MA DS Practice SET – 2
DS: Mixture & Alligation – Level 2
1.
Two salt solutions, A and B, are mixed. What is the concentration of the final mixture?
(1) Solution A is 10% salt and Solution B is 30% salt, mixed in a ratio of 1:1.
(2) The volume of Solution A is 50 liters.
(1) Solution A is 10% salt and Solution B is 30% salt, mixed in a ratio of 1:1.
(2) The volume of Solution A is 50 liters.
Solution (A):
(1) When two solutions are mixed, the final concentration depends ONLY on their individual concentrations and the ratio of mixing.
Avg = \(\frac{1(10) + 1(30)}{2} = 20\%\). The actual volume doesn’t matter for concentration. Sufficient.
(2) Gives volume but no concentrations. Not sufficient.
2.
In what ratio were Tea X ($5/lb) and Tea Y ($10/lb) mixed?
(1) The total cost of the mixture was $300.
(2) The cost per pound of the mixture was $8.
(1) The total cost of the mixture was $300.
(2) The cost per pound of the mixture was $8.
Solution (B):
(1) Total cost depends on total quantity. We can have 30lbs of Y or 60lbs of X. The ratio isn’t fixed. Not sufficient.
(2) Using Alligation: Target=8. X=5, Y=10.
\(|10-8|=2\). \(|5-8|=3\). Ratio is 2:3. Sufficient.
**Correction:** I labeled the answer as C in the prompt, but B is sufficient on its own. Let me fix the data-answer to B.
2.
In what ratio were Tea X ($5/lb) and Tea Y ($10/lb) mixed?
(1) The total cost of the mixture was $300.
(2) The cost per pound of the mixture was $8.
(1) The total cost of the mixture was $300.
(2) The cost per pound of the mixture was $8.
Solution (B):
(1) Total cost depends on quantity, not just ratio. Not sufficient.
(2) Using the Alligation rule with the unit prices ($5, $10, and mean $8), we can uniquely determine the ratio. Ratio = (10-8):(8-5) = 2:3. Sufficient.
3.
How much water must be added to 20 liters of milk to change the milk-to-water ratio to 2:1?
(1) The original mixture contains 10% water.
(2) The original mixture has 18 liters of milk.
(1) The original mixture contains 10% water.
(2) The original mixture has 18 liters of milk.
Solution (D):
To find the added water, we need the initial composition.
(1) 10% water means 90% milk. Milk = \(0.9 \times 20 = 18\). Water = 2.
Target M:W = 2:1. Milk stays 18. New Water = \(18/2 = 9\). Added = \(9 – 2 = 7\). Sufficient.
(2) Milk = 18. Since Total = 20, Water = 2. This gives the exact same info as (1). Sufficient.
Answer is D.
4.
Alloy X contains Copper and Zinc. Is the weight of Copper in Alloy X greater than 5 kg?
(1) The ratio of Copper to Zinc is 3:2.
(2) Alloy X is mixed with 10 kg of pure Zinc.
(1) The ratio of Copper to Zinc is 3:2.
(2) Alloy X is mixed with 10 kg of pure Zinc.
Solution (E):
(1) Ratio is 3:2. If Total = 1kg, Copper = 0.6kg (No). If Total = 100kg, Copper = 60kg (Yes). Not sufficient.
(2) Adding Zinc changes the composition but doesn’t tell us the original weight. Not sufficient.
Combined: We know the ratio, but we still don’t know the **Total Weight** of Alloy X. Not sufficient.
5.
A solution of alcohol and water is 20% alcohol. How many liters of pure alcohol must be added to create a 50% alcohol solution?
(1) The initial volume of the solution is 40 liters.
(2) The initial amount of water in the solution is 32 liters.
(1) The initial volume of the solution is 40 liters.
(2) The initial amount of water in the solution is 32 liters.
Solution (D):
Formula: \(\frac{\text{Alcohol}_{old} + x}{\text{Total}_{old} + x} = 0.50\).
(1) Gives Total = 40. Alcohol = 8. Eq: \(\frac{8+x}{40+x} = 0.5\). Solve for x. Sufficient.
(2) Water = 32. Since solution is 20% Alcohol, it is 80% Water.
\(0.80 \times \text{Total} = 32 \implies \text{Total} = 40\). This derives the same info as (1). Sufficient.
Score: 0 / 0