In Ethan’s piggybank there are nickels, dimes and quarters. There are 3 times as many quarters as nickels, and 5 more dimes than nickels. How many of each kind of coins are there? Ethan has saved a total of $3.20.
Q = 3N and D = 5N
.05N + .10D + .25Q = .05N + .10(5N) + .25(3N) = 3.20
Calculate for N, then Q and D.
To solve this problem, we can set up a system of equations based on the given information.
Let's represent the number of nickels, dimes, and quarters as "n," "d," and "q" respectively.
From the information given in the problem, we can write the following equations:
1) There are 3 times as many quarters as nickels: q = 3n
2) There are 5 more dimes than nickels: d = n + 5
3) The total value of the coins is $3.20. In dollars, nickels are worth 0.05, dimes are worth 0.10, and quarters are worth 0.25. So, we can write the equation:
0.05n + 0.10d + 0.25q = 3.20
Now, we have a system of three equations with three unknowns. Let's solve it to find the values of n, d, and q.
1) Substitute the value of q from equation 1 into equation 3:
0.05n + 0.10d + 0.25(3n) = 3.20
0.05n + 0.10d + 0.75n = 3.20
0.80n + 0.10d = 3.20
2) Substitute the value of d from equation 2 into the modified equation 3:
0.80n + 0.10(n + 5) = 3.20
0.80n + 0.10n + 0.50 = 3.20
0.90n + 0.50 = 3.20
3) Subtract 0.50 from both sides of the equation:
0.90n = 2.70
4) Divide both sides of the equation by 0.90:
n = 3
5) Substitute the value of n back into equation 1 to find the value of q:
q = 3(3) = 9
6) Substitute the value of n into equation 2 to find the value of d:
d = 3 + 5 = 8
Therefore, Ethan has 3 nickels, 8 dimes, and 9 quarters in his piggybank.