document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1113748>Question 1113748</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A box contains $7.15 in nickels, dimes, and quarters. There are 41 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #728837 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</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=728837\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Tutor ikleyn's solution is a perfectly good solution using formal algebra.<br><BR>\n" );
document.write( "Here is a solution that is basically the same, but using logical reasoning instead of the formal algebra.<br><BR>\n" );
document.write( "There are 41 coins in all; the number of nickels and dimes together is 3 less than the number of quarters.  So temporarily remove the 3 \"extra\" quarters, leaving a total of 38 coins, of which half are quarters.  That means 19 quarters, and a total of 19 nickels and dimes.  Then put back the 3 quarters; the number of quarters in the box is 19+3 = 22.<br><BR>\n" );
document.write( "The 22 quarters have a value of $5.50; so the total value of the 19 dimes and nickels is $7.15-$5.50 = $1.65.<br><BR>\n" );
document.write( "If all 19 of the remaining coins were dimes, the total value would be $1.90; but it is $1.65, which is 25 cents less than $1.90.  Replacing a dime with a nickel keeps the same total number of coins but reduces the total value by 5 cents.  Since all dimes gives us 25 cents more than we want, the number of times we need to replace a dime with a nickel is 25/5 = 5.<br><BR>\n" );
document.write( "So the number of nickels is 5, and the number if dimes is 19-5 = 14.<br><BR>\n" );
document.write( "Answer: 22 quarters, 14 dimes, 5 nickels.<br><BR>\n" );
document.write( "Check: 22(25)+14(10)+5(5) = 550+140+25 = 715 </SPAN>\n" );
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