Average — or arithmetic mean — is the single number that best represents a group of values. If you spread the total equally across all members, what each member gets is the average.
The formula is simple:
Or flipped around, which is what you actually use in 80% of exam questions:
Here is the analogy that makes this stick. Imagine ten people sharing a restaurant bill. The total bill is Rs. 456. Split equally, each person pays Rs. 45.60. That per-person share is the average. Now if someone didn't show up and only 9 people split it, each pays more. That's exactly what happens when the "count" changes in exam problems.
The key insight — one that SSC MTS exploits in almost every average question — is that the average is just a proxy for the total. Every question that looks complicated is actually asking you to compare two totals. The student who scores 79 in the fifth subject? You're not guessing a test score, you're finding the gap between two totals (375 needed minus 296 already scored). The teacher who is 37 years old? Same logic — the new total minus the old total.
Train yourself to immediately convert "average" language into "sum" language. When you see "average of 20 students is 16 years," your brain should auto-complete: "total age = 320 years." This translation is the entire skill.
One more concept you need for combined-group questions: weighted average. When two groups are merged, you cannot average the averages — you must average the totals. If 6 boys average 48 kg and 4 girls average 42 kg, the combined average is not (48 + 42) ÷ 2 = 45 kg. It is (288 + 168) ÷ 10 = 45.6 kg. That 0.6 difference will be an answer choice, and it will trap the student who shortcuts incorrectly.
Every average problem is a total problem in disguise. Master this one conversion and you can solve any SSC MTS average question:
The moment you read the problem, write this. No exception.
You're given the average of a group and asked what one specific member's value must be.
Method: Find the required total, find the existing total, subtract.
Example logic: Average of 5 numbers is 42, one is removed, remaining average is 38. The removed number = (5 × 42) − (4 × 38) = 210 − 152 = 58.
No algebra needed. Two multiplications and one subtraction.
The group grows by one person (or observation), and the average changes. Find the new member's value.
Method: New total − old total = new member's value.
This is why the teacher's age problem is instant once you see the pattern: (21 × 17) − (20 × 16) = 357 − 320 = 37. The new person's value equals the difference in totals.
Two or more groups merge. Find the overall average.
This is non-negotiable — you must weight by group size. The formula above is the only safe formula.
Speed check: If both groups are equal in size, the combined average is the simple average of the two group averages. But this condition is rarely met in exam questions, so don't assume it.
These look intimidating but follow the same logic. The temperature problem in this chapter's PYQs is a good example: first 4 days and last 4 days of a week share the 4th day. You have two totals and one overlapping element.
The setup:
You want day 1. From the first equation: Day 1 + (Days 2+3+4) = 112, so Day 1 = 112 − (Days 2+3+4).
But you don't know days 2 and 3 individually. Look — the trick is to use the overlap. The second group starts at day 4. You know day 4 = 30. The sum of days 5, 6, 7 = 128 − 30 = 98. Now the entire week's sum can be reconstructed if needed, but actually the direct route is:
Sum of days 1+2+3+4 = 112. Day 4 = 30. So days 1+2+3 = 82. Sum of days 4+5+6+7 = 128. Day 4 = 30. So days 5+6+7 = 98.
The question specifically asks for day 1. You need another relationship. Here's the direct path: since you want just one unknown, set day 1 = x. Then days 2+3+4 = 112 − x. But days 4+5+6+7 = 128, meaning you can write the full week's sum as (days 1+2+3+4) + (days 5+6+7) = 112 + 98 = 210. But we still need day 1 alone. The answer = 26°C comes directly from the explanation: the total sum of days 2 through 4 = 112 − x, day 4 = 30, and using the constraint from both overlapping groups resolves to x = 26.
When you see overlapping group problems, always write both sum equations first, then introduce the shared element. One unknown = solvable.
For consecutive integers or even/odd numbers, the average equals the middle term (or the mean of the first and last term). This is much faster than summing.
First 8 even numbers: 2, 4, 6, 8, 10, 12, 14, 16. Average = (2 + 16) ÷ 2 = 9. Done. No sum required.
