document.write( "Question 1099859: A box contains $7.10 in nickels, dimes, and quarters. There are 42 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there? \n" ); document.write( "

Algebra.Com's Answer #714340 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Let Q be the number of quarters.\r\n" );
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document.write( "Then the number of nickels and dimes is (42-Q), and the condition says that \r\n" );
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document.write( "this number (42-Q) is 2 less than Q.\r\n" );
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document.write( "It gives you an equation \r\n" );
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document.write( "42 - Q = Q - 2,  which implies\r\n" );
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document.write( "2Q = 42 - 2 = 40  and  then  Q =  = 20.\r\n" );
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document.write( "So, we have 20 quarters that sum up to 20*25 = 5 dollars.\r\n" );
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document.write( "Taking these 20 quarters away, we have 42-20 = 22 coins that are nickels and dimes and worth 710 - 500 = 210 cents.\r\n" );
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document.write( "If you think one minute on it, you will find the solution MENTALLY: 20 dimes and 2 nickels.\r\n" );
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document.write( "Answer.  20 quarters, 20 dimes and 2 nickels.\r\n" );
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