Average
~22 min read · AFCAT Numerical Ability
- Weight: About 1.25 marks per AFCAT paper; high-yield tier.
- Master identity: Change in sum = (Change in average) times (Number of members). Almost every AFCAT average item collapses to this one line.
- Top trap: Using (a+b)/2 for average speed when distances are equal — the correct formula is 2ab/(a+b).
Overview
Average appears about 1.3 times per paper across the last four AFCAT solved papers, placing it in the high yield band of Numerical Ability.
Average is one of the most reliable scoring topics in the AFCAT Numerical Ability section. Across the last four AFCAT solved papers, the topic shows up around five times in total, settling at roughly 1.25 questions per paper. That places it in the high-yield tier — not as dense as percentages or interest, but dense enough that every serious aspirant must clear it cold.
The good news is that AFCAT average questions almost never use deep statistics. The examiner stays inside arithmetic mean, with one or two weighted-mean items mixed in. What separates a quick solve from a sixty-second slog is whether you treat the question as a sum-recomputation problem (slow) or as a change-in-sum problem (fast). The single most useful identity in this chapter is: change in total sum equals change in average multiplied by number of members. Once you internalise that, age-of-class items, weight-replacement items, and even the cricket running-average format reduce to a single line.
This page builds the topic from the ground up. We start with the arithmetic-mean definition, walk through the algebraic properties that make shortcuts legal, list the consecutive-number formulas examiners love, develop the inclusion-exclusion identity, and then apply it to the AFCAT-favourite scenarios — class averages with a teacher added, member replacement, average speed for equal distances versus equal times, running batting averages, and mixtures via alligation. We finish with geometric and harmonic means (rare in AFCAT but useful to recognise), a list of trap patterns, and a recommended time budget.
Why average is a near-guaranteed marker
Numerical Ability carries 18 to 22 questions in each AFCAT paper, which is close to one-fifth of the entire 100-question test. Within that section, average sits in the high-yield band alongside time-and-work, time-speed-distance, ratio-mixtures, geometry, probability, number system, and algebra basics. It does not match the volume of percentages or interest, but its average of 1.25 questions per paper means you should expect at least one and sometimes two average items every sitting.
What makes it a near-guaranteed marker is the narrow style. The AFCAT examiner does not ask abstract statistics. The questions repeat across only a handful of templates: class average with addition or removal, weight replacement, weighted mean of two groups, average speed, average of consecutive numbers or progressions, and running batting average. If you have rehearsed each template once, the same answer surfaces in under a minute.
Negative marking on AFCAT is minus one per wrong answer against plus three per correct. That ratio means even one careful skip is cheap, but a guessed wrong answer costs you four marks of net swing. Average questions are short and pattern-driven enough that you should not be guessing — either you recognise the template and solve in fifty seconds, or you mark it for review and come back after the rest of the section is locked in.
Arithmetic mean — definition and base formula
The arithmetic mean, usually just called the average, of a set of numbers is the sum of those numbers divided by how many numbers there are. If the observations are x1, x2, up to xn, then:
Average A = (x1 + x2 + ... + xn) divided by n
The two quantities you almost always need are the average A and the sum S, related by S = n times A. Most AFCAT average questions are solved by going back and forth between these two — convert an average to a sum, do an addition or subtraction, then convert back to an average.
A small but high-leverage habit: as soon as you read an average problem, write down n, write down A, and immediately compute and write n times A. That number, the total sum, is what the question is really about. You will almost never use A directly; you will use S.
Properties of arithmetic mean
The arithmetic mean obeys several algebraic identities that justify the shortcuts you will use under exam pressure. Knowing them by name protects you from solving questions the long way.
- Sum identity: The sum of n observations equals n times their average. This is the most important single fact in the chapter.
- Deviation identity: The sum of deviations of all observations from the mean is zero. Formally, the sum of (xi minus A) over all i is exactly zero. This is the algebraic reason every change-in-average problem collapses to a single equation.
- Add a constant: If you add a constant k to every observation, the new average is A plus k. The whole distribution shifts.
- Subtract a constant: Similarly, if you subtract k from every observation, the new average is A minus k.
- Multiply by a constant: If you multiply every observation by k, the new average is k times A.
- Divide by a constant: If you divide every observation by k, the new average is A divided by k.
These transformation rules let you simplify ugly numbers before computing. For instance, if asked the average of 1003, 1007, 1011, 1015, 1019, subtract 1000 from each, take the average of 3, 7, 11, 15, 19 (which is 11), and add 1000 back to get 1011. Far faster than long addition.
