Averages, Ages & Number System
Averages, Ages & Number System is a frequently tested area in CUET General Test. Work through these free NTA-style sample questions with full answers and explanations, then attempt all 35 in a timed practice test to build exam-day speed.
Snapshot
- Averages, ages and number system bundle three fast-scoring families: the mean, age puzzles that are linear equations in disguise, and number facts such as divisibility, unit digits, HCF, LCM and remainders.
- One idea cracks the averages: work with totals, not averages. One habit cracks the ages: shift every person by the same number of years.
- This page gives the reasoning, a LaTeX formula sheet, shortcuts with the situations where they break, 15 worked examples, a traps table, 16 rapid-fire questions with answers and a one-week routine.
- Exam reality: +5 for a correct answer, −1 for a wrong one, and each question is one or two steps once you set it up correctly.
Part 1 — The core idea: an average is a fair share
Written and reviewed by Devendra Dubey, Mathematics & Commerce faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching). Prepared for the CUET UG General Test.
The average of a set is the value each member would hold if the total were shared out equally. That is why the useful form is the rearranged one:
$$\text{Sum} = \text{Average} \times \text{Number of items}$$
Why totals beat averages
Averages cannot be added, removed or replaced directly, because each one carries a different count. Totals can. When a person joins, leaves or is swapped, convert both averages into totals, compare the totals, and the answer is a simple subtraction. If the average of 10 numbers rises by 2 when one number is replaced, the total rose by $10 \times 2 = 20$, so the new number is 20 more than the old one. No equation needed.
Why the average lies between the smallest and the largest
An average balances the values around it: the amounts by which values exceed it equal the amounts by which others fall short. This balance is called the deviation principle. It makes the "assumed mean" shortcut work: pick a friendly number near the data, add up only the differences, and divide.
Why age problems are easy
Every age problem is a pair of linear equations. Two facts make them straightforward. First, everyone ages at the same rate, so "5 years ago" subtracts 5 from every person and "in 8 years" adds 8 to every person. Second, the difference between two ages never changes, which gives an immediate equation in many problems.
Why number-system questions are formulaic
Numbers repeat. Unit digits of powers follow a short cycle, remainders follow patterns, and every number has a unique prime factorisation. Once you know the pattern you read the answer off it, with no large calculation.
Pause and predict: a car goes from A to B at 40 km/h and returns at 60 km/h. Is the average speed for the whole trip more than, less than, or equal to 50 km/h?
Check your answer: less than 50. The car spends more time at the slower speed, so the average is pulled towards 40. The exact value is $\dfrac{2 \times 40 \times 60}{40 + 60} = 48$ km/h.
Part 2 — Formula sheet
Averages:
$$\text{Average} = \dfrac{\text{Sum of values}}{\text{Number of values}} \qquad \text{Weighted average} = \dfrac{n_1 a_1 + n_2 a_2}{n_1 + n_2}$$
$$\text{Average speed (equal distances)} = \dfrac{2ab}{a + b} \qquad \text{Average speed in general} = \dfrac{\text{Total distance}}{\text{Total time}}$$
Number system:
$$N = p^a q^b \Rightarrow \text{Number of factors} = (a + 1)(b + 1)$$
$$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$$
| Quantity | Formula | When to use |
|---|---|---|
| Sum of first n natural numbers | $\dfrac{n(n+1)}{2}$ | Average is $\dfrac{n+1}{2}$ |
| Sum of first n odd numbers | $n^2$ | Average is n |
| Sum of first n even numbers | $n(n+1)$ | Average is $n + 1$ |
| Average of an AP | $\dfrac{\text{First} + \text{Last}}{2}$ | Any evenly spaced list |
| Change in average on adding a value | $\dfrac{\text{New value} - \text{Old average}}{n + 1}$ | One item joins |
| New value after a rise of d in average | $\text{Old value} + nd$ | One item replaced in a set of n |
