<PRE><B>Question 997824</B>
.
You can solve this problem even without using equations.


Let us suppose for a moment that all &amp;nbsp;50 &amp;nbsp;coins are &amp;nbsp;10-cent. 

Then their value is &amp;nbsp;50*10 = $5.00. &amp;nbsp;It is less than &amp;nbsp;$5.30 in &amp;nbsp $30 cents. 

It is clear that the difference is due to presence of &amp;nbsp;25-cent coins that we intently counted as &amp;nbsp;10-cent coins.

It is also clear that the number of these &amp;nbsp;25-cent coins is &amp;nbsp;{{{30/(25-10)}}} = {{{30/15}}} = 2&amp;nbsp; to compensate the difference. 

So, &amp;nbsp;the answer is: &amp;nbsp;there are &amp;nbsp;2 &amp;nbsp;of &amp;nbsp;25-cent coins and &amp;nbsp;50-2 = 48 &amp;nbsp;of &amp;nbsp;10-cent coins.


The solution is completed.


If you want to see more problems solved in this way, &amp;nbsp;look into these lessons 

&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;- &lt;A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-two-wheels-and-three-wheels-bicycles-in-a-sale.lesson&gt;Problem on two-wheel and three-wheel bicycles&lt;/A&gt;, 

&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;- &lt;A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-animals-at-a-farm.lesson&gt;Problem on animals at a farm&lt;/A&gt; &amp;nbsp;and 

&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;- &lt;A HREF=http://www.algebra.com/algebra/homework/word/misc/Problem-on-tablets-in-containers.lesson&gt;Problem on pills in containers&lt;/A&gt; 

in this site.


See also the lesson &amp;nbsp;&lt;A HREF=http://www.algebra.com/algebra/homework/word/coins/Solving-coin-problem-without-equations.lesson&gt;Solving coin problems without using equations&lt;/A&gt;. 


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