MA DS Practice SET – 3
DS: Mixture & Alligation – Level 3
1.
From a tank of pure milk, 10 liters are removed and replaced with water. This is done twice. How much milk remains?
(1) The initial capacity of the tank is 50 liters.
(2) After the first replacement, the ratio of milk to water is 4:1.
(1) The initial capacity of the tank is 50 liters.
(2) After the first replacement, the ratio of milk to water is 4:1.
Solution (D):
Formula: \(Final = Initial (1 – \frac{10}{Initial})^2\). We need Initial.
(1) Gives Initial = 50. Sufficient.
(2) After 1st step: 10L removed/replaced.
Ratio 4:1 means Milk is \(4/5\) of total.
This implies \(\frac{Initial – 10}{Initial} = \frac{4}{5}\).
\(5(I – 10) = 4I \implies 5I – 50 = 4I \implies I = 50\).
We found Initial. Sufficient.
2.
A merchant mixes two grades of rice, A and B. What is the profit percentage if sold at $5/kg?
(1) Cost of A is $3/kg and Cost of B is $4/kg.
(2) They are mixed in a ratio of 2:1.
(1) Cost of A is $3/kg and Cost of B is $4/kg.
(2) They are mixed in a ratio of 2:1.
Solution (C):
To find Profit %, we need Cost Price (CP) of the mixture.
(1) Gives individual costs but not the ratio. CP could be anywhere between 3 and 4. Not sufficient.
(2) Gives ratio but not costs. Not sufficient.
Combined: \(CP = \frac{2(3) + 1(4)}{3} = \frac{10}{3} \approx 3.33\).
Now we have CP and SP ($5). We can calculate Profit %. Sufficient.
3.
A mixture contains chemical A, B, and C. Is the amount of A greater than the amount of C?
(1) The ratio A:B is 2:3 and B:C is 3:1.
(2) The total volume is 60 liters.
(1) The ratio A:B is 2:3 and B:C is 3:1.
(2) The total volume is 60 liters.
Solution (A):
(1) A:B = 2:3. B:C = 3:1.
Since B is ‘3’ in both, A:B:C = 2:3:1.
Is A (2 parts) > C (1 part)? Yes. Always true regardless of total volume. Sufficient.
(2) Gives Total only. No ratios. Not sufficient.
4.
How many kg of dried grapes can be obtained from \(x\) kg of fresh grapes?
(1) Fresh grapes contain 90% water.
(2) Dried grapes contain 20% water.
(1) Fresh grapes contain 90% water.
(2) Dried grapes contain 20% water.
Solution (E):
To solve, we balance the “pulp”.
\(0.10 \times \text{Fresh} = 0.80 \times \text{Dry}\).
\(\text{Dry} = \frac{0.10}{0.80} \times \text{Fresh} = \frac{1}{8} \times x\).
(1) Gives fresh water content.
(2) Gives dry water content.
Combined: We have the formula, but **we do not know the value of x**. The question asks “How many kg”, which implies a specific number. Since \(x\) is a variable and not defined, we cannot give a numerical answer. Not sufficient.
5.
Two alloys X and Y are mixed. Alloy X is 40% Gold. Alloy Y is 20% Gold. What is the ratio of Gold to non-Gold in the new mixture?
(1) The new mixture is 30% Gold.
(2) The total weight of the new mixture is 100g.
(1) The new mixture is 30% Gold.
(2) The total weight of the new mixture is 100g.
Solution (A):
The question asks for the **ratio of Gold to non-Gold** in the mixture. This depends solely on the **percentage of Gold** in the final mixture.
(1) New mixture is 30% Gold. This implies it is 70% non-Gold. Ratio = 30:70 = 3:7. Sufficient.
(2) Gives weight, but not composition. Not sufficient.
**Correction:** I labeled the answer as C in the prompt, but A is sufficient on its own to find the ratio *within* the mixture. I will fix the data-answer to A.
Score: 0 / 0