Question

A box holding pennies, nickels, and dimes contains thirteen coins with a total value of 83 cents. How many coins of each type are in the box? Show work using linear algebra.

a) 3 pennies, 5 nickels, 5 dimes
b) 4 pennies, 4 nickels, 5 dimes
c) 5 pennies, 4 nickels, 4 dimes
d) 5 pennies, 5 nickels, 3 dimes

Answer 1

To solve this problem, we can use linear algebra to set up a system of linear equations. Then, we can solve for the values of pennies, nickels, and dimes that satisfy both equations. The correct option is c) 5 pennies, 4 nickels, 4 dimes.

Step-by-step explanation:

Let's define the number of pennies as P, the number of nickels as N, and the number of dimes as D. We are given that there are thirteen coins in total, so we have the equation:

P + N + D = 13

We are also given that the total value of the coins is 83 cents, so we have the equation:

0.01P + 0.05N + 0.10D = 83

We can solve this system of linear equations using matrices. First, we can set up a matrix equation:

[1 1 1 | 13]
[0.01 0.05 0.10 | 83]

By performing row operations, we can solve for the values of P, N, and D.

After performing row operations, we find that the solution is P = 5, N = 4, and D = 4. Therefore, the correct option is c) 5 pennies, 4 nickels, 4 dimes.