Averages for CUET General Aptitude

Q: How do averages of ages change over time?

Every member gets a year older each year, so after t years the average age of the same group rises by exactly t, and the total by the number of members times t.

About 10 min to learn · 62 questions

Averages is one of the friendliest chapters in CUET's General Aptitude Test, because almost every question comes down to one idea: an average is a total shared out equally. This lesson covers simple averages, consecutive numbers, batting averages, combined and weighted averages, the alligation shortcut, adding, removing and replacing a value, ages over time, and the new-score questions, with worked examples and the traps beside them.

Syllabus area: General Mental Ability, Numerical Ability

On this page · 8 sections

01The core idea
02Simple averages
03Weighted averages
04When the group changes
05Exam shortcuts
06Common mistakes
07Quick revision
08Check your understanding

The core idea

An average is a total shared out equally. Turn that round and you get the one formula this chapter runs on: total = average × count.

Whenever a question gives you an average, multiply it by the count straight away and work with totals. Totals can be added and subtracted; averages cannot. This one habit solves nearly every question in under a minute, and it protects you from the chapter's biggest mistake.

Simple averages

What is an average and how do you calculate it?

The average, or arithmetic mean, is the sum of the values divided by how many there are. The scores 14, 19, 22 and 25 add up to 80, so their average is $80 \div 4 = 20$ .

Average and total

\text{Average} = \frac{\text{sum of values}}{\text{number of values}} \qquad \text{Sum} = \text{average} \times \text{number of values}

Example 1 · A missing value

The average of seven numbers is 52. Six of them add up to 305. What is the seventh?

Total of seven: $7 \times 52 = 364$ .
Seventh number: $364 - 305 = 59$ .

Answer

Watch out

Never add or average averages directly. A group of 6 averaging 30 and a group of 4 averaging 40 do not average 35 together. The combined average is $(6 \times 30 + 4 \times 40) \div 10 = 340 \div 10 = 34$ .

What is the average of consecutive or evenly spaced numbers?

For any evenly spaced list, the average is the first term plus the last, divided by 2. With an odd count it is also the middle term. The multiples of 6 from 12 to 96 average $(12 + 96) \div 2 = 54$ , however many there are.

List	Sum	Average
First $n$ natural numbers	$\tfrac{n(n + 1)}{2}$	$\tfrac{n + 1}{2}$
First $n$ even numbers	$n(n + 1)$	$n + 1$
First $n$ odd numbers	$n^2$	$n$
First $n$ multiples of $k$	$\tfrac{kn(n + 1)}{2}$	$\tfrac{k(n + 1)}{2}$

Example 2 · Consecutive even numbers

The average of nine consecutive even numbers is 64. What is the largest?

With nine numbers, the average is the 5th, the middle one: 64.
The largest is four even steps above it: $64 + 4 \times 2 = 72$ .

Answer

Watch out

Consecutive odd or even numbers go up in 2s, not 1s. And write down the real first and last terms of a range described in words. Two-digit numbers that leave remainder 2 on division by 4 run from 10 to 98, not from 2, so they average 54.

How does the average change if you change every value?

The average follows whatever you do to every value. Add 7 to every value and the average rises by 7; multiply every value by 4 and the average is multiplied by 4. If a set averages 15 and every value is doubled and then increased by 3, the new average is $2 \times 15 + 3 = 33$ .

Key idea

This holds only when the same change is made to every value. If one value changes, work through the total, as in the replacement section below.

What is a batting average in cricket?

A batting average is total runs divided by the number of times the batter was out, not by the number of innings. Not-out innings add runs but not to the divisor.

Example 3 · Batting average

Aditi plays 18 innings, is not out in 3 of them, and scores 870 runs. Find her batting average.

Times out: $18 - 3 = 15$ .
Average: $870 \div 15 = 58$ .

Answer

Exam shortcut

A bowling average is runs conceded divided by wickets taken, and for a bowler a lower average is better.

Weighted averages

How do you find the combined average of two groups?

Multiply each group's average by its size, add the totals, and divide by the total number of members. Bigger groups pull the result towards their own average.

Combined average

\bar{x} = \frac{n_1 a_1 + n_2 a_2}{n_1 + n_2}

Example 4 · Two batches

A coaching centre has a morning batch of 45 students averaging 68 marks and an evening batch of 15 averaging 76. Find the combined average.

Totals: $45 \times 68 = 3{,}060$ and $15 \times 76 = 1{,}140$ .
Combined total: 4,200 for 60 students.
Combined average: $4{,}200 \div 60 = 70$ .

Answer

It is much closer to 68 than to 76, because the morning batch is three times as big.

Exam shortcut

The same method prices a mixture: 10 kg of sugar at ₹42 a kg with 5 kg at ₹48 a kg costs $(420 + 240) \div 15 = 44$ rupees a kg. And a combined average always lies between the group averages, nearer the larger group.

What is the alligation shortcut for weighted averages?

Alligation gives the ratio in which two groups must combine to give a certain average. The distances of each group's average from the combined average, taken crosswise, give the ratio of their sizes.

Alligation

\text{Size of A} : \text{Size of B} = (b - m) : (m - a)

Here $a$ and $b$ are the two group averages and $m$ the combined average between them.

Example 5 · Mixing two oils

Oil at ₹130 a litre is mixed with oil at ₹160 a litre to get a blend worth ₹140 a litre. In what ratio are they mixed?

