Averages and mixtures are among the friendliest marks in CLAT Quantitative Techniques. The maths is class-10 level and the formulas are short. What trips students up is not the arithmetic — it is reading a data passage or chart, picking the right numbers, and choosing between a simple and a weighted average. Get those habits right and this topic becomes a reliable scoring zone.
📌 The one idea behind every average
An average is what each part would be if the total were shared equally. So average = sum ÷ count, and just as usefully, sum = average × count. That second form is the workhorse: in CLAT you are nearly always given an average and asked to recover or rebuild a total. Hold both versions in your head and most questions unlock themselves.
The average formula and what it really means
The average (or arithmetic mean) of a set of values is their sum divided by how many there are. If five students score 60, 70, 80, 90 and 100, the total is 400 and the average is 400 ÷ 5 = 80. The average is the single number that, repeated for every member, would give the same total.
- ✓Average = Sum ÷ Count — the definition. Add the values, divide by how many there are.
- ✓Sum = Average × Count — rearranged. This is the version CLAT questions usually need.
- ✓The average lies between the smallest and largest value — never below the minimum or above the maximum. A quick sanity check on any answer.
- ✓Adding the average itself doesn't change it — include one more value equal to the current average and the average stays put.
⚠️ The average is NOT the midpoint of the range
A common slip: students see values from 20 to 80 and guess the average is 50 — the middle of the range. That is only true if the values are evenly spread. The average is pulled towards wherever the numbers actually cluster. If most values sit near 20, the average sits near 20, however high the maximum is. Always compute sum ÷ count; never read the average off the endpoints.
How adding or removing a value shifts the average
CLAT loves the question 'the average of a group changes when one person joins or leaves — find the new value.' Don't rebuild the whole list. Work with totals instead. The total is average × count; adjust the total for the change, then divide by the new count.
- 1
Find the original totalTotal = old average × old count. This single number captures the whole group, so you never need the individual values.
- 2
Adjust for the changeAdding a value? Add it to the total. Removing one? Subtract it. Replacing one? Subtract the old and add the new.
- 3
Divide by the new countNew average = adjusted total ÷ new count. Watch the count carefully — adding a member raises it by one, removing one lowers it.
- 4
Sanity-check the directionA new value above the old average should pull the average up; below it, down. If your answer moves the wrong way, you have a sign error.
💡 Think in 'how far from the average'
A faster mental model: a new value only moves the average by how far it sits from the current average, spread across the group. If a team of 5 averages 40 and a sixth member scores 46, the extra 6 marks are shared over 6 people, lifting the average by 1, to 41. This 'surplus shared out' idea answers many add/remove questions without any long division.
🧩 Worked example
A coaching batch has 24 students with an average score of 62 in a mock test. A 25th student joins the batch. After including this student's score, the average of all 25 students rises to 63.
What did the 25th student score in the mock test?
A63
B85
C87
D88
▸ Show solution
Answer: C. Use totals. Original total = 24 × 62 = 1488. New total for 25 students = 25 × 63 = 1575. The 25th student's score is the difference: 1575 − 1488 = 87.
Quick check with the 'surplus' idea: the average rose by 1 across 25 people, so the new student is 25 marks above the new average of 63, i.e. 63 + 24 = 87 (the 1-mark lift also applies to the 24 existing students). Option C is correct.
Quick check with the 'surplus' idea: the average rose by 1 across 25 people, so the new student is 25 marks above the new average of 63, i.e. 63 + 24 = 87 (the 1-mark lift also applies to the 24 existing students). Option C is correct.
Weighted average — and why it differs from a simple average
A simple average treats every value as equally important. But often the groups behind the numbers are different sizes, and then you must weight each value by its group's size. The weighted average is the sum of (value × weight) divided by the sum of the weights.
Weighted average = (value₁ × weight₁ + value₂ × weight₂ + …) ÷ (weight₁ + weight₂ + …)
Suppose Section A has 40 students averaging 50, and Section B has 10 students averaging 80. The simple average of 50 and 80 is 65 — but that is wrong, because far more students sat in the lower-scoring section. The honest combined average weights each section by its size.
⚠️ Simple average ≠ weighted average
This is the single most exploited trap in CLAT averages. When you combine two groups of different sizes, you cannot just average the two averages. For the sections above, the true average = (40 × 50 + 10 × 80) ÷ (40 + 10) = (2000 + 800) ÷ 50 = 2800 ÷ 50 = 56, not 65. The combined average always leans towards the larger group. Average the averages only when the groups are the same size.
