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Mixtures and alligation for IBPS PO

Adding milk to change a ratio, mixing two grades to hit a price, replacing part of a mixture twice, combining two mixtures and selling at a profit. Mixture questions in IBPS PO, solved with alligation and a few rules, with six worked examples and eight practice questions.

9 Oct 2026 7 min read

In this guide
  1. Tool 1: keep the unchanged part fixed
  2. Tool 2: alligation
  3. Tool 3: repeated replacement
  4. Tool 4: total pure part ÷ total volume
  5. Worked examples
  6. Common mistakes
  7. Practice set
  8. What to do next

Mixture questions are ratio and percentage questions with a jug in them. The skill they test is noticing what stays fixed while something else changes. Add milk, and the water does not change. Replace part of a mixture, and a fixed fraction of the original liquid survives each time. Mix two grades, and the result is a weighted average of the two.

In IBPS PO, mixtures appear as standalone arithmetic in the prelims and as statements in quantity comparison. In the mains, they are often folded into a caselet with a profit twist. Four tools cover almost everything.

Tool 1: keep the unchanged part fixed

When only one component is added or removed, the other component's quantity stays the same. Write the new ratio using the fixed quantity and solve for the change.

Why: a ratio changes because one part changes. Anchoring on the part that did not move gives a one-step equation.

Tool 2: alligation

When two grades at prices (or strengths) c and d are mixed to get a mean m, with c < m < d:

  • Quantity of cheaper : quantity of dearer = (d − m) : (m − c).

Why it works: each kilogram of the cheaper grade is m − c below the mean, and each kilogram of the dearer grade is d − m above it. For the mixture to average exactly m, the total shortfall must balance the total excess. That happens when the quantities are in the inverse ratio of those distances. It is a see-saw: the heavier side sits closer to the middle.

Alligation works for anything that averages: price per kg, percentage strength, milk fraction, even average marks of two groups.

Tool 3: repeated replacement

If x litres are taken out of a V-litre container and replaced with water, and this is done n times, the original liquid left is:

  • V × (1 − x/V)ⁿ

Why: each round keeps the same fraction (1 − x/V) of whatever original liquid was there. Keeping the same fraction repeatedly is compounding, just like compound interest in reverse.

Tool 4: total pure part ÷ total volume

When two solutions are simply poured together, add the pure component from each and divide by the total volume. Use this when you are asked for the result. Use alligation when you are asked for the ratio.

You are asked forUse
The new ratio after adding one componentFixed-part rule
The ratio in which to mix two gradesAlligation
What is left after repeated replacementV × (1 − x/V)ⁿ
The strength of a mixture of two solutionsTotal pure ÷ total volume
The ratio in which to mix two mixturesAlligation on the fraction of one component

Worked examples

Example 1 (fixed part): A 60-litre mixture has milk and water in the ratio 7 : 5. How much milk must be added to make the ratio 3 : 2?

  • Milk = 35 L, water = 25 L. Water does not change.
  • (35 + x) ÷ 25 = 3 ÷ 2, so 35 + x = 37.5 and x = 2.5 litres.

Example 2 (alligation on price): Rice at ₹32 per kg and ₹45 per kg is mixed to get a mixture worth ₹40 per kg. In what ratio are they mixed?

  • Cheaper : dearer = (45 − 40) : (40 − 32) = 5 : 8.
  • Check: 5 kg × 32 + 8 kg × 45 = 160 + 360 = 520 for 13 kg, which is ₹40 a kg.

Example 3 (repeated replacement): A container holds 80 litres of milk. 8 litres are taken out and replaced with water. This is done twice more, three times in all. How much milk is left?

  • Milk left = 80 × (1 − 8/80)³ = 80 × (9/10)³ = 80 × 0.729 = 58.32 litres.
  • After two rounds it would be 80 × 0.81 = 64.8 litres. Each round keeps 90% of what was there.