The cheaper oil is $140 - 130 = 10$ below the blend price.
The costlier oil is $160 - 140 = 20$ above it.
Crosswise: cheaper to costlier is $20 : 10 = 2 : 1$ .

Answer

2 : 1

Check: $(2 \times 130 + 1 \times 160) \div 3 = 420 \div 3 = 140$ .

Watch out

Think of a see-saw: the group closer to the combined average is the bigger group. A ratio that puts more weight on the group further away has been flipped.

Why is the average price not always the average of the prices?

Because the average must be weighted by quantity, and equal spending does not mean equal quantity. Spending the same money at two prices buys more at the cheaper price, so the average price leans towards the cheaper one.

₹900 on cloth at ₹150 a metre buys 6 metres, and ₹900 at ₹100 a metre buys 9 metres. That is ₹1,800 for 15 metres, so ₹120 a metre, not ₹125.
When the quantities are equal, say 5 kg at each price, the average price is the plain average of the prices.

When the group changes

How do you find the new average when a value is added or removed?

Find the old total, add or remove the value, and divide by the new count. A faster way uses the excess: the amount by which the newcomer differs from the old average, shared among the new count.

Example 6 · Someone joins

The average age of 14 people is 30, and a 45-year-old joins. Find the new average. Then, separately: 19 members average 40, and one more person joins, raising the average to 41. How old is the newcomer?

The 45-year-old is 15 years above the average, shared among 15 people: the average rises by 1, to 31.
Second case: new total $20 \times 41 = 820$ , old total $19 \times 40 = 760$ , so the newcomer is $820 - 760 = 60$ .
Shortcut: newcomer = new average + old count × rise $= 41 + 19 \times 1 = 60$ .

Answer

31; 60 years

Watch out

Divide by the new count after someone joins or leaves, never the old one.

How do you handle a replacement or a wrongly entered value?

When one value is swapped for another, the count stays the same, so only the total changes, by the new value minus the old. The average changes by that difference divided by the count.

Example 7 · Replacement and correction

In a group of 12, a person weighing 70 kg is replaced and the average rises by 0.5 kg. How heavy is the new person? And the average of 25 readings is 60, but a reading of 47 was entered as 74. Find the correct average.

Replacement: the total rose by $12 \times 0.5 = 6$ kg, so the new person weighs $70 + 6 = 76$ kg.
Correction: the total was $74 - 47 = 27$ too high, so the average falls by $27 \div 25 = 1.08$ , to $60 - 1.08 = 58.92$ .

Answer

76 kg; 58.92

Exam shortcut

Ask whether the wrong value was too big or too small. That tells you whether the corrected average goes down or up before you calculate.

How do averages of ages change over time?

Every member gets a year older each year, so after $t$ years the average age of the same group rises by exactly $t$ , and the total by the number of members times $t$ .

Six years ago a family of 3 averaged 28 years, so they now total $3 \times (28 + 6) = 102$ . If the family of 4 now averages 26, their total is 104, so the youngest is 2 years old.
If a team of 10 has the same average age as 3 years ago because one member was replaced, the replacement is $10 \times 3 = 30$ years younger than the person replaced.

How do you solve "the average rises by x when a new score is added"?

The new score has to cover the old average and also lift all the innings, old and new, by the rise. So the old average is the new score minus the new count times the rise.

Example 8 · A big innings

After 14 innings a batter scores 110 in the 15th, and the average rises by 4. Find the new average.

Old average: $110 - 15 \times 4 = 50$ .
New average: $50 + 4 = 54$ .
Check: $14 \times 50 + 110 = 810 = 15 \times 54$ .

Answer

Exam shortcuts

Think in totals. Total = average × count.
Evenly spaced lists. Average = (first + last) ÷ 2.
Excess. A newcomer's difference from the average, shared by the new count, is the change.
Alligation. Crosswise distances give the ratio of sizes.
Sense checks. A combined average lies between the group averages. An older newcomer raises the average.

Common mistakes

Averaging the averages of groups of different sizes.
Dividing by the old count after someone joins or leaves.
A batting average over all innings. Divide by the times out.
Flipping the alligation ratio.
Forgetting that people age.
Counting an overlap once. If the first 6 and the last 6 of 11 numbers are given, the 6th appears in both totals.
Answering the wrong quantity. The old average or the total is often among the options.
The wrong first or last term. Multiples of 8 between 1 and 100 run from 8 to 96.
Spreading a change over too few values. One value up by 20 in a set of 10 lifts the average by 2.

Quick revision

Basics

Average = sum ÷ count. Sum = average × count
Evenly spaced: (first + last) ÷ 2
Same change to every value: the average follows it
Batting average: runs ÷ times out

Weighted averages

$\tfrac{n_1 a_1 + n_2 a_2}{n_1 + n_2}$
Alligation: A : B $= (b - m) : (m - a)$
Equal spending: weight by quantity bought

Changes

Join or leave: new total ÷ new count
Replace: change in average = (new − old) ÷ count
Ages: after $t$ years the average is $+t$
New score: old average = score − new count × rise

Median, mode and spread are worked in mean, median and mode, mixtures in mixtures and alligation, and average speed in time, speed and distance.

Next · 07Quick revision

Check your understanding

62 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Simple averageQ1 of 62Q1

Simple average

20 questions

Month	Bill (₹)
Jan	1,240
Feb	1,180
Mar	1,050
Apr	1,320
May	1,610
Jun	1,880

Check your understanding

Simple average

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Weighted average

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Change in average

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