🧩 Worked example
In a school, the boys' average height is 150 cm and the girls' average height is 140 cm. There are 30 boys and 20 girls in the group.
What is the average height of all 50 students, taken together?
A144 cm
B145 cm
C146 cm
D148 cm
▸ Show solution
Answer: C. Weight each average by its group size. Total height of boys = 30 × 150 = 4500 cm; of girls = 20 × 140 = 2800 cm. Combined total = 4500 + 2800 = 7300 cm over 50 students. Average = 7300 ÷ 50 = 146 cm. Note it sits closer to 150 (the boys) than to 140, because there are more boys — exactly what a weighted average should do. Option C is correct. (The naive simple average of 150 and 140 would give 145 — the trap option.)
The alligation rule — the cross method
Alligation is a shortcut for weighted-average problems where you know the two ingredient values and the final mixture value, and you want the ratio in which they were mixed. Instead of forming an equation, you take differences crosswise — the famous cross method. It is the fastest tool in this whole topic.
📌 The alligation rule in one line
When two ingredients at values cheaper and dearer are mixed to reach a mean value, the ratio of cheaper to dearer is:
(dearer − mean) : (mean − cheaper).
In words: each ingredient's share is proportional to the distance of the other ingredient from the mean. The mixture leans towards whichever ingredient you used more of — so the bigger ratio sits on the side nearer the mean.
(dearer − mean) : (mean − cheaper).
In words: each ingredient's share is proportional to the distance of the other ingredient from the mean. The mixture leans towards whichever ingredient you used more of — so the bigger ratio sits on the side nearer the mean.
Picture it as a cross. Write the cheaper value bottom-left, the dearer value top-left, the mean in the middle, and subtract along each diagonal. The two results, read across, give the ratio of the cheaper to the dearer ingredient.
| Cheaper value | Dearer value | |
|---|---|---|
| Place at one corner | → Mean (middle) ← | Place at the other corner |
| Take (Dearer − Mean) for the cheaper's share | Take (Mean − Cheaper) for the dearer's share | |
| Ratio cheaper : dearer = (Dearer − Mean) : (Mean − Cheaper) |
- 1
Identify the three valuesThe cheaper (lower) ingredient value, the dearer (higher) one, and the mean (the price or concentration the mixture must reach).
- 2
Cross-subtractCheaper's share = Dearer − Mean. Dearer's share = Mean − Cheaper. Always subtract the smaller from the larger so both shares are positive.
- 3
Write the ratioRatio of cheaper to dearer = (Dearer − Mean) : (Mean − Cheaper). Simplify the ratio if you can.
- 4
Scale to the quantity askedIf a total amount is given, split it in that ratio to find how much of each ingredient is needed.
🧩 Worked example
A tea merchant wants to mix two types of tea to sell at a cost of 90 rupees per kilogram. One variety costs 80 rupees per kilogram and the other costs 100 rupees per kilogram. He wants to prepare 50 kilograms of the blend.
In what ratio must the cheaper and dearer teas be mixed, and how many kilograms of the cheaper tea are needed?
A1 : 1, so 25 kg of the cheaper tea
B2 : 1, so about 33.3 kg of the cheaper tea
C1 : 2, so about 16.7 kg of the cheaper tea
D3 : 2, so 30 kg of the cheaper tea
▸ Show solution
Answer: A. Apply alligation. Cheaper = 80, dearer = 100, mean = 90.
Cheaper's share = Dearer − Mean = 100 − 90 = 10.
Dearer's share = Mean − Cheaper = 90 − 80 = 10.
So cheaper : dearer = 10 : 10 = 1 : 1. The mean (90) sits exactly midway between 80 and 100, so equal parts are needed. For 50 kg, that is 25 kg of each. Option A is correct. Whenever the target is the midpoint of the two prices, alligation gives a clean 1 : 1.
Cheaper's share = Dearer − Mean = 100 − 90 = 10.
Dearer's share = Mean − Cheaper = 90 − 80 = 10.
So cheaper : dearer = 10 : 10 = 1 : 1. The mean (90) sits exactly midway between 80 and 100, so equal parts are needed. For 50 kg, that is 25 kg of each. Option A is correct. Whenever the target is the midpoint of the two prices, alligation gives a clean 1 : 1.
ℹ️ Alligation is just weighted average, reversed
The cross method is not a separate theory — it is the weighted-average formula rearranged to solve for the ratio instead of the mean. Use the weighted-average equation when you want the final value; use alligation when you already know the final value and want the mixing ratio. Same idea, two directions.
