How many of each specific coins are in a bottle?

Question

Question

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Suppose you have a bottle that contains exactly twenty-two U.S. coins.
These coins only consist of nickels(\$0.05), dimes(\$0.10), and quarters(\$0.25).

In addition to the types of coins, you also know the following statements to be true:

1. When added together, the coins hold a total value of \$2.55.
2. The number of nickels and the number of dimes is greater than the number of quarters.
3. The number of nickels in the bottle is greater than five.

Using the information given, how can we calculate the quantity of each coin in the bottle?

Edit

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In efforts to clarify any vagueness or lack of information, please consider the following questions and their respective answers.

Clarification

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$(1)$ Can you provide a clarification of the 2nd constraint?

The number of nickels and the number of dimes is greater than the number of quarters.

Is it intended that $(n)$ and $(d)$ are each separately $(>q)$.
Furthermore, this means the constraint translates to $(n>q)$ and $(d>q)$.

Additional Information

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$(2)$ What is the source of the problem?

This problem originates from my undergraduate studies in a Linear Algebra I course (MA-237). While this problem is not located in the course textbook itself, a similar problem can be found on page 12 as "Exercise 1.37".

This is the free textbook (Author's Words):
LINEAR ALGEBRA - Fourth edition by Jim Hefferon:

https://hefferon.net/linearalgebra/

$(3)$ What theorems or previously solved problems or worked examples that led up to this problem do you think might be relevant?

When I originally posted this question, I intended to supply examples and relevant information in the answer's reference section. I see now that they should also be applied here.

Please find them listed below:

1. Constructing an Augmented Matrix
2. Gaussian-Elimination Operations
3. Insight into "Reducing a Matrix"
4. Brief Explanation of Free Variables
5. Parametric Vector Form Example

Background

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$(4)$ In general, what is your Math education/background?

Currently, I am an Undergraduate Student seeking a Bachelor of Science Degree with a minor in Mathematics.

$(5)$ What is your motivation for posting this question?

When I was a student in what we refer to at my university as MA-237 or Linear Algebra I, I distinctly remember struggling to work through this problem due to the variety of concepts that are prevalent in its solution.

Through the answer I provided, I hope that any future students also struggling to understand these concepts will find the same understanding that I did at completing this question.

Answer 1

In the solution parts, the answers are initially hidden.
This has been done to give you - the reader - the opportunity to work this question out on your own if desired.

First, we'll construct a system of linear equations - a.k.a: linear system - from the given information:

nickel + dime + quarter = total quantity $$n+d+q=22$$ (value)n + (value)d+ (value)q= total value $$5n+10d+25q=255$$

Second, we convert this system into Augmented Matrix Form:

Only use the coefficients of the linear equations
to build the augmented matrix $$\begin{bmatrix}1&1&1&|&22\\5&10&25&|&255\end{bmatrix}$$

Third, we begin the process of Gaussian-Elimination to reduce our matrix into RREF:

Note: Gaussian-Elimination allows us to perform the following row operations on a Matrix:

1. The swapping of two rows.
2. Multiplying a row by a non-zero number.
3. Adding a multiple of one row to another row.

Proceed:

Row 1: $R_1$
Row 2: $R_2$

Step 1:
1. Multiply $R_1$ by $-5$.
2. Add $R_1$ to $R_2$.
$$\xrightarrow[]{-5{R_1}+R_2}\begin{bmatrix}1&1&1&|&22\\5&10&25&|&255\end{bmatrix}=\begin{bmatrix}1&1&1&|&22\\0&5&20&|&145\end{bmatrix}=$$ Step 2:
1. Multiply $R_2$ by $\frac{1}{5}$. $$\xrightarrow[]{\frac{1}{5}{R_2}}\begin{bmatrix}1&1&1&|&22\\0&5&20&|&145\end{bmatrix}=\begin{bmatrix}1&1&1&|&22\\0&1&4&|&29\end{bmatrix}=$$ Step 3:
1. Multiply $R_2$ by $-1$.
2. Add $R_2$ to $R_1$.
$$\xrightarrow[]{-{R_2}+R_1}\begin{bmatrix}1&1&1&|&22\\0&1&4&|&29\end{bmatrix}=\begin{bmatrix}1&0&-3&|&-7\\0&1&4&|&29\end{bmatrix}$$ Reduced Row Echelon Form:$$\begin{bmatrix}1&0&-3&|&-7\\0&1&4&|&29\end{bmatrix}$$

