There are a total of 198 coins in a bag, all of which are nickels, dimes, and quarters.
The number of nickels is four times the number of quarters.
There are twice as many quarters as there are dimes.
If the total amount of money is $18.00, how many nickels, dimes, and quarters are in the bag?
There are nickels, dimes, and quarters.
I am not sure how to do this, can someone please explain?
sure - you will note that if you use my equations, you will get
q = 2d and n = 8d
x,y,z will also work, of course.
To solve this problem, we need to set up a system of equations based on the given information and then solve for the unknowns.
Let's assign variables to represent the number of nickels, dimes, and quarters.
Let's say the number of nickels is 'n', the number of dimes is 'd', and the number of quarters is 'q'.
Based on the given information, we can write the following equations:
1. The number of nickels is four times the number of quarters: n = 4q
2. There are twice as many quarters as there are dimes: q = 2d
3. The total number of coins is 198: n + d + q = 198
4. The total value of the coins is $18.00: 0.05n + 0.10d + 0.25q = 18.00
Now, we have a system of equations with four unknowns (n, d, q) and four equations. We can solve this system using substitution or elimination method.
Substitution method:
1. Rewrite equation 1 in terms of 'd': n = 4q
Substitute this into equation 2: 4q = 2d → d = 2q/2 → d = q
2. Substitute the values of d and n from the above steps into equation 3:
n + d + q = 198
(4q) + (2q) + q = 198
7q = 198
q = 198/7
q = 28
3. Substitute the value of q back into equations 1 and 2 to find n and d:
n = 4q = 4(28) = 112
d = q = 28
So, there are 112 nickels, 28 dimes, and 28 quarters in the bag.
To check if these values are correct, we can substitute them into equation 4 and see if the total value IS indeed 18.00:
0.05n + 0.10d + 0.25q = 18.00
0.05(112) + 0.10(28) + 0.25(28) = 5.60 + 2.80 + 7.00 = 18.00
The total value is indeed $18.00, thus confirming our solution.
just write the data in math
n+d+q = 198
n = 4q
q = 2d
That is enough to find how many of each coin there are.
The amount doesn't really matter, but its equation is
5n+10d+25q = 1800
hey oobleck, thanks for trying but, the correct way to solve it was different than what you advised... the correct way was..
let x=#of dimes; let 2x=#of quarters; and let 8x=#of nickels. That makes x+2x+8x=198 which then x=18. Therefore there was 18 dimes; 36 quarters; and 144 nickels. thanks for trying.