A bank teller has $255 in $5 bills and $10 bills. If the number of $10 bills is greater than...

1. Translate the problem requirements: We need to find the maximum total number of bills when we have \(\$255\) total, using only \(\$5\) and \(\$10\) bills, with the constraint that we must have more \(\$10\) bills than \(\$5\) bills.
2. Set up the constraint relationship: Express the value constraint (\(\$255\) total) and the quantity constraint (more \(\$10\) bills than \(\$5\) bills) using variables for each bill type.
3. Apply the maximization strategy: To maximize total bills, we want as many of the lower-value bills as possible while still satisfying both constraints.
4. Test the boundary condition: Find the optimal combination by working with the constraint that \(\$10\) bills must exceed \(\$5\) bills by at least 1, then verify this gives us exactly \(\$255\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in everyday language:

• We have exactly \(\$255\) total
• We can only use \(\$5\) bills and \(\$10\) bills
• We must have MORE \(\$10\) bills than \(\$5\) bills
• We want to find the MAXIMUM possible total number of bills

Think of this like packing a suitcase - we want to fit as many items as possible, but we have size and weight restrictions to follow.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical constraints

Let's call the number of \(\$5\) bills "f" (for five) and the number of \(\$10\) bills "t" (for ten).

Our constraints become:

• Value constraint: \(5\mathrm{f} + 10\mathrm{t} = 255\)
• Quantity constraint: \(\mathrm{t} > \mathrm{f}\)
• Goal: Maximize \(\mathrm{f} + \mathrm{t}\)

2. Set up the constraint relationship

Now let's think about this step by step. We have two main rules to follow:

Rule 1: The money must add up to exactly \(\$255\)
If we have f five-dollar bills and t ten-dollar bills, then:
\(5\mathrm{f} + 10\mathrm{t} = 255\)

We can simplify this by dividing everything by 5:
\(\mathrm{f} + 2\mathrm{t} = 51\)

This means: (number of \(\$5\) bills) + 2×(number of \(\$10\) bills) = 51

Rule 2: We need more \(\$10\) bills than \(\$5\) bills
This means: \(\mathrm{t} > \mathrm{f}\), or equivalently, \(\mathrm{t} \geq \mathrm{f} + 1\)

From Rule 1, we can express f in terms of t:
\(\mathrm{f} = 51 - 2\mathrm{t}\)

3. Apply the maximization strategy

Here's the key insight: to get the maximum number of bills total, we want as many of the smaller-value bills (\(\$5\) bills) as possible, while still satisfying our constraints.

The total number of bills is: \(\mathrm{f} + \mathrm{t}\)
Substituting \(\mathrm{f} = 51 - 2\mathrm{t}\):
Total bills = \((51 - 2\mathrm{t}) + \mathrm{t} = 51 - \mathrm{t}\)

This tells us something important: to maximize the total number of bills, we need to minimize t (the number of \(\$10\) bills).

But we can't make t too small because we need \(\mathrm{t} > \mathrm{f}\).

Process Skill: INFER - Drawing the non-obvious conclusion that minimizing \(\$10\) bills maximizes total bills

4. Test the boundary condition

We need \(\mathrm{t} > \mathrm{f}\), so let's find the smallest value of t that works.

Using our constraint \(\mathrm{f} = 51 - 2\mathrm{t}\) and the requirement \(\mathrm{t} > \mathrm{f}\):
\(\mathrm{t} > 51 - 2\mathrm{t}\)
\(\mathrm{t} + 2\mathrm{t} > 51\)
\(3\mathrm{t} > 51\)
\(\mathrm{t} > 17\)

Since t must be a whole number (we can't have fractions of bills), the smallest possible value is \(\mathrm{t} = 18\).

Let's check this works:

• If \(\mathrm{t} = 18\), then \(\mathrm{f} = 51 - 2(18) = 51 - 36 = 15\)
• Check constraint: Is \(\mathrm{t} > \mathrm{f}\)? Yes, \(18 > 15\) ✓
• Check value: \(5(15) + 10(18) = 75 + 180 = 255\) ✓

Total bills = \(\mathrm{f} + \mathrm{t} = 15 + 18 = 33\)

Process Skill: APPLY CONSTRAINTS - Ensuring all conditions are satisfied before concluding

4. Final Answer

The maximum possible number of bills is 33.

