document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1118429>Question 1118429</A>: <I><SPAN STYLE=\"word-break: break-all;\"> From his paper route, Brody collected $5.55 in nickels and dimes. The number of nickels was 6 more than the number of dimes. How many nickels were there? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #733765 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</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=733765\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Without algebra, using logical analysis....<br><BR>\n" );
document.write( "(1) Take away the \"extra\" 6 nickels, so that the remaining coins are equal numbers of nickels and dimes.  The amount left is $5.55 - 6($.05) = $5.55-$.30 = $5.25.<BR>\n" );
document.write( "(2) Pair the remaining coins up into groups of 1 nickel and 1 dime.  Each of those groups has a value of $0.15.<BR>\n" );
document.write( "(3) The number of those groups required to make the total of $5.25 is $5.25/$0.15 = 35.  So there are 35 nickels and 35 dimes.<BR>\n" );
document.write( "(4) Bring back the \"extra\" 6 nickels, giving you the final answer of 41 nickels nd 35 dimes.<br><BR>\n" );
document.write( "Of course, if an algebraic solution is required, then the solution from the other tutor is fine. </SPAN>\n" );
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