Whenever you hear "the average Filipino household," "her average grade," or "the average temperature this April," you are meeting one of the most useful ideas in all of mathematics: a single number that stands in for a whole group. Averages are easy to compute but surprisingly rich to reason about, and the Civil Service Exam loves testing them in clever, indirect ways. This chapter will make you fast and confident with all of them.
An average (also called the mean) is a "typical" or "central" value for a set of numbers. The basic formula is simple — add up the values and divide by how many there are — but the exam rarely asks it so plainly. Instead it hides the average inside problems about missing values, combining groups, a new member joining, replacing a member, and average speed.
On the Civil Service Exam, averages appear regularly in Numerical Reasoning and also underpin Data Interpretation and Statistics. Counting the disguised cases, this topic touches 4 to 7 items on a typical exam.
The difficulty is beginner to intermediate. The computation is light; what's tested is whether you can work backward from an average to a sum, handle weighted combinations correctly, and avoid the classic "average of two speeds" trap. Master the rearrangement Sum = Average × Count and most questions fall over instantly.
After completing this lesson you will be able to:
You should be comfortable with:
A quick refresher on the idea we depend on most: an average is really a fair redistribution. If you pooled everyone's amount together and shared it back out equally, each person's equal share is the average. That mental image — "level everything out" — explains every technique in this chapter, so keep it in mind.
Averages are how we summarize the world in a single number:
Understanding averages lets you read all these numbers critically — and spot when an "average" is being used to mislead — which is exactly the reasoning the exam rewards.
We build from the plain formula to the trickier exam patterns.
Average = (Sum of the values) ÷ (Number of values).
The scores 4, 8, 10, 6, and 12 have sum 40 and count 5, so the average is 40 ÷ 5 = 8.
Analogy: Imagine five glasses filled to different heights with water, then you pour them all together and share equally back into five glasses. The equal level each reaches is the average. Nothing is created or lost — the total is just spread evenly.
This is the key that unlocks most exam problems. If you know the average and the count, you can always recover the total:
Sum = Average × Count.
Five test scores average 82. Four of them are 78, 85, 90, and 76. Find the fifth. Total of all five = 82 × 5 = 410. Sum of the four known = 78 + 85 + 90 + 76 = 329. Fifth score = 410 − 329 = 81.
Whenever a problem gives you an average, immediately compute the sum — it turns a vague statement into a concrete number you can work with.
When two groups of different sizes are combined, you cannot simply average the two averages. Bigger groups pull the combined average toward themselves. You must weight by group size.
Class A (20 students) averages 85. Class B (30 students) averages 75. What is the combined average? Total score = (20 × 85) + (30 × 75) = 1,700 + 2,250 = 3,950. Combined average = 3,950 ÷ 50 = 79.
Notice 79 is closer to 75 than to 85 — because Class B has more students and pulls harder. If you had naïvely averaged 85 and 75 to get 80, you'd be wrong. Only when the groups are the same size does averaging the averages happen to work.
If a value joins a group and shifts the average, use: (old sum + new value) ÷ (old count + 1) = new average.
The average of 6 numbers is 30. A 7th number is added and the new average becomes 32. Find the 7th number. Old sum = 30 × 6 = 180. New sum = 32 × 7 = 224. Seventh number = 224 − 180 = 44.
Intuitively, the newcomer must not only be the new average but also lift everyone else up to it — which is why it's well above 32.
When a member is swapped for another and the average changes by a known amount, the change in the total equals (change in average) × (number of items).
The average weight of 10 people increases by 2 kg when a person weighing 60 kg is replaced by a newcomer. Find the newcomer's weight. Total increase = 2 × 10 = 20 kg. Newcomer's weight = 60 + 20 = 80 kg.
The single new person carries the entire increase for the whole group, which is why they weigh 20 kg more than the person they replaced.
Removing a member works the same way in reverse — recover both sums and subtract.
The average of 5 numbers is 20. When one number is removed, the average of the remaining 4 is 18. Find the removed number. Sum of 5 = 100; sum of 4 = 72. Removed = 100 − 72 = 28.
If every member of a group changes by the same amount, the average changes by exactly that amount — no more.
The average age of a family of 5 is 20 years today. What will it be in 3 years? Each person ages 3 years, so the average rises by exactly 3, to 23 — not by 15. A common trap is multiplying the 3 by the 5 people; don't.
