Question

: A vessel is full of a mixture of methanol and ethanol in which there is 20% ethanol. 10 litres of mixture are drawn off and filled with methanol. If the ethanol is now 15%, what is the capacity of the vessel?

Answer 1

Gɪᴠᴇɴ :-

• A vessel is full of a mixture of methanol and ethanol
• Initial ethanol = 20% .
• Mixture drawn out = 10 Litre .
• Quantity of Methanol filled = 10 Litre.
• Now % of Ethanol = 15% .

Tᴏ Fɪɴᴅ :-

• Capacity or vessel ?

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, Total Capacity of the vessel is 100x Litre.

Than ,

→ initial Ethanol = 20% of 100x = (20 * 100x) / 100 = 20x Litre.

→ Initial Methanol = Rest = 100x - 20x = 80x Litre.

So,

→ Ratio of Ethanol : Methanol = 20x : 80x = 1 : 4

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Now, 10 Litre of Mixture was drawn out.

So, we can say That, in This Mixture Ethanol & Methanol were in ratio of 1 : 4 .

Therefore,

→ Ethanol Drawn out = 10 * (1/5) = 2 Litre.

→ Methanol Drawn out = 10 * (4/5) = 8 Litre.

_________________

Hence,

→ Ethanol Left in vessel Now = (20x - 2) Litre .

→ Methanol Left in vessel Now = (80x - 8) Litre .

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Now, we Have Given That, it was Filled with methanol.

So,

→ Ethanol in Vessel Now = (20x - 2) Litre.

→ Methanol in Vessel Now = (80x - 8) + 10 = (80x + 2) Litre.

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Now, we Have Given That, % of Ethanol is 15% Now,

So,

→ [(20x - 2) * 100] / 100x = 15

→ 20x * 100 - 200 = 1500x

→ 2000x - 1500x = 200

→ 500x = 200

→ x = (2/5) .

Hence,

→ Capacity of vessel = 100x = 100(2/5) = 40 Litre (Ans.)

(Nice Question).