document.write( "Question 430647: Gina has a pile of 50 dimes and nickels worth $4.30. How many coins of each type does she have? \n" ); document.write( "

Algebra.Com's Answer #299041 by josmiceli(19441) $\"\"$ $\"About$

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She can't have an odd number of nickels, since
\n" ); document.write( "that would give her a 5 in the units position,
\n" ); document.write( "like $4.35.
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\n" ); document.write( "She can't have all dimes, since that would equal
\n" ); document.write( "$5.00, which is $.70 too much
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\n" ); document.write( "Suppose she had 40 dimes, making $4.00.
\n" ); document.write( "She needs $.30, which is 6 nickels.
\n" ); document.write( "46 coins isn't enough.
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\n" ); document.write( "If she has 30 dimes, that's $3.00, and
\n" ); document.write( "$1.30 must be made up with nickels.
\n" ); document.write( "That's 26 nickels, and 30 + 26 = 56,
\n" ); document.write( "which is too much
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\n" ); document.write( "36 dimes = $3.60, leaving $.70 woth of
\n" ); document.write( "nickels, or 14 nickels, and 14 + 36 = 50
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\n" ); document.write( "She has 36 dimes and and 14 nickels \n" ); document.write( "

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