# so embarrassing- math question.

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Alex dug out coins from the couches in his house. He was able to scavenge \$2.05 In Quarters and Nickels. Amazingly, there were 15 coins hidden in the folds of the cushions! How many quarters and nickels did he find?

Thanks!:tongue_smilie::blush:

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Alex dug out coins from the couches in his house. He was able to scavenge \$2.05 In Quarters and Nickels. Amazingly, there were 15 coins hidden in the folds of the cushions! How many quarters and nickels did he find?

Thanks!:tongue_smilie::blush:

Set this up as two equations in two unknowns.

Let A be the number of quarters, valued at .25 each.

Let B be the number of nickels, valued at .05 each.

A + B = 15

(because 15 is the total number of coins)

25A + 5B = 205

(because each A is worth 25 cents, each B is worth 5, and the total is 205 cents.)

Then I think you know what to do!

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(My embarrassing part was I had to take a moment to remember what nickels were. I them mixed up with dimes.)

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ok. That is what we did. But we can't get it to equal 205 :confused:

I get 25(7) + 5(8)= 175 + 40 =215

25 (6) + 5 (9) = 150 + 45= 195

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Yeah, I can't get it to equal 205 either.

If you have 8Q and 1N, you have 205, but not enough coins. If you exchange one quarter for 5 nickels, you still get 205 (7Q and 6N), but that's only 13 coins. Exchange another quarter (6Q and 11N) and you have too many coins. :confused:

Of course I wrote "communicate member" in an email to my pastor this morning instead of "communicant member" :svengo: so I might just be having a ditzy day.

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The problem has no solution.

a+b=15

25a+5b=205

solve first for b: b=15-a

put in 2nd:

25a+5(15-a)=205

25 a +75 -5a=205

20a = 130

has no integer solution for a, because 130:20=6.5, and you can not have half coins.

The \$2.05 must be a typo.

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Regentrude is right; I did it just that way and got exactly the same thing. That will teach me not to finish math problems! I thought the problem was the set up, but not this time.

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Well, technically it doesn't say that he ONLY found quarters and nickels. Maybe he found eight quarters, one nickel (that's the \$2.05), and six other random coins? :-)

Was the original question from a book?

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The answer is nine or 13.

There are only two ways that you can get \$2.05 out of quarters and nickels without going over 15 coins. Eight quarters and one nickle, or seven quarters and six nickles.

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Thank you! That is what we were thinking - so now I do not feel so dumb :001_smile:

Well, technically it doesn't say that he ONLY found quarters and nickels. Maybe he found eight quarters, one nickel (that's the \$2.05), and six other random coins? :-)

Was the original question from a book?

This is what my husband is thinking could be it also. It is a question from ds's math class @ co-op.

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Well, technically it doesn't say that he ONLY found quarters and nickels. Maybe he found eight quarters, one nickel (that's the \$2.05), and six other random coins? :-)

But then there would still be no unique solution, because he could also have found 7 quarters and six nickels and two random coins.

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But then there would still be no unique solution, because he could also have found 7 quarters and six nickels and two random coins.

I thought sometimes math problems were like that? More than one solution?

ETA- I don't like it though and am hoping it was a typo.

Edited by kwg

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