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Mixture and Alligation — GMAT Focus Questions with Explanations

Questions & explanations

1. A chemist has 200 liters of a 40% acid solution. She evaporates some water, then adds 50 liters of a 60% acid solution. The final mixture is 50% acid. How many liters of water were evaporated?

20 liters
30 liters
40 liters
50 liters
60 liters

Answer: 30 liters

Initial acid: 0.4*200 = 80 L. Let x = liters evaporated. After evaporation, volume = 200 - x, acid remains 80 L. Then add 50 L of 60% acid: added acid = 0.6*50 = 30 L. Final volume = (200 - x) + 50 = 250 - x. Final acid = 80 + 30 = 110 L. Final concentration = 50% = 0.5, so 110 / (250 - x) = 0.5 → 110 = 0.5(250 - x) → 110 = 125 - 0.5x → 0.5x = 15 → x = 30 liters.

2. A coffee blend is made by mixing three types of beans: Arabica, Robusta, and Liberica. The blend contains a total of 120 kg of beans. The cost per kg of Arabica is $12, Robusta is $8, and Liberica is $6. The total cost of the blend is $1000. If the amount of Arabica is twice the amount of Robusta, how many kg of Liberica are in the blend?

20 kg
30 kg
40 kg
50 kg
60 kg

Answer: 40 kg

Let R = kg of Robusta, then Arabica = 2R, Liberica = L. Total: 2R + R + L = 120 → 3R + L = 120. Cost: 12(2R) + 8R + 6L = 1000 → 24R + 8R + 6L = 1000 → 32R + 6L = 1000. Multiply first eq by 6: 18R + 6L = 720. Subtract from second: (32R+6L) - (18R+6L) = 1000-720 → 14R = 280 → R = 20. Then Arabica = 40, total 60, so L = 120 - 60 = 40 kg.

3. A solution is made by mixing two liquids: Liquid A (concentration 30%) and Liquid B (concentration 60%). The resulting mixture has a concentration of 45%. What is the ratio of the volume of Liquid A to the volume of Liquid B in the mixture?

Answer: 1:1

Using alligation: The difference between the mean (45%) and the lower concentration (30%) is 15%, and the difference between the higher concentration (60%) and the mean is 15%. The ratio of volumes is inversely proportional to these differences: A:B = (60-45):(45-30) = 15:15 = 1:1.

4. A 50-liter solution of alcohol and water contains 40% alcohol. How many liters of pure alcohol must be added to increase the alcohol concentration to 60%?

10 liters
15 liters
20 liters
25 liters
30 liters

Answer: 25 liters

Let x = liters of pure alcohol added. Initial alcohol = 0.40*50 = 20 liters. Final volume = 50 + x. Final alcohol = 20 + x. Concentration: (20+x)/(50+x) = 0.60. Solve: 20+x = 0.60(50+x) → 20+x = 30 + 0.6x → 0.4x = 10 → x = 25. So 25 liters of pure alcohol must be added.

5. A store sells three types of nuts: almonds at $10 per kg, cashews at $14 per kg, and peanuts at $6 per kg. The owner wants to create a 50 kg mixture that costs $9.20 per kg. If the mixture contains twice as many almonds as cashews, and the weight of peanuts is 10 kg more than the weight of cashews, how many kilograms of almonds are in the mixture?

Answer: 20

Let cashews = x kg. Then almonds = 2x kg, peanuts = x + 10 kg. Total weight: 2x + x + (x+10) = 4x+10 = 50 → 4x=40 → x=10. So almonds = 20 kg. Check cost: almonds: 20*10=200, cashews: 10*14=140, peanuts: 20*6=120, total cost=460, per kg=460/50=9.20. Correct.

6. A 20-liter mixture of milk and water contains 60% milk. Some of the mixture is replaced with pure milk, resulting in a mixture that is 80% milk. How many liters of the original mixture were replaced?

4 L
5 L
6 L
8 L
10 L

Answer: 10 L

The original mixture has 12 L milk (60% of 20 L). Replacing y L of mixture removes 0.6y L milk and adds y L pure milk. New milk = 12 - 0.6y + y = 12 + 0.4y. New total is still 20 L. Set (12 + 0.4y)/20 = 0.8 → 12 + 0.4y = 16 → 0.4y = 4 → y = 10 L.

7. A tank contains 100 liters of a 10% salt solution. 10 liters of the solution are removed and replaced with pure water. This process is repeated once more. What is the final percentage of salt in the solution?

7.2%
8.1%
8.5%
9.0%
9.5%

Answer: 8.1%

After each removal, salt left = initial salt * (1 - fraction removed). Fraction removed each time = 10/100 = 0.1. After first: salt = 10 * 0.9 = 9 liters. After second: salt = 9 * 0.9 = 8.1 liters. Final percentage = (8.1/100)*100% = 8.1%.

8. A 20-liter solution of acid and water contains 30% acid. How many liters of pure water must be added to obtain a 12% acid solution?

10 liters
15 liters
20 liters
25 liters
30 liters

Answer: 30 liters

Let x = liters of water added. Initial acid = 0.30*20 = 6 liters. Final volume = 20 + x. Final acid concentration = 6/(20+x) = 0.12. Solve: 6 = 0.12(20+x) → 6 = 2.4 + 0.12x → 3.6 = 0.12x → x = 30. So 30 liters of water must be added.

9. A mixture of nuts contains almonds, cashews, and peanuts. The price per pound of almonds is $8, cashews $12, and peanuts $4. The mixture costs $7 per pound. If the mixture contains twice as many pounds of peanuts as almonds, and the weight of cashews is equal to the weight of almonds, what is the ratio of the weight of peanuts to the total weight?

Answer: 1/2

Let weight of almonds = a, then cashews = a, peanuts = 2a. Total weight = a + a + 2a = 4a. Cost: 8a + 12a + 4(2a) = 8a+12a+8a = 28a. Cost per pound = 28a/4a = $7, which matches. Ratio of peanuts to total = 2a/4a = 1/2.

10. A chemist mixes two solutions: Solution A contains 20% acid, and Solution B contains 50% acid. The chemist wants a mixture that is 30% acid. Using the alligation method, what is the ratio of Solution A to Solution B in the mixture?

Answer: 2:1

Alligation: Place the desired concentration (30%) in the middle. Higher concentration (50%) on top, lower (20%) on bottom. Differences: 50-30=20, 30-20=10. Ratio of A (lower) to B (higher) is 20:10 = 2:1. So A:B = 2:1.

11. A coffee blend is made by mixing a premium coffee costing $12 per pound with a standard coffee costing $8 per pound. The blend costs $9 per pound. According to the alligation rule, what is the ratio of premium to standard coffee in the blend?

Answer: 1:3

Alligation: Mean = 9, lower = 8, higher = 12. Differences: mean - lower = 1, higher - mean = 3. Ratio of higher (premium) to lower (standard) = (mean - lower): (higher - mean) = 1:3. So premium:standard = 1:3.

12. A chemist needs to create a 40% acid solution. She has a 25% acid solution and a 60% acid solution. How many liters of the 60% solution must she add to 2 liters of the 25% solution to obtain the desired concentration? (Assume volumes are additive.)

1.0 L
1.2 L
1.5 L
1.8 L
2.0 L

Answer: 1.5 L

Let x be liters of 60% solution. Total acid = 0.25*2 + 0.6x = 0.5 + 0.6x. Total volume = 2 + x. Desired concentration = 0.4, so (0.5 + 0.6x)/(2+x) = 0.4 => 0.5 + 0.6x = 0.8 + 0.4x => 0.2x = 0.3 => x = 1.5 L.