Question 1080827: Zachary had nickels, dimes, and quarters. He has 58 coin in total, value is 6.50. There are twice as many dimes as nickels. How many of each does he have.
Answer by ikleyn(52797)   (Show Source):
You can put this solution on YOUR website! .
Let N be the number of nickels and Q be the number of quarters.
Then the number of dimes is 2N.
Nickels contribute 5N cents to the total.
Dimes contribute 10*(2N) = 20N cents to the total.
Quarter contribute 25Q cents.
Hence, the "value" equation is
5N + 20N + 25Q = 650 cents, or
25N + 25Q = 650, or
N + Q = 26,
The "coins" equation is
N + 2N + 5Q = 58, or
3N + 5Q = 58.
So, you have this system
N + Q = 26, (1)
3N + 5Q = 58. (2)
The setup is done.
Solve the system by any method you know (substitution, elimination).
|
|
|
