SOLUTION: You have $24.05 with 210 coins in quarters, nickles, and dimes, such that twice the number of quarters is 13 more than the number of nickels. Find how many of each coin you have.
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: You have $24.05 with 210 coins in quarters, nickles, and dimes, such that twice the number of quarters is 13 more than the number of nickels. Find how many of each coin you have.
Log On
Question 36424: You have $24.05 with 210 coins in quarters, nickles, and dimes, such that twice the number of quarters is 13 more than the number of nickels. Find how many of each coin you have. Answer by Paul(988) (Show Source):
You can put this solution on YOUR website! Let the dimes be x
Let the nickels be y
Let the quarters be z
Equation 1 :
"twice the number of quarters is 13 more than the number of nickels"
2(z)=13+y
y=2z-13 (subsitution 1)
Equation 2:
x+y+z=210 (subsitute for y)
x+(2z-13)+z=210
x=210+13-2z-z
x=223-3z (subsitution 2)
Equation 3:
25z+5y+10x=2405
Subsitute for y:
25z+5(2z-13)+10x=2405
subsitute for x:
25z+10z-65+10(223-3z)=2405
35z-30z=2405-2230+65
5z=240
z=48
(subsitution 1)-- y=2(48)-13
y=83
(subsitution 2)-- x=223-3(48)
x=79
Hence, there are 48 quarters, 79 dimes and 83 nickels.
Paul.
(