View Full Version : Is there a way to figure this algebraically?

designermamajessica

06-24-2009, 06:36 PM

Jorge claims that he has a certain combination of U.S. coins and he cannot make change for a dollar, half dollar, quarter, dime, or nickel. What is he greatest amount of money he could have and what coins would they be? He does not have any dollar coins.

This just blows me away. I have no idea where to even begin.

Thank you for the help.

Loren

06-24-2009, 06:57 PM

Maybe there is a better way, but this is what I would do.

If he has 4 cents, he can't make change for a nickel.

If he has 4 cents and 1 nickel, he can't make change for a dime and still can't for a nickel.

If he has 4 cents, 1 nickel, and 2 dimes, he can't make change for a quarter and still can't for a nickel or dime.

etc.

designermamajessica

06-24-2009, 07:08 PM

Can you explain this further? I am looking at

"If he has 4 cents, 1 nickel, and 2 dimes, he can't make change for a quarter and still can't for a nickel or dime."

and seeing that you could make change for the quarter by using two dimes and a nickel. What am I doing wrong?

Loren

06-24-2009, 08:04 PM

Nothing. I goofed. I guess we would eliminate the nickel leaving 4 cents and 2 dimes won't make change for a quarter and we still can't make change for a dime or a nickel.

galactus

06-24-2009, 09:33 PM

It would be easy to miscount here. Get a bunch of change and try figuring it out.

If we have 2 quarters, then we have change for 50 cents.

2 nickels give change for a dime.

How about 1 quarter, 4 dimes, and 4 pennies.

That is 69 cents.

Does that work?. Did I miss something?.

Denis

06-25-2009, 12:17 AM

Here I come to save the day.....

4 pennies, 4 dimes, 1 quarter, 1 half dollar = 1.19

galactus

06-25-2009, 05:35 AM

I came up with 1.19, but with the wrong coins. Good, D

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