document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1127229>Question 1127229</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a box contains 77 coins, only dimes and nickels. The amount of money in the box is $5.25. How many dimes and how many nickles are in the box? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #743619 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13206)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=743619\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Here is a common sense way to solve problems like this, without the formal algebra.<br><BR>\n" );
document.write( "If all 77 coins were nickels, the total value would be 77*5 = 385 cents or $3.85.<BR>\n" );
document.write( "The actual total value is $5.25, which is $1.40 (140 cents) more than $3.85.<BR>\n" );
document.write( "Replacing 1 nickel with 1 dime raises the total value by 5 cents (while keeping the total number of coins the same).<BR>\n" );
document.write( "The number of dimes needed to replace nickels to make up the other 140 cents is 140/5 = 28.<br><BR>\n" );
document.write( "So there are 28 dimes and 77-28 = 49 nickels. </SPAN>\n" );
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