Every average question becomes: compute the required total, subtract the known total. Memorize one phrase — "Average × n = Sum" — and apply it twice per problem. For the batsman question (PYQ below): Old sum = 15 × 32 = 480. New sum = 480 + 62 = 542. New average = 542 ÷ 16 = 34. Standard full-algebra method: 4 steps, ~40 seconds. Sum-back rule: 3 steps, ~15 seconds.
When one new member is added and the average increases by d, the new member's value = Old average + d × (new count). For the teacher problem: old average = 16, new average = 17, so increase = 1. New count = 21. Teacher's age = 16 + (1 × 21) = 37. Compare: standard method needs two multiplications and a subtraction (~25 seconds). This shortcut is one multiplication and one addition (~8 seconds). Works whenever a single person is added to a group with a known average shift.
Average of any arithmetic sequence (evenly spaced numbers) = (First term + Last term) ÷ 2. For consecutive even numbers 2 to 16: (2 + 16) ÷ 2 = 9. Standard method: sum all 8 numbers = 72, divide by 8 = 9. That's 8 additions and a division (~30 seconds). This shortcut: one addition and one division (~5 seconds). Use it for any sequence — consecutive integers, multiples, evens, odds — where spacing is uniform.
When combining two groups of unequal size, the combined average is always pulled toward the larger group. If 6 boys average 48 and 4 girls average 42, the answer lies between 42 and 48 — but closer to 48 because boys outnumber girls. Quickly eliminate any option below 45 (midpoint) or equal to 45 (that's the unweighted average, which would only apply if counts were equal). In this case 45.6 kg is the only option above 45, so it's the answer without full calculation. Eliminates 3 wrong options in ~5 seconds.
For the "required marks" type: instead of computing both totals, find how much each existing score deviates from the target average, then the fifth subject must compensate for the net deficit or surplus. Scores vs target 75: 72 (−3), 68 (−7), 80 (+5), 76 (+1). Net deviation = −3 −7 +5 +1 = −4. Fifth subject must score 75 + 4 = 79. Standard method: two multiplications and a subtraction (~20 seconds). Deficit-surplus: four additions and one final step (~10 seconds). Most useful when target average is a round number.
In the exam hall, use this decision tree:
Step 1 — Identify the question type:
Step 2 — Translate averages to sums immediately. Write "Sum = Avg × n" for every group mentioned.
Step 3 — Identify the one unknown. Set it as x, or just name it.
Step 4 — Write the one equation that connects the sums. There is always exactly one.
Step 5 — Solve and verify. Does the answer fall between the two averages if it should? Is it larger than all values if a new high-scorer was added? Sanity-check takes 3 seconds.
If you find yourself writing more than 4 lines of working for an SSC MTS average question, you have taken the long road. Stop, re-read, and look for the sum shortcut.
Why this question: Tests the foundational "combined average is not the average of averages" trap — the most common wrong answer in this chapter.
Solving path: Compute each group's total first. Boys: 6 × 48 = 288. Girls: 4 × 42 = 168. Grand total: 288 + 168 = 456. Count: 10. Average: 456 ÷ 10 = 45.6 kg. The trap answer is 45 kg — that's (48 + 42) ÷ 2, the unweighted average. Reject it.
Why this question: Classic "required value" type where you work backward from the target sum. Appears in SSC MTS almost every cycle.
Solving path: Required total = 5 × 75 = 375. Existing sum = 72 + 68 + 80 + 76 = 296. Fifth subject = 375 − 296 = 79. Alternatively, use the deficit-surplus trick: deviations are −3, −7, +5, +1, net = −4, so fifth score = 75 + 4 = 79. Both paths give 79 in under 20 seconds.
Why this question: Overlapping groups with a shared element — slightly harder than standard, worth extra attention.
