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Quantitative Aptitude · Chapter 10

Mixture and alligation: balance quantities by value.

Understand weighted averages, use alligation to find mixing ratios, and track concentration when part of a mixture is repeatedly removed and replaced.

Illustration of Rahul Gautam guiding learners through mixture calculations
“Every mixture question is a weighted balance hiding inside a story.”— Rahul Gautam

Concept 1

A mixture combines quantities with different values

The “value” may be price per kilogram, percentage concentration, purity, marks, speed or any comparable rate. The mean value of a mixture must lie between the component values when all quantities are positive.

Balance rule: total value = sum of (quantity × component value). Mean value = total value ÷ total quantity.

Concept 2

Use a weighted average when quantities are known

Mix 6 kg rice at ₹40/kg with 4 kg at ₹55/kg. Mean price = (6×40 + 4×55)/(6+4) = ₹46/kg.

A simple average works only for equal quantities. Unequal quantities must carry unequal weights.

Concept 3

Alligation finds the required mixing ratio

To mix a cheaper value C and dearer value D to obtain mean M, the quantity ratio is cheaper:dearer = (D−M):(M−C).

ComponentValueOpposite difference
Cheaper₹3050−42 = 8
Dearer₹5042−30 = 12

Thus the ratio of ₹30 material to ₹50 material for a ₹42 mean is 8:12 = 2:3.

Concept 4

Track pure substance, not just total liquid

A 20-litre solution containing 30% acid has 6 litres of acid and 14 litres of water. Adding 5 litres of water leaves acid at 6 litres but makes total volume 25 litres, so the new concentration is 24%.

When two solutions are mixed, treat their percentage concentrations as component values and their volumes as weights.

Concept 5

One replacement removes both ingredients proportionally

From a well-mixed vessel of volume V, removing x litres leaves the fraction (1−x/V) of every original ingredient. Replacing the removed amount with a new liquid restores total volume but not the removed ingredient.

For a 40-litre milk vessel, remove 10 litres and replace with water. Original milk remaining = 40×(1−10/40) = 30 litres.

Concept 6

Repeated equal replacements use a retention factor

If x is removed and replaced n times from a vessel of fixed volume V, original substance remaining = initial amount × (1−x/V)ⁿ.

From 64 litres of pure milk, replace 16 litres with water twice. Milk remaining = 64×(3/4)² = 36 litres; water = 28 litres.

Worked methods

Examples with reasoning

Known quantities and prices

8 kg at ₹35 and 12 kg at ₹45 cost ₹820 total. Mean = 820/20 = ₹41/kg.

Find a mixing ratio

Mix 20% and 50% solutions to obtain 32%. Ratio low:high = (50−32):(32−20) = 18:12 = 3:2.

Find an unknown component value

5 kg at ₹24 and 3 kg at ₹x average ₹30. Then 120+3x=240, so x=₹40/kg.

Dilution with water

15 litres at 40% contains 6 litres pure substance. To make it 30%, total volume must be 6/0.30=20 litres, so add 5 litres water.

Repeated replacement

From 50 litres, replace 10 litres twice. Original liquid remains 50×(4/5)²=32 litres.

Exam traps

Errors that upset the balance

Taking a simple mean

Average component values directly only when their quantities are equal.

Writing the alligation ratio backward

Use opposite differences and keep the component order consistent.

Mean outside the range

A positive mixture mean cannot be below both values or above both values.

Removing only one ingredient

A sample removed from a uniform mixture contains ingredients in the same ratio.

Adding percentages

Convert percentages into actual pure quantities before changing volume.

Using n instead of an exponent

Repeated replacement multiplies the retention factor, producing a power.

Original practice

Check understanding after learning

Original syllabus-aligned questions—not claimed previous-year questions.

1. Six kilograms at ₹40/kg are mixed with four kilograms at ₹55/kg. Find the mean price.
2. In what ratio should 20% and 50% solutions be mixed to obtain 32%?
3. From 64 litres of milk, 16 litres are removed and replaced with water twice. How much milk remains?

Rapid revision

The one-minute sheet

  • Weighted mean = Σ(quantity × value) ÷ Σquantity.
  • Cheaper:dearer = (dearer−mean):(mean−cheaper).
  • The mean must lie between the component values.
  • Pure substance = total mixture × concentration fraction.
  • After one replacement, original fraction remaining = 1−x/V.
  • After n equal replacements, original amount = initial × (1−x/V)ⁿ.

Source ledger

Official foundations

Explanations, worked examples and practice are original. Official references are credited and not reproduced. Drafted 25 September 2026.

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