| Father-son type age | Set ages as $kx$ and $x$ | One age is a multiple of the other |
| Age ratio $p : q$ | Write $px$ and $qx$ | Ratio given for the present |
| Age change | Add or subtract the same years to all | Past or future statements |
| Sum of factors of $p^a q^b$ | $\dfrac{p^{a+1} - 1}{p - 1} \times \dfrac{q^{b+1} - 1}{q - 1}$ | Sum of all factors |
| HCF of fractions | $\dfrac{\text{HCF of numerators}}{\text{LCM of denominators}}$ | Fractions in lowest terms |
| LCM of fractions | $\dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}$ | Fractions in lowest terms |
| Least number leaving remainder r | $\text{LCM} + r$ | Same remainder with every divisor |
| Greatest divisor leaving remainders | HCF of the (number − remainder) values | Divisor with fixed remainders |
Divisibility tests
| Divisor | Test |
|---|---|
| 2 | Last digit is even |
| 3 | Digit sum is a multiple of 3 |
| 4 | Last two digits form a multiple of 4 |
| 5 | Last digit is 0 or 5 |
| 6 | Divisible by both 2 and 3 |
| 8 | Last three digits form a multiple of 8 |
| 9 | Digit sum is a multiple of 9 |
| 11 | (Sum of digits in odd places) − (sum in even places) is 0 or a multiple of 11 |
Unit digit cycles
| Last digit of the base | Cycle of the unit digit as the power rises | Cycle length |
|---|---|---|
| 0, 1, 5, 6 | The same digit always | 1 |
| 4 | 4, 6 | 2 |
| 9 | 9, 1 | 2 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
For the cycles of length 4, divide the power by 4 and use the remainder. A remainder of 0 means the last digit of the cycle. For 4 and 9 the odd powers give the first digit and the even powers give the second.
Part 3 — Shortcut methods and when they fail
| Shortcut | How it works | When it fails |
|---|---|---|
| Total method | Convert averages to totals, then add or subtract | Always valid, but you must use the right count (old or new) for each total |
| Assumed mean | Pick a base near the data, add the deviations, divide by the count | A poorly chosen base, such as 0, gives no saving |
| Middle value | The average of consecutive numbers is the middle one | Only for evenly spaced lists. For a general set you must compute |
| Weighted average cross | Distances from the overall average are inverse to the group sizes | Works only for two groups with a known overall average |
| Harmonic mean $2ab/(a+b)$ | Equal distances at two speeds | Fails for equal times, where the plain mean is right, and for three or more legs, where you need total distance over total time |
| Age difference is constant | Use the gap as an equation | Does not help with ratios of ages at one moment, where you need two equations |
| Unit digit cycle | Only the power mod 4 matters | It gives the last digit only. It does not give the remainder on division by any other number |
| HCF × LCM = product | Recover a missing number | Valid for two numbers. It is false for three numbers |
Habit for departures. When someone leaves, the total falls by that person's value and the count falls by one. When the average rises after a departure, the person who left was below the group average.
Sanity checks. The average must lie between the smallest and largest value. HCF is never larger than the smaller number, and LCM is never smaller than the larger number. These three checks eliminate many wrong options immediately.
Part 4 — Worked examples: averages
Worked example 1: missing value. Five tests average 50 marks. The first four add up to 190. What was the fifth?
Total of five = $50 \times 5 = 250$. Fifth = $250 - 190 = 60$.
Worked example 2: a number leaves. The average of 6 numbers is 30. When one number is removed the average of the rest is 32. Which number was removed?
| Step | What we do | Result |
|---|---|---|
| 1 | Old total | $30 \times 6 = 180$ |
| 2 | New total of 5 numbers | $32 \times 5 = 160$ |
| 3 | Removed number | $180 - 160 = 20$ |
Sanity check: the average rose, so the removed number should be below 30, and 20 is.
Worked example 3: a number is replaced. The average of 10 numbers is 15. One number, 15, is replaced and the average becomes 17. What is the new number?
The total rose by $10 \times 2 = 20$, so the new number is $15 + 20 = 35$.