Drill averages, mixtures & alligation now
10 drills, 150 questions — CLAT-style data passages with simple-vs-weighted traps and full step-by-step working in every answer.
Concentration, replacement and dilution
Mixture questions are not only about price. They also ask about concentration — how much milk in a milk-water mix, how much acid in a solution, how much pure metal in an alloy. Alligation works identically here: treat the percentage concentration as the 'value', and the cross method gives the ratio of the two solutions you must blend to hit a target concentration.
- ✓Concentration — the fraction (or percentage) of the key ingredient in the whole. Pure water is 0% milk; pure milk is 100%.
- ✓Mixing two solutions — alligation on their concentrations gives the ratio to combine them in to reach the target concentration.
- ✓Dilution (adding water) — adding water is mixing with a 0%-concentration ingredient. The quantity of the key ingredient stays the same; only the total grows.
- ✓Replacement — removing some mixture and topping up with water lowers the concentration each round; the key ingredient that remains is the focus.
💡 In dilution, track what stays constant
When you add water to a solution, the amount of the key ingredient does not change — only the total volume rises, so the concentration falls. The cleanest way to solve a dilution question is to fix the quantity that stays constant (the acid, the milk, the pure metal), then express the new concentration as that fixed amount over the new larger total. Chasing the water around is where students go wrong.
🧩 Worked example
A 40-litre solution contains acid and water in the ratio 3 : 1, so it is 75 percent acid. A chemist wants to dilute it down to a solution that is 60 percent acid by adding pure water only.
How many litres of water must be added?
A8 litres
B10 litres
C12 litres
D15 litres
▸ Show solution
Answer: B. Fix what stays constant — the acid. In 40 litres at 75%, acid = 0.75 × 40 = 30 litres. Adding water does not change this 30 litres of acid; it only raises the total. We need the 30 litres of acid to become 60% of the new total, so new total = 30 ÷ 0.60 = 50 litres. Water added = 50 − 40 = 10 litres. Option B is correct. Notice we never tracked the water already present — only the constant acid.
Averages in data interpretation
Because CLAT Quant is passage and chart based, averages most often appear inside a table or bar chart: 'find the average sales across the four years', or 'which month was above the yearly average.' The maths is unchanged — sum ÷ count — but the skill is reading the right cells and not miscounting.
- 1
Read the question's exact scopeAverage of which rows or years? 'Average monthly sales' over a year means divide by 12; 'average of the three branches' means divide by 3. Misreading the scope is the top error.
- 2
Pull only the needed valuesCharts pack in extra data to distract you. List just the values the question asks about before adding anything.
- 3
Add carefully, then divide by the countSum the chosen values, divide by how many you pulled. Keep units consistent (don't mix thousands with lakhs).
- 4
Use the average as a benchmarkFollow-up parts often ask 'how many years were above average?' — compute the average once, then compare each value against it.
⚠️ Count the categories, not the bars you see
On a grouped bar chart a year may show three bars (one per product), but if the question asks for the average per year you divide by the number of years, not the number of bars. Equally, a 'total of all years' divided by years gives a yearly average, while dividing by the number of products gives something meaningless. Always match your divisor to what the question is averaging over.
🧩 Worked example
A company's annual revenue, in crore rupees, was: 2019 — 40, 2020 — 30, 2021 — 50, 2022 — 60, 2023 — 70. A reader studies the five-year figures shown in the chart.
In how many of the five years was the company's revenue above its own five-year average?
A1 year
B2 years
C3 years
D4 years
▸ Show solution
Answer: B. First find the average. Sum = 40 + 30 + 50 + 60 + 70 = 250 crore over 5 years, so average = 250 ÷ 5 = 50 crore. Now compare each year to 50: 2019 (40) below, 2020 (30) below, 2021 (50) equal — not above, 2022 (60) above, 2023 (70) above. Only 2022 and 2023 exceed the average, so the answer is 2 years — option B. The year that exactly equals the average does not count as 'above' it: read such stems precisely.
ℹ️ 'Above average' excludes 'equal to average'
When a question asks how many values are above the average, a value sitting exactly on the average does not count. If it had said 'at least the average' or 'not below', the equal value would count. CLAT writes these stems deliberately — one careful reading saves a mark.
A repeatable method for the exam screen
Under time pressure, don't decide afresh each time. Run the same short routine and the right tool — plain average, weighted average or alligation — picks itself.
