SOLUTION: A jar with pennies, nickels, and dimes contains a total of 55 coins worth $3.23. How many of each are there if the number of pennies is 10 less than the sum of nickels and dimes to
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: A jar with pennies, nickels, and dimes contains a total of 55 coins worth $3.23. How many of each are there if the number of pennies is 10 less than the sum of nickels and dimes to
Log On
Question 181055: A jar with pennies, nickels, and dimes contains a total of 55 coins worth $3.23. How many of each are there if the number of pennies is 10 less than the sum of nickels and dimes together? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! YOU NEED TO RECHECK YOUR PROBLEM. UNLESS I MADE A MISTAKE, WE ARE COMING OUT WITH FRACTIONS OF COINS AND WE CAN'T HAVE FRACTIONS OF COINS.
Let x=number of pennies
y=number of nickels
z=number of dimes
Then:
x+y+z=55-----------------eq1
Lets deal in pennies
x+5y+10z=323------------------eq2
and
x+10=y+z or
x-y-z=-10 or
x-(y+z)=-10----------------------eq3
but from eq1, y+z=55-x. Substitute this into eq3
x-(55-x)=-10 or
x-55+x=-10 add 55 to each side
2x=45-----------------------------------------NO GOOD!!!!
LET'S TRY ANOTHER APPROACH:
subtract eq1 from eq2 and we get:
4y+9z=268-----------------------------------eq1a
subtract eq3 from eq2 and we get:
6y+11z=333------------------------------------eq2a
Multiply eq1a by 3:
12y+27z=804----------------------eq1b
Multiply eq2a by 2:
12y+22z=666--------------------------eq2b
subtract eq2b from eq1b:
5z=138-------------------------------------------NO GOOD!!!!!
Hope this helps---ptaylor
