Question 1092135
<br>Here is a solution that combines informal algebra with logical analysis... instead of using purely formal algebra.<br>
Suppose all 200 coins were quarters.  The total value would be $50.<br>
But the total is only $41.  That's $9 short of the total we would get with all quarters.<br>
So let's start taking away quarters and replacing them with dimes. Each time we do that, the total value of the coins is reduced by 15 cents -- because the quarter was worth 25 cents and we replaced it with a dime worth only 10 cents.<br>
But if we need to get the $50 down to $41, we need to reduce the value of the coins by $9, or 900 cents.  If each time we replace a quarter with a dime we reduce the total value by 15 cents, then the number of quarters we need to replace is 900/15 = 60.<br>
And when we have replaced 60 quarters with dimes, we find we have 140 quarters and 60 dimes.