Worked example 4: weighted average. 30 boys average 45 kg and 20 girls average 40 kg. Find the class average.
$$\dfrac{30 \times 45 + 20 \times 40}{50} = \dfrac{1350 + 800}{50} = \dfrac{2150}{50} = 43 \text{ kg}$$
Common wrong step: taking $(45 + 40)/2 = 42.5$. The groups are unequal, so the average sits closer to the larger group (boys).
Worked example 5: average speed. A bus covers equal distances at 30 km/h and 20 km/h. Find its average speed.
$$\dfrac{2 \times 30 \times 20}{30 + 20} = \dfrac{1200}{50} = 24 \text{ km/h}$$
Check with numbers: a 60 km leg each way takes 2 h and 3 h, so 120 km in 5 h gives 24 km/h. The simple mean of 25 km/h would be wrong.
Worked example 6: batting average. A batsman scores 85 in his 17th innings and his average rises by 3. What is his average after 17 innings?
| Step | What we do | Result |
|---|---|---|
| 1 | Let the old average be x, so 16 innings total $16x$ | Old total $16x$ |
| 2 | New total gives new average x + 3 | $16x + 85 = 17(x + 3)$ |
| 3 | Expand and solve | $85 - 51 = x$, so $x = 34$ |
| 4 | New average | $34 + 3 = 37$ |
Shortcut: the last score must exceed the new average by $16 \times 3 = 48$ (3 for each of the 16 earlier innings), so the new average is $85 - 48 = 37$.
Worked example 7: assumed mean. Find the average of 82, 87, 91, 79, 86.
Assume 85. Deviations: −3, +2, +6, −6, +1. Their sum is 0, so the average is 85. Direct check: $425 \div 5 = 85$.
Worked example 8: family average. The average age of a family of 5 is 24 years. The youngest member is 4. What was the average age of the other four members when the youngest was born?
Total now = $24 \times 5 = 120$. The other four have a total of $120 - 4 = 116$. Four years ago each was 4 years younger, so their total was $116 - 4 \times 4 = 100$ and the average was $100 \div 4 = 25$ years.
Part 5 — Worked examples: ages and number system
Worked example 9: multiple and a shift. A father is three times as old as his son. In 12 years he will be twice as old. Find their present ages.
| Step | What we do | Result |
|---|---|---|
| 1 | Son = x, father = 3x | Present ages |
| 2 | After 12 years | $3x + 12 = 2(x + 12)$ |
| 3 | Solve | $x = 12$ |
| 4 | Ages | Son 12, father 36 |
Check: in 12 years they are 24 and 48.
Worked example 10: ratio and a shift. The present ages of two people are in the ratio 5 : 7. After 6 years the ratio becomes 3 : 4. Find their ages.
$$\dfrac{5x + 6}{7x + 6} = \dfrac{3}{4} \Rightarrow 20x + 24 = 21x + 18 \Rightarrow x = 6$$
The ages are 30 and 42. Check: after 6 years they are 36 and 48, and the ratio 36 : 48 is 3 : 4.
Worked example 11: past and future. Five years ago A was three times as old as B. Ten years from now A will be twice as old as B. Find their present ages.
A − 5 = 3(B − 5) gives A = 3B − 10. A + 10 = 2(B + 10) gives A = 2B + 10. Equate: 3B − 10 = 2B + 10, so B = 20 and A = 50. Check: five years ago they were 45 and 15, and in ten years they will be 60 and 30.
Worked example 12: sum and a past ratio. The sum of a father's and son's ages is 60. Six years ago the father was five times as old as the son. Find the present ages.
F + S = 60 and F − 6 = 5(S − 6), so F = 5S − 24. Substitute: 6S − 24 = 60, so S = 14 and F = 46. Check: six years ago they were 40 and 8.
Worked example 13: unit digits. Find the unit digit of $7^{123}$ and of $2^{100}$.
