Mixtures and alligation for IBPS Clerk

Mixture questions give you two things of different price or strength and ask what happens when you combine them. Two kinds of rice make a blend at an in-between price. A strong solution and a weak one make a medium solution. Milk gets diluted with water, or some of it is drawn out and replaced.

In the clerk exam these usually come as single arithmetic questions, and they also sit inside DI and profit questions. The good news is that one idea, the weighted average, handles almost all of them. Once you see that, alligation stops being a trick you memorise and becomes a quick way of writing an equation you already understand.

The alligation rule, and why it works

Suppose x units of a cheaper item at price c are mixed with y units of a dearer item at price d, and the mixture is worth m per unit. The total cost does not change when you mix, so:

• c × x + d × y = m × (x + y)
• Rearranging: x(m − c) = y(d − m)
• So x : y = (d − m) : (m − c)

That is the whole rule. Cheaper : Dearer = (Dearer − Mean) : (Mean − Cheaper). People draw it as a cross, with c and d on top, m in the middle and the two differences at the bottom, but the cross is only a picture of this equation.

Two checks come free with it:

• The mean must lie between c and d. If it does not, the question has a trap or you have misread it.
• The closer the mean is to one price, the more of that item there is in the mixture. If m is near c, most of the blend is the cheaper item.

The same rule works for anything that averages by quantity: prices per kg, percentage strength of a solution, marks per student, speed over equal times.

Adding one component

When only water (or only one ingredient) is added, the other quantity stays the same. Build the equation around the quantity that does not change. If a mixture has 30 L of milk, it will still have 30 L of milk after any amount of water is poured in.

Removal and replacement

When x litres are drawn from a vessel holding V litres and replaced with water, the removed liquid has the same ratio as the whole mixture. Each round keeps a fraction (1 − x/V) of the original liquid. After n rounds:

Original liquid left = V × (1 − x/V)ⁿ

This works because each draw takes the same fraction of whatever is in the vessel at that moment, so the fractions multiply.

Mixing two mixtures

If vessel A has milk and water 5 : 3 and vessel B has 2 : 3, convert each to the fraction of milk (5/8 and 2/5). Then alligate those fractions against the target fraction. It is the same rule with fractions in place of prices.

Worked examples

Example 1. Rice at ₹40/kg is mixed with rice at ₹60/kg to get a blend worth ₹45/kg. In what ratio are they mixed? If 24 kg of the cheaper rice is used, how much of the dearer rice is needed?

1. Cheaper : dearer = (60 − 45) : (45 − 40) = 15 : 5 = 3 : 1.
2. 3 parts = 24 kg, so 1 part = 8 kg. Dearer rice = 8 kg.
3. Check: (24 × 40 + 8 × 60) ÷ 32 = (960 + 480) ÷ 32 = 1,440 ÷ 32 = 45.

Example 2. A trader mixes tea at ₹180/kg with tea at ₹280/kg and sells the blend at ₹264/kg, making a 20% profit. Find the mixing ratio.

1. Cost price of the blend = 264 ÷ 1.2 = ₹220/kg.
2. Cheaper : dearer = (280 − 220) : (220 − 180) = 60 : 40 = 3 : 2.

If you had used 264 as the mean, you would have got (280 − 264) : (264 − 180) = 16 : 84, or 4 : 21, which is wrong.

Example 3. A 40-litre mixture has milk and water in the ratio 3 : 1. How much water must be added to make the ratio 3 : 2?

1. Milk = 30 L, water = 10 L. Milk stays 30.
2. 30 ÷ (10 + x) = 3 ÷ 2, so 60 = 30 + 3x and x = 10 litres.

Example 4. A 60-litre mixture has milk and water in the ratio 2 : 1. 15 litres are removed and replaced with water. Find the new ratio.

1. Milk 40 L, water 20 L. The 15 L removed is a quarter of the mixture, so it takes 10 L milk and 5 L water.
2. Left: milk 30, water 15. Add 15 L water: milk 30, water 30. Ratio 1 : 1.

Example 5. A vessel holds 80 litres of pure milk. 8 litres are drawn out and replaced with water. This is done twice. How much milk is left?

1. Each round keeps 1 − 8/80 = 9/10 of the milk.
2. Milk left = 80 × 9/10 × 9/10 = 64.8 litres. Water = 80 − 64.8 = 15.2 litres.

Example 6. Vessel A has milk and water in the ratio 5 : 3; vessel B has them in the ratio 2 : 3. In what ratio should they be mixed to get milk and water in equal amounts?

1. Milk fraction: A = 5/8, B = 2/5, target = 1/2.
2. A : B = (1/2 − 2/5) : (5/8 − 1/2) = 1/10 : 1/8 = 4 : 5.
3. Check with 4 litres of A and 5 of B: milk = 2.5 + 2 = 4.5, water = 1.5 + 3 = 4.5. Equal.

Common mistakes

• Taking the selling price of the blend as the mean price.
• Writing the ratio the wrong way round: the difference next to the dearer price belongs to the cheaper item.
• In "add water" questions, changing the milk quantity as well.
• In replacement questions, subtracting the same number of litres of milk each round. Each round removes a fraction, not a fixed amount of milk.

Practice

Set a six-minute timer.

1. Tea at ₹200/kg and ₹300/kg is mixed to get tea worth ₹240/kg. Find the ratio.
2. A 30-litre mixture has milk and water in the ratio 4 : 1. How much water must be added to make it 2 : 1?
3. A 10% salt solution and a 40% salt solution are mixed to get a 20% solution. Find the ratio.
4. Wheat at ₹24/kg and ₹32/kg is mixed in the ratio 3 : 1. Find the price of the mixture per kg.
5. How many kg of rice at ₹50/kg must be mixed with 30 kg of rice at ₹35/kg so that the mixture is worth ₹40/kg?
6. A container holds 50 litres of milk. 5 litres are drawn out and replaced with water, and this is done twice. How much milk is left?
7. Sugar at ₹40/kg is mixed with sugar at ₹50/kg. The mixture is sold at ₹52.80/kg for a 20% profit. Find the ratio.
8. A 20-litre solution is 15% alcohol. How much water must be added to make it 12% alcohol?

Answers:

1. 3 : 2. (300 − 240) : (240 − 200) = 60 : 40.
2. 6 litres. Milk 24, water 6. 24 ÷ (6 + x) = 2 gives x = 6.
3. 2 : 1. (40 − 20) : (20 − 10) = 20 : 10.
4. ₹26/kg. (3 × 24 + 1 × 32) ÷ 4 = 104 ÷ 4.
5. 15 kg. Cheaper : dearer = (50 − 40) : (40 − 35) = 2 : 1, so dearer = 30 ÷ 2. Check: (1,050 + 750) ÷ 45 = 40.
6. 40.5 litres. 50 × 0.9 × 0.9.
7. 3 : 2. Cost price = 52.80 ÷ 1.2 = ₹44. (50 − 44) : (44 − 40) = 6 : 4.
8. 5 litres. Alcohol = 3 L stays fixed. 3 ÷ (20 + x) = 0.12 gives 20 + x = 25.

What to do next

• Derive the alligation rule once on paper from the total-cost equation, so you trust it.
• Do ten mixed questions and label each with its type from the table above before solving.
• Revise ratio and proportion and profit and loss, since both feed into mixture questions.
• Try alligation on averages questions with two groups. It is the same idea.