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Bunuel
A tank contains a mixture of 200 litres of wine and water. 20% of the mixture is water and the remaining is wine. How many litres of water should be added to the mixture to increase the percentage of water to 25% in the new mixture?

A 13 1/3
B 16 1/3
C 19 1/3
D 22 1/3
E 25 1/3
Solution:

We see that currently 200 x 0.2 = 40 liters of the mixture is water. If w let x = the number of liters of water should be added to increase the percentage of water to 25% of the mixture, we can create the equation:

(40 + x) / (200 + x) = 1/4

4(40 + x) = 200 + x

160 + 4x = 200 + x

3x = 40

x = 40/3 = 13 1/3

Answer: A
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A tank contains a mixture of 200 litres of wine and water. 20% of the mixture is water and the remaining is wine. How many litres of water should be added to the mixture to increase the percentage of water to 25% in the new mixture?

Amount of water in 200 l mixture = 20% of 200 = 40

Assume that the new amount of water to be added to the mixture be x and the percentage of water is increased to 25 %.

25/100 = (40+x)/(200+x)

1/4 = (40+x)/(200+x)

200 + x = 160 + 4x

3x = 40

x = 40/3 = 13 1/3

Option A is the answer.

Thanks,
Clifin J Francis
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