1. Rule of alligation
When two items of cost c₁ (cheaper) and c₂ (dearer) are mixed to get mean price m:
Cheaper : Dearer = (c₂ − m) : (m − c₁)
c₁ c₂
m
c₂ − m m − c₁
It works for prices, percentages (acid, milk), speeds and averages — anything that is a weighted average.
2. Mean price
Mean = (q₁c₁ + q₂c₂)/(q₁ + q₂).
3. Replacement formula
A vessel has x litres of pure liquid. y litres are taken out and replaced by water, n times. Pure liquid left = x (1 − y/x)ⁿ.
4. Mixture types
- Adding water to change a ratio: the milk stays the same; find new water from the new ratio.
- Two vessels mixed: convert each to fractions of milk, add, then compare. Never add ratios directly.
- Mixture sold at a profit: first find the cost price of the mixture (SP × 100/(100 + gain%)), then use alligation.
5. Solved examples
Rice ₹30 and ₹45 mixed to get ₹40 → (45 − 40) : (40 − 30) = 1 : 2.
60 L milk : water = 2 : 1 (40 : 20). Water to add for 1 : 2 → water needed 80 → add 60 L.
80 L milk; 8 L taken out and replaced, twice → 80 × (9/10)² = 64.8 L.
Vessels 3 : 1 and 1 : 3 mixed equally → milk 3/4 + 1/4 = 1, water 1 → 1 : 1.
Tea ₹62 and ₹72 mixed and sold at ₹80.50 for 15% gain → CP = 70 → (72 − 70) : (70 − 62) = 1 : 4.
Average of a class is 50; boys average 45, girls 60 → boys : girls = (60 − 50) : (50 − 45) = 2 : 1.
Solved examples by question type
Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.
Basic level
Example 1. In what ratio must rice at ₹60/kg be mixed with rice at ₹90/kg so that the mixture costs ₹70/kg?
(a) 2 : 1 (b) 60 : 90 (c) 1 : 2 (d) 3 : 1
Answer: (a) 2 : 1
Solution: Alligation: cheaper : dearer = (90 − 70) : (70 − 60) = 2 : 1.
Example 2. 26 kg of rice at ₹90/kg is mixed with 13 kg at ₹67/kg. The price of the mixture per kg is:
(a) ₹78.5 (b) ₹82.33 (c) ₹80.33 (d) ₹83.33
Answer: (b) ₹82.33
Solution: Total cost = 26×90 + 13×67 = 3211. Per kg = 3211/39 = ₹82.33.
Example 3. In what ratio must tea at ₹18/kg be mixed with tea at ₹51/kg so that the mixture is worth ₹43/kg?
(a) 25 : 8 (b) 6 : 17 (c) 8 : 25 (d) 9 : 25
Answer: (c) 8 : 25
Solution: Cheaper : dearer = (51 − 43) : (43 − 18) = 8 : 25 = 8 : 25.
Example 4. In what ratio must tea at ₹45/kg be mixed with tea at ₹57/kg so that the mixture is worth ₹54/kg?
(a) 3 : 1 (b) 15 : 19 (c) 1 : 3 (d) 4 : 9
Answer: (c) 1 : 3
Solution: Cheaper : dearer = (57 − 54) : (54 − 45) = 3 : 9 = 1 : 3.
ASO exam level
Example 5. 60 litres of a mixture has milk and water in the ratio 2 : 1. How much water must be added to make the ratio 1 : 2?
(a) 60 litres (b) 30 litres (c) 20 litres (d) 65 litres
Answer: (a) 60 litres
Solution: Milk = 40, water = 20. Milk stays the same: water needed = 40 × 2/1 = 80. Add 60 litres.
Example 6. From a vessel of 80 litres of milk, 4 litres is taken out and replaced with water. This is done 2 times. How much milk is left?
(a) 69.2 litres (b) 72.2 litres (c) 74.2 litres (d) 72 litres
Answer: (b) 72.2 litres
Solution: Milk left = 80 × (1 − 4/80)² = 72.2 litres.
Example 7. How many kg of sugar at ₹20/kg must be mixed with 21 kg of sugar at ₹30/kg so that the mixture costs ₹27/kg?
(a) 49 kg (b) 9 kg (c) 21 kg (d) 20 kg
Answer: (b) 9 kg
Solution: Cheaper : dearer = (30 − 27) : (27 − 20) = 3 : 7. Cheaper = 21 × 3/7 = 9 kg.
