A tank contains a mixture of 200 litres of wine and water. 20% of the

Question

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

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

Dashingnightmare · Answer

In 200 litres since 20% is water => 40 litre is water
Let x litres of water is added to increase % to 25%
now, 40+x / 200+x=25/100 => x=40/3 =>x=13 1/3 Hence A.

TestPrepUnlimited · Answer

By adding water, we are keeping wine constant. Originally we had  80\% * 200 = 160 L  of wine, we want that to be 75% of the new mixture.

Thus  \frac{160}{75\%} = 160*\frac{4}{3} = \frac{640}{3} = 200 + \frac{40}{3}  is the new mixture's volume. Then we added  \frac{40}{3}  which is choice A.

Ans: A

ScottTargetTestPrep · Answer

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

CrackverbalGMAT · Answer

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

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

A tank contains a mixture of 200 litres of wine and water. 20% of the

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