<PRE><B>Question 1117666</B>
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Here are a couple of ways to work this kind of problem that, if you can do some mental arithmetic, get you to the answer much faster than the traditional algebraic approach shown by the other tutor.&lt;br&gt;
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Of course, if for some reason an algebraic approach is required, the solution shown by the other tutor is fine.&lt;br&gt;
You can solve the problem using a little logical reasoning.&lt;br&gt;
Suppose all 100 coins were nickels; the total value would be only $5 (500 cents) instead of $8 (800 cents).  The total would be 300 cents short of the actual total.&lt;br&gt;
Exchanging one of the nickels for a dime keeps the number of coins at 100 but increases the total value by 5 cents.&lt;br&gt;
To make up the 300 cents that you were short using all nickels, the number of nickels you have to exchange for dimes is 300/5 = 60.&lt;br&gt;
So there are 60 dimes, leaving 40 nickels.&lt;br&gt;
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And here is the method I use on problems like this, because it is even faster (for me...) than any other method.&lt;br&gt;
The total value using all nickels would be $5; the total using all dimes would be $10.
The actual total, $8, is 3/5 of the way from $5 to $10.
So 3/5 of the coins have to be dimes.
3/5 of 100 is 60; so there are 60 dimes and 40 nickels.</PRE>