Let the number of nickels be N
let the number of dimes be D, so D = 2N
let the number of quarters be Q
but Q would then be 30-N-D
= 30 - N - 2N
= 30 - 3N
now form an equation using the "value" of those coins:
5(N) + 10(2N) + 25(30-3N) = 550
solve for N, then backsubstitute for the others.
Let me know what you got.
A collection of 30 coins worth $5.50 consists of nickels, dimes, and quarters. There are twice as many dimes as nickels. How many quarters are there?
4 answers
so i did...
5n+10(2n)+750-75n=550
5n+20n-75n=-200
-50n=-200
n=4
four nickels
eight dimes
12 quarters
.. that's not right! D:
5n+10(2n)+750-75n=550
5n+20n-75n=-200
-50n=-200
n=4
four nickels
eight dimes
12 quarters
.. that's not right! D:
4 nickels
8 dimes, and
30-4-8 = 18 quarters, (how did you get 12?)
check: 4(5) + 8(10 + 18(25) = 550
8 dimes, and
30-4-8 = 18 quarters, (how did you get 12?)
check: 4(5) + 8(10 + 18(25) = 550
ohh!
for some reason i multiplied 4 and three to get 12 quarters..
wow silly mistake!
thanks so much for the help!!
for some reason i multiplied 4 and three to get 12 quarters..
wow silly mistake!
thanks so much for the help!!