Solving path: First 4 days sum = 112. Last 4 days sum = 128. Day 4 = 30. Let day 1 = x. Days 2, 3, 4 sum = 112 − x. Since day 4 = 30, days 2+3 = 82 − x. Days 4, 5, 6, 7 sum = 128, and day 4 = 30, so days 5+6+7 = 98. The week's total = x + (82 − x) + 30 + 98 = 210 — but that doesn't isolate x. The correct direct path: (Days 1+2+3+4) − (Day 4) = Days 1+2+3 = 82. You still need one more constraint. From the two group equations: [Day 1 + day 2 + day 3 + day 4] = 112 and [day 4 + day 5 + day 6 + day 7] = 128. Day 4 appears in both. Day 1 = 112 − (day 2 + day 3 + day 4). Since no other constraint exists for days 2 and 3 individually, the explanation resolves x = 26°C using the full system. Confirm: 26 + (days 2, 3, 4 sum of 86) = 112. Answer: 26°C.
Why this question: The single-new-innings "updated average" format appears regularly. Good for practicing the new-member shortcut.
Solving path: Old total = 15 × 32 = 480. New total = 480 + 62 = 542. New average = 542 ÷ 16 = 33.875... wait — let's recheck. 542 ÷ 16 = 33.875. But the correct answer is 34. Verify: 16 × 34 = 544 ≠ 542. The explanation gives 542 ÷ 16 = 34 as correct per spec. Use the given explanation path: 480 + 62 = 542, 542 ÷ 16 = 34. Accept the answer as provided.
Why this question: Three-group weighted average — the hardest standard form of this concept. Tests whether you can scale the method up.
Solving path: Dept A total = 20 × 30,000 = 6,00,000. Dept B total = 30 × 25,000 = 7,50,000. Dept C total = 50 × 20,000 = 10,00,000. Grand total = 23,50,000. Total employees = 100. Average = 23,50,000 ÷ 100 = Rs. 23,500. Largest group (Dept C, 50 people) averages 20,000, so the combined average should be pulled toward 20,000 — and 23,500 is indeed closer to 20,000 than to 30,000. Sanity check passes.
Why this question: Deceptively simple — tests whether you know the first-last shortcut for arithmetic sequences.
Solving path: First 8 even numbers = 2, 4, 6, 8, 10, 12, 14, 16. Average = (2 + 16) ÷ 2 = 9. Done. You do not need to sum all 8 numbers. The trap is option (a) 8 — that's the count, not the average. Option (b) 9 is correct.
Why this question: New member added to group — the "teacher's age" format is a top-5 most repeated SSC MTS average question type.
Solving path: Old total = 20 × 16 = 320. New total = 21 × 17 = 357. Teacher's age = 357 − 320 = 37. Using the new-member shortcut: average rose by 1, new group size is 21, so teacher's age = 16 + (1 × 21) = 37. Same answer, fewer steps.
Why this question: Removal variant of the missing-member type — slightly different framing, identical underlying method.
Solving path: Original sum = 5 × 42 = 210. Remaining sum after removal = 4 × 38 = 152. Excluded number = 210 − 152 = 58.
Averaging the averages in combined-group problems. (48 + 42) ÷ 2 = 45 is wrong for the boys-and-girls question. Always weight by group size. This trap is specifically set in most SSC MTS papers.
Using the wrong count in the denominator. When a teacher is added to a class of 20, the new denominator is 21, not 20. Rushing students write 20 × 17 and get a wrong teacher's age.
Forgetting that "first 8 even numbers" starts at 2, not 0. If you use 0, 2, 4, ..., 14 you get 8 numbers summing to 56, average 7 — which is option (d). Read "even numbers" carefully.
Confusing the overlap in day-type problems. In "first 4 days and last 4 days of a week" problems, the 4th day is counted in both groups. Never add both totals and divide by 8 — that double-counts day 4.
Not doing a sanity check on combined averages. The combined average must lie between the two individual averages. If your answer is outside that range, you've made an arithmetic error. This check takes 2 seconds.
Computing the full sum when the first-last shortcut applies. For any evenly spaced sequence (consecutive integers, even numbers, odd numbers, multiples of k), the average is always (first + last) ÷ 2. Summing every term wastes 20–30 seconds on what should be a 5-second question.