Question 1041472
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robert has 50 coins, all in nickels and dimes, amounting to $3.50. how many nickels does he have?
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Similar problems were solved so many times, that I'd like to offer you a non-traditional solution.


<pre>
Let us suppose for a moment that all &nbsp;50 &nbsp;coins are &nbsp;dimes. 

Then their value is &nbsp;50*10 = 500 cents = $5. &nbsp;It is more than &nbsp;$3.50 exactly in &nbsp $1.50 = 150 cents. 

It is clear that the difference is due to presence of &nbsp;5-cent nickels that we intently counted as 10-cent dimes.

It is also clear that the number of these &nbsp;5-cent coins is &nbsp;{{{150/(10-5)}}} = {{{150/5}}} = 30&nbsp; to compensate the difference. 

So, the answer is: &nbsp;there are &nbsp;30 &nbsp;of &nbsp;5-cent nickels and &nbsp;50-30 = 20 &nbsp;of &nbsp;10-cent dimes.

<U>Answer</U>. There are 30 nickels and 20 dimes.
</pre>

For more traditional solutions see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/Coin-problems.lesson>Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson>More Coin problems</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Kevin-and-Randy-Muise-have-a-jar.lesson>Kevin and Randy Muise have a jar containing coins</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/coins/Typical-coin-problems-from-the-archive.lesson>Typical coin problems from the archive</A>

in this site.