Mixtures

1. Understanding mixtures and choosing the correct base

A mixture combines two or more components, such as milk and water, salt and water, or two grades of rice. Arithmetic questions usually ask for an unknown quantity, concentration, mixing ratio or average cost. The same weighted-average principle applies to marks and group averages, although these are not physical mixtures. Before calculating, identify the total quantity and the particular constituent or value being tracked.

If a 40-litre solution contains 25% acid by volume, its acid content is 10 litres and the remaining 30 litres is the other constituent. A ratio of acid to water of 1:3 also means 25% acid, because acid accounts for one of four total parts. Confusing constituent-to-constituent ratios with constituent-to-total fractions is a frequent source of error.

The units define the calculation. A 10% salt solution by mass contains 10 grams of salt in 100 grams of solution, not in 100 grams of water. Do not combine litres and kilograms without density information. Unless otherwise specified, aptitude questions involving liquid volumes assume that volumes are additive; real liquids can behave differently.

• Constituent quantity = total mixture quantity × constituent fraction.
• If A:B = a:b, the fraction of A is a/(a + b).
• For a two-component mixture, the constituent percentages add to 100%.

2. Weighted averages and the conservation equation

The most reliable method is to account for the constituent before and after mixing. If quantities q1 and q2 have concentrations c1 and c2, the final concentration is c = (q1c1 + q2c2)/(q1 + q2). Concentrations may be entered consistently as decimals or percentage numbers. However, when calculating an actual constituent quantity from a percentage, divide the percentage by 100.

For example, combine 20 litres of 30% acid solution with 30 litres of 50% acid solution. The acid quantities are 6 litres and 15 litres. The resulting 50 litres contains 21 litres of acid, so its concentration is 42%. Simply averaging 30% and 50% gives 40%, which is wrong because the two volumes are unequal.

The same reasoning determines average purchase cost. Mixing 12 kg of rice costing Rs 40 per kg with 8 kg costing Rs 55 per kg gives a total cost of Rs 920 for 20 kg. The mixture therefore costs Rs 46 per kg. If sold at Rs 50 per kg, profit is Rs 4 per kg, and the profit percentage is 4/46 × 100, approximately 8.70%.

For three or more components, add all constituent quantities or costs and divide by the combined quantity. A conservation equation is generally safer than an elaborate shortcut when a question includes several additions, withdrawals or unknown quantities.

• Equal quantities give the simple arithmetic mean; unequal quantities require weights.
• A combined average is closer to the value associated with the larger quantity.
• Profit percentage normally uses cost price, not selling price, as its denominator.

3. Alligation: finding the mixing ratio

Alligation reverses the weighted-average calculation. Suppose a lower-strength mixture has concentration L, a higher-strength mixture has concentration H, and the required concentration is M. Provided L < M < H, the quantity ratio of lower-strength mixture to higher-strength mixture is (H − M):(M − L). The differences cross over: the difference involving H belongs to the quantity of the lower-strength mixture.

To prepare a 35% solution from 20% and 50% solutions, the required ratio is (50 − 35):(35 − 20), or 1:1. For a 30% target using the same source solutions, the ratio becomes 20:10, or 2:1. If the final quantity is 60 litres, use 40 litres of the 20% solution and 20 litres of the 50% solution.

Alligation also works for prices when both prices use the same unit, such as rupees per kilogram. It does not directly use a selling price as the target average cost when a profit condition is present. First convert the selling price into the required cost price. At 25% profit on cost, a selling price of Rs 100 corresponds to a cost price of Rs 80.

A target outside the two source concentrations cannot be reached merely by mixing them in positive quantities. A target equal to one endpoint requires only that source, unless another operation changes the composition. This feasibility check can eliminate options before any detailed calculation.

• Write source values and target value in the same units.
• Preserve the order: lower quantity:higher quantity = higher difference:lower difference.
• Convert the resulting ratio into actual quantities only when required.

4. Dilution, evaporation and repeated replacement

Adding pure solvent changes the total quantity but not the amount of solute. Thus, for dilution, initial quantity × initial concentration = final quantity × final concentration. If 30 litres of a 40% solution must become 25%, its unchanged 12 litres of solute requires a final volume of 48 litres. Add 18 litres of solvent.

Adding pure solute is different because both the solute amount and the mixture total increase. If x kg of salt is added to 20 kg of a 10% salt solution, the new concentration is (2 + x)/(20 + x). To reach 25%, solve 2 + x = 0.25(20 + x), giving x = 4 kg. Assume complete dissolution and no loss.

Evaporation questions commonly assume that only the solvent leaves. For example, 50 kg of a 20% salt solution contains 10 kg of salt. To reach 25% salt, the final solution must weigh 40 kg, so 10 kg of water must evaporate. This model is inappropriate if the solute also evaporates or material is otherwise lost.

For repeated replacement, a homogeneous container of fixed volume V has v withdrawn and replaced with a liquid containing none of the tracked constituent. After n operations, the constituent amount remaining is A0(1 − v/V)^n. Starting with 40 litres of pure milk, withdrawing and replacing 10 litres with water twice leaves 40 × (3/4)^2 = 22.5 litres of milk. Each withdrawal removes some water as well as milk after the first mixing.

• If withdrawal volumes differ, multiply the successive retention fractions rather than using one power.
• If replacement liquid has concentration r, final concentration after n equal operations is r + (c0 − r)(1 − v/V)^n.
• Mix thoroughly before every withdrawal; otherwise the homogeneous-mixture model may not apply.

5. A practical CSAT solving strategy

Begin by classifying the operation: combining mixtures, adding solvent, adding solute, evaporating solvent, or withdrawing and replacing. Then write the relevant quantity balance. Pure solvent has 0% of the tracked solute, while a pure constituent has 100%. Represent an unspecified total by a convenient number, such as 100 units, only when the question is scale-independent.

Use fractions when they simplify calculation: 12.5% is 1/8, 20% is 1/5 and 37.5% is 3/8. For replacement, compute the fraction retained rather than separately calculating every withdrawn constituent. Nevertheless, a short two-step table is useful when the replacement quantities or liquids change.

Check direction, bounds and units before marking an answer. Adding water cannot increase salt concentration; removing a homogeneous portion alone does not change concentration; and combining two solutions cannot produce a concentration beyond both source values. Use option substitution when it is faster than algebra, but avoid treating a rough estimate as exact when options are close.

• Separate withdrawal from replacement: these are two distinct operations.
• State the final ratio in the order requested, such as milk:water rather than water:milk.
• Do not assume that equal volumes imply equal masses or equal costs.

Previous year questions

No UPSC question has been asked directly on this micro-topic yet. Use the practice questions below.

Practice questions