document.write( "Question 648794: liviana has $ 2.20 in coins in her change purse. if there is a total of 15 coins how many quarters,dimes,nickles,and pennies does she have? \n" ); document.write( "
Algebra.Com's Answer #406825 by josmiceli(19441)![]() ![]() You can put this solution on YOUR website! There are 4 unknowns but only 2 possible \n" ); document.write( "equations, so this can't be solved directly, \n" ); document.write( "but it still can be figured out. \n" ); document.write( "Let \n" ); document.write( "Let \n" ); document.write( "Let \n" ); document.write( "Let \n" ); document.write( "The equations are: \n" ); document.write( "(1) \n" ); document.write( "(2) \n" ); document.write( "------------------------------------------ \n" ); document.write( "The logic steps are: \n" ); document.write( "(a) \n" ); document.write( "There doesn't have to be any pennies, \n" ); document.write( "although there might be, but they will be in \n" ); document.write( "multiples of 5 or 10 \n" ); document.write( "(b) \n" ); document.write( "I know there must be less than 9 quarters, since \n" ); document.write( "9 x 25 = 225, which is too much \n" ); document.write( "(c) \n" ); document.write( "I know there must be more than 2 quarters since \n" ); document.write( "2 x 25 = 50, that leaves 13 coins to make $1.70 \n" ); document.write( "and even if I have 13 dimes, that is only $1.30 \n" ); document.write( "(d) \n" ); document.write( "Likewise, there must be more that 3 quarters \n" ); document.write( "3 x 25 = 75 and 220 - 75 = 145, and \n" ); document.write( "12 x10 = 120 which doesn't make it \n" ); document.write( "(e) \n" ); document.write( "Also, there must be more than 4 quarters \n" ); document.write( "4 x 25 = 100 and 220 - 100 = 120 \n" ); document.write( "There are 11 coins left to make $1.20 \n" ); document.write( "and 11 x 10 = 110, too small \n" ); document.write( "(f) \n" ); document.write( "It is possible with 5 quarters since \n" ); document.write( "5 x 25 = 125 and 220 - 125 = 95 \n" ); document.write( "and 10 x 10 = 100, so it's possible \n" ); document.write( "-------------------------------- \n" ); document.write( "5q = 125 which is 5 coins leaving 10 \n" ); document.write( "220 - 125 = 95 \n" ); document.write( "9d = 90 leaving 1 coin \n" ); document.write( "1 x 5 = 5 \n" ); document.write( "So, the answer is \n" ); document.write( "5q \n" ); document.write( "9d \n" ); document.write( "1n \n" ); document.write( "But if I have to use pennies it doesn't work \n" ); document.write( "-------------------------- \n" ); document.write( "I'll try \n" ); document.write( "6q \n" ); document.write( "5d \n" ); document.write( "4n \n" ); document.write( "This works, but still no pennies \n" ); document.write( "--------------------------- \n" ); document.write( "I'll try \n" ); document.write( "7q \n" ); document.write( "1d \n" ); document.write( "This leaves 7 coins to make 35 cents \n" ); document.write( "and no room for pennies \n" ); document.write( "--------------------- \n" ); document.write( "8q \n" ); document.write( "1d \n" ); document.write( "Leaves 6 coins to make 10 cents \n" ); document.write( "that's \n" ); document.write( "1n \n" ); document.write( "5p \n" ); document.write( "That's it \n" ); document.write( "------------ \n" ); document.write( "8 quarters, 1 dime, 1 nickel, and 5 pennies \n" ); document.write( "are 15 coins that make $2.20\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |