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Alligation or Mixture Questions

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Thirty litres of water is added to 150 litres of 20% solution of alcohol in water.The resulting strenght of alcohol is


A16.67%.

B17.2%.

C18.3%.

DNone of these

Answer: Option A

Explanation:

As per question, 20% solution of alcohol in water => 30 litres of alcohol & 120 litres of water.
Then 30 litres of water is added, so the solution now contains 150 litres of water and 30 litres of alcohol.
Therefore, the resulting strength of alcohol = 30*100/180 = 16.67%.

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A 10 Liter mixture of milk and water contains 30 percent water. Two liters of this mixture is taken away. How many liters of water should now be added so that the amount of milk in the mixture is double that of water?


A1.4

B0.8

C0.4

D0.7

Answer: Option A

Explanation:

Two liters were taken away So we have the only 8L of the mixture.
Amount of milk in 8 L of mixture = 8 * 70% = 5.6 liters
Amount of water in 8 L of mix = (8 - 5.6) = 2.4 L.
Half of milk i.e half of 5.6 = 2.8 L.
We need (2.8 - 2.4)L water more = 0.4 L

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One type of liquid contains 25% of Kerosene, the other contains 30% of Kerosene. P can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of the Kerosene in the new mixture.


A28%

B23%

C30%

D27%

Answer: Option D

Explanation:

Let P be filled by 60 lts of 1st liquid and 40 lts. of 2nd liquid.
Amount of kerosene = (25*60/100) + (30*40/100) = 27 lts.
% of kerosene = 27 %.

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Two liquids are mixed in the proportion of 3:2 and the mixture is sold at $11 per litre at a 10% profit. If the first liquid costs $2 more per litre than the second, what does it cost per litre?


A$10.30

B$10.80

C$10.40

D$10.20

Answer: Option B

Explanation:

Given two liquids proportion as 3:2
from the mixtures:
Suppose second liquid cost = $x, then first liquid = $(x+2)
(x-10)/(10-x-2) = 3/2
2x-20 = 24-3x
5x = 44
x=8.8
so first liquid cost is x+2 = 8.80+2 = $10.80

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In what proportion must walnut at Rs. 62/kg be mixed with walnut at Rs. 72/kg in order to obtain a mixture worth Rs. 65/ kg ?


A3:1

B5:3

C6:7

D7:3

Answer: Option D

Explanation:

The required ratio is (72-65)/ (65-62) = 7/3.

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In two alloys, copper and tin are related in the ratios of 4:1 and 1:3. 10kg of Ist alloy, 16 kg of 2nd alloy and some pure copper melted together. An alloy obtained in which the ratio of copper and tin was 3:2. Find the weight of the new alloy.


A34 kg.

B35 kg.

C36 kg.

D30 kg.

Answer: Option B

Explanation:

Suppose x kg of pure copper is melted with 10kg of Ist alloy and 16 kg of 2nd alloy , then
Pure cooper in new alloy = 10*4/5 + 16*1/4 +x = 12+x kg
Pure tin in new alloy = 10*1/5+ 16*3/4 = 14 kg
As per given condition
=>(12+x)/14 = 3/2
=> 24+2x= 42
=> x=9 kg.
Total weight of new alloy = (10+16+9) = 35 kg.

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Two solutions of 90% and 97% purity are mixed resulting in 21 liters of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture?


A7.5 liters

B6 liter

C7 liters

D9 liters

Answer: Option D

Explanation:

lets assume firstsolution mixture = x liters
90x+(21-x)*97=21*94
(97x-90x)=21(97-94)
=> 7x=21*3
=>x= 9 liters

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A container of production line solvent is mixed at 22% strength and another is mixed at 53% strength.to obtain 12 liter of a mixture that is 25% strength. How many liter should be used from the container that is mixed at 53% strength?


A4 liters

B1 5/31 liters

C2 16/25 liters

D5 35/53 liters

Ee)4/59

Answer: Option B

Explanation:

container-of-production-line-solvent-is-mixed

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An alloy of zinc and copper contains the metals in the ratio 5 : 3. The quantity of zinc to be added to 16 kg of the alloy so that the ratio of the metal may be 3 : 1 is:


A2 Kg.

B4 Kg.

C8 Kg.

D3 Kg.

Answer: Option C

Explanation:

Lets assume quantity of zinc to added = x kg.
zinc quantity in alloy = 5/8*16=10 Kg.
and copper quantity = 3/8*16=6 kg.
Alloy new ratio of zinc and copper = 3:1
Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg.

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A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?


A1/4

B1/3

C1/2

D2/3

Answer: Option B

Explanation:

Consider a solution = 1 ltr
X is the certain quantity which has to be replaced
Now,
40%of (1-x)+(25%of x)=35/100
=> 40/100 * (1-x) + 25x/100 = 35/100
=> 40/100 - 40x/100 + 25x/100 = 35/100
=> 15x/100 = 5/100
=> x = 1/3

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