Average is the central value of a group — the single number that would replace every item in a set without changing their total. Formally:
Think of it like distributing water equally into containers. You have 5 containers holding different amounts. Pour them all into one tank, then split equally back into 5 — each now holds the average amount. The total water never changed.
This single idea drives every average question on SSC CGL:
Here's why average problems feel easy but trap candidates: the formula is simple, but questions layer two or three group changes on top of each other. The examiner doesn't want to see if you know the formula — they want to see if you can track the total sum correctly under pressure.
Analogy: Imagine a class where everyone has to "donate" marks to a common pool and take back an equal share. If one student with unusually high marks is removed, everyone gets a smaller share. Your job in most SSC questions is to find exactly how high (or low) that one outlier's marks were.
The core principle you must internalize:
When the average increases or decreases after a change, the difference in totals tells you what was added or removed. This is the backbone of the "teacher's age," "new member joins," and "one number excluded" question families — which together account for a large share of SSC CGL average questions.
Never work with the average directly in multi-group problems. Always convert to sums immediately.
Then add or subtract sums, and divide by the new count. This is the one habit that prevents every calculation error in this topic.
Standard structure: Average of n items is A. One item is added/removed and the new average is A'. Find the added/removed item.
Formula:
Use + when someone joins, - when someone leaves.
Example: Average of 30 students = 14 years. Teacher joins, new average = 15 years.
The shortcut version: The average went up by 1 for 31 people. So someone contributed 31 × 1 = 31 extra above the old average of 14. Teacher's age = 14 + 31 = 45. Same answer, fewer multiplications.
Structure: Average of n items is A. Sub-group 1 has average a₁, sub-group 2 has average a₂. Find the average of sub-group 3.
Approach:
There is no shortcut here. Compute each sum, subtract, divide. The numbers are usually small and manageable.
If every number in a set is multiplied by k, the new average = k × old average.
If a constant c is added to every number, the new average = old average + c.
This is a direct consequence of how sums work:
SSC uses this type to test if you know the property or if you'll waste time recomputing everything.
First n natural numbers: Average =
First n multiples of k: Average =
This comes from the sum formula for natural numbers and for multiples of k.
For any evenly spaced (arithmetic) sequence: Average = (First term + Last term) / 2
So first 10 multiples of 6: first = 6, last = 60, average = (6 + 60)/2 = 33. Done in 4 seconds.
Look — this is the most commonly mishandled average problem in SSC CGL. The instinct is to average the speeds directly: (60 + 40)/2 = 50. That is wrong when the distances (not times) are equal.
When equal distances are covered at speeds a and b:
This is the harmonic mean of the two speeds. It is always less than the arithmetic mean. The reason: you spend more time at the slower speed, so it drags the average down.
When equal times are spent at speeds a and b:
Average speed = (a + b)/2 — the arithmetic mean. This case is rarer in SSC.
Always check: does the problem say equal distance or equal time? The word "half of the journey" almost always means equal distance — use the harmonic mean formula.
When groups of different sizes have different averages:
This is just the sum-first approach applied to multiple groups simultaneously. SSC doesn't often use the word "weighted average" but every multi-group combination problem is exactly this.
When one person joins and the average changes, the new person's value = New average + (change in average × new total count).
Example: 30 students, average 14. Teacher joins, average becomes 15. Rise = 1. New count = 31. Teacher's age = 15 + (1 × 31) — wait, that's wrong. Correct: Teacher's age = Old average + rise × new count = 14 + (1 × 31) = 45.
More cleanly: the new member must cover the old average for themselves AND lift every one of the 31 people by 1 year. = 14 (their own share at old average) + 31 × 1 (the lift they provide) = 45.
Standard method: two multiplications + subtraction = 6 steps. This method: 1 multiplication + addition = 3 steps.
For any arithmetic progression (evenly spaced list), average = (first + last) / 2.
First 10 multiples of 7: first = 7, last = 70. Average = (7 + 70)/2 = 77/2 = 38.5. First 50 natural numbers: first = 1, last = 50. Average = 51/2 = 25.5.
Standard method: compute sum using n(n+1)/2 formula, then divide by n — 3 steps. This method: identify first and last, add, halve — 2 steps. Saves 15-20 seconds on these questions.
For two speeds a and b covering equal distances: Average speed = 2ab/(a+b).
Quick recall: it's the harmonic mean. A number trick — the harmonic mean is always less than (a+b)/2. If your answer equals (a+b)/2, you've made the classic error.
For 60 and 40: 2×60×40 / 100 = 4800/100 = 48. Check: (60+40)/2 = 50 ≠ 48. Correct.
Step count — standard (computing total distance and total time): 5-6 steps. Formula: 2 multiplications + 1 addition + 1 division = 4 operations, completable in 20 seconds.
If every number is multiplied by k: new average = k × old average. No need to touch individual numbers.
