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CAT Exam  >  CAT Tests  >  Level-wise Tests for CAT  >  Test Level 2: Mixtures and Alligations - 1 - CAT MCQ

Test Level 2: Mixtures and Alligations - 1 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 2: Mixtures and Alligations - 1

Test Level 2: Mixtures and Alligations - 1 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 2: Mixtures and Alligations - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 2: Mixtures and Alligations - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 2: Mixtures and Alligations - 1 below.
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Test Level 2: Mixtures and Alligations - 1 - Question 1
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Two liquids A and B are in the ratio 5 : 1 in container X and in the ratio 1 : 3 in container Y. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?  

Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 1

Let X units of liquid be taken from container X.
So, liquid A taken = (5/6)X
Liquid B taken = (1/6)X
Let Y units of liquid be taken from container Y.
So, liquid A taken = (Y/4)
Liquid B taken = (3/4)Y
So, as the ratio is 1 : 1, we get
(5/6)X + (Y/4) = (X/6) + (3/4)Y
Or X/Y = 3/4
Hence, the required ratio is 3 : 4.
Thus, option d is correct.

Test Level 2: Mixtures and Alligations - 1 - Question 2
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A giant watermelon weighs 30 kg, of which 90% is water. After remaining in the sun for one hour, some of the water evaporates such that water makes up 80% of the weight. What is the weight of the watermelon now?  

Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 2

Water portion of the watermelon weighs 90% of 30 = 27 kg.
Solid portion of the watermelon weighs 30 - 27 = 3 kg.
When water is 80%, solid portion = 20% 
As 20% = 3 kg; 
Total weight = 3 x 5 = 15 kg

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Test Level 2: Mixtures and Alligations - 1 - Question 3
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In what ratio must water be mixed with milk to gain on selling the mixture at cost price?

Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 3

Let the C.P. of 1 litre milk be Re. 1. 
S.P. of 1 litre of mixture = Re. 1
Gain = 
C.P. of 1 litre of mixture = (100 x 3/350 x 1) = Rs. 6/7 
So, 6/7 litres of milk is present in 1 litre of mixture. 
Hence, ratio of water to milk = 1/7 : 6/7 = 1 : 6 
Alternate Solution:
To make a profit of 16.67% we need to add 16.67% water.
i.e. we need to add water in milk in the ratio 1 : 6.

Test Level 2: Mixtures and Alligations - 1 - Question 4
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In what proportion must wheat at Rs. 1.8 per kg be mixed with wheat at Rs. 2.5 per kg to get a mixture worth Rs. 2.0 per kg?
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 4

Applying the rule of alligation:

⇒ Required ratio = 5 : 2

Test Level 2: Mixtures and Alligations - 1 - Question 5
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A milkman has 48 litres of mixture of milk and water in the ratio 5 : 1, respectively. He withdraws some amount of the mixture from the vessel and replaces it with milk. The ratio of the mixture of milk and water now becomes 7 : 1. What amount of the mixture is withdrawn?
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 5

Capacity = 48 litres
Amount of milk = 5/6 x 48 = 40 litres
Amount of water =1/6 x 48 = 8 litres
Let the amount withdrawn be x.
Thus, 
⇒  x = 12 litres

Test Level 2: Mixtures and Alligations - 1 - Question 6
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Puneet bought a mobile and a computer. Then he sold these to his friend Manjeet. He made a loss of 10% on the mobile and an overall profit of 15%. If the ratio of the costs of mobile and computer was 2 : 5, what was the profit made on the computer?
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 6


By the rule of Alligation,

⇒ 5x - 75 = 50
⇒ 5x = 125
⇒ x = 25
Profit made on the computer = 25%

Test Level 2: Mixtures and Alligations - 1 - Question 7
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How many litres of a 77% solution of alcohol must be added to 20 litres of a 65% alcoholic drink to obtain a 68% alcoholic drink?
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 7

Let the amount of solution having 77% solution of alcohol be x litres.
Now, apply the alligation formula.

Test Level 2: Mixtures and Alligations - 1 - Question 8
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The ratios of spirit to wine in two flasks labelled as A and B, containing spirit and wine mixture, are 7 : 3 and 3 : 1, respectively. Some amounts of the mixtures from the two flasks are poured into a third flask. The new mixture now contains spirit and wine in the ratio of 11 : 4. Find the ratio in which the mixtures from the flask A and flask B are mixed together to get this new mixture.
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 8

Fraction of wine in flask A =
Fraction of wine in flask B =
Fraction of wine in final mixture = 
By alligation method,

Therefore, ratio of amount of mixture taken from A to that taken from B

Test Level 2: Mixtures and Alligations - 1 - Question 9
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A sum of Rs. 39,00,000 is allotted for providing compensation to the victims of a train accident. The authorities have decided to sanction an amount of Rs. 30,000 to the injured and Rs. 80,000 to the families of the dead passengers. If there are total 65 victims, then find the number of injured passengers.  
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 9

Total amount allotted for compensation to the victims = Rs. 39,00,000
Total victims = 65
Average amount of compensation =
By alligation method,

Therefore,
Ratio of injured to dead passengers = 2 : 3
Number of injured passengers = 
This is our required solution.

Test Level 2: Mixtures and Alligations - 1 - Question 10
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One alloy contains 90% copper and 10% tin. Another alloy contains 93% copper and 4% tin. If they are mixed so that the mixture may contain 9% of tin, then what percent of copper will it contain?
Detailed Solution for Test Level 2: Mixtures and Alligations - 1 - Question 10

Apply the rule of alligation,
Quantity of tin

The two alloys should be mixed in the ratio 5 : 1.
Now, amount of copper in the mixture if mixture is 6 units = = 5.43 units
% of copper in the mixture = 

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