Weighted mean — when groups are not the same size
The plain arithmetic mean treats every observation equally. The weighted mean lets you handle observations that carry different weights — typically because they represent groups of different sizes, or measurements with different reliability.
Weighted mean = (w1 x1 + w2 x2 + ... + wn xn) divided by (w1 + w2 + ... + wn)
The most common AFCAT-style use is combining two group averages into a single overall average. If group one has m members with average A1, and group two has n members with average A2, then the combined average is:
Combined average = (m times A1 plus n times A2) divided by (m plus n)
This is just the weighted mean with the group sizes as weights. The derivation is easy: the total sum of group one is m times A1, the total sum of group two is n times A2, and the combined sum divided by the combined head-count gives the overall average. Always write the sums explicitly before combining — the mistake of averaging two averages without weighting them is the single most common slip on this template.
Average of consecutive numbers and progressions
AFCAT loves asking for averages of standard number sequences because the answer is a one-formula computation. Memorise this table.
| Sequence | Sum | Average |
|---|---|---|
| First n natural numbers (1 to n) | n(n+1)/2 | (n+1)/2 |
| First n odd natural numbers | n2 | n |
| First n even natural numbers | n(n+1) | n+1 |
| First n squares (12 to n2) | n(n+1)(2n+1)/6 | (n+1)(2n+1)/6 |
| First n cubes (13 to n3) | [n(n+1)/2]2 | n(n+1)2/4 |
| Arithmetic progression, first term a, last term l | n(a+l)/2 | (a+l)/2 |
The last row is the most useful single line in the whole topic. For any arithmetic progression — any sequence with a constant common difference — the average is simply the mean of the first and last terms. This means the average of 17, 19, 21, 23, 25 is just (17+25)/2 = 21, no addition required. The average of any block of consecutive integers from a to b is (a+b)/2.
Inclusion-exclusion identity — the master technique
This is the single most powerful idea in the chapter and the technique that almost every AFCAT average question rewards. Suppose a group of n observations has average A, so the total sum is S = n times A. Now suppose the group changes in one of three ways: a new member is added, an existing member is removed, or one member is replaced by another. In every case, you can compute the new average without ever recomputing the full sum from scratch.
Add a member with value x: The new sum is S + x, and the new count is n + 1. The new average is (n times A + x) divided by (n + 1). Rearranged, if the average changes by an amount delta when the new member joins, then x = A + delta times (n + 1).
Remove a member with value x: The new sum is S minus x, the new count is n minus 1, and the new average is (n times A minus x) divided by (n minus 1). If the average changes by delta on removal, then x = A minus delta times (n minus 1).
Replace one member by another: The head-count stays at n, so the new average is A plus (change in sum divided by n). Equivalently, change in average times n equals the new value minus the old value. This is the formula behind every weight-replacement and age-replacement question.
| Change | Identity | Solve for unknown |
|---|---|---|
| Add a member with value x | New average = (nA + x)/(n+1) | x = New average times (n+1) minus nA |
| Remove a member with value x | New average = (nA - x)/(n-1) | x = nA minus New average times (n-1) |
| Replace old value by new value | Change in average times n = New value minus Old value | New value = Old value + n times (change in average) |
Carry the sign of the change with care. If the average goes up by 1.5 kg, delta is +1.5; if it falls by two years, delta is minus 2. Most wrong answers on this template come from sign mistakes, not concept failures.
Age problems — class-averages and teacher additions
Age problems are the single most common AFCAT average template. The standard setup is a class of students with a known average age, and then either a teacher is added, a student leaves, or one student is replaced by another. Every one of these reduces to the inclusion-exclusion identity in the previous section.
The class-with-teacher format is so common it deserves its own line. If a class of n students has average age A, and the teacher's age is included so the new average becomes B, then the teacher's age T is:
T = (n + 1) times B minus n times A
This is just the add-a-member identity dressed up. Treat it as a single line, write it down once, and you will recognise the format on sight.
The replacement format also appears. If a member of age x leaves and is replaced by a member of age y, the new average changes by (y minus x) divided by n. Many AFCAT items give you the change in average and ask for either x or y; solve for the unknown directly.