Fourth, we will convert our augmented matrix into Parametric Vector Form to find the solution to the Linear System:

Step 1:
1. Covert the augmented matrix back in to a linear system.
$$\begin{equation}\begin{bmatrix}1&0&-3&|&-7\\0&1&4&|&29\end{bmatrix} = \begin{cases}n- 3q=-7\\d+4q=29\\\end{cases}\end{equation}$$ Step 2:
1. Determine Free Variables (q does not have a leading row while in RREF).
2. Write each unknown variable in terms of Free Variables (q).
$$\left\{\begin{array}{cc}n-3q=-7\\d+4q=29\end{array}\right\}=\left\{\begin{array}{cc}n=-7+3q\\d=29-4q\\q=q\end{array}\right\}$$ Step 3:
Write all three equations as one equation using Vectors.
1. First vector contains all of the variables.
2. Second vector contains the equations the variables are equal to. $$\left(\begin{array}{cc}n\\d\\q\end{array}\right)=\left(\begin{array}{cc}-7+3q\\29-4q\\q\end{array}\right)$$ 3. Expand the second vector into multiple vectors that consist of Free Variable Terms and Constant Terms.
$$\left(\begin{array}{cc}-7+3q\\29-4q\\q\end{array}\right)=\left(\begin{array}{cc}-7\\29\\0\end{array}\right)+\left(\begin{array}{cc}3q\\-4q\\q\end{array}\right)$$ 4. Factor out the Free Variable(q) from the vector containing the Free Variable (q).
$$\left(\begin{array}{cc}3q\\-4q\\q\end{array}\right)=q\left(\begin{array}{cc}3\\-4\\1\end{array}\right)$$ 5. Combine findings.

Parametric Vector Form:
$$\left(\begin{array}{cc}n\\d\\q\end{array}\right)=\left(\begin{array}{cc}-7\\29\\0\end{array}\right)+q\left(\begin{array}{cc}3\\-4\\1\end{array}\right)$$

Fifth, we will use our Linear System solution in Parametric Vector Form to determine how many quarters are in the bottle:

Since q is a Free Variable in our solution and we know that q must be a non-negative whole number, we can plug in any real number into our solution for q.

In doing so, we will determine which value of q matches the information given in the question.

Testing q values:
1. q $=1$:$$\left(\begin{array}{cc}n\\d\\q\end{array}\right)=\left(\begin{array}{cc}-7\\29\\0\end{array}\right)+(1)\left(\begin{array}{cc}3\\-4\\1\end{array}\right)=\left\{\begin{array}{cc}n=-7+3(1)\\d=29-4(1)\\q=(1)\end{array}\right\}=\left\{\begin{array}{cc}n=-4\\d=25\\q=1\end{array}\right\}$$We find the following:
1. Number of nickels is negative.
2. Number of dimes greater than the overall number of coins in the bottle.
So, we can conclude that the solution set for when q $=1$ is not possible.

Intuitive Findings:
1. In order to have a non-negative number of nickels, q must be greater than $2$.
2. In order for the number of dimes to be less than $21$, q must be less than $8$.
So, we can exclude the testing for q values less than $3$ and greater than $8$ since they will yield impossible solutions.

These testing values for q are shown through the the following table and calculated the same way as when q $=1$.

