A Paradox: solving "solution" exercise gives impossible ans.

John Whitaker

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Text: Write an equation based on the following problem, solve the equation, and explain why the problem has no solution:

How mucn 30% acid should be mixed with 15 L of 50% acid to obtain a mixture that is 60% acid?

I know the described procedure can only dilute the mixture, but I tried a conventional approach to come up with an equation: .30x + .50(15) = .60(x+15). Working this results in a "5 L" answer. But because this is not possible, I'm not sure how to explain "why the problem has no solution" when it appears (5 L) that it does.

Thank you.
John Whitaker
 
Re: A Paradox

Check your arithmetic. The answer to your equation is \(\displaystyle x = -5L\).
 
Re: A Paradox

royhass Thank you. I got the -5L, but adding a negative (amount of liquid) is not possible. Can that be the answer I was seeking: "It is not possible to 'add' a negative amount of liquid?"
John Whitaker
 
Re: A Paradox

John Whitaker said:
royhass Thank you. I got the -5L, but adding a negative (amount of liquid) is not possible. Can that be the answer I was seeking: "It is not possible to 'add' a negative amount of liquid?"....yes
John Whitaker
 
John Whitaker said:
How mucn 30% acid should be mixed with 15 L of 50% acid to obtain a mixture that is 60% acid?
You have a thirty-percent mix and a fifty-percent mix. You need to get a sixty-percent mix. But even if you used ONLY the fifty-percent mix, you would never get anything more than a fifty-percent mix. If you dilute with a lesser concentration, you can only end up with exactly that: a lesser concentration. There is no way to add less of the acid and somehow arrive at more of it. :oops:

It would be like asking "How much white ting should be added to a pink paint in order to get a red mix?" You simply can't go in the desired direction with the given inputs; it's physically impossible. :shock:

Eliz.
 
Hey, how ya doin' John? Haven't heard from you since Dec 16th :shock:
 
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