Question 1095347
<br>You can solve this kind of problem informally, with logical analysis.  The actual calculations you will need to do are nearly identical to what you do in the formal algebraic solution, as provided by another tutor.<br>
The method I use is first to suppose what the total amount would have been if all 100 coins had been dimes.  That's easy: $10.00.<br>
But the actual total was only $7.85; simple arithmetic shows that is $2.15 short of $10.00.<br>
But the difference between the value of a dime and a nickel is 5 cents.  So the number of nickels there must be the $2.15, divided by 5 cents:
{{{2.15/0.05 = 215/5 = 43}}}<br>
So there are 43 nickels, leaving 57 for the number of dimes.