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A vessel contains 80 litres of pure alcohol. A certain quantity of water is added to make the ratio of alcohol and water 4 : 1. Some quantity of mixture was removed and 10 liters of water were added to it. If the quantity of water is 20 less than that of alcohol in the resultant mixture then alcohol is what percent of the water in the resultant mixture?

A vessel contains 80 litres of pure alcohol. A certain quantity of water is added to make the ratio of alcohol and water 4 : 1. Some quantity of mixture was removed and 10 liters of water were added to it. If the quantity of water is 20 less than that of alcohol in the resultant mixture then alcohol is what percent of the water in the resultant mixture? Correct Answer 200%

Given:

Pure alcohol in the vessel = 80 l

Quantity of water added = 10 l

Calculation:

Quantity of water added = (80/4) = 20 l

Let x be the quantity of mixture taken out

20 – (x/100) × 20 + 10 + 20 = 80 – (x/100) × 80

⇒ 20 – (x/5) + 30 = 80 – (4x/5)

⇒ (4x/5) – (x/5) = 80 – 50

⇒ (3x/5) = 30

⇒ x = {(30 × 5)/3} = 50

Quantity of alcohol after 50 liters was removed

⇒ 80 – (50/100) × 80

⇒ New quantity of alcohol = 40 l

Quantity of water after 50 liters was removed

⇒ 20 – (50/100) × 20 + 10

⇒ 20 – 10 + 10 = 20

⇒ Required percentage = (40/20) × 100 = 200%

∴ The required percentage is 200%

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