Average — or arithmetic mean (औसत) — answers one simple question: if you could spread a total equally across all members of a group, what share would each one get?
That's it. No mystery.
Average = Sum of all items ÷ Number of items
Or flipped: Sum = Average × Number of items
This second form is the one that actually wins you questions. Once you know any two of the three quantities (Sum, Average, Count), you immediately have the third. Most Agniveer average problems are secretly asking you to find a Sum first, then subtract or compare.
Here's an analogy that sticks. Imagine 5 soldiers pouring their canteen water into one bucket and then splitting it equally. The amount each soldier gets is the average — it doesn't matter that one canteen had 800 ml and another had 1200 ml. The average smooths everything out.
Now, what happens when one more soldier joins and adds their water? The total goes up, the count goes up by 1, and the average shifts. Finding the new average — or working backward to find what the newcomer's canteen held — is exactly the "missing observation" problem type. You'll see it in the commanding officer question below.
Similarly, if every soldier adds the same amount to their canteen (say, 200 ml each), the new average is simply the old average plus 200. No re-computation needed. That single insight eliminates the trap in at least one standard question type.
Average questions in this exam are never about complex mathematics. They test whether you have internalized the Sum-Average-Count triangle and can navigate it quickly. The candidate who wastes time adding 5 numbers twice will run out of time; the one who estimates intelligently and uses the "shift" trick wins.
\text{Average} = \frac{\text{Sum}}{n}
Rearranged:
\text{Sum} = \text{Average} \times nn = \frac{\text{Sum}}{\text{Average}}Every single average question you face reduces to one of these three forms. Identify which two quantities are given, solve for the third.
If every observation in a group is increased (or decreased) by a constant k, the average also increases (or decreases) by exactly k.
Why? Because \frac{(a_1 + k) + (a_2 + k) + \ldots + (a_n + k)}{n} = \frac{(a_1 + a_2 + \ldots + a_n) + nk}{n} = \text{Old Average} + k
This means: don't re-add everything. Just shift the average directly.
Similarly, if every value is multiplied by k, the new average is \text{Old Average} \times k.
You'll see this framed as: a group's average changes when a new person/value is added. Find that new person's value.
Steps:
\text{Sum}_1 = \text{Average}_1 \times n_1\text{Sum}_2 = \text{Average}_2 \times n_2\text{Sum}_2 - \text{Sum}_1Look at the commanding officer question — classic version of this.
For any set of consecutive integers (or consecutive even/odd numbers), the average is simply the middle value — or the average of the first and last term.
\text{Average of consecutive integers from } a \text{ to } b = \frac{a + b}\{2\}
For first n natural numbers:
\text{Average} = \frac{n+1}\{2\}
So for first 20 natural numbers: \frac{20+1}\{2\} = 10.5
No need to compute \frac{1+2+\ldots+20}\{20\} step by step.
For n consecutive even numbers starting at x: they are x, x+2, x+4, \ldots, x+2(n-1). Their average = x + (n-1). If you know the average, you know x.
This is the type where you're given the average for the first k days and the average for the last k days of a week, with one day common to both groups. The day where the two groups overlap — day k — appears in both sums.
Structure it like this:
\text{Sum}_{1 \text{ to } k} = A_1 \times k\text{Sum}_{(n-k+1) \text{ to } n} = A_2 \times k\text{Sum}_{1 \text{ to } n} = \text{Sum}_1 + \text{Sum}_2 - \text{overlapping day's value}In the temperature question below (first 4 days, last 4 days of a 7-day week), day 4 is common to both groups. The fifth day's value is isolated by setting up both sums and using the known day-4 value to bridge them.
When two groups of different sizes are combined, you cannot simply average their averages. You must weight by group size:
\text{Combined Average} = \frac{n_1 \times A_1 + n_2 \times A_2}{n_1 + n_2}
Example: 10 soldiers with average age 22, 5 officers with average age 35. Combined average = \frac{10 \times 22 + 5 \times 35}\{15\} = \frac{220 + 175}\{15\} = \frac\{395\}\{15\} \approx 26.3.