Example 8. How many kg of sugar at ₹22/kg must be mixed with 125 kg of sugar at ₹48/kg so that the mixture costs ₹47/kg?
(a) 68 kg (b) 125 kg (c) 5 kg (d) 3,125 kg
Answer: (c) 5 kg
Solution: Cheaper : dearer = (48 − 47) : (47 − 22) = 1 : 25. Cheaper = 125 × 1/25 = 5 kg.
Example 9. Two vessels of equal size have milk and water in the ratios 1 : 2 and 3 : 8. Both are emptied into a third vessel. The ratio of milk to water is:
(a) 3 : 16 (b) 23 : 10 (c) 2 : 5 (d) 10 : 23
Answer: (d) 10 : 23
Solution: Milk = 1/3 + 3/11 = 0.61; water = 1.39. Ratio = 10 : 23. (Adding the ratios directly is wrong.)
Example 10. Two vessels of equal size have milk and water in the ratios 3 : 2 and 7 : 8. Both are emptied into a third vessel. The ratio of milk to water is:
(a) 21 : 16 (b) 8 : 7 (c) 1 : 1 (d) 7 : 8
Answer: (b) 8 : 7
Solution: Milk = 3/5 + 7/15 = 1.07; water = 0.93. Ratio = 8 : 7. (Adding the ratios directly is wrong.)
Example 11. A 81-litre mixture has milk and water in the ratio 4 : 5. How much milk must be added to make the ratio 8 : 5?
(a) 45 litres (b) 37 litres (c) 36 litres (d) 72 litres
Answer: (c) 36 litres
Solution: Milk 36, water 45. Water stays the same: milk needed = 45 × 8/5 = 72. Add 36 litres.
Example 12. A 170-litre mixture has milk and water in the ratio 9 : 8. How much milk must be added to make the ratio 11 : 8?
(a) 40 litres (b) 30 litres (c) 90 litres (d) 20 litres
Answer: (d) 20 litres
Solution: Milk 90, water 80. Water stays the same: milk needed = 80 × 11/8 = 110. Add 20 litres.
Advanced level
Example 13. In what ratio must a 10% acid solution be mixed with a 40% acid solution to get a 25% acid solution?
(a) 10 : 40 (b) 3 : 1 (c) 2 : 1 (d) 1 : 1
Answer: (d) 1 : 1
Solution: Alligation: (40 − 25) : (25 − 10) = 15 : 15 = 1 : 1.
Example 14. Two alloys contain gold and copper in the ratios 2 : 1 and 1 : 2. If equal weights of both are melted together, the ratio of gold to copper is:
(a) 1 : 1 (b) 3 : 3 (c) 2 : 1 (d) 3 : 1
Answer: (a) 1 : 1
Solution: Gold per unit: 2/3 + 1/3 = 1; copper = 1. Ratio = 1 : 1. (Adding ratios directly is wrong.)
Example 15. Two alloys contain gold and copper in the ratios 3 : 2 and 1 : 4. If equal weights of both are melted together, the ratio of gold to copper is:
(a) 2 : 3 (b) 3 : 2 (c) 4 : 6 (d) 1 : 1
Answer: (a) 2 : 3
Solution: Gold per unit: 3/5 + 1/5 = 0.8; copper = 1.2. Ratio = 2 : 3. (Adding ratios directly is wrong.)
Example 16. A container has 160 litres of milk. 32 litres is taken out and replaced with water, and this is done 2 times in all. Milk now in the container is:
(a) 134.4 litres (b) 128 litres (c) 102.4 litres (d) 96 litres
Answer: (c) 102.4 litres
Solution: Milk left = 160 × (1 − 32/160)^2 = 102.4 litres.
Example 17. A grocer mixes tea at ₹21/kg and ₹41/kg and sells the mixture at ₹52.5/kg, gaining 50%. In what ratio did he mix them?
(a) 2 : 5 (b) 7 : 3 (c) 4 : 7 (d) 3 : 7
Answer: (d) 3 : 7
Solution: Cost of mixture = 52.5 × 100/150 = ₹35. Alligation: (41 − 35) : (35 − 21) = 3 : 7. (Do not use the selling price in alligation.)
Topic-wise practice for Mixture and alligation (Quant): Alligation rule, repeated replacement, profit on mixtures. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.
Question types covered
- Alligation rule: mixing two varieties
- Mean price of a mixture
- Milk and water ratio problems
- Repeated replacement (taking out and refilling)
- Profit or loss on a mixture
- Mixing at a given ratio of three items