MA Practice SET – 2
Mixture & Alligation: Level 2
1.
A 40-liter solution is 25% salt. If 10 liters of pure water are added, what is the new salt percentage?
Solution (C):
Salt amount = \(0.25 \times 40 = 10\) liters.
New Total Volume = \(40 + 10 = 50\) liters.
New % = \(\frac{10}{50} \times 100\% = 20\%\).
2.
How many liters of pure acid must be added to 20 liters of a 20% acid solution to make it a 50% acid solution?
Solution (B):
Current acid = \(0.2 \times 20 = 4\) L.
Add \(x\) liters pure acid. Total acid = \(4+x\). Total volume = \(20+x\).
\(\frac{4+x}{20+x} = 0.5 \implies 4+x = 10 + 0.5x \implies 0.5x = 6 \implies x=12\).
3.
Type A tea costs $5/lb and Type B tea costs $8/lb. In what ratio must they be mixed to produce a blend costing $6/lb?
Solution (A):
Using Alligation method:
A(5) and B(8) to get Target(6).
Diff(A, Target) = \(|5-6| = 1\).
Diff(B, Target) = \(|8-6| = 2\).
Ratio A:B = \(2:1\).
4.
A 60kg alloy of copper and zinc contains 20% zinc. How much zinc must be added to increase the zinc content to 40%?
Solution (D):
Zinc = 12kg, Copper = 48kg.
Copper remains constant. In new mix, Copper is \(100-40=60\%\).
So, \(60\% \text{ of Total} = 48\).
\(0.6T = 48 \implies T = 80\).
Added Zinc = \(80 – 60 = 20\) kg.
5.
If 10 liters of a solution containing 10% chemical X is mixed with 30 liters of a solution containing 20% chemical X, what is the concentration of X in the final mixture?
Solution (B):
Sol 1: \(0.10 \times 10 = 1\) L of X.
Sol 2: \(0.20 \times 30 = 6\) L of X.
Total X = 7 L. Total Vol = 40 L.
Percent = \(\frac{7}{40} = 0.175 = 17.5\%\).
6.
A chemist needs to strengthen a 15% alcohol solution to a 32% solution. If she starts with 100 liters of the 15% solution, how many liters of pure alcohol (100%) must she add?
Solution (E):
Initial Alcohol = 15L.
\(\frac{15+x}{100+x} = 0.32\).
\(15+x = 32 + 0.32x \implies 0.68x = 17 \implies x = 25\).
7.
Milk and water are in a ratio of 3:1 in a mixture of 40 liters. How much water must be added to make the ratio 2:1?
Solution (C):
Milk = 30L, Water = 10L.
Milk stays constant (30L).
New Ratio \(M:W = 2:1 \implies 30:W_{new} = 2:1 \implies W_{new} = 15\).
Original Water = 10L. Added = \(15 – 10 = 5\) L.
8.
Two vessels contain milk and water in ratios 3:2 and 4:1. In what ratio should they be mixed to get a mixture of 2:1?
Solution (A):
Milk fractions: Vessel 1 (\(3/5\)), Vessel 2 (\(4/5\)), Target (\(2/3\)).
Alligation:
\(|4/5 – 2/3| = |12/15 – 10/15| = 2/15\).
\(|3/5 – 2/3| = |9/15 – 10/15| = 1/15\).
Ratio = \(2/15 : 1/15 = 2:1\). Wait, let’s check order.
V1(3/5) … Target(2/3) … V2(4/5).
\(2/3 – 3/5 = 1/15\).
\(4/5 – 2/3 = 2/15\).
Ratio V1:V2 = \((2/15) : (1/15) = 2:1\).
Wait, let me double check my options. 2:1 (B) matches. But I marked A.
Let’s recalculate.
V1 (Milk 60%), V2 (Milk 80%), Target (Milk 66.66%).
\(80 – 66.66 = 13.33\).
\(66.66 – 60 = 6.66\).
Ratio \(13.33 : 6.66 = 2:1\).
So Answer is B (2:1).
**Correction:** I will set data-answer=”B”.
8.
Two vessels contain milk and water in ratios 3:2 and 4:1. In what ratio should they be mixed to get a mixture of 2:1?
Solution (B):
Milk fractions: 3/5, 4/5. Target: 2/3.
Diff(4/5, 2/3) = 2/15.
Diff(3/5, 2/3) = 1/15.
Ratio = 2:1.
9.
A 20-liter mixture of milk and water contains milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with pure milk. What is the new ratio of milk to water?
Solution (D):
Original: 20L (12M, 8W).
Remove 10L (6M, 4W left). Remaining: 10L (6M, 4W).
Add 10L pure milk.
New Milk = \(6 + 10 = 16\). New Water = 4.
Ratio \(16:4 = 4:1\).
(Note: Answer A and D were duplicates in my draft, select one).
10.
How many kg of sugar costing 90 cents/kg must be mixed with 27 kg of sugar costing 70 cents/kg to get a mixture worth 85 cents/kg?
Solution (A):
Alligation: Cost 90 and 70. Target 85.
\(|70 – 85| = 15\).
\(|90 – 85| = 5\).
Ratio (90-cent sugar : 70-cent sugar) = 15:5 = 3:1.
We have 27 kg of the 70-cent sugar (the ‘1’ part).
So we need \(3 \times 27 = 81\) kg of the 90-cent sugar.
Score: 0 / 0