Average speed — equal distances versus equal times
Average speed is the textbook example where intuition fails. The most common AFCAT trap on this topic is using the simple average of two speeds when it does not apply. The correct rule depends on what is held constant — distance or time.
| Scenario | Correct average speed | Why |
|---|---|---|
| Equal distances at speeds a and b | 2ab/(a+b) | Total distance is 2d; total time is d/a + d/b. Divide to cancel d. |
| Equal times at speeds a and b | (a+b)/2 | Total distance is at + bt; total time is 2t. |
| Three speeds, equal distances | 3abc/(ab+bc+ca) | Harmonic-mean generalisation. |
Almost every AFCAT speed-averaging item is the equal-distances scenario — phrased as a return journey, a round trip, or a half-and-half route. The signature word to look for is the phrase covering the same stretch at two different speeds. When you see it, reach for 2ab/(a+b) immediately. Use the simple average (a+b)/2 only when the question explicitly tells you the time spent at each speed is the same.
Running averages — the cricket batting template
The running-average format is the inclusion-exclusion identity applied innings by innings, and AFCAT examiners enjoy phrasing it as a cricket question. A batsman has played n innings with batting average A. After scoring r in the next innings, the new average is some new value B. Find r, or n, or A — depending on what is missing.
The identity is identical to add-a-member:
r = (n + 1) times B minus n times A
The variation that catches candidates out is when the question says the batting average rises or falls by a fixed amount after the next innings. If the average rises by delta after the (n+1)th innings, then the runs scored in that innings equal the new average B plus n times delta, or equivalently A plus (n+1) times delta. If the average falls by delta, the runs scored in that innings equal A minus (n+1) times delta. These two lines handle every running-average AFCAT item.
Averages of fractions and mixed numbers
When the observations are fractions or mixed numbers, the arithmetic does not change but the bookkeeping does. Two practical habits.
- Convert mixed numbers to improper fractions before averaging. An average of 2 and 1/2, 3 and 3/4, 4 and 1/8 is much easier as 5/2, 15/4, 33/8 — bring to a common denominator (eighths in this case), add, then divide.
- Use the deviation trick on near-equal fractions. If the observations all sit close to a round number, subtract that round number from each, average the small leftover deviations, and add the round number back. This is the same property as adding a constant.
AFCAT rarely sets a pure fraction-average question, but fractional observations appear inside larger problems — for example, an average score expressed in marks-per-question when each question is worth 2.5 marks. The conversion-to-improper-fraction habit prevents most arithmetic slips.
Weighted average for mixtures and alligation
Mixture problems are weighted-average problems in disguise. Two ingredients with different costs, concentrations, or strengths are mixed in some ratio to produce a final mixture with an average value somewhere between the two. The combined cost or strength is just the weighted mean of the two ingredient values, with the quantities of each ingredient as the weights.
The alligation cross is a graphical shortcut for the inverse problem: given two ingredient values and a target average, find the ratio in which to mix them. Place the two ingredient values on a line, the target average between them, and take the absolute differences crosswise. The ratio of the differences is the ratio of the two quantities — with the larger ingredient quantity paired with the smaller difference.
Quantity of cheaper / Quantity of dearer = (Dearer price minus Mean price) / (Mean price minus Cheaper price)
If a shopkeeper mixes rice at 30 rupees per kg with rice at 45 rupees per kg to sell at 36 rupees per kg, the cross gives (45 minus 36) to (36 minus 30) = 9 to 6 = 3 to 2. So the cheaper rice is mixed with the dearer rice in the ratio 3 to 2 by quantity. Read the labels carefully — the side with the smaller difference always belongs to the larger quantity.
Geometric and harmonic means — when each applies
The arithmetic mean is the only mean AFCAT asks about in nine out of ten papers. But the geometric mean (GM) and harmonic mean (HM) appear occasionally, and recognising them protects you from misapplying the arithmetic formula.
| Mean | Formula for two values a and b | Use case |
|---|---|---|
| Arithmetic mean (AM) | (a + b) / 2 | Additive quantities — marks, ages, weights, prices. |
| Geometric mean (GM) | Square root of (a times b) | Multiplicative quantities — compound growth rates, ratios over time. |
| Harmonic mean (HM) | 2ab / (a + b) | Rates over equal amounts — average speed over equal distances, average price per unit when equal money is spent. |
The classic inequality, for any two positive numbers, is AM is greater than or equal to GM is greater than or equal to HM, with equality only when the two numbers are equal. The 2ab/(a+b) formula you already used for average speed is the harmonic mean — recognise the family resemblance and you will not confuse the two formulas under pressure.
AFCAT trap patterns to recognise
- Trap 1 — Averaging two averages without weighting. If section A has 30 students averaging 70 marks and section B has 45 students averaging 80 marks, the combined average is not 75. Use the weighted mean with sizes as weights.