If the same distance is covered at two different speeds, more time is spent at the slower speed, so the average speed is pulled below the simple average. For equal distances at speeds a and b:
Average speed = (2 × a × b) ÷ (a + b) — the "harmonic mean."
A cyclist rides out at 20 km/h and returns the same route at 30 km/h. Average speed for the round trip = (2 × 20 × 30) ÷ (20 + 30) = 1,200 ÷ 50 = 24 km/h — not 25.
For numbers that are evenly spaced (consecutive integers, an arithmetic sequence), the average is simply the middle value, or equivalently (first + last) ÷ 2.
The average of 1, 2, 3, …, 9 is the middle number, 5. The average of 10, 20, 30, 40 is (10 + 40) ÷ 2 = 25.
This turns a long addition into a two-second glance.
With the ideas in hand, let's picture, tabulate, and drill.
| Formula | Meaning |
|---|---|
| Average = Sum ÷ Count | The mean of a data set |
| Sum = Average × Count | Recover the total from an average |
| Weighted avg = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂) | Combine groups of different sizes |
| New value = (new avg × new count) − old sum | Value that joined a group |
| Total change = (Δaverage) × count | Effect of replacing a member |
| Average speed (equal distance) = 2ab ÷ (a + b) | Round trip / equal legs |
| Average of evenly spaced = (first + last) ÷ 2 | Arithmetic sequences |
Why the average-speed formula isn't (a + b)/2: you travel each leg for a different amount of time. Since more time passes on the slow leg, that speed gets more "weight," dragging the true average below the midpoint. The 2ab/(a+b) form accounts for this automatically.
| When the problem says… | Do this… |
|---|---|
| "average of these numbers" | Sum ÷ Count |
| "find the missing score/value" | Sum = Average × Count, then subtract |
| "two classes/groups combined" | weighted average (never average the averages) |
| "a new value is added and the average becomes…" | new value = new sum − old sum |
| "one value is replaced and the average changes by…" | Δtotal = Δaverage × count |
| "in X years / everyone gains…" | average shifts by that same amount |
| "goes at speed a, returns at speed b" | 2ab ÷ (a + b) |
| "consecutive / evenly spaced" | middle value = (first + last) ÷ 2 |
Step 1 — Convert every average into a sum using Sum = Average × Count. This is almost always the first move. ↓ Step 2 — Identify what's unknown — a missing value, a combined average, a newcomer, a replacement. ↓ Step 3 — Choose the tool: plain formula, weighted average, join/leave/replace, or average speed. ↓ Step 4 — Solve, keeping counts exact (mind whether the count went up or down). ↓ Step 5 — Sanity-check. A combined average must lie between the two group averages; a value that raises the average must be above it. If not, recheck.
Why Step 1 matters: averages are hard to manipulate directly but sums add and subtract freely. Turning averages into sums is the master key of this topic.
Example 1. Find the average of 12, 15, 18, and 15. Solution: Sum = 60; count = 4; average = 60 ÷ 4 = 15. Difficulty: ★☆☆☆☆
Example 2. The average of 6 numbers is 25. What is their sum? Solution: Sum = 25 × 6 = 150. Difficulty: ★☆☆☆☆
Example 3 (missing value). The average of five numbers is 40. Four of them are 35, 42, 38, and 50. Find the fifth. Solution: Total = 40 × 5 = 200. Known sum = 165. Fifth = 200 − 165 = 35. Difficulty: ★★☆☆☆
Example 4 (weighted average). Section 1 has 25 pupils averaging 88; Section 2 has 15 pupils averaging 80. Find the combined average. Solution: Total = 25 × 88 + 15 × 80 = 2,200 + 1,200 = 3,400. Combined = 3,400 ÷ 40 = 85. Check: 85 lies between 80 and 88, nearer the bigger section's 88. ✓ Difficulty: ★★★☆☆
Example 5 (evenly spaced shortcut). Find the average of the first 20 counting numbers (1 to 20). Solution: (1 + 20) ÷ 2 = 10.5. Difficulty: ★★☆☆☆
Example 6 (new member). The average of 8 numbers is 15. A ninth number is added and the average rises to 16. Find the ninth number. Solution: Old sum = 120; new sum = 16 × 9 = 144; ninth = 144 − 120 = 24. Difficulty: ★★★☆☆
