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Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers. Quantity I: A vessel contains 50 litres of milk but the milkman is planning to replace 5 litres of milk with water and he will repeat this process total 3 times. Find the quantity of milk left in the container. Quantity II: There is a mixture of 33 litres of honey and water in the ratio of 5 : 6. Find the amount which should be added to both honey and water to make the ratio of water to honey 53 : 50.

Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers. Quantity I: A vessel contains 50 litres of milk but the milkman is planning to replace 5 litres of milk with water and he will repeat this process total 3 times. Find the quantity of milk left in the container. Quantity II: There is a mixture of 33 litres of honey and water in the ratio of 5 : 6. Find the amount which should be added to both honey and water to make the ratio of water to honey 53 : 50. Correct Answer Quantity I > Quantity II

Quantity I:

Initial quantity of milk = 50 litres

Milk to be replaced with water = 5 litres

Process repeated = 3 times

So, Final quantity = Initial quantity n

Where “ n ” = Number of times volume taken out

So, Final quantity = 50 3

⇒ 50 × (9/10)3

⇒ (50 × 9 × 9 × 9)/(10 × 10 × 10)

⇒ 36450/1000

⇒ 36.45 litres

∴ Final quantity of milk is 36.45 litres

Quantity II:

Total mixture of honey and water = 33 litres

Ratio of honey and water = 5 : 6

Honey quantity = 5/11 × 33

⇒ 15 litres

Water quantity = (33 – 15) litres

⇒ 18 litres

Now, let quantity to be added in the honey and water be x litres

Final quantity of Honey = 15 + x

Final quantity of water = 18 + x

And, ratio of water and honey = 53 : 50

So, (18 + x)/(15 + x) = 53/50

⇒ 900 + 50x = 795 + 53x

⇒ 3x = 105

⇒ x = 35 litres

∴ Quantity I > Quantity II

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