If a constant c is added to every number: new average = old average + c.
SSC question type: "Average of 8 numbers is 21. Each number multiplied by 8. New average?" Trap answer: 21 + 8 = 29 (wrong — they added instead of multiplied). Correct: 8 × 21 = 168.
Recognition time: under 5 seconds. Computation: 1 multiplication. Vs. standard approach of recomputing sum: 2 steps. Saves ~30 seconds on this question type.
When a set is split into three sub-groups and you know the averages of two of them, find the third.
Total sum = n × A. Sum₁ = n₁ × A₁. Sum₂ = n₂ × A₂. Sum₃ = Total − Sum₁ − Sum₂. Average₃ = Sum₃ / n₃.
The elimination is: don't guess, don't work backwards from options. Compute all three sums first, then subtract. This approach makes decimal-heavy questions (like 3.95, 3.4, 3.85) mechanical — no estimation needed.
Step count vs. back-substituting options: back-substitution requires checking up to 4 answer choices (4×3 = 12 multiplications). Forward computation: 3 multiplications + 2 subtractions + 1 division = 6 operations.
In the exam hall, classify the question in the first 5 seconds:
If the question doesn't fit cleanly into one of these five types within 10 seconds, default to: Total = Average × Count for each group mentioned, then set up the equation. This brute-force approach solves every average question on SSC CGL, just slower. Use the typed shortcuts when you can recognize the pattern; fall back to totals when you can't.
Never start with the answer options on average questions — forward computation is fast enough and back-substitution adds confusion risk.
Why this question: The "excluded number" format is the most frequently repeated average question in SSC CGL. It tests whether you can work backwards from average to sum.
Solving path: Sum of 5 numbers = 5 × 27 = 135. Sum of remaining 4 = 4 × 25 = 100. Excluded number = 135 − 100 = 35. The question is purely about tracking two totals. No algebra required.
Why this question: The "teacher joins the class" format is structurally identical to any "new member added" problem. SSC repeats this template with different contexts — factory workers, cricket teams, committees.
Solving path: Old sum = 30 × 14 = 420. New sum = 31 × 15 = 465. Teacher's age = 465 − 420 = 45. Deviation method check: 14 + 31 × 1 = 45. Both give the same answer — use whichever you computed faster.
Why this question: Average of a specific multiple sequence. Tests knowledge of the arithmetic sequence shortcut vs. brute-force summation.
Solving path: First 10 multiples of 6 form an AP: 6, 12, ..., 60. Average = (6 + 60)/2 = 33. Alternatively: average of multiples of 6 = 6 × (average of 1 to 10) = 6 × 5.5 = 33. Same result, two valid routes.
Why this question: Three-subgroup split with decimals. A classic trap — candidates panic at 3.95, 3.4, 3.85 and try to estimate. Don't. Compute mechanically.
Solving path: Total = 6 × 3.95 = 23.7. Sum of group 1 = 2 × 3.4 = 6.8. Sum of group 2 = 2 × 3.85 = 7.7. Remaining sum = 23.7 − 6.8 − 7.7 = 9.2. Average = 9.2 / 2 = 4.6. The decimals look scary but every calculation here is manageable — the numbers are designed to work out cleanly.
Why this question: Average speed with equal distances. The most common conceptual trap on SSC CGL — many candidates write 50 km/h and move on.
Solving path: Equal distances at 60 and 40 → use harmonic mean formula: 2 × 60 × 40 / (60 + 40) = 4800 / 100 = 48 km/h. The answer 48 is less than the arithmetic mean 50 — this is always true for unequal speeds over equal distances, which serves as a quick sanity check.
Averaging the averages directly. When sub-groups have different sizes, you cannot add their averages and divide by the number of groups. You must weight by group size. (3.4 + 3.85 + x)/3 ≠ 3.95 in the three-subgroup problem.)
Using arithmetic mean for equal-distance speed problems. "First half at 60, second half at 40" — the average is NOT 50. It is 48. The word "half" refers to distance, not time, which is why the harmonic mean applies.
Off-by-one on group size. When a teacher or new member is added, the denominator becomes n+1, not n. Writing 30 × 15 instead of 31 × 15 is the single most common arithmetic error in this topic.
Multiplying when you should add (scalar operations). "Each number multiplied by 8" → new average = 8 × 21 = 168. "8 added to each number" → new average = 21 + 8 = 29. Mixing these up costs you the question.
Not identifying the sequence type before summing. For first 50 natural numbers, many candidates compute 50 × 51/2 = 1275 then divide by 50. That is correct but takes longer than (1 + 50)/2 = 25.5 directly.
Back-calculating from wrong total. In multi-step problems (like three sub-groups), some candidates compute the total as n × A and then misread n or A from the question. Always re-read the question after writing down your total — one digit wrong at this step cascades into a wrong final answer.