document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=952187>Question 952187</A>: <I><SPAN STYLE=\"word-break: break-all;\"> My piggy bank contains 29 coins, each of which is either a penny, nickel, dime, or quarter. The total value of all the coins is $2.86, and the number of quarters is equal to the sum of the nickels and dimes, and is less than the number of pennies. How many nickels 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 #581568 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=581568\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> p = no.of pennies<BR>\n" );
document.write( "n = no. of nickles<BR>\n" );
document.write( "d = no. of dimes<BR>\n" );
document.write( "q = no. of quarters<BR>\n" );
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document.write( "write an equation for each statement<BR>\n" );
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document.write( " My piggy bank contains 29 coins, each of which is either a penny, nickel, dime, or quarter.<BR>\n" );
document.write( "p + n + d + q = 29<BR>\n" );
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document.write( "The total value of all the coins is $2.86,<BR>\n" );
document.write( ".01p + .05n + .10d + .25q = 2.86 <BR>\n" );
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document.write( " and the number of quarters is equal to the sum of the nickels and dimes, <BR>\n" );
document.write( "q = n + d<BR>\n" );
document.write( "d = q - n<BR>\n" );
document.write( "n = q - d <BR>\n" );
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document.write( "and is less than the number of pennies.<BR>\n" );
document.write( "q < p<BR>\n" );
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document.write( "In the 1st equation replace n + d with q and you have<BR>\n" );
document.write( "p + 2q = 29<BR>\n" );
document.write( "p = -2q + 29<BR>\n" );
document.write( "From the above equation and the total$ equation we know the number of pennies has to be 21 or 11 pennies which would make it 4 or 9 quarters<BR>\n" );
document.write( ".01p + .05n + .10d + .25q = 2.86<BR>\n" );
document.write( "lets say 9 quarters and 11 pennies<BR>\n" );
document.write( ".11 + .05n + .10d + .25(9) = 2.86<BR>\n" );
document.write( ".11 + .05n + .10d + 2.25 = 2.86<BR>\n" );
document.write( ".05n + .10d + 2.36 = 2.86<BR>\n" );
document.write( ".05n + .10d = 2.86 - 2.36<BR>\n" );
document.write( ".05n + .10d = .50<BR>\n" );
document.write( "simplify, divide by .05<BR>\n" );
document.write( "n + 2d = 10<BR>\n" );
document.write( "we know the d = q - n, d = 4 - n<BR>\n" );
document.write( "n + 2(9-n) = 10<BR>\n" );
document.write( "n + 18 - 2n = 10<BR>\n" );
document.write( "-n = 10 - 18<BR>\n" );
document.write( "-n = -8<BR>\n" );
document.write( "n = 8 nickels<BR>\n" );
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document.write( "Check this find the dimes<BR>\n" );
document.write( "8 + 2d = 10<BR>\n" );
document.write( "2d = 2<BR>\n" );
document.write( "d = 1 dime<BR>\n" );
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document.write( "11 + 8 + 1 + 9 = 29<BR>\n" );
document.write( "and<BR>\n" );
document.write( ".01(11) + .05(8) + .10(1) + .25(9) = 2.86<BR>\n" );
document.write( ".11 + .40 + .10 + 2.25 = 2.86\r<BR>\n" );
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document.write( "--------------------Response to Comment-----------------------<BR>\n" );
document.write( "Tried sending this to the given email address, came back as \"Milbox not available\"<BR>\n" );
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document.write( "Yes, well to start with, we have 4 unknowns and only 3 meaningful equations, which means we will have to apply some logical assumptions<BR>\n" );
document.write( "Assuming you followed me Ok up to the equation: <BR>\n" );
document.write( "p = –2q + 29<BR>\n" );
document.write( "Since we have a total of 2.86, and the total of the rest of the coins have to be a multiple of 5, the number of pennies have to end in a 1 or a 6<BR>\n" );
document.write( "and looking at the above equation, subtract 2q from 29 will give you an odd number of pennies less than 29 of course, leaves us with a choice of 11 or 21 pennies<BR>\n" );
document.write( "This narrows it down to a manageable, try it and see” situation<BR>\n" );
document.write( "We can find the number of quarters with either p=11 or p=21<BR>\n" );
document.write( "-2q + 29 = 11<BR>\n" );
document.write( "-2q = 11 – 29<BR>\n" );
document.write( "-2q = –18<BR>\n" );
document.write( "q = –18/-2<BR>\n" );
document.write( "q = 9 quarters<BR>\n" );
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document.write( "The other alternative, when p = 21<BR>\n" );
document.write( "-2q + 29 = 21<BR>\n" );
document.write( "-2q = 21 – 29<BR>\n" );
document.write( "q = –8/-2<BR>\n" );
document.write( "q = 4 quarters<BR>\n" );
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document.write( "Hence the statement,<BR>\n" );
document.write( "”From the above equation and the total$ equation we know the number of pennies has to be 21 or 11 pennies which would make it 4 or 9 quarters”<BR>\n" );
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document.write( "Hopefully it made made sense to you after that. If not, email me your concerns about this,  email me at ankor@att.net<BR>\n" );
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