Test: Alligation & Mixture- 1 - CAT MCQ

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

► 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:

► 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

► 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

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

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.

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

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.

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.