This corresponds to:

• 15 five-dollar bills
• 18 ten-dollar bills
• Total value: \(\$75 + \$180 = \$255\) ✓
• More \(\$10\) bills than \(\$5\) bills: \(18 > 15\) ✓

The answer is (B) 33.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint "number of \(\$10\) bills is greater than number of \(\$5\) bills"
Students often confuse this with "the VALUE of \(\$10\) bills is greater than the VALUE of \(\$5\) bills" or interpret it as "greater than or equal to" instead of strictly greater than. This leads to setting up \(\mathrm{t} \geq \mathrm{f}\) instead of \(\mathrm{t} > \mathrm{f}\), which gives an incorrect maximum.

2. Confusion about the optimization objective
Students may think that to maximize the number of bills, they should maximize the number of higher-value bills (\(\$10\) bills) since they're worth more. This is backwards - to get the maximum COUNT of bills with a fixed total value, you actually want to minimize the higher-value bills and maximize the lower-value bills.

3. Setting up the wrong equation from the constraint
When translating "\(\mathrm{t} > \mathrm{f}\)" combined with "\(\mathrm{f} = 51 - 2\mathrm{t}\)", students often make errors in the substitution or inequality direction, writing "\(\mathrm{t} < 51 - 2\mathrm{t}\)" instead of "\(\mathrm{t} > 51 - 2\mathrm{t}\)".

Errors while executing the approach

1. Arithmetic errors when solving the inequality \(3\mathrm{t} > 51\)
Students may incorrectly calculate \(51 \div 3\), getting 16 instead of 17, or forget that since we need \(\mathrm{t} > 17\) (not \(\mathrm{t} \geq 17\)), the minimum integer value is \(\mathrm{t} = 18\), not \(\mathrm{t} = 17\).

2. Sign errors when rearranging the inequality \(\mathrm{t} > 51 - 2\mathrm{t}\)
When moving terms across the inequality, students often make sign errors, writing "\(\mathrm{t} - 2\mathrm{t} > 51\)" instead of "\(\mathrm{t} + 2\mathrm{t} > 51\)", leading to incorrect solutions.

Errors while selecting the answer

1. Failing to verify that all constraints are satisfied
Students may arrive at \(\mathrm{t} = 17, \mathrm{f} = 17\) (giving 34 total bills) and select this without checking that it violates the constraint \(\mathrm{t} > \mathrm{f}\). Since 17 is not greater than 17, this solution is invalid, but students often miss this verification step.

Alternate Solutions

Smart Numbers Approach

Step 1: Set up the foundational relationship
From the constraint that we have \(\$255\) in \(\$5\) and \(\$10\) bills:
\(5\mathrm{x} + 10\mathrm{y} = 255\)
Dividing by 5: \(\mathrm{x} + 2\mathrm{y} = 51\)

Step 2: Choose smart numbers based on the constraint \(\mathrm{y} > \mathrm{x}\)
Since we want to maximize total bills (\(\mathrm{x} + \mathrm{y}\)), let's work with the boundary condition where y is just slightly greater than x. The smallest difference would be \(\mathrm{y} = \mathrm{x} + 1\).

Step 3: Substitute our smart number relationship
If \(\mathrm{y} = \mathrm{x} + 1\), then substituting into \(\mathrm{x} + 2\mathrm{y} = 51\):
\(\mathrm{x} + 2(\mathrm{x} + 1) = 51\)
\(\mathrm{x} + 2\mathrm{x} + 2 = 51\)
\(3\mathrm{x} + 2 = 51\)
\(3\mathrm{x} = 49\)
\(\mathrm{x} = \frac{49}{3} = 16.33...\)

Step 4: Adjust to the nearest valid integers
Since we need whole numbers of bills, we test \(\mathrm{x} = 16\) and \(\mathrm{x} = 17\):

If \(\mathrm{x} = 16\): \(\mathrm{y} = \frac{51 - 16}{2} = \frac{35}{2} = 17.5\) (not a whole number)
If \(\mathrm{x} = 17\): \(\mathrm{y} = \frac{51 - 17}{2} = \frac{34}{2} = 17\)

Step 5: Verify our smart numbers work
With \(\mathrm{x} = 17\) and \(\mathrm{y} = 17\):
- Check constraint \(\mathrm{y} > \mathrm{x}\): \(17 > 17\)? No, this violates our constraint.

Let's try \(\mathrm{x} = 15\): \(\mathrm{y} = \frac{51 - 15}{2} = \frac{36}{2} = 18\)
- Check constraint \(\mathrm{y} > \mathrm{x}\): \(18 > 15\)? Yes ✓
- Check value: \(5(15) + 10(18) = 75 + 180 = 255\) ✓

Step 6: Calculate maximum bills
Total bills = \(\mathrm{x} + \mathrm{y} = 15 + 18 = 33\)

Answer: (B) 33