Mixtures and Alligation
CSAT · Quantitative Aptitude
Introduction
Mixture problems – blending grades of rice, diluting milk, combining acid solutions – are weighted averages in disguise. The alligation rule turns them into a two-step subtraction.
Key concepts
Alligation
When two ingredients at prices (or concentrations) c (cheaper) and d (dearer) are mixed to give a mean m, the quantities are in the ratio:
cheaper : dearer = (d − m) : (m − c).
Example: rice at ₹30 and ₹45 mixed to make ₹40 → (45 − 40) : (40 − 30) = 5 : 10 = 1 : 2.
Weighted concentration
Mixing V₁ litres at p₁% with V₂ litres at p₂% gives (V₁p₁ + V₂p₂)/(V₁ + V₂) %. The result always lies between p₁ and p₂.
Repeated replacement
From a vessel of V litres of liquid, x litres are removed and replaced by water n times. Liquid left = V(1 − x/V)ⁿ.
Adding water to change concentration
The amount of pure substance stays fixed. If 40 L at 20% acid is diluted to 10%, acid = 8 L, so the new total must be 80 L → add 40 L of water.
Profit on mixtures
A milkman who mixes water free of cost and sells at cost price gains the water's share: 1 part water in 4 parts milk = 25% profit.
- Cheaper : dearer = (d − m) : (m − c)
- Mixture % = (V₁p₁ + V₂p₂)/(V₁ + V₂)
- After n replacements: V(1 − x/V)ⁿ
- Pure quantity is conserved on dilution
Solved examples
Example 1
In what ratio must rice at ₹30/kg be mixed with rice at ₹60/kg so that the mixture costs ₹38/kg?
(a) 30 : 60 (b) 4 : 11 (c) 12 : 4 (d) 11 : 4
- Alligation: (dearer − mean) : (mean − cheaper).
- = (60 − 38) : (38 − 30) = 22 : 8 = 11 : 4.
Answer: (d) 11 : 4
Example 2
40 litres of a 20% acid solution are mixed with 60 litres of a 70% acid solution. What is the acid concentration of the mixture?
(a) 55% (b) 50% (c) 45% (d) 46%
- Acid = 40 × 20% + 60 × 70% = 8 + 42 = 50 litres.
- Concentration = 50 ÷ 100 × 100 = 50%.
Answer: (b) 50%
Example 3
40 litres of a 40% acid solution are mixed with 60 litres of a 60% acid solution. What is the acid concentration of the mixture?
(a) 48% (b) 57% (c) 50% (d) 52%
- Acid = 40 × 40% + 60 × 60% = 16 + 36 = 52 litres.
- Concentration = 52 ÷ 100 × 100 = 52%.
Answer: (d) 52%
Example 4
10 litres of a 40% acid solution are mixed with 20 litres of a 80% acid solution. What is the acid concentration of the mixture?
(a) 60% (b) 66 2/3% (c) 71 2/3% (d) 62 2/3%
- Acid = 10 × 40% + 20 × 80% = 4 + 16 = 20 litres.
- Concentration = 20 ÷ 30 × 100 = 66 2/3%.
Answer: (b) 66 2/3%
Example 5
10 litres of a 30% acid solution are mixed with 30 litres of a 60% acid solution. What is the acid concentration of the mixture?
(a) 52 1/2% (b) 45% (c) 57 1/2% (d) 48 1/2%
- Acid = 10 × 30% + 30 × 60% = 3 + 18 = 21 litres.
- Concentration = 21 ÷ 40 × 100 = 52 1/2%.
Answer: (a) 52 1/2%
Example 6
In what ratio must rice at ₹40/kg be mixed with rice at ₹60/kg so that the mixture costs ₹57/kg?
(a) 3 : 17 (b) 17 : 3 (c) 4 : 17 (d) 40 : 60
- Alligation: (dearer − mean) : (mean − cheaper).
- = (60 − 57) : (57 − 40) = 3 : 17 = 3 : 17.
Answer: (a) 3 : 17
Example 7
In what ratio must rice at ₹32/kg be mixed with rice at ₹50/kg so that the mixture costs ₹48/kg?
(a) 1 : 8 (b) 2 : 8 (c) 32 : 50 (d) 8 : 1
- Alligation: (dearer − mean) : (mean − cheaper).
- = (50 − 48) : (48 − 32) = 2 : 16 = 1 : 8.
Answer: (a) 1 : 8
Example 8
A container holds 100 litres of milk. 5 litres are taken out and replaced with water. This is done 3 times in all. How much milk is left?
(a) 90 1/4 litres (b) 85 litres (c) 85 59/80 litres (d) 87 59/80 litres
- Milk left = V × (1 − x/V)^n.
- = 100 × (1 − 5/100)^3 = 100 × (19/20)^3 = 85 59/80 litres.
Answer: (c) 85 59/80 litres
- Draw the alligation cross: write c and d at the top, m in the middle, subtract diagonally.
- Check that the mean lies between the two values – otherwise the question has a trick.
- For dilution, track only the pure part.
- Reversing the ratio (it is (d − m) for the cheaper ingredient).
- Using simple subtraction for repeated replacement.
