MA Practice SET – 3
Mixture & Alligation: Level 3
1.
A container contains 40 liters of milk. From this, 4 liters are removed and replaced with water. This process is repeated two more times (3 times total). How much milk is now contained in the container?
Solution (B):
Formula: \( \text{Final} = \text{Initial} (1 – \frac{\text{Removed}}{\text{Total}})^n \).
\( 40 (1 – \frac{4}{40})^3 = 40 (0.9)^3 \).
\( 40 \times 0.729 = 29.16 \) liters.
2.
Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruit?
Solution (D):
Pulp remains constant.
Fresh: 32% pulp (100-68). \(0.32 \times 100 = 32\) kg pulp.
Dry: 80% pulp (100-20). \(0.80 \times \text{Weight} = 32\).
Weight = \(32 / 0.8 = 40\) kg.
3.
Two alloys contain gold and silver in the ratio 3:1 and 5:3. A goldsmith mixes these to create a new alloy with ratio 2:1. In what ratio did he mix them?
Solution (C):
Gold fractions: \(3/4\) and \(5/8\). Target: \(2/3\).
Diff 1: \(|5/8 – 2/3| = |15/24 – 16/24| = 1/24\).
Diff 2: \(|3/4 – 2/3| = |9/12 – 8/12| = 1/12 = 2/24\).
Ratio = \(1/24 : 2/24 = 1:2\).
4.
A mixture of 120 liters of milk and water contains 20% water. How much water must be added to make the water 40% of the new mixture?
Solution (A):
Milk is constant.
Initial Milk = 80% of 120 = 96 L.
In new mixture, Water = 40%, so Milk = 60%.
\(0.60 \times \text{NewTotal} = 96 \implies \text{NewTotal} = 160\) L.
Added Water = \(160 – 120 = 40\) L.
5.
A merchant has 1000 kg of sugar. He sells part at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is:
Solution (E):
Alligation: 8 and 18. Target 14.
\(|18 – 14| = 4\).
\(|8 – 14| = 6\).
Ratio (8% part : 18% part) = 4:6 = 2:3.
Total 5 parts = 1000 kg. 1 part = 200 kg.
18% part (3 parts) = \(3 \times 200 = 600\) kg.
6.
From a cask of milk containing 30 liters, 6 liters are drawn out and replaced with water. If the same process is repeated a second time, what is the amount of milk left?
Solution (B):
\( \text{Final} = 30 (1 – 6/30)^2 \).
\( 30 (0.8)^2 = 30 \times 0.64 = 19.2 \) liters.
7.
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. What fraction of whisky was replaced?
Solution (D):
Orig (40%), Replaced Part (19%), Target (26%).
Alligation:
\(|19 – 26| = 7\) (Proportion of Orig remaining).
\(|40 – 26| = 14\) (Proportion of Replaced part).
Ratio Rem:Rep = 7:14 = 1:2.
Total = 3 parts. Replaced = 2 parts.
Fraction replaced = 2/3.
8.
The ratio of acid to water in solutions A and B are 2:7 and 4:5 respectively. They are mixed in ratio 4:3. What is the ratio of acid to water in the new solution?
Solution (C):
Take 36L total (LCM of 9, 9, 7). No, simpler:
Acid fractions: \(2/9\) and \(4/9\).
Mixed in 4:3 ratio.
Weighted Avg Acid = \( \frac{4(2/9) + 3(4/9)}{7} = \frac{8/9 + 12/9}{7} = \frac{20/9}{7} = \frac{20}{63} \).
So Acid is 20, Total is 63. Water = \(63-20=43\).
Ratio = 20:43.
9.
Solution X is 20% acid. Solution Y is 60% acid. They are mixed to form 100L of 30% acid. How many liters of X were used?
Solution (A):
Alligation: 20, 60. Target 30.
\(|60-30| = 30\).
\(|20-30| = 10\).
Ratio X:Y = 30:10 = 3:1.
Total 100L. X = \( \frac{3}{4} \times 100 = 75 \) L.
10.
A vessel contains 80L of milk. 8L milk is taken out and replaced by water. This process is repeated one more time. What is the ratio of milk to water in the final mixture?
Solution (B):
Rem Milk = \( 80(1 – 8/80)^2 = 80(0.9)^2 = 80(0.81) = 64.8 \) L.
Water = \( 80 – 64.8 = 15.2 \) L.
Ratio \( 64.8 : 15.2 = 648 : 152 \).
Divide by 8: \( 81 : 19 \).
Score: 0 / 0