Practice Questions
Concept Explanation
Mixed alligation problems combine multiple concepts: price mixtures, concentration, replacement, and ratio finding in multi-step scenarios typical of placement exams.
Key Formulas
- Alligation: cheaper/dearer = (d − m)/(m − c)
- Mean price = (C1Q1 + C2Q2)/(Q1 + Q2)
- Replacement: pure = V(1 − x/V)n
- Concentration mix: C1V1 + C2V2 = Cm(V1 + V2)
Example
Mix Rs. 30/kg and Rs. 50/kg tea for Rs. 38/kg, then add water (Rs. 0/L). Use alligation first, then dilution if needed.
Common Trap
Break multi-step problems into stages: mix first, then replace or dilute.
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Q1 EasyRs. 30/kg and Rs. 50/kg tea mixed for Rs. 38/kg. Ratio cheaper : dearer is:ExplanationAlligation: cheaper : dearer = (50 − 38) : (38 − 30) = 12 : 8 = 3 : 2. Hence the answer is 3 : 2.
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Q2 MediumA mixture of 40 litres of milk and water is in the ratio 3 : 1. How much water should be added so that milk becomes 50% of the mixture?ExplanationMilk = (3/4) × 40 = 30 L; water = 10 L. Milk stays 30 L. Need 30 / (40 + x) = 1/2 ⇒ 60 = 40 + x ⇒ x = 20 L. Hence the answer is 20 L.
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Q3 MediumMix 20% and 50% acid for 35%, then dilute with an equal volume of water. Final acid concentration is:ExplanationFirst mixture is 35% acid. Diluting with equal volume of water halves the concentration to 17.5%. Hence the answer is 17.5%.
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Q4 MediumRs. 24/kg and Rs. 36/kg for Rs. 30/kg. In a 60 kg mix, the cheaper quantity is:ExplanationAlligation ratio = (36 − 30) : (30 − 24) = 1 : 1. Cheaper quantity = 30 kg. Hence the answer is 30 kg.
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Q5 Medium64 L pure milk: 8 L replaced with water twice. Milk left is:ExplanationPure = 64 × (7/8)^2 = 64 × 49/64 = 49 L. Hence the answer is 49 L.
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Q6 MediumA 50 kg mix at Rs. 32/kg uses goods at Rs. 26/kg and Rs. 38/kg. Ratio cheaper : dearer is:ExplanationAlligation: (38 − 32) : (32 − 26) = 6 : 6 = 1 : 1. Hence the answer is 1 : 1.
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Q7 HardFrom a vessel of 40 litres of pure milk, 4 litres is replaced by water repeatedly 3 times. The quantity of pure milk left is:ExplanationPure milk left = V(1 − x/V)^n = 40 × (1 − 4/40)^3 = 40 × (0.9)^3 = 40 × 0.729 = 29.16 L. Hence the answer is 29.16 L.
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Q8 Hard100 litres of milk: 10 litres is replaced by water, thrice. Milk left is:ExplanationPure milk left = 100 × (1 − 10/100)^3 = 100 × (0.9)^3 = 100 × 0.729 = 72.9 L. Hence the answer is 72.9 L.
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Q9 Hard30 kg mix costs Rs. 840. After adding 10 kg water (Rs. 0/kg), the new mean price per kg is:ExplanationTotal cost remains Rs. 840 for 40 kg. New mean = 840 / 40 = Rs. 21/kg. Hence the answer is Rs. 21.00/kg.
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Q10 HardRs. 15/kg and Rs. 45/kg for Rs. 27/kg. Cheaper share in 60 kg is:ExplanationCheaper : dearer = (45 − 27) : (27 − 15) = 18 : 12 = 3 : 2. Cheaper = (3/5) × 60 = 36 kg. Hence the answer is 36 kg.
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Q11 HardMix 10 L of 20% acid with 20 L of 40% acid, then add 20 L water. Final acid concentration is:ExplanationAcid = 0.20 × 10 + 0.40 × 20 = 2 + 8 = 10 L. Total volume after adding water = 10 + 20 + 20 = 50 L. Concentration = 10 / 50 × 100 = 20%. Hence the answer is 20%.
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Q12 Hard80 L vessel of pure liquid: 20 L removed and replaced by water, 3 times. Pure liquid left is:ExplanationPure = 80 × (3/4)^3 = 80 × 27/64 = 33.75 L. Hence the answer is 33.75 L.
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Q13 HardMilk : water = 5 : 3 in 40 L. How much milk should be added so that the ratio becomes 2 : 1?ExplanationMilk = (5/8) × 40 = 25 L; water = 15 L. Water stays 15 L. Need (25 + x) : 15 = 2 : 1 ⇒ 25 + x = 30 ⇒ x = 5 L. Hence the answer is 5 L.
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Q14 HardRs. 22/kg and Rs. 58/kg for Rs. 34/kg. Dearer share in 90 kg is:ExplanationCheaper : dearer = (58 − 34) : (34 − 22) = 24 : 12 = 2 : 1. Dearer = (1/3) × 90 = 30 kg. Hence the answer is 30 kg.
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DS1 Easy
How much pure milk remains after three replacements?
Statement IFrom 40 L pure milk, 4 L is replaced by water three times.
Statement IIPure left equals 40 × (0.9)^3 litres.
ExplanationStatement I: pure = 40 × (0.9)^3 = 29.16 L. Statement II states the same computed quantity. Each alone is sufficient. Answer: D. -
DS2 Medium
How much water must be added so that milk becomes 50%?
Statement IA 40 L mixture of milk and water is in the ratio 3 : 1.
Statement IIMilk quantity in the mixture is 30 L and total is 40 L.
ExplanationStatement I: milk = 30 L; 30/(40+x)=1/2 ⇒ x=20 L. Statement II: same data ⇒ x=20 L. Each alone is sufficient. Answer: D.
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DS3 Medium
What is the cheaper quantity in the mix?
Statement IRs. 24/kg and Rs. 36/kg give mean Rs. 30/kg.
Statement IITotal mixture is 60 kg.
ExplanationStatement I alone gives ratio 1 : 1 but not absolute quantity. Statement II alone lacks the ratio. Together: cheaper = 30 kg. Both are needed. Answer: C. -
DS4 Hard
Can the final concentration after mixing and diluting be found?
Statement I20% and 50% solutions are mixed to get 35%.
Statement IIThe resulting 35% solution is then diluted with an equal volume of water.
ExplanationStatement I alone gives 35% but not the diluted result. Statement II alone needs the pre-dilution concentration. Together: final = 17.5%. Both are needed. Answer: C.