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anonymous
 5 years ago
How many ounces of a 22% alcohol solution must be mixed with 23 ounces of a 5% solution to produce a 10% alcohol solution? Round to the nearest tenth of an ounce. Can somebody explain how to do this problem?
anonymous
 5 years ago
How many ounces of a 22% alcohol solution must be mixed with 23 ounces of a 5% solution to produce a 10% alcohol solution? Round to the nearest tenth of an ounce. Can somebody explain how to do this problem?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i can do it; but splaining it might be a challenge :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1a + 23 = total amount of liquid a(.22) +23(.05) = (a+23)(.10) .22a + 1.15 = .10a + 2.3 .12a = 2.30  1.15 = 1.15 a = 1.15/.12 = 115/12 = 9.5833....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That got the right answer...and your equation does make sense to me...I got the stuff on the left side of the equal sign right, but had messed up on the right side of the equal sign. I had .10x on the right side. Can you explain why this was wrong, and I needed your a+23 times .1Ox? :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0times just .10, that is

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1a+23 is the total amont of the , im assuming liguid, that the system holds. so (total amount) * (.10) gives us the volume of solution and % content

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah. I guess I just had the % content on the right side, and did not have the amount times .1 to give me the volume...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the amount of 23 was stated inte problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know....I hope I can remember your equations for these type of problems, for my math test tomorrow. :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the are all of a basic form: amount(%) of one item + amount(%) of another item = total amount(%) of ending item

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm copying this now and saving it...:)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if amount is money, we get interest

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.120(.05) + 30(.10) = 50(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm actually good at simple interest problems...:)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1then consider these simple interests problems :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1$a(.22) + $23(.05) = $(a+23)(.10)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Uh...the problem of relating simple interest to mixture problems for me...is that I actually have a little diagram circle that I write and label to help me with simple interest problems...your above sentence is kinda confusing...without a story problem to go with it, that is...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1story problem then: if $23 is invested at 5%, how much money needs to be invested at 22% to get as much interest at a single account at 10%

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That makes your equation make more sense to me..:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But now the a+23 does not seem like it would equal a 'single account'...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe I'm thinking too hard...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess I'm stuck on, just doing the .1 times something, would equal a 'single account' or the 'final solution'...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that tends to be my train of thought right now, for some reason..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Bu t thanks anyway. :)
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