Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
All Direct removal and addition 448 Alligation cross rule 270 Ratios, quantities, and percentages 233 Weighted mean and concentration 195 Equal quantity replacement 189 Three or more components 174 Repeated equal-volume replacement 154 Profit-based alligation 153

Mixtures and Alligation Questions

Multiple choice

In a 135 liter of mixture, ratio of water and milk is 2:7. If water is added in the above mixture in such a way so that final amount of the water in the mixture becomes twice of the initial amount which was in the mixture then what is the ratio of water and milk in the resulting mixture?
  1. $7: 15$
  2. $4:7$
  3. $14: 25$
  4. $8: 13$
  5. $9: 14$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial mixture has water:milk ratio of 2:7 in 135L, so water = 30L, milk = 105L. Water is doubled to 60L while milk remains 105L. New ratio of water:milk = 60:105 = 4:7. Option A gives an incorrect ratio. Options C, D, and E don't match the simplified ratio after calculation.

Multiple choice

In a can filled with solution, percentage of water is 25%. When 15 liter mixture is taken out and replaced with water new percentage of water in mixture becomes 40%. Find the capacity of can in liters. विलयन से भरे एक केन में पानी 25% है। जब 15 लीटर मिश्रण निकाला जाता है और पानी के साथ प्रतिस्थापित किया जाता है तो मिश्रण में पानी का नया प्रतिशत 40% हो जाता है। केन की क्षमता लीटर में ज्ञात कीजिये।

  1. 60 liter/ लीटर

  2. 75 liter/ लीटर

  3. 48 liter/ लीटर

  4. 36 liter/ लीटर
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let can capacity be x liters. Initially: 25% water = 0.25x, 75% solution = 0.75x. After removing 15L (which has 25% water): water removed = 15 × 0.25 = 3.75L, solution removed = 15 × 0.75 = 11.25L. Remaining water = 0.25x - 3.75, remaining solution = 0.75x - 11.25. After adding 15L water: new water = (0.25x - 3.75 + 15) = 0.25x + 11.25. This equals 40% of x: 0.25x + 11.25 = 0.4x, so 0.15x = 11.25, x = 75 liters.

Multiple choice

Ramesh mixed 20% water in 56 liter milk and sold mixture at 15% more price than cost price of milk. Find his overall profit percent. रमेश ने 56 लीटर दूध में 20% पानी मिलाया और मिश्रण को दूध की कीमत से 15% अधिक कीमत पर बेचा। उसका कुल लाभ प्रतिशत ज्ञात कीजिये।

  1. 37.5%

  2. 41.4%

  3. 38%

  4. 32.25%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Ramesh adds 20% water to 56 L milk, so water added = 56 × 0.20 = 11.2 L. Total mixture = 56 + 11.2 = 67.2 L. He sells at 15% above milk's cost price, so selling price is 115% of milk CP per unit. Effective CP includes water cost (0), so CP = milk CP. Profit = (115% × 56)/56 - 1 = 0.15 from markup + 0.20 from dilution = 38% total. Using formula: (1.15 × 1.20 - 1) × 100 = (1.38 - 1) × 100 = 38%.

Multiple choice

A mixture of 56 litre milk and 30 litre water is in a container. X litre milk and X litre water are added in it to form a new mixture. If 20% of the new mixture is 36 litre, then find the value of x.

  1. 47 litre

  2. 32 litre

  3. 25 litre

  4. 64 litre

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

New mixture total = 86 + 2X liters. 20% of new mixture = 36L, so 0.2 × (86 + 2X) = 36. Solving: 17.2 + 0.4X = 36, so 0.4X = 18.8, giving X = 47. Option A matches.

Multiple choice

A shopkeeper sells a mixture of X and Y at a rate of Rs 400 per kg. Find the profit percentage of the shopkeeper I: He bought X at the rate of Rs 350 per kg II: He bought Y at Rs 10 higher than the rate of X per kg. III: He mixed Y, 40 kg more than that of X.

  1. Only I and II

  2. Only I and III

  3. Any two of the three

  4. All three together are sufficient

  5. All three together are not sufficient
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let X bought at Rs 350/kg, Y at Rs 360/kg. Let quantity of X = q kg, Y = (q+40) kg. Selling price is Rs 400/kg for the mixture. But we don't know the total quantity or what the mixture ratio is. We cannot determine cost price per kg of the mixture without knowing the total weight and selling price structure.

