 76
- JJ. 79
Question 3
- AA. 37.5
- BB. 30
- CC. 33.33
- DD. 40
Question 4
- FF. 106
- GG. 96
- HH. 100
- JJ. 110
Question 5
- AA. 3.6
- BB. 3.5
- CC. 3.7
- DD. 3.4
Question 6
- FF. 45
- GG. 50
- HH. 40
- JJ. 42.5
Question 7
- AA. 23
- BB. 19
- CC. 15
- DD. 25
Question 8
- FF. 3.4
- GG. 3.5
- HH. 3.6
- JJ. 3.3
Question 9
- AA. 51.43
- BB. 52.5
- CC. 50
- DD. 55
Question 10
- FF. 9
- GG. 15
- HH. 12
- JJ. 6
Question 11
- AA. 78
- BB. 85
- CC. 92
- DD. 80
Question 12
- FF. 45
- GG. 48
- HH. 50
- JJ. 35
Question 13
- AA. −3
- BB. 3
- CC. 5
- DD. 1
Question 14
- FF. 83
- GG. 78
- HH. 81
- JJ. 85
Question 15
- AA. 25
- BB. 28
- CC. 22
- DD. 31
Question 16
- FF. 3.0
- GG. 3.2
- HH. 3.4
- JJ. 2.8
Question 17
- AA. 50
- BB. 44
- CC. 52
- DD. 48
Question 18
- FF. 15
- GG. 12
- HH. 18
- JJ. 9
Question 19
- AA. 82.5
- BB. 77.5
- CC. 80
- DD. 85
Question 20
- FF. 88
- GG. 84
- HH. 90
- JJ. 86.8
▸ Answers and Explanations
Show the answers and explanations
1: D. Total = 15×4 = 60; 60 − (10+14+19) = 60 − 43 = 17. Choice A reports the given average itself instead of solving for the missing value.
2: J. 0.40(70) + 0.60(85) = 28 + 51 = 79. Choice F computes the simple average of the two exam scores instead of the weighted average.
3: C. Time: 100/50=2 hr and 100/25=4 hr; average speed = 200/6 ≈ 33.33. Choice A averages the two speeds directly instead of using total distance divided by total time.
4: F. Total = 90×5 = 450; 450 − (88+92+79+85) = 450 − 344 = 106. Choice G comes from an arithmetic slip while recovering the total or subtracting the known values.
5: A. (3×4.0 + 2×3.0) ÷ (3+2) = (12+6) ÷ 5 = 3.6. Choice B computes the simple average of the two GPAs instead of the weighted average by credit hours.
6: H. Time: 60/30=2 hr and 60/60=1 hr; average speed = 120/3 = 40. Choice F averages the two speeds directly instead of using total distance divided by total time.
7: A. Total=19×4=76; 76−(14+22+17)=23. Choice B reports the given average itself instead of solving for the missing value.
8: G. (3×3.0+5×3.8)÷8=(9+19)÷8=3.5. Choice F computes the simple average of the two GPAs instead of the weighted average by credit hours.
9: A. Time: 180/60=3 hr and 180/45=4 hr; average speed=360/7≈51.43. Choice B averages the two speeds directly instead of using total distance divided by total time.
10: F. Total=12×5=60; 60−(8+10+15+18)=9. Choice G comes from an arithmetic slip while recovering the total or subtracting the known values.
11: B. 0.50(78)+0.50(92)=39+46=85. Choice A reports one exam score alone instead of computing the weighted average of both.
12: G. Time: 90/45=2 hr and 150/50=3 hr; average speed=240/5=48. Choice F comes from an arithmetic slip while computing the total distance or total time.
13: B. (−5−2+8+15−1)÷5=15÷5=3. Choice A comes from a sign error while summing the values.
14: H. (25×75+15×91)÷40=(1,875+1,365)÷40=3,240÷40=81. Choice F computes the simple average of the two class averages instead of the weighted average by class size.
15: B. Total=22×4=88; 88−(16+25+19)=28. Choice A comes from an arithmetic slip while recovering the total or subtracting the known values.
16: G. (4×2.8+4×3.6)÷8=(11.2+14.4)÷8=3.2. Choice F comes from an arithmetic slip while combining the two weighted GPA components.
17: D. Time: 240/60=4 hr and 240/40=6 hr; average speed=480/10=48. Choice A averages the two speeds directly instead of using total distance divided by total time.
18: F. Total=15×5=75; 75−(10+12+18+20)=15. Choice G comes from an arithmetic slip while recovering the total or subtracting the known values.
19: C. 0.60(70)+0.40(95)=42+38=80. Choice A computes the simple average of the two exam scores instead of the weighted average.
20: J. (18×82+12×94)÷30=(1,476+1,128)÷30=2,604÷30=86.8. Choice F computes the simple average of the two class averages instead of the weighted average by class size.