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 nickels
8 dimes, and
30-4-8 = 18 quarters, (how did you get 12?)
check: 4(5) + 8(10 + 18(25) = 550
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:
ohh!
for some reason i multiplied 4 and three to get 12 quarters..
wow silly mistake!
thanks so much for the help!!
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the number of nickels as x.
Then the number of dimes will be 2x, as it is twice the number of nickels.
The number of quarters can be calculated using the total number of coins, which is 30, and subtracting the number of nickels and dimes, i.e., 30 - x - 2x = 30 - 3x.
The value of the nickels will be 5 cents, dimes will be 10 cents, and quarters will be 25 cents.
Now, based on the given information, we can create an equation for the value of the coins:
5x + 10(2x) + 25(30 - 3x) = 550
Simplifying the equation, we have:
5x + 20x + 750 - 75x = 550
-50x + 750 = 550
-50x = -200
x = 4
So, there are 4 nickels, 8 dimes (twice the number of nickels), and the number of quarters is 30 - 4 - 8 = 18.
Therefore, there are 18 quarters in the collection.
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.