Question 1203240
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The problem is more easily solved if you do a bit of logical reasoning and mental arithmetic before plunging into solving the problem with a system of 3 equations.<br>
The total number of bills is 71, and the number of $20 bills is 7 less than the total number of $5 and $10 bills.  A bit of mental arithmetic (or formal algebra, if you want) tells us that the number of $20 bills is 32 and the total number of $5 and $10 bills is 39.<br>
So now we have 32($20) = $640 in $20 bills and $925-$640 = $285 in $5 and $10 bills.<br>
For me, the easiest way to finish solving the problem is to first consider the case where there are 39 $5 bills, then see how many of those need to be replaced with $10 bills to make the required total of $285.<br>
39 $5 bills make a total of $195
The actual total is $285, which is $90 more than $195
Each time we replace a $5 bill with a $10 bill, the total increases by $5.  So the number of times we need to do that is $90/$5 = 18.  So the number of $10 bills is 18, leaving 39-18 = 21 $5 bills.<br>
ANSWER: 32 $20 bills, 18 $10 bills, and 21 $5 bills<br>
CHECK: 32($20)+18($10)+21($5)= $640+$180+$105 = $925<br>