Average — or arithmetic mean — is the single number that represents a group of numbers equally. You find it by dividing the total sum by the count of numbers.
Flip this around and you get the most useful formula in the chapter:
That second form is the one you will use in 90% of SSC CHSL questions. The question rarely asks you to compute an average directly — it usually gives you averages of sub-groups and asks for a missing value. Your job is to convert averages into sums, then add or subtract.
Here is a physical analogy that makes this concrete. Imagine 5 friends share a restaurant bill of ₹1,200. The average per person is ₹240. Now one more person joins and the new bill average drops to ₹220 — so the total is now ₹1,320. The new person's share is ₹1,320 − ₹1,200 = ₹120. You never needed to think about "what is the average" — you worked with sums throughout.
This is the mental shift that separates a slow solver from a fast one. Average is just a compressed version of the sum. Whenever you see the word "average" in a question, immediately translate it:
"Average of 11 numbers is 50" → "Sum of 11 numbers is 550"
Do this translation in the margin before you read the rest of the question. You will find that most questions then become simple addition and subtraction.
The three question types you will see in CHSL are:
Each type has a clean two-step path. Let's go through all of them.
Every average problem reduces to this:
You never need to divide unless the question explicitly asks for a new average. Work in sums until the last step.
Example: Average of 5 students is 75. Four scores are 70, 80, 65, 90. Find the fifth.
5 × 75 = 37570 + 80 + 65 + 90 = 305375 − 305 = 70Done in two lines.
This is the "middle number" trap that SSC loves. The structure looks like this:
Average of 11 numbers is X. First 6 have average A. Last 6 have average B. Find the 7th number.
The 7th number appears in both the "first 6" and "last 6" groups — but it should only appear once in the total. So:
This works because the middle element is counted twice (once in each sub-group) but should be counted only once. Subtracting the total removes the extra count.
Generalisation: If you have n total numbers, and the two overlapping groups together cover m numbers with one common middle element, the formula above always holds.
When two groups of different sizes merge, the combined average is NOT the simple average of the two group averages. You must weight by group size.
Example: 15 students average 80, 10 students average 60. Class of 25:
Note that 72 is closer to 80 than to 60, because the group with average 80 is larger. This is your sanity check — the combined average always lies between the two sub-averages, and leans toward the larger group.
These are weighted average problems in disguise. The trick: when a new person joins or leaves, the total changes by that person's age, and the count changes by 1.
If a person of age a joins a group with current average A and count n:
Rearranged: a = New Average × (n+1) − A × n
Most CHSL age-average questions give you two group averages (e.g., children alone vs. children + parents) and ask for a specific age. Use the sum technique, not the formula above.
If the numbers form an evenly spaced sequence (AP), the average equals the middle term — no addition needed.
For the sequence 12, 18, 24, 30, 36: the middle term is 24. Average = 24. Done in 3 seconds.
This works because in any AP, values above and below the middle cancel out symmetrically.
Before touching the calculator or doing arithmetic, rewrite every "average" statement as a "sum" statement in your margin. "Average of 20 matches is 35" becomes "Total runs = 700" written immediately. This single habit prevents the most common CHSL mistake (dividing when you should be multiplying). Standard approach: re-read the question 2–3 times to track averages — 60s. Translation-first: write sums, subtract — 15s. That's a 4× speed gain.
For overlapping-group questions, memorise this one line: Middle = Left group sum + Right group sum − Grand total. No algebra needed. Example: First 7 average 65 (sum 455), last 7 average 75 (sum 525), all 13 average 70 (total 910). Middle (7th) = 455 + 525 − 910 = 70. Standard method (setting up equations): 5 steps, ~50s. This pattern: 3 multiplications and 1 subtraction, ~18s.
When numbers are equally spaced (an arithmetic progression), the average is simply the middle term. Spot the equal gap: 12, 18, 24, 30, 36 — gap is 6, middle term is 24. You skip addition entirely. Standard method (add all 5, divide by 5): ~20s. AP recognition: 3s. Use this whenever the numbers in the question look "evenly spread" — check the gap first, not last.
Before computing the weighted average, check which group is larger. The answer must be closer to that group's average. If 15 students average 80 and 10 average 60, the answer must be between 60 and 80 and closer to 80. This eliminates wrong options before you do any arithmetic. In a 4-option MCQ, this often reduces choices from 4 to 2 in under 5 seconds, cutting calculation risk in half.
When one unknown person is added to/removed from a group and the average shifts, use: Unknown age = New average × New count − Old average × Old count. Example: 5 children average 15 (sum 75), with parents the 7 members average 25 (sum 175). Parents' combined age = 175 − 75 = 100. No algebra setup required. Standard approach (two equations, substitute): 40s. Direct subtraction: 10s.