For 7 the cycle is 7, 9, 3, 1. Divide 123 by 4: the remainder is 3, so take the third term: 3. For 2 the cycle is 2, 4, 8, 6. Divide 100 by 4: the remainder is 0, so take the last term of the cycle: 6.
Common wrong step: treating a remainder of 0 as the first term of the cycle.
Worked example 14: factors and HCF/LCM. (a) How many factors does 72 have? (b) The HCF of two numbers is 6 and their LCM is 60. One number is 12. Find the other.
(a) $72 = 2^3 \times 3^2$, so the count is $(3 + 1)(2 + 1) = 12$. (b) Other number = $\dfrac{6 \times 60}{12} = 30$. Check: HCF of 12 and 30 is 6, and LCM is 60.
Worked example 15: remainders. (a) The greatest number that divides 245 and 1029 leaving remainder 5 in each case. (b) The least number that leaves remainder 3 when divided by 8, 12 and 15. (c) Remainder when $2^{10}$ is divided by 7.
(a) Subtract the remainder: 240 and 1024. Their HCF is 16, since $240 = 16 \times 15$ and $1024 = 16 \times 64$ with 15 and 64 sharing no factor. (b) LCM of 8, 12, 15 is 120, so the number is $120 + 3 = 123$. (c) $2^3 = 8$ leaves remainder 1 when divided by 7, so $2^{10} = (2^3)^3 \times 2$ leaves $1 \times 2 = 2$. Check: 1024 = 7 × 146 + 2.
Part 6 — Common traps
| Trap | Why it is tempting | The truth |
|---|---|---|
| Averaging two group averages directly | The numbers are right there | Use a weighted average when the group sizes differ |
| Average speed as $(a + b)/2$ | It looks natural | For equal distances use $2ab/(a + b)$, which is always lower |
| Aging only one person | The question mentions one person last | A time shift applies to every age in the problem |
| Forgetting that the count changes | The average appears to stay | After a departure divide by $n - 1$, after an arrival by $n + 1$ |
| Remainder 0 treated as first of the cycle | The cycle starts at the first power | A remainder of 0 selects the last value in the cycle |
| HCF × LCM = product used for three numbers | It worked for two | Valid for two numbers only |
| Counting factors as the exponent | The exponents are visible | Add 1 to each exponent before multiplying |
| 11-test with absolute sums | Both sums look useful | The difference of the odd-place and even-place sums must be 0 or a multiple of 11 |
Part 7 — Rapid-fire practice with answers
| No. | Question | Answer |
|---|---|---|
| 1 | Average of 12, 18, 24, 30, 36 | 24 |
| 2 | Average of the first 10 natural numbers | 5.5 |
| 3 | Average of the first 15 odd numbers | 15 |
| 4 | Average of 5 numbers is 20. A sixth number, 32, is added. New average | 22 |
| 5 | Average of 8 numbers is 15. One of them, 29, is removed. Average of the rest | 13 |
| 6 | A walker goes at 30 km/h and returns at 20 km/h over the same road. Average speed | 24 km/h |
| 7 | A father is 4 times his son's age. Their ages add up to 50. Son's age | 10 |
| 8 | Two ages are in the ratio 3 : 5 and total 64. The elder | 40 |
| 9 | Unit digit of $7^{45}$ | 7 |
| 10 | Unit digit of $8^{22}$ | 4 |
| 11 | Is 345678 divisible by 3? By 9? | Yes, no (digit sum 33) |
| 12 | Number of factors of 100 | 9 |
| 13 | HCF 12, LCM 180, one number 36. The other | 60 |
| 14 | Smallest number divisible by 6, 8 and 12 | 24 |
| 15 | Remainder when $2^{20}$ is divided by 3 | 1 |
| 16 | Sum of the first 20 natural numbers | 210 |
Part 8 — A timed-practice routine
| Day | Task | Time |
|---|---|---|
| 1 | Learn the divisibility tests and the unit-digit cycle table until you can write them from memory | 15 min |
| 2 | Re-solve worked examples 1 to 8 closed book, writing the total of each average first | 25 min |
| 3 | Re-solve worked examples 9 to 15 closed book, and define the variable on every age problem before you write an equation | 25 min |
| 4 | Attempt the rapid-fire table against a clock: 16 questions in 12 minutes | 12 min |
| 5 | Mixed set of 10 questions at 60 seconds each, marked as solved, guessed or skipped | 10 min |
| 6 | Redo every miss and write the one rule that would have saved it | 20 min |
| 7 | Timed test of 15 questions in 15 minutes, aiming for 80% accuracy | 15 min |
Exam strategy. Age and average questions are short, so do them early. If an age question is taking more than a minute, you have probably chosen the wrong variable. Restart with the youngest person as x.
