A mixture of 125 gallons of wine and water contains 20% water. How muc

In 125 gallons of the solution there are \(0.2*125=25\) gallons of water. We want to add \(w\) gallons of water to 125 gallons of solution so that \(25+w\) gallons of water to be 25% of new solution: \(25+w=0.25(125+w)\) --> \(w=\frac{25}{3}\).

Answer: C.

Given: A mixture of 125 gallons of wine and water contains 20% water.
Asked: How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?

In 125 gallons mixture: -
Wine = 80%*125 = 100 gallons
Water = 20%*125 = 25 gallons

Let the water to be added be x gallons
(x+25)/(125+x) = 25% = 1/4
4(x+25) = 125 + x
4x + 100 = 125 + x
3x = 25
x = 25/3 = 8 1/3gallons

IMO C

Kinshook
Hey could you explain what did you do after reaching 1:15 ratio?

I am subtracting new amount 1/16*125 (amount of water in new solution) - old amount of water 25 L
This is not giving the answer.

Jaychoudhary

Please see the revised solution.

All the other posted solutions get to the finish line by introducing a variable, setting up an algebraic equation, and solving for the variable. If that's ever what I want to do, I always see if PITA (Plugging In The Answers). In this case, we can probably knock it out by doing the math in our heads...with very little risk of making a silly mistake.

I like trying B and D. D looks easier to work with, so let's try D.
We start with 125 gallons, 20% of which is water. So we start with 125 gallons, 25 of which is water.
Add 8 gallons of water. Now we have 133 gallons total and 33 gallons of water.
33*2 = 132. So close! But not quite...

Okay, let's try B.
Add 8.5 gallons of water. Now we have 133.5 gallons total and 33.5 gallons of water.
33.5*2 = 134. So close! But not quite...

D was too low. B was too high. Only one answer is between them: C.

Answer choice C.


ThatDudeKnowsPITA

just consider water constant equation
let x litre of water is added

water initial + added =final
125* 20% + x = 25% ( 125+ x)

x = 25/3

ans C

20% of 125 = \(\frac{1}{5}\) * 125 = 25.
Therefore, 25 gallons of the 125 gallon mixture is water and hence the remaining 100 gallons is wine.

Water is being added to this mixture in such a way that the percentage of the water in the new mixture is 25%. Note that no addition or removal of wine is being carried out.

Let the volume of water to be added be x gallons. Then,

Total volume of the new mixture = (125 + x) gallons
Volume of water in the new mixture = (25 + x) gallons
Volume of wine in the new mixture = 100 gallons

Percentage of water in the new mixture = 25%

Therefore, \(\frac{(25 + x) }{ (125 + x)}\) * 100 = 25

Simplifying, \(\frac{(25 + x) }{ (125 + x)}\) = \(\frac{1 }{ 4}\)

Further simplification gives 100 + 4x = 125 + x

Solving for x, we have x = \(\frac{25 }{ 3}\), which is equivalent to 8 \(\frac{1}{3}\).
Therefore, 8 \(\frac{1}{3}\) gallons of water should be added to the original mixture.

The correct answer option is C.

Hi, ThatDudeKnows! I have a question about the method used, with option D, 8gal is added and 133gal is compared to what?
"33*2 = 132." what does this line mean? could you please elaborate here?

Thanks!