Mark bought a set of 6 flower pots of different sizes at a total cost of $8.25. Each pot cost...

1. Translate the problem requirements: We have 6 flower pots where each pot costs \(\$0.25\) more than the next smaller pot, the total cost is \(\$8.25\), and we need to find the cost of the largest (most expensive) pot.
2. Set up the cost pattern using the smallest pot as our base: If we call the smallest pot's cost 'x', then we can express all 6 pot costs in terms of x using the \(\$0.25\) increment pattern.
3. Create and solve the total cost equation: Add up all 6 pot costs (expressed in terms of x) and set equal to \(\$8.25\) to find the value of x.
4. Calculate the largest pot's cost: Once we know x (smallest pot cost), add 5 increments of \(\$0.25\) to get the largest pot's cost.

Execution of Strategic Approach

1. Translate the problem requirements

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

• Mark bought 6 flower pots of different sizes
• The total cost for all 6 pots was \(\$8.25\)
• Each pot costs \(\$0.25\) more than the next smaller pot
• We need to find the cost of the largest (most expensive) pot

Think of it like this: imagine you're looking at 6 pots lined up from smallest to largest. As you move from left to right, each pot costs exactly 25 cents more than the one before it. This creates a step-by-step increase in price.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Set up the cost pattern using the smallest pot as our base

Let's use plain English first. If we call the cost of the smallest pot our starting point, then:

• Smallest pot costs: our starting amount
• Second pot costs: starting amount + \(\$0.25\)
• Third pot costs: starting amount + \(\$0.50\) (that's \(2 \times \$0.25\))
• Fourth pot costs: starting amount + \(\$0.75\) (that's \(3 \times \$0.25\))
• Fifth pot costs: starting amount + \(\$1.00\) (that's \(4 \times \$0.25\))
• Largest pot costs: starting amount + \(\$1.25\) (that's \(5 \times \$0.25\))

Now let's use 'x' to represent our starting amount (the smallest pot's cost):

• Pot 1 (smallest): \(\mathrm{x}\)
• Pot 2: \(\mathrm{x} + 0.25\)
• Pot 3: \(\mathrm{x} + 0.50\)
• Pot 4: \(\mathrm{x} + 0.75\)
• Pot 5: \(\mathrm{x} + 1.00\)
• Pot 6 (largest): \(\mathrm{x} + 1.25\)

3. Create and solve the total cost equation

Since all 6 pots together cost \(\$8.25\), we add up all the individual costs:

In plain English: (smallest pot) + (second pot) + (third pot) + (fourth pot) + (fifth pot) + (largest pot) = \(\$8.25\)

Substituting our expressions:

\(\mathrm{x} + (\mathrm{x} + 0.25) + (\mathrm{x} + 0.50) + (\mathrm{x} + 0.75) + (\mathrm{x} + 1.00) + (\mathrm{x} + 1.25) = 8.25\)

Let's collect like terms:

\(6\mathrm{x} + (0.25 + 0.50 + 0.75 + 1.00 + 1.25) = 8.25\)

\(6\mathrm{x} + 3.75 = 8.25\)

Subtracting 3.75 from both sides:

\(6\mathrm{x} = 8.25 - 3.75\)

\(6\mathrm{x} = 4.50\)

Dividing both sides by 6:

\(\mathrm{x} = 0.75\)

So the smallest pot costs \(\$0.75\).

4. Calculate the largest pot's cost

Now that we know the smallest pot costs \(\$0.75\), we can find the largest pot's cost.

Remember: the largest pot costs \(\$1.25\) more than the smallest pot (since it's 5 steps of \(\$0.25\) each).

Largest pot cost = \(\$0.75 + \$1.25 = \$2.00\)

Let's verify our answer by checking that all 6 pots total \(\$8.25\):

• Pot 1: \(\$0.75\)
• Pot 2: \(\$0.75 + \$0.25 = \$1.00\)
• Pot 3: \(\$0.75 + \$0.50 = \$1.25\)
• Pot 4: \(\$0.75 + \$0.75 = \$1.50\)
• Pot 5: \(\$0.75 + \$1.00 = \$1.75\)
• Pot 6: \(\$0.75 + \$1.25 = \$2.00\)

Total: \(\$0.75 + \$1.00 + \$1.25 + \$1.50 + \$1.75 + \$2.00 = \$8.25\) ✓

Final Answer

The cost of the largest pot is \(\$2.00\), which matches answer choice C.

Common Faltering Points

Errors while devising the approach

• Misinterpreting the cost progression direction: Students might assume each pot costs \(\$0.25\) less than the next larger pot (reading "next one below it" as referring to physical position rather than size hierarchy), leading them to set up the equation with the wrong increment direction.
• Starting with the wrong base variable: Some students might choose to use the largest pot's cost as their variable 'x' instead of the smallest, making the algebra unnecessarily complex with negative increments and increasing chances of sign errors.
• Misunderstanding the increment pattern: Students may think each pot costs \(\$0.25\) more than ALL smaller pots combined, rather than understanding it's a step-by-step increase where each pot is exactly \(\$0.25\) more than just the next smaller one.

Errors while executing the approach

• Arithmetic errors in calculating the increment sum: When adding \(0.25 + 0.50 + 0.75 + 1.00 + 1.25 = 3.75\), students often make basic addition mistakes, commonly getting 3.50 or 4.00, which throws off the entire solution.
• Algebraic manipulation errors: Students frequently make mistakes when solving \(6\mathrm{x} + 3.75 = 8.25\), either incorrectly subtracting (getting \(6\mathrm{x} = 4.50\) wrong) or incorrectly dividing by 6, leading to wrong values for the smallest pot.
• Incorrect increment calculation: Some students miscalculate how much more the largest pot costs compared to the smallest pot, forgetting it's 5 increments of \(\$0.25\) (= \(\$1.25\)) and instead using 4 or 6 increments.

Errors while selecting the answer

• Selecting the smallest pot's cost instead of the largest: After correctly calculating that the smallest pot costs \(\$0.75\), students might mistakenly select an answer choice close to this value rather than adding \(\$1.25\) to get the largest pot's cost of \(\$2.00\).
• Rounding errors in final calculation: Students who get slight arithmetic errors in earlier steps might arrive at values like \(\$1.75\) or \(\$2.25\) and select these as their final answer without double-checking their work against the total cost constraint.