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Kamikaze_Badger

Math Question

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I never cared about math in high school and only did the bare essentials of introductory algebra and geometry. Now that I'm in college and every single penny counts, I have the option of passing an Algebra 1 level assessment to get on the nursing program wait list, or wait for Spring 2013 to take the next available prealgebra and algebra one classes. Obviously I've decided to get off my butt and learn some Algebra with an instructor's textbook from the school library and lots of caffeine. So far all has gone well, then I ran into this problem:

 

How many liters of a 60% acid solution would need to be mixed with a 75% acid solution to get 20L of a 72% acid solution?

 

All the other similar problems leading up to this one I haven't had trouble with, but this one and others like it where you're only given the amount of the end product and not of the initial ones, like "how many liters of x% solution need to be mixed into 5L of y% solution to get a z% solution?", are just kicking my butt. The book says the answer is "4 Liters," but doesn't explain the formula for such a problem in the text, and doesn't show any work. Everything I've tried has given me high or low numbers and often decimals, so I'm thinking I'm not setting up the formula properly or I'm not simplifying properly. Here's what I've pulled out so far:

 

Rate | Amount | Amt of Pure Acid
60% | x           | .60(x)
75% | (x-20)    | .75(x-20)
72% | 20         | .72*20 = 14.4 L

x(.60) + (x-20) = 14.4
+20 to both sides
x(.60) + x = 34.4
Divide by .6
2x = 57 1/3
Divide by 2
28 3/5

 

Where am I going wrong here? I'm about ready to pull my hair out.

Edited by Kamikaze_Badger

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well firstly the amount of 75% acid is 20-x, not x-20

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and the first line should be x(0.6)+(20-x)(0.75)=(20)(0.72)

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Adjusted accordingly. I'm sure now it's just me screwing up in simplifying the formula:

 

x(0.6) + (20-x)(.75) = 14.4

-20 Both sides

x(0.6) -x(.75) = -5.6

-x(.15) = -5.6

-x = 37 1/3

Rage

 

Where are my painfully obvious mistakes at here?

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I did this not too long ago, and if I remember correctly, it would go something like this:

 

x = 60%

y = 75%

 

x + y = 20

 

.60x + .75y = 20(.72) -------------------- 20 * .72 = 14.4

 

The first equation states that both solutions will equal to a combined 20 liters.

The second equation states that 60% solution plus the 75% solution should equal to 20 liters of the 72% solution.

 

You would then have to solve by elimination or by substitution.

 

If this does not make ANY sense to you, or you do not even know what I'm talking about, just disregard what I wrote; I do not want to confuse you if you already have a method that you are comfortable with.

 

I solved the problem on paper using elimination, and I got y = 16L of the 75% solution and x = 4L of the 60% solution. Is that the answer your book has?

 

Hope this helps!!!

Edited by saywhut

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Adjusted accordingly. I'm sure now it's just me screwing up in simplifying the formula:

 

x(0.6) + (20-x)(.75) = 14.4

-20 Both sides

x(0.6) -x(.75) = -5.6

-x(.15) = -5.6

-x = 37 1/3

Rage

 

Where are my painfully obvious mistakes at here?

 

because expanding the first line results in x(0.6)+20(.75)-x(.75)=14.4

which then goes to 0.6x-0.75x+15=14.4 (expanded more)

then -0.15x=-0.6 (cancelled)

then 0.15x=0.6 (times both sided by -1)

then x=4 (divided both by .15)

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Starting to make sense now. Thanks so much! I can finally sleep at night!

 

Just don't give up even if you get really frustrated and feel like crumbling your paper. If you ever get stuck again, go to www.khanacademy.org and they have videos of every type of math problems all the way up to calculus.

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Here's how I do such math fast in my head , though i would advise against it. 75 - 72 = 3% change in the final solution. 72-60 e 12% change in the final solution. 12% = 4 x 3% there fore you need 4 times more of solution 1 than you need of solution 2 to get 72% there fore x + 4x = 20 thus x = 4; 4 liters + 16 liters. I do most of my statistical math intuitively like this and even though the math is correct i still go 38/100 on my discrete math exam :lol:

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I'm gonna check out Kahn academy because I'm terrible at math lol.

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