A coin bank contains nickels, dimes, and quarters totaling $5.45. If there ae twice as many quarters as dimes and 11 more nickels than quarters, how many of each coin are in the bank?
My set up:
x = number of dimes
2x = number of quarters
2x + 11 = number of nickels
Is this correct so far?
Yes, it's correct. Now you only need to multiply the quantities with the values for Dimes, quarters, and nickels.
Originally Posted by magentarita
(0,10)(x) + (0,25)(2x) + (0,05)(2x + 11) = 5,45
Now solve for x.
I find x = 7, and then you simply have to plug the values in.