Help on another ques. plz =]??

Question

How many quarts of water must be added to 5qt of an 80% antifreeze solution to make a 50% antifreeze solution?

......Answer is 3qt of water..... how do I get that answer?

Answer 1

The original solution has 4 qts of antifreeze and 1 qt of water

For 4qts of antifreeze to be 50% of the solution, you need to add 3qts of water to make the solution a total of 8 quarts

Here's a chart if it helps any

................Antifreeze......Water.... of anti.....total

Old..............4qt..............qt.....

New............4qt..............qt+x.....

4qt = qt+x

3qt = x

Answer 2

If we have 5qt of 80% antifreeze, then the amount of raw antifreeze we have will be:

5 * 80% = 4qts

and of course, the other 1qt will be water. If we don't add any more antifreeze, that same 4qts will still be the only antifreeze in the solution, we just need to put water in so that it matches the amount of antifreeze (to make it 50%). So, obviously, we need 3qts of water.

Answer 3

0.8 X 5 = 4.0 gallons - that is how much antifreeze is in the original solution.  to make a 50% solution, you want equal amounts of water and antifreeze- so the total amount is 8 gallons - you are starting with 5 gallons of solution so 8-5=3 gallons of water needed to make it 50%...

Answer 4

The first thing to think about is volume of water.  There is one variable.  Let W = the amount added.  (Note:)  There will be a total of W + 5 quarts at the end of the mixing process..

Next we have to think about the volume of antifreeze.  For example, the current solution has 80% of 5 quarts and that is .8 x 5 = 4

So:

0W + (.8 x 5) = (W + 5) x 50%  (which is .5, just as 80% = .8)

0 + 4 = (.5 x W) + (.5 x 5)

4 = (W/2) + 2.5

W/2 = 4 - 2.5

W/2 = 1.5

Multiply both sides of the equation by 2.

W = 3   <<<<