Solve this problem using matrices: A large piggy bank contains a total of 2258 coins in pennies, nickels, dimes, and quarters. The total amount of money in the piggy bank is $55.77. If the number of pennies and quarters in the piggy bank remained the same but the number of nickels and dimes doubled, then the amount of money in the piggy bank would be $67.12. If the number of pennies, nickels, and dimes remained the same but the number of quarters were tripled, then the amount of money in the piggy bank would be $104.27. Find the number of pennies, nickels, dimes, and quarters in the piggy bank.
Let's create a system of equations using the information given in the problem.
Let x be the number of pennies, y be the number of nickels, z be the number of dimes, and w be the number of quarters.
From the first sentence of the problem, we know:
x + y + z + w = 2258 (equation 1)
0.01x + 0.05y + 0.1z + 0.25w = 55.77 (equation 2)
From the second sentence of the problem, we know that if the number of nickels and dimes doubles, then:
x + 2y + 2z + w = 2258 (number of pennies and quarters remain the same)
0.01x + 0.1y + 0.2z + 0.25w = 67.12 (equation 3)
From the third sentence of the problem, we know that if the number of quarters triples, then:
x + y + z + 3w = 2258 (number of pennies, nickels, and dimes remain the same)
0.01x + 0.05y + 0.1z + 0.75w = 104.27 (equation 4)
We can write these equations in matrix form as:
\begin{bmatrix}1 & 1 & 1 & 1 \\ 0.01 & 0.05 & 0.1 & 0.25 \\ 1 & 2 & 2 & 1 \\ 0.01 & 0.1 & 0.2 & 0.25 \\ 1 & 1 & 1 & 3 \\ 0.01 & 0.05 & 0.1 & 0.75\end{bmatrix}
\begin{bmatrix}x \\ y \\ z \\ w\end{bmatrix}
=
\begin{bmatrix}2258 \\ 55.77 \\ 2258 \\ 67.12 \\ 2258 \\ 104.27\end{bmatrix}
We can solve this matrix equation using row reduction:
\begin{bmatrix}1 & 1 & 1 & 1 & 2258 \\ 0.01 & 0.05 & 0.1 & 0.25 & 55.77 \\ 1 & 2 & 2 & 1 & 2258 \\ 0.01 & 0.1 & 0.2 & 0.25 & 67.12 \\ 1 & 1 & 1 & 3 & 2258 \\ 0.01 & 0.05 & 0.1 & 0.75 & 104.27\end{bmatrix}
R2 = R2 - 0.01R1
R3 = R3 - R1
R4 = R4 - 0.01R1
R5 = R5 - R1
R6 = R6 - 0.01R1
\begin{bmatrix}1 & 1 & 1 & 1 & 2258 \\ 0 & 0.04 & 0.09 & 0.24 & 33.19 \\ 0 & 1 & 1 & 0 & 1124 \\ 0 & 0.09 & 0.19 & 0.24 & 12.35 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0.74 & 92.69\end{bmatrix}
R2 = R2/0.04
R4 = R4/0.09
R6 = R6/0.74
\begin{bmatrix}1 & 1 & 1 & 1 & 2258 \\ 0 & 1 & 2.25 & 6 & 829.75 \\ 0 & 1 & 1 & 0 & 1124 \\ 0 & 1 & 2.11 & 2.67 & 137.22 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 & 125.3\end{bmatrix}
R1 = R1 - R2
R3 = R3 - R2
R4 = R4 - R2
R4 = R4 - R3
R6 = R6 - R4
\begin{bmatrix}1 & 0 & -1.25 & -5 & 1408.25 \\ 0 & 1 & 2.25 & 6 & 829.75 \\ 0 & 0 & -1.25 & -6 & 294.25 \\ 0 & 0 & -0.14 & -3.33 & -692.96 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 409.89\end{bmatrix}
From the last row, we see that 409.89 = 0, which is nonsensical. This means that the system of equations has no solution. Therefore, there is no way to find the number of pennies, nickels, dimes, and quarters that fit the given conditions. Something has gone wrong with the problem statement or calculations.
AAAaannndd the bot gets it wrong yet again!
These equations are bogus, because the number of coins would change!
x + 2y + 2z + w = 2258
x + y + z + 3w = 2258
Of course, they are not used anyway.
Getting rid of the decimal stuff and just using integers, we have
x + y + z + w = 2258
x + 5y + 10z + 25w = 5577
x + 10y + 20z + 25w = 6712
x + 5y + 10z + 75w = 10427
That gives the solution (x,y,z,w) = (2017,61,83,97)
You are correct, I apologize for the mistake in my earlier response. Thank you for providing the correct equations and solution.
Let's define the variables as follows:
- Let P represent the number of pennies in the piggy bank.
- Let N represent the number of nickels in the piggy bank.
- Let D represent the number of dimes in the piggy bank.
- Let Q represent the number of quarters in the piggy bank.
We can create a system of linear equations based on the given information.
Equation 1: The total amount of money in the piggy bank is $55.77.
0.01P + 0.05N + 0.10D + 0.25Q = 55.77
Equation 2: If the number of pennies and quarters remained the same, but the number of nickels and dimes doubled, then the amount of money in the piggy bank would be $67.12.
0.01P + 0.05(2N) + 0.10(2D) + 0.25Q = 67.12
Equation 3: If the number of pennies, nickels, and dimes remained the same, but the number of quarters were tripled, then the amount of money in the piggy bank would be $104.27.
0.01P + 0.05N + 0.10D + 0.25(3Q) = 104.27
Simplifying this equation, we get:
0.01P + 0.05N + 0.10D + 0.75Q = 104.27
We can represent these equations in the matrix form:
[0.01 0.05 0.10 0.25] [P] [55.77]
[0.01 0.10 0.20 0.25] [N] = [67.12]
[0.01 0.05 0.10 0.75] [D] [104.27]
To solve this matrix equation, we need to find the inverse of the coefficient matrix [A]. Let [X] represent the inverse of [A].
[A] = [0.01 0.05 0.10 0.25]
[0.01 0.10 0.20 0.25]
[0.01 0.05 0.10 0.75]
[X] = [P]
[N]
[D]
[Q]
To find [X], we can multiply both sides of the matrix equation by the inverse of [A].
[X] = [A]^-1 * [B]
Where [B] is the matrix on the right-hand side representing the constants in the equations.
Solving this matrix equation will give us the values of P, N, D, and Q, which represent the number of pennies, nickels, dimes, and quarters in the piggy bank.