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.
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document.write( " How many of each of Samy's pieces does he have? \n" );
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Algebra.Com's Answer #798740 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \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( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \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. \n" ); document.write( " \n" ); document.write( " |