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8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16:65. How much wine did the cask originally hold?

  • a)
    30 litres

  • b)
    26 litres

  • c)
    24 litres

  • d)
    32 litres

Correct answer is option 'C'. Can you explain this answer?
Verified Answer
8 litres are drawn from a cask full of wine and is then filled with wa...
8 litres are drawn from a cask full of wine and is then filled with water.


This operation is performed three more times.


The ratio of the quantity of wine now left in cask to that of the water is 16 ∶ 65

Calculations:


Let the quantity of the wine in the cask originally be x litres.


After first operation, wine left in the cask = (x - 8) litres So, the ratio of the quantity of wine now left in cask to that of filled cask = 


This operation is performed three more times. Then, Wine : Water = 16 : 65 or wine : water = 16 : (16 + 65) = 81



Correct answer is 24.
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Most Upvoted Answer
8 litres are drawn from a cask full of wine and is then filled with wa...
Let initial quantity of wine = x litre
After a total of 4 operations, quantity of wine = x(1−yx)n=x(1−8x)4 
Given that after a total of 4 operations, the ratio of the quantity of wine left in cask 
to that of water = 16 : 65
Hence we can write as x(1−8x)4x=1681⇒(1−8x)4=(23)4⇒(1−8x)=23⇒(x−8x)=23⇒3x−24=2x⇒x=24
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Community Answer
8 litres are drawn from a cask full of wine and is then filled with wa...
Given:
- 8 litres of wine is drawn from a cask full of wine and is then filled with water.
- This operation is performed three more times.
- The ratio of the quantity of wine now left in the cask to that of the water is 16:81.

To find:
- The original amount of wine in the cask.

Solution:
Let's assume that the cask originally held 'x' litres of wine.

After the first operation, 8 litres of wine is drawn and replaced with water. So the cask now contains:
- Wine: (x-8) litres
- Water: 8 litres

After the second operation:
- Wine: (x-8) * (x-8)/x litres
- Water: 8 * (x-8)/x + 8 litres

After the third operation:
- Wine: (x-8) * (x-8)/x * (x-8)/x litres
- Water: 8 * (x-8)/x * (x-8)/x + 8 * (x-8)/x + 8 litres

After the fourth operation:
- Wine: (x-8) * (x-8)/x * (x-8)/x * (x-8)/x litres
- Water: 8 * (x-8)/x * (x-8)/x * (x-8)/x + 8 * (x-8)/x * (x-8)/x + 8 * (x-8)/x + 8 litres

Given that the ratio of wine to water is 16:81, we can express this as:
- Wine: (x-8) * (x-8)/x * (x-8)/x * (x-8)/x = 16k
- Water: 8 * (x-8)/x * (x-8)/x * (x-8)/x + 8 * (x-8)/x * (x-8)/x + 8 * (x-8)/x + 8 = 81k

Simplifying the above equations, we get:
- (x-8)^4 = 16kx^3
- 8(x-8)^3 = 81kx^3

Dividing the above equations, we get:
- (x-8)/2 = 3/4
- x-8 = 6
- x = 14

Therefore, the original amount of wine in the cask was 14 litres.

Answer: Option C) 24 litres.
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8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16:65. How much wine did the cask originally hold?a)30 litresb)26 litresc)24 litresd)32 litresCorrect answer is option 'C'. Can you explain this answer?
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