document.write( "Question 1136557: A cashier box has 3 types of coins (nickels, dimes and quarters). There are a total of 88 coins with
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Algebra.Com's Answer #754349 by ikleyn(52782)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let N be the number of nickels.\r\n" );
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document.write( "Then the number of quarters is 2N, according to the condition,\r\n" );
document.write( "and the number of dimes is the rest (88 - N - 2N) = 88-3N.\r\n" );
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document.write( "The \"money\" equation then is\r\n" );
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document.write( "    5N + 10*(88-3N) + 25*(2N) = 1455 cents.\r\n" );
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document.write( "Simplify and solve for N\r\n" );
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document.write( "    5N + 880 - 30N + 50N = 1455\r\n" );
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document.write( "    25N = 1455 - 880\r\n" );
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document.write( "    25N = 575  =============>  N = \"575%2F25\" = 23.\r\n" );
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document.write( "ANSWER.  23 nickels,  2*23 = 46 quarters and  88-23-46 = 19 dimes.\r\n" );
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document.write( "CHECK.   23*5 + 46*25 + 19*10 = 1455 cents.    ! Correct !\r\n" );
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document.write( "    This problem is to be solved by reduction to one single equation in one unknown.\r\n" );
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