...

IMO A,

presently - 30 gallons of mix - 20% water -> 6 gallons water - 24 gallon milk
If we add 90 gallon of milk in this mixture -

90+30 -> 120 gallon of mixture

water remains 6 gallons

6/120 * 100 -> 5%

So 90 gallons of milk needed to be added - So A.

what is OA?

Tricky question - I wouild rate it at 700 becuase it doesn't ask how much water you need to replace, nor does it ask how much milk you need to add, as the other explanations did. it asks how many gallons of the MIXTURE should be replaced.

So at 30 gallons - we have 6 g water and 24 g milk. Fo 5% water we need 1.5g water and 28.5g milk.

But for each gallon of the mixture we remove we take out .2gwater and .8g milk. and add in one gallon pure milk. I am sure there is some algebra to be done here but i found it easier to simiply use the answers.

First I tried 15 gal which would mean taking out 15(.2) or 3g water and 15(.8) or 12 gal Milk and adding 15g milk - but even without adding the milk we know this won't be enough water taken out because we are looking for 1.5 g water in the final.

So Then I realized you simply need to look at the water you take out and I multiplied my best guess at an answer (22.5) and .2 for the concentration of water to get (22.5)(.2)=4.5 if we take out 4.5 g of water we are left with 1.6g of water in the mix.

Therefore the answer is C.

ahhh...Thanks Becky ..that was a trap and i missed that !!

GoBlue wrote:30 gallons of milk and water contains 20 % water in it. How many gallons of the mixture should be replaced with pure milk to bring down the percentage of water in mixture to 5 %.

A. 90
B. 15
C. 22.5
D. 18
E. 30

Answer with explaination. Also, according to your experience rate it as 500 level, 600 level or 700 level.

Here's an efficient approach that requires very little arithmetic:

We currently have .2*30 = 6 gallons of water in the mixture.
We want the amount of water to be reduced to .05 * 30 = 1.5.
So we need to lose 6 - 1.5 = 4.5 gallons of water.

Which answer choice (the amount of pure milk to be added) will reduce the amount of water by 4.5 gallons?
Since the mixture is 20% water, 20% of each answer choice will go toward replacing the amount of water currently in the mixture.
So let's take 20% of each answer choice (the amount of pure milk to be added) to see which gives us 4.5 (the amount of water we need to lose):

Answer choice C:
.2 * 22.5 = 4.5 Success!

The correct answer is C.

Alternatively, if we look into the final percentage composition of milk, we see that milk should now be 95% of the solution (as water is only 5% of the final solution). So algebraically,

If we draw x litres of solution and replace it with x litres of milk, total milk before was 80% (30-x) and after is x litres (as I am adding x litres of milk) = 80%(30-x) + x.

So equation is ::

(80%(30-x) + x)
______________ * 100 = 95, solving for x results in 22.5 litres.

30

Alternatively, if we look into the final percentage composition of milk, we see that milk should now be 95% of the solution (as water is only 5% of the final solution). So algebraically,

If we draw x litres of solution and replace it with x litres of milk, total milk before was 80% (30-x) and after is x litres (as I am adding x litres of milk) = 80%(30-x) + x.

So equation is ::

(80%(30-x) + x)
______________ * 100 = 95, solving for x results in 22.5 litres.

30

Thank you all. Sorry, I forgot to post the answer.. its "C". Thank you all for the detailed explanation. I think this is a 700 + question and I had a real tough time with this one.

It is a pleasure to see so many different solutions for a problem. apporaches from all angles.