Part 9 — Where this connects
Weighted averages are alligation in disguise, so revisit Percentage, Ratio and Alligation. Equal-distance average speed returns in Time, Speed and Work, and averages are used in almost every question set in Data Interpretation. The sum formulas used here come from Sequences and Series, and the full treatment of the mean and spread is in Statistics. Pattern-based number questions continue in Series.
One-line revision: sum = average × count, equal-distance average speed is 2ab/(a + b), a time shift moves every age equally, test 3 and 9 by digit sum, read unit digits from the cycle of four, and HCF × LCM equals the product for two numbers.
Practice questions
Now test yourself. 8 free sample questions with explanations. 27 more in the timed practice test.
Q1. The average score of a cricketer in $10$ matches was $38.9$ and in $5$ other matches it was $50$. What is his average score in all $15$ matches?
▸ Show answer & explanation
Answer: A
Total $= 10\times38.9 + 5\times50 = 389+250 = 639$. Average over $15 = \frac{639}{15} = 42.6$.
Q2. The average of $a, b, c$ is $20$ and the average of $b, c, d$ is $25$. If $d = 30$, then what is the value of $a$?
▸ Show answer & explanation
Answer: B
$a+b+c = 60$ and $b+c+d = 75$. Subtracting, $d-a = 15$, so $a = d-15 = 30-15 = 15$.
Q3. The average of $9$ observations is $45$. The average of the first $5$ is $42$ and that of the last $5$ is $48$. What is the value of the $5$th observation?
▸ Show answer & explanation
Answer: A
Total of $9 = 9\times45 = 405$. Sum of first $5 = 210$, sum of last $5 = 240$. The $5$th is counted twice, so it equals $210+240-405 = 45$.
Q4. Simplify: $12+8\div2\times3-4$.
▸ Show answer & explanation
Answer: A
$8\div2 = 4$, then $4\times3 = 12$. So $12+12-4 = 20$.
Q5. The largest $4$-digit number exactly divisible by $88$ is:
▸ Show answer & explanation
Answer: A
$9999\div88 = 113$ remainder $55$ (since $88\times113 = 9944$). So the largest $4$-digit multiple is $9944$.
Q6. What is the greatest number that divides $43$, $91$ and $183$ leaving the same remainder in each case?
▸ Show answer & explanation
Answer: A
The required number is the HCF of the differences: $91-43 = 48$, $183-91 = 92$, $183-43 = 140$. HCF of $48, 92, 140$ is $4$.
Q7. The ratio of the present ages of A and B is $4:5$. After $6$ years the ratio becomes $5:6$. What is the present age of A?
▸ Show answer & explanation
Answer: A
Let ages be $4x$ and $5x$. Then $\frac{4x+6}{5x+6} = \frac{5}{6}$, so $6(4x+6) = 5(5x+6)$, giving $24x+36 = 25x+30$, so $x = 6$. A's age $= 4\times6 = 24$ years.
Q8. The average weight of $8$ persons increases by $2.5$ kg when a new person replaces one of them who weighs $65$ kg. What is the weight of the new person?
▸ Show answer & explanation
Answer: C
Total increase $= 8\times2.5 = 20$ kg. New person's weight $= 65+20 = 85$ kg.
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