Tables not supported in spoilers :(
| q = | 1 | 3 | 4 | 5 | 6 | 7 |
| (n,d) =| (-4,25) | (2,17) | (5,13) | (8,9) | (11,5) | (14,1) |

Finally, we can draw our final conclusion:

We first consider the earlier truths that:
1. The number of nickels and the number of dimes is greater than the number of quarters.
2. The number of nickels in the bottle is greater than five.

Using the table we created for q values, we find that when q $=3$, $4$, $6$, or $7$, the truths are violated.

Therefore, the only possible amount of quarters in the bottle is $\textbf{5}$, meaning there are $\textbf{8}$ nickels and $\textbf{9}$ dimes.

References:

1. Constructing an Augmented Matrix
2. Gaussian-Elimination Operations
3. Insight into "Reducing a Matrix"
4. Brief Explanation of Free Variables
5. Parametric Vector Form Example

Answer 2

Alternative approach:

Since the number of nickels must be greater than 5, and since $~\displaystyle \frac{255}{22} \approx 10$, a reasonable first guess is that there are $6$ nickels and $16$ dimes.

This totals $190$, which leaves a deficit of $65$.

It is clear, from the constraints of the problem that this deficit must be cleared by converting some of the dimes to quarters and converting some of the dimes to nickels.

Each dime --> quarter reduces the deficit by 15.

Each dime --> nickel increases the deficit by 5.

Since $65$ is not a multiple of $15$, and since you are not allowed to change nickels to dimes (since there must be at least $6$ nickels), it is clear that the conversion of dimes to quarters must focus on multiples of $15$ that are greater than $65$.

Further, the only (other) constraint to fade is that the number of dimes + the number of nickels must exceed the number of quarters.

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Edit
The above paragraph may represent a misinterpretation of the 2nd constraint. This issue is discussed more fully, later in this answer.

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So, the obvious play is to explore converting 5 dimes to quarters, and 2 dimes to nickels.

This results in 8 nickels, 9 dimes and 5 quarters, which clearly satisfies all of the constraints. However, it remains to determine whether this solution is unique.

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One way of investigating this is to return to the baseline distribution of $(n,d,q) = (6,16,0)$ and its deficit of 65.

Consider what happens if you aim for the dime --> quarter conversion of $6$, rather than $5$. You have changed the deficit from $(+65)$ to $(-25)$ which implies that you then have to convert a further $5$ dimes to nickels.

This results in $(n,d,q) = (11,5,6)$, which (arguably - see later in this answer) also satisfies the constraints.

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At this point, one must question how the following constraint is to be interpreted:

• The number of nickels and the number of dimes is greater than the number of quarters.

I was interpreting this to mean that $(n + d) > q$.

However, the alternative interpretation is that

• $n > q$ and $d > q$.

Under this alternative interpretation, it is clear that $(n,d,q) = (11,5,6)$ must be excluded, as well as any conversion of a greater number of dimes to quarters.

This leaves as unique, the original answer of $(n,d,q) = (8,9,5).$

Meta-cheating, one might guess that this is the problem composer's intent, since it is customary for the solution to be unique. Then, one simply surmises that the pertinent (2nd) constraint is poorly written.

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It only remains to explore, what other solutions, besides $(n,d,q) \in \{(8,9,5), (11,5,6)\}$ are possible, under the alternative assumption that the 2nd constraint is to be interpreted as $(n + d) > q$.

Returning again to the baseline, of $(n,d,q) = (6,16,0)$, the conversion of $7$ dimes to quarters, changes the deficit from $(+65)$ to $(-40)$. This means that you then have to convert $8$ (other) dimes to nickels.

This results in $(n,d,q) = (14,1,7)$ which also works. It is then clear that any conversion of a larger number of dimes to quarters, must be excluded, because this would trigger also converting dimes to nickels. So, the result would be a negative number of dimes.

So, under the alternative interpretation, in addition to the answer of $(n,d,q) = (8,9,5)$, the only other permissible answers are $(11,5,6)$, or $(14,1,7)$.