When you see an averages question in the exam hall, run through this decision tree in under 10 seconds:
Step 1 — Translate all averages to sums. Write "Sum = Average × Count" for every group mentioned. Do this before reading the question a second time.
Step 2 — Is there an overlapping group? If first-K and last-K both appear, the shared middle element = Sum(left) + Sum(right) − Grand total.
Step 3 — Is it weighted average? Two different group sizes, one combined group? Use (n₁A₁ + n₂A₂)/(n₁+n₂). Apply the lean test to check your answer direction.
Step 4 — Is there a missing individual? Missing value = Total sum − Sum of known values.
Step 5 — AP check. If the list looks evenly spaced, spot the middle term. Done.
Most CHSL average questions fit exactly one of these five paths. Do not mix them up. If you have converted all averages to sums and the answer still isn't obvious, recheck whether one value is being double-counted (overlapping group pattern).
Why this question: The classic overlapping-groups structure that appears almost every year. Tests whether you know the "middle = left + right − total" formula.
Solving path: Total of 11 numbers = 11 × 50 = 550. Sum of first 6 = 6 × 49 = 294. Sum of last 6 = 6 × 52 = 312. The 6th number appears in both groups, so: 6th number = 294 + 312 − 550 = 56. No equations, no variables — three multiplications and one subtraction.
Why this question: Standard weighted average. Tests whether you know NOT to simply average 80 and 60 (which would give 70, a trap option that is not even listed here but which students compute instinctively).
Solving path: Total score = 15 × 80 + 10 × 60 = 1200 + 600 = 1800. Average = 1800 ÷ 25 = 72. Lean test confirms: 72 is closer to 80 (larger group). Answer: 72.
Why this question: Tests your ability to work with two nested groups — children alone vs. children + parents — and then use a simple linear relation to split two unknowns.
Solving path: Total runs = 20 × 35 = 700. First-12 runs = 12 × 45 = 540. Last-8 runs = 700 − 540 = 160. Average of last 8 = 160 ÷ 8 = 20. Clean three-step path.
Why this question: Introduces a third sub-group. You must account for all three parts summing to the grand total. Straightforward but mistakes happen if you forget to subtract both groups.
Solving path: Total age = 16 × 20 = 320. Sum of 5 students = 5 × 20 = 100. Sum of 10 students = 10 × 20.4 = 204. 16th student = 320 − 100 − 204 = 16. Note: the 16th student's age (16) equals their position number — a coincidence that makes the answer look suspicious, but the arithmetic is clean.
Why this question: Another overlapping-group question, identical structure to the first PYQ. Seeing both helps you recognise the pattern regardless of surface-level changes.
Solving path: Total = 13 × 70 = 910. First-7 sum = 7 × 65 = 455. Last-7 sum = 7 × 75 = 525. 7th result = 455 + 525 − 910 = 70. The answer (70) equals the overall average — this is a pattern worth remembering: when sub-group averages are symmetric around the total average, the middle term equals the total average.
Why this question: The simplest missing-value structure. If you cannot solve this in under 20 seconds using the sum method, go back and drill the translation step.
Solving path: Total = 5 × 75 = 375. Known sum = 70 + 80 + 65 + 90 = 305. Fifth score = 375 − 305 = 70. Two operations.
Why this question: Tests AP recognition. If you add all five numbers you waste 15 seconds on a question that can be answered in 3.
Solving path: The sequence 12, 18, 24, 30, 36 has a common difference of 6 — it is an AP. The middle term (3rd of 5) is 24. Average = 24. No addition needed. If you didn't spot the AP: sum = 120, average = 120 ÷ 5 = 24 — same answer, more time spent.
Averaging the averages. In a weighted average question, you cannot add the two group averages and divide by 2 unless both groups are the same size. This is the single most common error. Always multiply average by count first to get the sum.
Forgetting to subtract the total in overlapping groups. Students correctly compute the two sub-group sums but then add them and report that as the answer. The middle element is double-counted — you must subtract the grand total to un-double it.
Treating "first 6 and last 6 out of 11" as non-overlapping. First 6 = positions 1–6. Last 6 = positions 6–11. Position 6 appears in both. Recognise overlap whenever first-K + last-K > total-N.
Not doing the AP check. When numbers look evenly spread, students jump straight to addition. Spend 2 seconds checking for equal gaps. If the AP check passes, the middle term is the answer — no arithmetic required.
Misreading "older by 4 years" in age problems. Students set up father = mother + 4 correctly but then solve for the wrong variable. After getting the combined age of two unknowns, write out both equations clearly before solving — don't do this in your head.
Dividing too early. Once you have the missing sum or missing value, the question is done. Do not divide by the count again unless the question explicitly asks for a new average. Re-read the question's final ask before computing anything.