- 1
Ask: average, or mixing?If you are summarising a set of numbers, it is an average. If you are blending two ingredients to a target, it is a mixture — reach for alligation.
- 2
Check the group sizesCombining groups of different sizes? You need a weighted average, never the simple average of the two averages.
- 3
Work with totalsFor add/remove/replace questions, convert averages to totals (average × count), adjust, then divide by the new count.
- 4
For mixtures, cross-subtractKnow both ingredient values and the mean? Use alligation: ratio = (dearer − mean) : (mean − cheaper), then scale to the quantity.
🎯 Averages, mixtures & alligation in a nutshell
- Average = sum ÷ count, and sum = average × count — the second form solves most CLAT questions; work with totals.
- The average always lies between the smallest and largest value — and it is not the midpoint of the range.
- Adding, removing or replacing a value: adjust the total, then divide by the new count.
- Weighted average (value × weight, over total weight) is needed when groups differ in size — never just average the averages.
- Alligation (cross method): ratio of cheaper to dearer = (dearer − mean) : (mean − cheaper) — the fast way to find a mixing ratio.
- Dilution keeps the key ingredient constant while the total grows; concentration problems use alligation on percentages.
Common mistakes to stop making
- ✓Taking a simple average of two averages when the underlying groups are different sizes — use weights.
- ✓Reading the average as the midpoint of the range instead of computing sum ÷ count.
- ✓Getting the count wrong after a value is added or removed — it changes by one.
- ✓In dilution, chasing the water instead of fixing the constant ingredient (acid, milk, pure metal).
- ✓On a chart, dividing by the number of bars rather than the number of years or categories the question averages over.
- ✓Counting a value equal to the average as 'above' it when the stem says 'above'.
🔢
Quantitative Techniques hub
All CLAT Quant chapters, the section strategy and every data-interpretation drill in one place.
Section guide
📈
Next: Profit, Loss, Interest & Time-Speed-Work
Percentages in action — margins, simple and compound interest, and the classic rate problems CLAT recycles.
Guide + 10 drills
Ready for the next chapter?
Profit, Loss, Interest & Time-Speed-Work turns percentages into marks — cost-price tricks, interest formulas, and rate problems straight from CLAT data passages.
Frequently asked questions
What is the difference between a simple average and a weighted average in CLAT?
A simple average treats every value as equally important: sum divided by count. A weighted average multiplies each value by the size of the group behind it, then divides by the total of those sizes. You need the weighted version whenever you combine groups of different sizes — averaging two averages directly is wrong unless the groups are equal in size.
What is the alligation rule and when do I use it?
Alligation is a shortcut for finding the ratio in which two ingredients must be mixed to reach a target value. The ratio of cheaper to dearer equals (dearer − mean) : (mean − cheaper). Use it whenever you know both ingredient values and the final mixture value and want the mixing ratio. It is simply the weighted-average formula solved for the ratio instead of the mean.
Why is the average not the midpoint between the smallest and largest values?
The average depends on where all the numbers actually cluster, not just the endpoints. It equals the midpoint of the range only when the values are evenly spread. If most numbers sit near the low end, the average sits near the low end too, however high the maximum is. Always compute sum divided by count rather than guessing from the extremes.
How do I solve a question where one value is added to or removed from a group?
Work with totals, not the individual values. Find the original total as average times count, then add the new value or subtract the removed one. Divide the adjusted total by the new count to get the new average. Check the direction: a value above the old average should raise it, and a value below should lower it.
How do dilution questions work when water is added?
Adding pure water does not change the amount of the key ingredient — only the total volume grows, so the concentration falls. Fix the constant quantity, such as the acid or milk, then set it equal to the target percentage of the new larger total to find that total. The extra water is the new total minus the original volume. Never track the water itself.
How are averages tested inside CLAT data interpretation passages?
Quant is chart and passage based, so averages usually appear inside a table or bar graph: average sales across years, average across categories, or how many values beat the average. The maths is still sum divided by count. The skill is reading the right cells and matching your divisor to what is being averaged — years, months or categories — not the number of bars on screen.
Do I need anything beyond class-10 maths for CLAT averages and mixtures?
No. CLAT Quantitative Techniques uses only class-10 level arithmetic — averages, ratios, percentages and basic algebra. There is no calculus, trigonometry or advanced statistics. The challenge is comprehension and speed: extracting the right numbers from a passage and applying short, well-practised formulas under the clock, which is exactly what the drills train.