Example 7 (replacement). The average age of a team of 11 players increases by 6 months when a 30-year-old is replaced by a new player. Find the new player's age. Thinking: Work in years: 6 months = 0.5 year. Total increase = 0.5 × 11 = 5.5 years. Solution: New player's age = 30 + 5.5 = 35.5 years. Difficulty: ★★★★☆
Example 8 (average speed). A jeepney covers the first half of a route at 40 km/h and the second half at 60 km/h. Find the average speed. Solution: 2 × 40 × 60 ÷ (40 + 60) = 4,800 ÷ 100 = 48 km/h (not 50). Difficulty: ★★★☆☆
Example 9 (removal). The average of 6 numbers is 45. When one number is removed, the average of the remaining 5 becomes 42. Find the removed number. Solution: Sum of 6 = 270; sum of 5 = 210; removed = 270 − 210 = 60. Check: removing a number above the mean (60 > 45) lowers the average, as observed. ✓ Difficulty: ★★★★☆
Example 10 (weighted → find the ratio). In a room, the men average 70 kg and the women average 55 kg; everyone together averages 64 kg. What is the ratio of men to women? Thinking: Use alligation — the distances from the combined mean give the ratio, crossed over. Solution: Men are 6 above (70 − 64); women are 9 below (64 − 55). Ratio men:women = 9:6 = 3:2. Check: (3 × 70 + 2 × 55) ÷ 5 = (210 + 110) ÷ 5 = 320 ÷ 5 = 64. ✓ Difficulty: ★★★★★
Example 11 (time shift trap). The average age of 4 members of a family is 25 years. Five years from now, what will their average age be? Solution: Each ages 5 years, so the average rises by exactly 5, to 30 — not by 20. Difficulty: ★★★☆☆
Example 12 (two-stage average). A student's average over 4 tests is 78. To raise the average to 80 after a 5th test, what must the 5th score be? Solution: Current sum = 312; required sum for 5 tests = 80 × 5 = 400; needed = 400 − 312 = 88. Difficulty: ★★★★☆
An average is the "leveled-out" value of a set: Sum ÷ Count. The master move is the rearrangement Sum = Average × Count, which converts every stated average into a concrete total you can add and subtract. When combining groups of different sizes, use a weighted average — never average the averages — and remember the result always lands between the parts. For a member joining, leaving, or being replaced, work through the sums; the whole group absorbs the change. A uniform shift (like "in X years") moves the average by exactly that shift. And for equal-distance trips at two speeds, use 2ab/(a+b), not the plain average. Convert to sums, respect the weights, and dodge the two famous traps — that's mastery of averages.
| Topic | Key point |
|---|---|
| Average | Sum ÷ Count |
| Master move | Sum = Average × Count |
| Combine groups | weighted: (n₁a₁ + n₂a₂) ÷ (n₁ + n₂) |
| Between-ness | combined avg lies between the two |
| Add/remove a value | recover sums, then subtract |
| Replace a value | Δtotal = Δaverage × count |
| Time shift | average moves by the same shift |
| Round trip speed | 2ab ÷ (a + b), not (a+b)/2 |
| Evenly spaced | average = (first + last) ÷ 2 |
Why can't I just average the two class averages? Because a bigger class contributes more scores and pulls the combined average toward itself. Averaging the averages only works when the groups are the same size.
What's the difference between mean, median, and mode? The mean is this lesson's average (sum ÷ count). The median is the middle value when sorted; the mode is the most frequent value. The exam sometimes tests all three — see the Probability & Statistics lesson.
Why is the round-trip average speed less than the middle of the two speeds? Because you spend more time at the slower speed, so it counts for more. The harmonic mean 2ab/(a+b) builds this in.
How do I handle "the average increased by X"? Multiply X by the number of items to get the total change, then apply it to whatever changed (the replaced member, or the newcomer's excess).
Is the average always one of the numbers in the set? No. The average of 2 and 4 is 3, which isn't in the set. The average is a representative value, not necessarily a member.
Tick them all and averages become fast, reliable points — and a foundation for Data Interpretation and Statistics.
Put it to the test with 1,389 practice questions on this topic.