document.write( "Question 659025: you have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have \n" ); document.write( "

Algebra.Com's Answer #852487 by n2(19) $\"\"$ $\"About$

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\n" ); document.write( "You have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have ?
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document.write( "Let x be the number of pennies, \r\n" );
document.write( "    y be the number of nickels,\r\n" );
document.write( "    z be the number of dimes,\r\n" );
document.write( "    w be the number of quarters.\r\n" );
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document.write( "We have two equations\r\n" );
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document.write( "    x +  y +   z +   w =  25,   (1)    for the total number of coins\r\n" );
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document.write( "    x + 5y + 10z + 25w = 100    (2)    for the total value.\r\n" );
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document.write( "From equation (2), subtract equation (1).  You will get\r\n" );
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document.write( "        4y + 9z  + 24w = 75,\r\n" );
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document.write( "        4y +       24w = 75 - 9z.    (3)\r\n" );
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document.write( "In the last equation, left side value is always a multiple of 4.\r\n" );
document.write( "Hence, right side must be a multiple of 4.\r\n" );
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document.write( "z = 0  gives for the right side of (3)  the value  75 - 9*0 = 75, which is not a multiple of 4.\r\n" );
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document.write( "z = 1  gives for the right side of (3)  the value  75 - 9*1 = 66, which is not a multiple of 4.\r\n" );
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document.write( "z = 2  gives for the right side of (3)  the value  75 - 9*2 = 57, which is not a multiple of 4.\r\n" );
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document.write( "z = 3  gives for the right side of (3)  the value  75 - 9*3 = 48, which is a multiple of 4.\r\n" );
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document.write( "At z = 3,  from (3), we have\r\n" );
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document.write( "        4y +       24w = 48.    (3)\r\n" );
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document.write( "One possible solution for (3) is  (y,w) = (0,2).  Then   x + y + z + w = x + 0 + 3 + 2 = 25  implies  x = 20.\r\n" );
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document.write( "    So, one possible solution for the problem is  20 pennies, 0 nickels, 3 dimes and 2 quarters.\r\n" );
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document.write( "Second possible solution for (3) is  (y,w) = (6,1).  Then   x + y + z + w = x + 6 + 3 + 1 = 25  implies  x = 15.\r\n" );
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document.write( "    So, second possible solution for the problem is  15 pennies, 6 nickels, 3 dimes and 1 quarters.\r\n" );
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document.write( "Third possible solution for (3) is  (y,w) = (12,0).  Then   x + y + z + w = x + 12 + 3 + 0 = 25  implies  x = 10.\r\n" );
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document.write( "    So, third possible solution for the problem is  10 pennies, 12 nickels, 3 dimes and 0 quarters.\r\n" );
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document.write( "Thus the problem has three possible solutions:\r\n" );
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document.write( "    (1)  20 pennies,  0 nickels,  3 dimes and 2 quarters, \r\n" );
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document.write( "    (2)  15 pennies,  6 nickels,  3 dimes and 1 quarters, \r\n" );
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document.write( "    (3)  10 pennies, 12 nickels,  3 dimes and 0 quarters.\r\n" );
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document.write( "It is clear that there no other solutions.\r\n" );
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