- Averaging concentrations without weighting by volume.
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. 40 litres of a 20% acid solution are mixed with 30 litres of a 50% acid solution. What is the acid concentration of the mixture?
(a) 32 6/7% (b) 35% (c) 37 6/7% (d) 28 6/7%
Q2. A container holds 80 litres of milk. 10 litres are taken out and replaced with water. This is done 3 times in all. How much milk is left?
(a) 50 litres (b) 55 19/32 litres (c) 61 1/4 litres (d) 53 19/32 litres
Q3. In what ratio must rice at ₹25/kg be mixed with rice at ₹30/kg so that the mixture costs ₹29/kg?
(a) 4 : 1 (b) 25 : 30 (c) 1 : 4 (d) 2 : 4
Q4. 30 litres of a 10% acid solution are mixed with 60 litres of a 60% acid solution. What is the acid concentration of the mixture?
(a) 48 1/3% (b) 43 1/3% (c) 35% (d) 39 1/3%
Q5. In what ratio must rice at ₹25/kg be mixed with rice at ₹50/kg so that the mixture costs ₹35/kg?
(a) 3 : 2 (b) 25 : 50 (c) 4 : 2 (d) 2 : 3
Q6. A container holds 100 litres of milk. 5 litres are taken out and replaced with water. This is done 2 times in all. How much milk is left?
(a) 92 1/4 litres (b) 90 litres (c) 90 1/4 litres (d) 95 litres
Q7. In what ratio must rice at ₹20/kg be mixed with rice at ₹50/kg so that the mixture costs ₹29/kg?
(a) 20 : 50 (b) 8 : 3 (c) 3 : 7 (d) 7 : 3
Q8. 30 litres of a 30% acid solution are mixed with 20 litres of a 50% acid solution. What is the acid concentration of the mixture?
(a) 43% (b) 40% (c) 34% (d) 38%
Q9. In what ratio must rice at ₹20/kg be mixed with rice at ₹45/kg so that the mixture costs ₹28/kg?
(a) 20 : 45 (b) 8 : 17 (c) 17 : 8 (d) 18 : 8
Q10. A container holds 40 litres of milk. 8 litres are taken out and replaced with water. This is done 2 times in all. How much milk is left?
(a) 24 litres (b) 25 3/5 litres (c) 32 litres (d) 27 3/5 litres
Q11. In what ratio must rice at ₹36/kg be mixed with rice at ₹40/kg so that the mixture costs ₹39/kg?
(a) 1 : 3 (b) 3 : 1 (c) 2 : 3 (d) 36 : 40
Q12. In what ratio must rice at ₹40/kg be mixed with rice at ₹60/kg so that the mixture costs ₹56/kg?
(a) 40 : 60 (b) 1 : 4 (c) 2 : 4 (d) 4 : 1
Q1 – (a) 32 6/7%. Acid = 40 × 20% + 30 × 50% = 8 + 15 = 23 litres. Concentration = 23 ÷ 70 × 100 = 32 6/7%.
Q2 – (d) 53 19/32 litres. Milk left = V × (1 − x/V)^n. = 80 × (1 − 10/80)^3 = 80 × (7/8)^3 = 53 19/32 litres.
Q3 – (c) 1 : 4. Alligation: (dearer − mean) : (mean − cheaper). = (30 − 29) : (29 − 25) = 1 : 4 = 1 : 4.
Q4 – (b) 43 1/3%. Acid = 30 × 10% + 60 × 60% = 3 + 36 = 39 litres. Concentration = 39 ÷ 90 × 100 = 43 1/3%.
Q5 – (a) 3 : 2. Alligation: (dearer − mean) : (mean − cheaper). = (50 − 35) : (35 − 25) = 15 : 10 = 3 : 2.
Q6 – (c) 90 1/4 litres. Milk left = V × (1 − x/V)^n. = 100 × (1 − 5/100)^2 = 100 × (19/20)^2 = 90 1/4 litres.
Q7 – (d) 7 : 3. Alligation: (dearer − mean) : (mean − cheaper). = (50 − 29) : (29 − 20) = 21 : 9 = 7 : 3.
Q8 – (d) 38%. Acid = 30 × 30% + 20 × 50% = 9 + 10 = 19 litres. Concentration = 19 ÷ 50 × 100 = 38%.
Q9 – (c) 17 : 8. Alligation: (dearer − mean) : (mean − cheaper). = (45 − 28) : (28 − 20) = 17 : 8 = 17 : 8.
Q10 – (b) 25 3/5 litres. Milk left = V × (1 − x/V)^n. = 40 × (1 − 8/40)^2 = 40 × (4/5)^2 = 25 3/5 litres.
Q11 – (a) 1 : 3. Alligation: (dearer − mean) : (mean − cheaper). = (40 − 39) : (39 − 36) = 1 : 3 = 1 : 3.
Q12 – (b) 1 : 4. Alligation: (dearer − mean) : (mean − cheaper). = (60 − 56) : (56 − 40) = 4 : 16 = 1 : 4.
- Alligation cross: (d − m) : (m − c)
- Weighted average of %
- Replacement: V(1 − x/V)^n
- Pure part stays constant
