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20 L of pure water was added to a vessel containing 80 L of pure milk. 36 L of the resultant mixture was then sold and some more pure milk and pure water was added to the vessel in the respective ratio of 7 : 2. If the final quantity of water was 3 liters less than the initial quantity of water in the vessel, what was the quantity of pure milk that was added to the vessel? (in liters)

20 L of pure water was added to a vessel containing 80 L of pure milk. 36 L of the resultant mixture was then sold and some more pure milk and pure water was added to the vessel in the respective ratio of 7 : 2. If the final quantity of water was 3 liters less than the initial quantity of water in the vessel, what was the quantity of pure milk that was added to the vessel? (in liters) Correct Answer 14.7

GIVEN:

Initial quantity of pure water = 20 L

Initial quantity of pure milk = 80 L

36 L of the resultant mixture was then sold and some more pure milk and pure water was added to the vessel in the respective ratio of 7 : 2.

CONCEPT:

Mixing of 2 entities concept.

CALCULATION:

Ratio of pure milk and pure water initially = 8 : 2 = 4 : 1

Total quantity of mixture initially = 20 + 80 = 100 L

36 L of the mixture is sold:

⇒ Remaining quantity of mixture = 100 - 36 = 64 L

∵ Ratio of pure milk and pure water will remain same as 4 : 1.

⇒ Quantity of pure milk = 64 × 4 / 5 = 51.2 L

⇒ Quantity of pure water = 64 - 51.2 = 12.8 L

Suppose the quantities of pure milk and pure water that was added to the vessel are 7x and 2x respectively.

⇒ Quantity of pure milk now in the mixture = (51.2 + 7x) L

⇒ Quantity of pure water now in the mixture = (12.8 + 2x) L

According to the question,

(12.8 + 2x) = 20 - 3

⇒ 2x = 4.2

⇒ x = 2.1

∴ Quantity of pure milk that was added to the vessel = 7x = 7 × 2.1 = 14.7 L

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