Multiple choice

A mixture of two types of rice costing Rs. 32 per Kg and Rs 48 per kg mixed in ratio 5 : 7 if 24 kg of mixture is replaced by 8 kg of first type of rice the ratio in the mixture becomes 6 : 7 . Find the amount of first type of rice originally.

  1. 1200

  2. 1800

  3. 1500

  4. 2250

  5. None of these
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let original amounts be 5x kg and 7x kg. When 24 kg mixture (in 5:7 ratio) is removed, 10 kg of first type and 14 kg of second type are taken out. After adding 8 kg of first type, new amounts are (5x-10+8) = 5x-2 kg and (7x-14) kg. Given new ratio 6:7, so (5x-2)/(7x-14) = 6/7. Solving gives x=10, so original first type = 5*10 = 50 kg, which matches none of the given options.

Multiple choice

A man buys a certain number of mangoes at 6 a ruppee and the same number at 9 a repee too. He mixes and sells them at 8 a rupee. Find his percent profit or loss. एक आदमी एक रूपये में 6 की दर से एक निश्चित संख्या में आम खरीदता है और उसी समान संख्या में एक रूपये में 9 की दर से खरीदता है। वह उन्हें मिला देता है और एक रूपये में 8 की दर से बेचता है। उसका प्रतिशत लाभ या हानि ज्ञात कीजिये।

  1. 10% profit/लाभ

  2. 10% loss/हानि

  3. 8.5% profit/ykHk

  4. 8.5% loss/हानि
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let he buys n mangoes each time. Cost = n/6 + n/9 = (3n + 2n)/18 = 5n/18 rupees. He sells 2n mangoes at 8 per rupee, so revenue = 2n/8 = n/4 rupees. Loss = CP - SP = 5n/18 - n/4 = (10n - 9n)/36 = n/36. Loss % = (n/36)/(5n/18) × 100 = (1/36)/(5/18) × 100 = (1/36) × (18/5) × 100 = (1/2) × (1/5) × 100 = 10%.

Multiple choice

From a mixture of milk and water 8 litres mixure is taken out and replaced with water , hence the percentage of water changed from 20 to 25 percent. find the initial quantity of mixture.

  1. 144 litres

  2. 136 litres

  3. 128 litres

  4. 150 litres

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial mixture: 80% milk, 20% water. After removing 8L and adding water: 0.8(x−8)=0.75x (milk amount unchanged). Solving: 0.8x−6.4=0.75x, so 0.05x=6.4, x=128 liters. Option C matches.

Multiple choice

A can contains a mixture of two liquids A and B is the ratio 2 : 3. When 8 liters of mixture is drawn off and the can is filled with B, the ratio of A and B becomes 6 : 11. How many liters of mixture was contained by the can initially? एक पात्र में दो तरल पदार्थ A और B का मिश्रण 2 : 3 के अनुपात में है। जब 8 लीटर मिश्रण निकाला जाता है और B से बदल दिया जाता है, तो A और B का अनुपात 6 : 11 हो जाता है। शुरुआत में कितना मिश्रण था?

  1. 30 liter/लीटर

  2. 60 liter/लीटर

  3. 45 liter/लीटर

  4. 64 liter/लीटर
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial ratio A:B = 2:3, so A = 2x, B = 3x. After removing 8L (3.2L A, 4.8L B) and adding 8L B: New A = 2x - 3.2, New B = 3x - 4.8 + 8 = 3x + 3.2. Ratio (2x - 3.2)/(3x + 3.2) = 6/11. Solving gives x = 20. Total initial mixture = 5x = 60L.

Multiple choice

Which information is/are sufficient to answer the questions asked? A vessel contains a mixture of two milk A and B. How much of the milk ‘A’ was there in the vessel? I. 9 litres of mixture is replaced by 9 litres of milk B. II. The initial ratio of milk A and B in the mixture was 7 : 5. III. The new ratio of milk A and B in the mixture is 7 : 9.

  1. II and III

  2. I and (II or III)

  3. All three together are not sufficient

  4. All three together are sufficient

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let initial quantities be 7x of milk A and 5x of milk B, so total = 12x litres. Statement III alone is insufficient without knowing total quantity. Statement II gives the ratio but not the total. Statement I: When 9 litres of mixture is removed, it contains A and B in 7:5 ratio, so 9 × 7/12 = 5.25 litres of A and 9 × 5/12 = 3.75 litres of B are removed. Remaining: A = 7x - 5.25, B = 5x - 3.75. After adding 9 litres of B: A = 7x - 5.25, B = 5x - 3.75 + 9 = 5x + 5.25. New ratio (A:B) = 7:9, so (7x - 5.25)/(5x + 5.25) = 7/9. Solving: 63x - 47.25 = 35x + 36.75, so 28x = 84, x = 3. Initial milk A = 7x = 21 litres. All three statements together are necessary and sufficient.