- Trap 2 — Simple-average speed. When distances are equal, never use (a+b)/2. Reach for 2ab/(a+b) instead.
- Trap 3 — Wrong head-count after change. When a teacher is added, the new average has n+1 in the denominator, not n. When a member leaves, the new count is n-1. Write the count down before computing.
- Trap 4 — Sign of the change in average. If the average falls, the change in sum is negative. Carry the sign through.
- Trap 5 — Misreading runs versus average. In cricket-style running-average items, do not confuse the next-innings runs with the new batting average. Read the question twice before equating.
- Trap 6 — Including or excluding the endpoints in a consecutive-numbers question. Average of integers from 10 to 50 inclusive is (10 + 50)/2 = 30 only when both ends are counted; check the wording.
- Trap 7 — Mixed units. If some values are in metres and others in centimetres, or some times are in hours and others in minutes, convert everything before averaging.
Time budget on average items
The AFCAT total budget is 120 minutes for 100 questions. After lighter sections take their share, you typically have around 35 to 45 minutes for the full Numerical Ability section. With 18 to 22 questions in the section, that gives you roughly 90 to 120 seconds per item as a ceiling — but average items should beat the average pace.
Set yourself a target of 50 to 70 seconds per average item. Anything beyond 90 seconds is a signal to mark and move on. The reason the budget is tight is precisely because every average question reduces to a single identity once you spot the template. If you find yourself doing more than three arithmetic steps, you have probably missed the shortcut — pause, re-read the question for the change-in-average phrase, and switch tactics.
In the order in which you attack Numerical Ability under exam pressure, average is one of the first three topics you should clear, alongside simplification and percentage. Lock in the easy marks before spending time on geometry or probability.
Worked AFCAT-style examples
The average age of 30 students in a class is 13.5 years. When the class teacher's age is included, the average rises to 14.2 years. How old is the teacher?
Sum of students = 30 times 13.5 = 405. Sum including teacher = 31 times 14.2 = 440.2. Teacher's age = 440.2 minus 405 = 35.9 years. Alternatively, use the shortcut T = (n+1) times new average minus n times old average = 31 times 14.2 minus 30 times 13.5.
The average weight of a rowing crew of 8 people increases by 2 kg when one member weighing 60 kg is replaced by a new member. What is the new member's weight?
Change in sum = 8 members times 2 kg per member = 16 kg. New member's weight = old member's weight plus change in sum = 60 + 16 = 76 kg.
Section A of a coaching batch has 24 students with average score 62. Section B has 36 students with average score 72. Find the combined average score of both sections.
Sum for A = 24 times 62 = 1488. Sum for B = 36 times 72 = 2592. Combined sum = 4080. Combined head-count = 60. Combined average = 4080 divided by 60 = 68. The trap answer 67 comes from averaging the two averages directly.
A car covers a distance of 120 km at 40 km/h and returns the same distance at 60 km/h. Find the average speed for the entire journey.
Equal distances, so use 2ab/(a+b) = 2 times 40 times 60 divided by 100 = 4800/100 = 48 km/h. The trap answer 50 comes from using the simple average (40+60)/2 — wrong because times spent at each speed are different.
A pilot flies for 2 hours at 500 km/h and for the next 3 hours at 400 km/h. Find the average speed across the full 5-hour flight.
Distances are different but the times are stated, so compute directly. Total distance = 2 times 500 + 3 times 400 = 1000 + 1200 = 2200 km. Total time = 5 hours. Average speed = 2200/5 = 440 km/h.
The average of the first 20 odd natural numbers is what?
Average of first n odd natural numbers is n. So the answer is 20. As a check, the first 20 odd numbers run from 1 to 39, and (1 + 39)/2 = 20 by the arithmetic progression rule.
Find the average of all even numbers between 11 and 49.
Even numbers between 11 and 49 are 12, 14, 16, up to 48 — an arithmetic progression with first term 12 and last term 48. Average of an arithmetic progression equals (first + last) divided by 2 = (12 + 48)/2 = 30.
A batsman's average across 16 innings is 36. In his 17th innings he scores 70 runs. What is his new average?
Sum after 16 innings = 16 times 36 = 576. Sum after 17 innings = 576 + 70 = 646. New average = 646 divided by 17 = 38. Alternatively, the runs (70) exceed the old average (36) by 34, which spread over 17 innings raises the average by 34/17 = 2.
