Solve the problem using matrices.
John has a jarful of quarters and nickels. There are 84 coins in the jar. The value of the coins is $13.00. How many of each type of coin are there?
A) 49 quarters ; 35 nickels
B) 44 quarters; 40 nickels
C) 79 quarters; 5 nickels
D) 40 quarters; 44 nickels
It can't be A because that's $12.25 in quarters.
Check out the others to see which fits.
The answer is B.
44 quarters+40 nickels=84 coins
(0.25*44)=$11
(0.05*40)=$2
11+2=$13
Tony has a collection of coins valued at $5.45. In this collection there are only quarters and nickels. If he has 79 Nichols in this collection create any equation to find how many quarters, Q, he has
To solve this problem using matrices, we can set up a system of equations. Let's denote the number of quarters as "q" and the number of nickels as "n".
From the given information, we can create two equations:
1. The total number of coins equation: q + n = 84
2. The total value equation: 0.25q + 0.05n = 13.00
To represent these equations in matrix form, we can create a coefficient matrix and a constant matrix:
Coefficient matrix:
| 1 1 |
| 0.25 0.05 |
Constant matrix:
| 84 |
| 13 |
Now, we can represent the system of equations using matrix multiplication:
Coefficient matrix * | q | = Constant matrix
| n |
| 1 1 | * | q | = | 84 |
| n |
| 0.25 0.05 | | 13 |
To find the values of "q" and "n", we can solve the equation by multiplying the inverse of the coefficient matrix to the constant matrix:
Inverse of the coefficient matrix * Coefficient matrix * | q | = Inverse of the coefficient matrix * Constant matrix
| n |
| q | = Inverse of the coefficient matrix * Constant matrix
| n |
Now, let's calculate:
Inverse of the coefficient matrix:
| 0.05 -1 |
|-0.25 1 |
Inverse of the coefficient matrix * Constant matrix:
| 0.05 -1 | * | 84 | = | 49 |
|-0.25 1 | | 13 | | 35 |
Therefore, the solution to the system of equations is q = 49 and n = 35.
So, the answer is A) 49 quarters; 35 nickels.