document.write( "Question 1139825: A stack of $5 and $20 bills was counted by the treasurer of an organization. The total value of the money was $1710 and there were 47 more $5 bills than $20 bills. Find the number of each type of bill. \n" ); document.write( "

Algebra.Com's Answer #760318 by greenestamps(13206) $\"\"$ $\"About$

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\n" ); document.write( "Let x = number of $20 bills
\n" ); document.write( "Then x+47 = number of $5 bills

\n" ); document.write( "The total value of the bills was $1710:

\n" ); document.write( " $\"20%28x%29%2B5%28x%2B47%29+=+1710\"$

\n" ); document.write( "Solve using basic algebra....

\n" ); document.write( "You can also solve the problem without the formal algebra, using logical reasoning and some mental arithmetic.

\n" ); document.write( "(1) Count the \"extra\" 47 $5 bills and subtract that total from the $1710.
\n" ); document.write( "(2) The rest of the money is equal numbers of $20 and $5 bills; one $20 bill and one $5 bill have a total value of $25. So divide the remaining amount by $25 to see how many of each bill there are left.
\n" ); document.write( "(3) Then of course add back in those \"extra\" 47 $5 bills that you counted earlier.

\n" ); document.write( "With the informal solution method, you do virtually the same calculations as with the formal algebra. If your mental arithmetic is good, you can get to the answer faster because you aren't burdened with the written equations. \n" ); document.write( "

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