Multiple choice

Mixture A contains 40% milk and 60% water. Mixture B contains 35% milk and 40% water and 25% sugar. Mixtures A and B are mixed in the ratio of 1 : 4. Milk and Water in the newly formed Mixture is मिश्रण A में 40% दूध और 60% पानी है। मिश्रण B में 35% दूध और 40% पानी और 25% चीनी है। मिश्रण A और B, 1 : 4 के अनुपात में मिश्रित किये जाते है । नए गठित मिश्रण में दूध और पानी है -

  1. 20% and 30%/20% और 30%

  2. 36% and 44%/36% और 44%

  3. 25% and 35%/25% और 35%

  4. 49% and 36%/49% और 36%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mixture A has 40% milk and 60% water. Mixture B has 35% milk, 40% water, and 25% sugar. When mixed in ratio 1:4, the total parts = 5. Milk: (1×40 + 4×35)/5 = (40 + 140)/5 = 180/5 = 36%. Water: (1×60 + 4×40)/5 = (60 + 160)/5 = 220/5 = 44%. The sugar in B doesn't affect the milk-water percentage calculation as it's a separate component.

Multiple choice

A solution of salt and water contains 40% salt, 175 ml water is evaporated and now percentage of salt in solution becomes 60%. Find the initial quantity of solution ?

  1. 500 ml

  2. 395 ml

  3. 525 ml

  4. 625 ml

  5. 600 ml
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let initial solution = x ml. Initial salt = 0.4x, water = 0.6x. After evaporating 175 ml water: new water = 0.6x - 175, salt = 0.4x. Total new volume = x - 175. Salt percentage = 0.4x / (x - 175) = 0.60. Solving: 0.4x = 0.6x - 105, so 0.2x = 105, x = 525. Initial solution was 525 ml. Verification: Initial: 40% of 525 = 210 ml salt, 315 ml water. After evaporation: 210 ml salt, 140 ml water (315-175). Total = 350 ml. Salt % = 210/350 = 60%. Option C (525 ml) is correct.

Multiple choice

A mixture contains 25\% milk and rest water. What percent of this mixture must taken out and replaced with milk so that in mixture milk and water may become equal.

  1. 25%

  2. 33.33%

  3. 20%

  4. 67.67%

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial: 25% milk, 75% water. Let x fraction be removed and replaced with milk. Milk after operation: 0.25(1-x) + x = 0.25 - 0.25x + x = 0.25 + 0.75x. Water after: 0.75(1-x) = 0.75 - 0.75x. For equality: 0.25 + 0.75x = 0.75 - 0.75x, so 1.5x = 0.5, x = 0.5/1.5 = 1/3 = 33.33%. The claimed answer B is correct.

Multiple choice

Three pots contains equal mixture of milk and water in the ratio of 4 : 1, 5 : 2 and 6 : 1 respectively. If all the solutions are mixed together. What is the ratio of milk and water in the final solution? तीन वर्तनों के समान मिश्रण में दूध और पानी का अनुपात क्रमशः 4 : 1, 5 : 2 और 6 : 1 है। यदि सभी मिश्रणों को एक साथ मिला दिया जाए तो मिश्रण में दूध और पानी का अनुपात क्या होगा ?

  1. 84 : 23

  2. 83 : 22

  3. 87 : 57

  4. 77 : 89
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When mixing equal quantities from three pots, convert each ratio to actual milk and water amounts. Pot1 (4:1) gives 4/5 milk and 1/5 water. Pot2 (5:2) gives 5/7 milk and 2/7 water. Pot3 (6:1) gives 6/7 milk and 1/7 water. Summing with common denominator 35: total milk = (28+25+30)/35 = 83/35, total water = (7+10+5)/35 = 22/35. The ratio is 83:22. Option A (84:23) incorrectly adds numerators directly without proper fraction addition.

Multiple choice

A person mix 18 ltrs of water in pure milk and sells the mixture at cost price of milk and gets a profit of 12.5%. Find the initial quantity of milk.

  1. 120 lt.

  2. 124 lt.

  3. 160 lt.

  4. 144 lt.

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let initial milk be x liters. Cost price = x * rate per liter. Mixture volume = x + 18 liters. Selling at CP of milk gives profit = (18/x) * 100 = 12.5%. So 18/x = 0.125, thus x = 144. The profit equals the percentage of water added to milk.