After his 12th innings a batsman has a batting average of 47 runs. If his average increased by 4 runs because of his 12th-innings score, how many runs did he score in that innings?
Average before the 12th innings = 47 minus 4 = 43. Sum before = 11 times 43 = 473. Sum after = 12 times 47 = 564. Runs in the 12th innings = 564 minus 473 = 91.
Five years ago the average age of a husband and wife was 23 years. Today the average age of the husband, wife, and a child is 20 years. How old is the child?
Five years ago, husband + wife age sum = 2 times 23 = 46. Today, husband + wife age sum = 46 + 2 times 5 = 56. Today, total age of all three = 3 times 20 = 60. Child's age = 60 minus 56 = 4. Re-check with a careful read — and note that the typical AFCAT version sets the numbers so the child age comes out as a clean small integer.
A trader mixes two varieties of tea costing 80 rupees per kg and 120 rupees per kg to produce a blend that sells at cost for 92 rupees per kg. In what ratio are the two varieties mixed?
Use alligation. Cheaper-to-dearer ratio = (120 minus 92) to (92 minus 80) = 28 to 12 = 7 to 3. So the trader mixes 7 parts of the 80-rupee tea with 3 parts of the 120-rupee tea.
The average of 9 observations is 35. If the average of the first five observations is 32 and the average of the last five is 39, what is the value of the fifth observation?
Sum of all nine = 9 times 35 = 315. Sum of first five = 5 times 32 = 160. Sum of last five = 5 times 39 = 195. The fifth observation is counted in both groups, so sum of first five + sum of last five = sum of all nine + fifth observation. So fifth observation = 160 + 195 minus 315 = 40.
Exam-day strategy
- Convert every stated average to a sum immediately by multiplying by the head-count. You will use the sum, not the average.
- Recognise the change-in-average template on sight — any phrase like 'rises by', 'falls by', 'is replaced by', 'is added' triggers the inclusion-exclusion identity.
- For two-group combined-average questions, write the two sums explicitly with their head-counts as weights. Never average two averages directly.
- For average-speed questions with equal distances, use 2ab/(a+b); for equal times, use (a+b)/2. The signature phrase 'same stretch' or 'returns by the same route' marks the equal-distance case.
- For consecutive-number and arithmetic-progression averages, use (first + last)/2 — never add up the whole list.
- Carry the sign of the change in average. If the average falls, treat the change as negative; if it rises, treat it as positive.
- Aim for 50 to 70 seconds per average item. Anything beyond 90 seconds is a sign you have missed the shortcut — re-read the question or skip.
- Lock in average items in the first pass through Numerical Ability, alongside simplification and basic percentage items.
Practise Average for AFCAT
Practise AFCAT-pattern average drills with the inclusion-exclusion identity, weighted averages of two groups, equal-distance speed averages and cricket running-average items — all timed to the 50-to-70-second AFCAT pace.
Start free AFCAT practiceFrequently asked questions
How many average questions does AFCAT ask per paper?
About 1.25 questions per paper on average, based on the last four AFCAT solved papers — so expect one most papers and two occasionally.
What is the most common AFCAT average template?
The change-in-average format. A class adds a teacher, a crew replaces a member, a batsman plays another innings. Every version of this template reduces to the inclusion-exclusion identity: change in sum equals change in average times head-count.
When should I use 2ab/(a+b) versus (a+b)/2 for average speed?
Use 2ab/(a+b) when equal distances are covered at the two speeds — for example, a round trip or a return journey along the same route. Use (a+b)/2 only when the two speeds are maintained for equal lengths of time. The default AFCAT pattern is equal distance, so 2ab/(a+b) is the more useful formula.
Do I need to memorise the geometric and harmonic mean formulas for AFCAT?
GM and HM rarely appear by name in AFCAT. But the harmonic mean is hidden inside the equal-distance average-speed formula, so knowing the relationship helps you recognise the question. Do not spend extra preparation time on GM and HM beyond the table on this page.
What is the fastest way to find the average of an arithmetic progression?
Take the mean of the first and last terms. The formula is (a + l) divided by 2, where a is the first term and l is the last term. This works for any sequence with a constant common difference — including all natural numbers, all odd numbers, all even numbers, and any block of consecutive integers.
How is AFCAT average different from CDS or NDA average?
AFCAT average sits between NDA and CDS in depth. The calculations are mostly integer-and-percentage arithmetic; the test rewards pattern recognition and time control over algebraic depth. SSC CGL Tier-1 quant questions on averages are a close benchmark.