Example 4 (two solutions): 20 litres of a 30% acid solution are mixed with 30 litres of a 50% acid solution. Find the strength of the mixture.

  • Acid = 6 + 15 = 21 litres in 50 litres, so the strength is 42%.

Example 5 (two mixtures): Vessel A has milk and water in the ratio 5 : 3. Vessel B has them in the ratio 3 : 1. In what ratio should the two be mixed to get milk and water in the ratio 7 : 3?

  • Milk fractions: A = 5/8 = 0.625, B = 3/4 = 0.75, target = 7/10 = 0.7.
  • A : B = (0.75 − 0.7) : (0.7 − 0.625) = 0.05 : 0.075 = 2 : 3.
  • Check: 16 L of A has 10 milk and 6 water; 24 L of B has 18 milk and 6 water. Together 28 : 12 = 7 : 3.

Example 6 (mixture with profit): A shopkeeper mixes tea costing ₹180 per kg with tea costing ₹240 per kg and sells the mixture at ₹253 per kg, making a 10% profit. In what ratio were they mixed?

  • Cost price of the mixture = 253 ÷ 1.1 = ₹230 per kg.
  • Cheaper : dearer = (240 − 230) : (230 − 180) = 10 : 50 = 1 : 5.

Common mistakes

  • Changing the part that should stay fixed. In Example 1, adding milk does not change the 25 L of water.
  • Writing the alligation ratio the wrong way round. The cheaper grade's share is the dearer price minus the mean.
  • Using simple subtraction for repeated replacement. Removing 8 L three times does not remove 24 L of milk, because later rounds remove a mixture.
  • Alligating on ratios instead of fractions. For two mixtures, convert each to the fraction of one component (Example 5) before you alligate.
  • Using the selling price as the mean when a profit is involved.

Practice set

  1. 40 litres of a mixture have milk and water in the ratio 3 : 1. How much water must be added to make the ratio 3 : 2?
  2. Tea at ₹200 and ₹260 per kg is mixed to get tea worth ₹240 per kg. In what ratio?
  3. A 50-litre container full of milk has 5 litres replaced with water, twice. How much milk is left?
  4. 10 litres of a 20% solution are mixed with 15 litres of a 40% solution. Find the strength of the mixture.
  5. A 64-litre mixture has wine and water in the ratio 5 : 3. How much water must be added to make the ratio 1 : 1?
  6. A milk vendor mixes water into milk and sells the mixture at the cost price of milk, gaining 25%. Find the ratio of water to milk.
  7. A 30-litre mixture contains 20% alcohol. How much pure alcohol must be added to make it 40% alcohol?
  8. Two vessels have milk and water in the ratios 4 : 1 and 2 : 3. Equal quantities from both are mixed. Find the new ratio.

Answers:

  1. 10 litres. Milk 30, water 10; 30 ÷ (10 + x) = 3/2, so 10 + x = 20.
  2. 1 : 2. (260 − 240) : (240 − 200) = 20 : 40.
  3. 40.5 litres. 50 × (9/10)² = 50 × 0.81.
  4. 32%. Acid 2 + 6 = 8 litres in 25 litres.
  5. 16 litres. Wine 40, water 24; water must rise to 40.
  6. 1 : 4. The vendor pays for milk only, so the gain comes from the water. Gain = water ÷ milk = 25% = 1/4.
  7. 10 litres. Alcohol 6 L. (6 + x) ÷ (30 + x) = 0.4, so 6 + x = 12 + 0.4x and x = 10.
  8. 3 : 2. Take 5 L of each: milk 4 + 2 = 6, water 1 + 3 = 4.

What to do next

  • Solve six mixture questions a day for five days. Before each, write which of the four tools it needs.
  • Redo Examples 5 and 6 without looking. Mixing two mixtures and profit-plus-mixture are the mains favourites.
  • Revise profit and loss and ratio and partnership; mixture questions borrow from both.

A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Institute of Banking Personnel Selection website .

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