document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=233465>Question 233465</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There were 35 nickels, dimes, and silver dollars in the pile, and their value was $12.25. How many coins of each kind were there if there were twice as many silver dollars as nickels? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #172455 by </B><A HREF=/tutors/aboutme.mpl?userid=edjones><B>edjones(8007)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=edjones><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=edjones><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=172455\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> divide 2.05 (2 dollars and a nickel) into 12.25.<BR>\n" );
document.write( "12.25/2.05=5.9...<BR>\n" );
document.write( "So, there can't be more than 5 nickels and since 12.25 ends in a five there can only be 1, 3 or 5 nickels.<BR>\n" );
document.write( "5 seems the most likely.<BR>\n" );
document.write( "5=.25<BR>\n" );
document.write( "5*$2=10<BR>\n" );
document.write( "12.25-20*.1=2<BR>\n" );
document.write( "=12.25 with 10 dollars, 20 dimes and 5 nickels.<BR>\n" );
document.write( "There is probably a more elegant way to solve this but I don't know what it is.<BR>\n" );
document.write( ".<BR>\n" );
document.write( "Ed </SPAN>\n" );
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