document.write( "Question 1173505: Samy counts his change. He has twice as many 10 ¢ coins as he has 5 ¢ and 6 25 ¢ coins more than 10 ¢. He also has 4 $ 1 coins and 7 $ 2 coins. In all, he has $ 29.25.
\n" ); document.write( " How many of each of Samy's pieces does he have?
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Algebra.Com's Answer #798740 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "let x be the number of 5 cent coins
\n" ); document.write( "then 2x is the number of 10 cent coins (twice as many as 5 cent)
\n" ); document.write( "then 2x+6 is the number of 25 cent coins (6 more than the number of 10 cent)

\n" ); document.write( "The total value of the coins, including 4 $1 coins and 7 $2 coins, is $29.25:

\n" ); document.write( "\"5%28x%29%2B10%282x%29%2B25%282x%2B6%29%2B4%28100%29%2B7%28200%29+=+2925\"

\n" ); document.write( "\"5x%2B20x%2B50x%2B150%2B400%2B1400+=+2925\"
\n" ); document.write( "\"75x%2B1950+=+2925\"
\n" ); document.write( "\"75x+=+975\"
\n" ); document.write( "\"x+=+13\"
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\n" ); document.write( "ANSWERS:
\n" ); document.write( "5 cent coins: x = 13
\n" ); document.write( "10 cent coins: 2x = 26
\n" ); document.write( "25 cent coins: 2x+6 = 32

\n" ); document.write( "CHECK: 13(5)+26(10)+32(25)+4(100)+7(200) = 65+260+800+400+1400 = 2925

\n" ); document.write( "A solution using logical reasoning can follow nearly the same path of calculations.

\n" ); document.write( "(1) Count the 4 $1 coins and the 7 $2 coins first. That is $18, leaving $11.25.
\n" ); document.write( "(2) Count the \"extra\" 6 quarters next; that is $1.50, leaving $9.75.
\n" ); document.write( "(3) What is left is x 5 cent coins, 2x 10 cent coins, and 2x quarters. Group those coins into groups of 1 5 cent coin, 2 10 cent coins, and 2 25 cent coins.
\n" ); document.write( "(4) The value of each of those groups is 5+20+50 = 75 cents. The number of groups, at 75 cents each, needed to make the remaining $9.75 is 975/75 = 13.

\n" ); document.write( "So there are 13 5 cent coins, 2*13=26 10 cent coins, and 2*13+6=32 25 cent coins.

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