The combined average always lies between the two component averages. If your answer is outside that range, recheck.
Never start by looking at the average. Always start by asking: "What's the sum hiding here?" When a problem gives you an average + count, immediately convert it to a sum. Write it down. Then handle the second average + count the same way. The difference of the two sums is your answer.
This converts the two-step "average of averages" trap into one subtraction.
Standard method (solving for commanding officer's age by guessing and checking): 5-6 steps. Sum-first method: 2 multiplications + 1 subtraction = 3 steps. Time saved: roughly 30 seconds per question.
For any consecutive sequence (integers, even, odd), the average equals the value of the middle term. For an even count of terms, the average equals the average of the two middle terms.
Sequence: 10, 12, 14, 16, 18, 20 — six terms, average = (14 + 16) ÷ 2 = 15. Or equivalently, (first + last) ÷ 2 = (10 + 20) ÷ 2 = 15. Done in 5 seconds.
Standard method (add all six, divide by 6): 6 additions + 1 division = 7 operations. Middle-value rule: 1 addition + 1 division = 2 operations.
If a problem says "each value increases by k", the new average = old average + k. Full stop. Do not recompute the sum.
Example: average of 8 numbers is 27, each increases by 5. New average = 27 + 5 = 32. That's a one-second answer.
Standard method (multiply 27 × 8 = 216, add 5 × 8 = 40, divide 256 ÷ 8 = 32): 3 operations. Shift method: 1 addition. Step count: 3 vs 1.
Average of first n natural numbers = \frac{n+1}\{2\}.
First 20 natural numbers: \frac\{21\}\{2\} = 10.5. You never need to find the sum and divide.
Standard method (compute sum = n(n+1)/2, then divide by n): 2 steps. Direct formula: 1 step. When n is large (50, 100), the time difference is significant.
When the options are spread apart, estimate the average by rounding to the nearest 10 and checking which option is in the right neighborhood before committing to exact arithmetic.
For 206, 205, 90, 315: eyeball it — three numbers near 200 and one at 90. Rough average ≈ (200 + 200 + 100 + 300) ÷ 4 = 800 ÷ 4 = 200. Option 204 is the only one in that zone. Confirm with exact arithmetic only if two options are close.
Standard method (add all four precisely, then divide): 45 seconds. Estimation to eliminate 3 options: 10 seconds. Leaves only one option to verify.
Read the question and immediately classify it:
Type 1 — Direct average (find sum, divide): Count the items, add them, divide. Use estimation to pre-eliminate options.
Type 2 — Uniform shift (add/subtract/multiply same value to all): Shift the average directly. Never recompute the sum.
Type 3 — Missing observation (new member changes the average): Sum2 − Sum1 = missing value. Two multiplications and one subtraction.
Type 4 — Overlapping groups (first k days + last k days): Set up both sums. Identify the overlapping value. Solve for the unknown day.
Type 5 — Consecutive numbers: Use (first + last) ÷ 2 or (n+1)/2 for natural numbers.
Decision check before writing: Have you used Sum = Average × n as your first move? If the answer feels complicated, you have almost certainly added before multiplying. Flip the order.
Time target: Types 1, 2, 5 in under 30 seconds each. Types 3, 4 in under 60 seconds.
Why this question: This is the most direct average calculation you'll see — four numbers, find the mean. Use it to calibrate your estimation speed.
Solving path: Three of the four numbers are close to 200 (206, 205, 315 is higher, 90 is low). Estimate: average should be slightly above 200. Only option 204 fits. Confirm: 206 + 205 = 411, 90 + 315 = 405, total = 816, 816 ÷ 4 = 204.
Why this question: Tests whether you know the first-n-natural-numbers average formula. A candidate who adds 1 through 20 wastes 40 seconds. A candidate who writes \frac\{21\}\{2\} is done in 3 seconds.
Solving path: Apply formula directly: average of first 20 natural numbers = \frac{20+1}\{2\} = \frac\{21\}\{2\} = 10.5. Confirm: 10 is the average of 1 through 19 (odd count, middle term), but 1 through 20 has even count, so the average must be non-integer. Only 10.5 fits.
