document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=426689>Question 426689</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A student had 100 coins totalling $7.00.<BR>\n" );
document.write( "He had no nickels.<BR>\n" );
document.write( "What coins did he have if 60% of the coins were of one kind? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #312096 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=312096\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assuming no half-dollars as well as no nickels, all the student has are pennies, dimes, and quarters. He can't have 60 quarters, because that's too much.  Similarly, dimes won't work out.  He must therefore have 60 pennies.<BR>\n" );
document.write( "$7.00-$.60=6.40<BR>\n" );
document.write( "x+y=40<BR>\n" );
document.write( "10x+25y=640<BR>\n" );
document.write( "10x+10y=400<BR>\n" );
document.write( "15y=240<BR>\n" );
document.write( "y=16<BR>\n" );
document.write( "x=24<BR>\n" );
document.write( "Thus, our student has 60 pennies, 24 dimes, and 16 quarters.. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );