Averages is one of the most reliable scoring areas in AFCAT Numerical Ability — almost every paper carries one or two direct questions. The good news is that a handful of formulas and two or three speed tricks let you solve them mentally. This Cavalier guide builds the concept from scratch, then hands you the shortcuts that save precious seconds in the exam hall.
Why Averages is a must-score topic
The average (or arithmetic mean) is simply the single value that could replace every number in a set without changing their total. Examiners love it because the idea is intuitive yet the questions can be dressed up as ages, marks, cricket scores, salaries or temperatures.
For AFCAT, Averages overlaps with Percentage, Ratio and Mixtures, so mastering it pays off across the whole Numerical Ability section. Most questions are one-step or two-step — if you know which quantity to find (total, count or mean), you are already halfway done.
At The Cavalier we tell our defence aspirants to treat Averages as a guaranteed two-mark booster: the calculations are light, the data is friendly, and a calm reader rarely loses marks here. The only enemies are careless arithmetic and reading the question in a hurry. Once you internalise the three forms of the formula and the deviation shortcut below, you will start solving most items without writing more than two lines of working. That speed is exactly what frees up time for the heavier Profit-Loss and Time-Distance sums later in the paper.
An average always lies between the smallest and the largest value of the set. If your answer falls outside that range, you have made an arithmetic slip.
The core formula and its three forms
Everything in this topic flows from one relationship between three quantities: the sum of observations, the number of observations and the average.
Average = Sum of observations ÷ Number of observations
Rearranged, the same equation gives you:
- Sum = Average × Number
- Number = Sum ÷ Average
Roughly 70% of AFCAT average questions are solved just by switching between these three forms. Always ask: which one is unknown? If the question gives you the average and the count, you want the sum. If it gives you the sum and the average, you want the count. Identifying the missing piece before you touch the numbers stops you from solving the wrong thing under exam pressure.
It also helps to remember why the formula works. The average spreads the entire total evenly across all members, so each member is treated as if it held exactly the mean value. That is why multiplying the average back by the count always rebuilds the original sum. Keep this picture in mind and the rearranged forms will never feel like formulas to memorise — they are just the same equation read from a different angle.
The average of 5 numbers is 28. What is their sum?
The deviation (assumed-mean) shortcut
When you must average several large numbers that are close together, adding them all is slow. Instead, pick a convenient assumed mean and average only the small deviations.
Real average = Assumed mean + (Average of deviations from the assumed mean). Choose a round number near the middle as your assumed mean.
Find the average of 102, 98, 105, 95, 100.
You never had to add 102 + 98 + ... — the deviations cancelled out. This trick is gold for closely-grouped data such as daily temperatures, monthly incomes or a batsman's recent scores, where every value hovers near a common figure. The bigger and tighter the numbers, the more time the deviation method saves over straight addition.
One practical tip: choose your assumed mean as a number that already appears in the list, or a clean multiple of ten near the middle. A well-chosen assumed mean keeps the deviations tiny and the signs easy to track, which is the whole point of the shortcut.
Averages of evenly spaced numbers
For any list that increases by a fixed step (an arithmetic progression), the average equals the middle term — or the mean of the first and last terms.
Average of an evenly spaced list = (First term + Last term) ÷ 2
- Average of first n natural numbers = (n + 1) ÷ 2
- Average of first n odd numbers = n
- Average of first n even numbers = n + 1
Find the average of all integers from 21 to 39.
No need to count how many numbers there are — the endpoints alone give the answer instantly. This works because in any evenly spaced set the values are perfectly balanced around the centre: every number above the middle is matched by an equal distance below it, so the mean sits exactly at the midpoint. Examiners often disguise this as “average of all multiples of 5 between 10 and 90” or “average of the first 50 even numbers” — recognise the evenly spaced pattern and you skip all the addition.
Replacement: when one member is swapped
A classic AFCAT setup: a person in a group is replaced by a new person, and the average changes. The shift tells you the new person's value.
Change in total = Number of members × Change in average.
New member's value = Old member's value ± (n × change in average).
The average weight of 8 men increases by 1.5 kg when one man of 65 kg is replaced by a new man. Find the new man's weight.
Do not multiply the change in average by the wrong count. Use the total number of members in the group, not the number replaced.
Adding or removing a member
When a new member joins (or one leaves) and the average shifts, work with totals.
New value = New total − Old total. Build each total from count × average, then subtract.
The average age of 10 students is 15 years. When the teacher's age is added, the average becomes 16. Find the teacher's age.
