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Test: Alligation & Mixture- 1 - CAT MCQ


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10 Questions MCQ Test Quantitative Aptitude (Quant) - Test: Alligation & Mixture- 1

Test: Alligation & Mixture- 1 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Test: Alligation & Mixture- 1 questions and answers have been prepared according to the CAT exam syllabus.The Test: Alligation & Mixture- 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Alligation & Mixture- 1 below.
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Test: Alligation & Mixture- 1 - Question 1

A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 1

Assume that a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid


Hence milk now contained by the container = 40

Test: Alligation & Mixture- 1 - Question 2

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 2

► Since first and second varieties are mixed in equal proportions.So, their average price = Rs.
= Rs. 130.50

► So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.

By the rule of alligation, we have:

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Test: Alligation & Mixture- 1 - Question 3

A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 3

► Let x and (12-x) litres of milk be mixed from the first and second container respectively

► Amount of milk in x litres of the the first container = .75x

► Amount of water in x litres of the the first container = .25x

► Amount of milk in (12-x) litres of the the second container = .5(12-x)

► Amount of water in (12-x) litres of the the second container = .5(12-x)

► Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)] = 3 : 5

⇒ (.25x+6-5x)/(.75x+6-.5x) =3/5

⇒(6−.25x)/(.25x+6) =3/5

⇒30−1.25x=.75x+18

⇒2x=12

⇒x=6

► Since x = 6, 12-x = 12-6 = 6

Hence 6 and 6 litres of milk should mixed from the first and second container respectively

Test: Alligation & Mixture- 1 - Question 4

Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 4

► Spirit in 1 litre mix of A = 5/7 litre.
► Spirit in 1 litre mix of B = 7/13 litre.
► Spirit in 1 litre mix of C = 8/13 litre. 
By rule of alligation we have required ratio X:Y

Therefore required ratio = 1/13 : 9/91
= 7:9

Test: Alligation & Mixture- 1 - Question 5

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16:65. How much wine did the cask originally hold?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 5

Test: Alligation & Mixture- 1 - Question 6

A jar is full of whiskey contains 40% alcohol. A part of this whiskey is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whiskey replaced is:

Detailed Solution for Test: Alligation & Mixture- 1 - Question 6

By the rule of alligation we have 
► So ratio of 1st and 2nd quantities = 7:14
= 1:2
∴ Required quantity replaced = 2/3

solution

Test: Alligation & Mixture- 1 - Question 7

A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 7

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Test: Alligation & Mixture- 1 - Question 8

Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. 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: Alligation & Mixture- 1 - Question 8

Fraction of A in contained 1 = 5/6
Fraction of A in contained 2 = 1/4
Let the ratio of liquid required from containers 1 and 2 be x:1-x

=> Ratio = 3:4

Test: Alligation & Mixture- 1 - Question 9

There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

Detailed Solution for Test: Alligation & Mixture- 1 - Question 9

Let the volume of the cup be V.
Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V
Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.
So, in each cup, the amount of water contained is

Amount of alcohol contained in the second container is 

So, the required proportion of water in the first container and alcohol in the second container are equal.

Test: Alligation & Mixture- 1 - Question 10

DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

What is the maximum amount of sucrose (to the nearest gram) that can be added to one-gram of saccharin such that the final mixture obtained is atleast 100 times as sweet as glucose?

Detailed Solution for Test: Alligation & Mixture- 1 - Question 10

For the mixture to be 100 times as sweet as glucose, its sweetness relative to the mixture should be at least 74.

1 gm of saccharin = 675

Let the number of grams of sucrose to be added be N. Thus, the total weight of the mixture = N + 1.

So, (675 + N) / (N+1) = 74

=> 675 + N = 74N + 74

=> 601 = 73N => N = 8.23

When N=9, sweetness will be S = (675+9)/10 = 684/10 = 68.4

When N=8, sweetness will be S = (675+8)/9 = 683/9 = 75.8

So, option b) is the correct answer.

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