Test Level 1: Mixtures and Alligations - CAT MCQ

Amount of mixture = 60 litres
Ratio of milk to water = 2 : 1

Let the amount of added water be x litres to make the ratio of milk to water 1 : 2.

⇒ x = 60 litres
Amount of water to be added = 60 litres

Amount of alcohol in the resulting mixture = 20 × 0.15 + 30 × 0.18 = 8.4 litres
Total amount of the resulting mixture = 50 litres
Concentration of alcohol in the resulting mixture = 8.4 × 100/50 = 16.8%

Total mixture = 100 litres
Quantity of milk = 80 litres
Quantity of water = 20 litres
¼ of the mixture removed = 25 litres
Quantity of milk removed = 25 × 4/5  = 20 litres
Quantity of water removed = 25 × 1/5 = 5 litres
After adding 25 litres of water, total quantity of milk = 80 - 20 = 60 litres
Total quantity of water = 20 - 5 + 25 = 40 litres
Hence, required ratio of water to milk = 2 : 3

Let the total volume of vessel be 'x' litres.
Initially, the %age of liquor in the mixture = 18%
After the replacement of 8 litres of the mixture with water, liquor %age = 15%
∴ 0.18x - 8 18/100 = 0.15x
⇒ 0.03x = 1.44
⇒ x = 48 litres
Hence, volume of vessel = 48 litres

where, A = Quantity of wine in the final mixture,
Q = Volume of cask,
q = Quantity removed, and
n = Number of times the operation is repeated

q = 6
n = 2

Q = 60 gallons

Suppose, we mix 1 litre of each solution.
So, the quantity of milk in the new solution would be 1/4 + 2/5 = 13/20
And, the quantity of water in the new solution would be 3/4 + 3/5 = 27/20
The ratio of milk and water would be (13/20) : (27/20) = 13 : 27

Let the weight of gold be x g and weight of the other metal be (18 - x) g.
Let the price of the other metal be Rs. y
Now, according to the question
7.20x + y(18 - x) = 74
7.20(18 - x) + yx = 60.10
Solving these two equations simultaneously,
x = 10 and y = 0.25
Quantity of the other metal = (18 - 10) g = 8 g

Suppose, we mix 1 litre of each solution.
So, the quantity of milk in the new solution would be 1/4 + 2/5 = 13/20
And, the quantity of water in the new solution would be 3/4 + 3/5 = 27/20
The ratio of milk and water would be (13/20) : (27/20) = 13 : 27

Let the weights of first and second alloys be 7x and 7y, respectively.
According to the question, both are mixed in equal ratios.
So, x = y
In first alloy, zinc and tin are 3x and 4x, respectively.
In second alloy, zinc and silver are 4x and 3x, respectively.
After mixing these two alloys, new alloy contains:
Zinc = 7x, tin = 4x and silver = 3x
Hence, ratio of zinc and tin = 7 : 4

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.