Quick rule: the new member's value = new average + (extra people already there) × change. Here 16 + 10 × 1 = 26 — same answer in one line.
Weighted average of two groups
When two groups of different sizes are combined, you cannot simply average the two averages. You must weight each by its count.
Combined average = (n1A1 + n2A2) ÷ (n1 + n2)
where n is the count and A is the average of each group.
A class of 20 boys averages 60 marks; 30 girls average 70 marks. Find the class average.
Averaging 60 and 70 to get 65 is wrong — the girls group is larger, so the answer pulls toward 70, giving 66.
The alligation shortcut for two groups
When you know the two group averages and the combined average, alligation gives the ratio of the counts in seconds — perfect for back-calculation questions.
Ratio of group 1 to group 2 = (A2 − Acombined) : (Acombined − A1). Take the distance of each group's average from the mean and cross them.
Some boys averaging 60 and some girls averaging 70 give a class average of 66. Find the ratio of boys to girls.
This matches the earlier 20:30 split — alligation reaches it without solving an equation. The logic is that the group whose average is farther from the combined mean must be the smaller one, because a small group can only shift the overall average so much. So the ratio of counts is the inverse of the ratio of distances. Whenever an AFCAT question gives all three averages and asks for the proportion of two categories, alligation is your fastest route to the answer.
Average speed and run-rate problems
Two sub-types appear regularly. They look like the basic formula but hide a trap.
Average speed
If equal distances are covered at speeds x and y, the average speed is the harmonic mean, not the simple average.
Average speed (equal distances) = 2xy ÷ (x + y)
Run rate
Required run rate uses total runs over total overs — a weighted-average idea.
A car covers half a journey at 40 km/h and the other half at 60 km/h. Find the average speed.
The simple average of 40 and 60 is 50, but the correct answer is 48. Equal distances always pull the average below the midpoint.
Previous-year style question
Q. The average age of a cricket team of 11 players is 28 years. If the captain's age and the wicketkeeper's age (who is 3 years older than the captain) are excluded, the average of the remaining 9 players drops by 1 year. Find the captain's age.
Answer: Total of 11 players = 11 × 28 = 308. Average of remaining 9 = 27, so their total = 9 × 27 = 243. Captain + wicketkeeper = 308 − 243 = 65. Let captain = C, then wicketkeeper = C + 3, so 2C + 3 = 65, giving C = 31 years.
Traps and time-savers to keep in mind
- Averaging two averages without weighting by their counts.
- Using the number replaced instead of the total group size in replacement problems.
- Forgetting that average speed over equal distances is the harmonic mean.
- Mixing up units — convert months to years or minutes to hours first.
If the average increases by k when a new member of n+1-strong group joins, the newcomer's value = new average + n × k. Memorise this one-liner — it appears almost every cycle.
Don't confuse mean with median or mode
AFCAT sometimes tests whether you mix up the arithmetic mean with the other measures of central tendency. The mean is sum ÷ count; the median is the middle value when the data is sorted; the mode is the most frequent value. For a symmetric, evenly spaced set all three coincide, which is exactly why the average of consecutive numbers equals the middle term. But skewed data — a few very high salaries, say — pulls the mean up while leaving the median almost untouched, so read the wording carefully to see which one the question wants.
Quick revision
- Average = Sum ÷ Number; rearrange to find Sum or Number.
- Use the deviation trick for closely-grouped large numbers.
- Evenly spaced list: average = (first + last) ÷ 2.
- Replacement: change in total = group size × change in average.
- Two groups: combined average = (n1A1 + n2A2) ÷ (n1 + n2); use alligation to get the ratio.
- Equal-distance average speed = 2xy ÷ (x + y).
Frequently asked questions
How many average questions appear in AFCAT?
Typically one or two direct questions per paper, and the concept also supports Mixtures, Ratio and Data Interpretation items. It is a high-return topic for the time invested.
Why can't I just average two group averages?
Because the groups usually have different sizes. You must weight each average by its number of members, otherwise the larger group's pull is ignored and the answer is wrong.
What is the fastest way to average large, close numbers?
Use the deviation method: pick a round assumed mean, average only the small plus and minus deviations, then add that to the assumed mean. It avoids long addition entirely.
Is average speed the same as the average of two speeds?
Only when equal time is spent at each speed. For equal distances, use the harmonic mean 2xy/(x+y), which is always less than the simple average.
What's a quick one-liner for the new-member problem?
When one member joins an existing group of n and the average rises by k, the newcomer's value equals the new average plus n times k.
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