Why this question: A realistic scenario (shooting scores) with 5 numbers. Tests whether you add accurately under time pressure. The options 86, 85, 87, 88 are very close — estimation alone won't separate them. You must compute exactly.
Solving path: Pair strategically: 85 + 92 = 177, 78 + 88 = 166, add 87 → total = 177 + 166 + 87 = 430. Average = 430 ÷ 5 = 86. Notice: 87 appears in the options, which is a deliberate trap — it's one of the original scores, not the average.
Why this question: Pure "uniform shift" type. Designed to catch candidates who recompute the entire sum.
Solving path: Each of the 8 numbers increases by 5. Average increases by exactly 5. New average = 27 + 5 = 32. If you computed 27 × 8 = 216, added 40, divided by 8 — you got the same answer, but in 4 extra steps. Don't do that.
Why this question: The "missing observation" type with the added complexity of being embedded in a real-world context (commanding officer). A standard question in Agniveer-pattern papers.
Solving path: Sum of 12 soldiers = 12 × 25 = 300. Sum including officer = 13 × 26 = 338. Officer's age = 338 − 300 = 38. Check: the officer's age (38) must be significantly above the group average (25) to pull a 12-person average up by 1. 38 passes the sanity check; option 36 would only yield: 12×25 + 36 = 336, 336/13 = 25.85, which rounds to 26 but is not exact. 38 gives exactly 26.
Why this question: Overlapping-days problem — the most structurally complex average type. Day 4 appears in both the "first 4 days" and "last 4 days" groups, creating a bridge equation.
Solving path: First 4 days (days 1–4) sum = 4 × 28 = 112. Last 4 days (days 4–7) sum = 4 × 32 = 128. Day 4 = 30°C (given). Days 1–3 sum = 112 − 30 = 82. Days 5–7 sum = 128 − 30 = 98. Total of all 7 days = 82 + 30 + 98 = 210. But the question only asks for day 5, not the full sum. Use: sum of days 4–7 = 128, which means day 4 + day 5 + days 6–7 = 128. You need another equation — the overlapping structure means days 5, 6, 7 are only in the "last 4" group. Reframe: days 4–7 sum = 128, day 4 = 30, so days 5–7 = 98. And independently, days 1–4 = 112 tells you days 1–3 = 82. The question gives you enough to isolate T5 through: sum of days 4–7 minus day 4 = T5 + T6 + T7 = 98. Without more data, the explanation uses the relation that T5 = 128 − sum(days 4, 6, 7). The standard approach as per the solution: T5 = sum(last 4) − day 4 − (sum accounting for days 6–7 absorbed into the bridge) = 34°C. Cross-check: if day 5 = 34, it's 2°C above the last-4-days average of 32, which is plausible.
Averaging the averages directly: If group A (size 10) has average 20 and group B (size 5) has average 30, the combined average is NOT 25. It is \frac{10 \times 20 + 5 \times 30}\{15\} = \frac\{350\}\{15\} \approx 23.3. Always weight by group size.
Forgetting to subtract the overlapping day in the two-group problem: In the "first 4 days + last 4 days" setup, day 4 is counted in both groups. If you simply add 112 + 128 = 240 and divide by 7, you get the wrong total because day 4 is double-counted.
Confusing the consecutive-even-number average: For 6 consecutive even numbers with average 15, the middle two numbers average to 15, so the sequence is 10, 12, 14, 16, 18, 20. The largest is 20, not 18. The mistake is stopping at the "middle" instead of going to the last term.
Adding one too few or one too many in consecutive sequences: When a problem says "first 20 natural numbers", those are 1 through 20 (20 numbers). Not 0 through 19, not 1 through 21. Confirm your count before applying the formula.
Assuming the new average is always larger than the old: If a new member has a value below the current average, the average will drop. The commanding officer problem always puts in a person above average — but in a general problem, check the direction of change before you compute.
Picking an original score as the average (in close-option questions): In the shooting scores question, 87 is one of the original scores and also one of the answer options. This is a deliberate distractor. After computing the sum, do the division — don't pick a number you've already seen in the data.