At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is...

1. Translate the problem requirements: Clarify what "average price" means in terms of total cost, and identify what we need to find - how many oranges to remove to change the average from 56 cents to 52 cents
2. Determine the initial fruit composition: Use the given average price of 56 cents to find how many apples and oranges Mary initially selected
3. Calculate the target scenario: Determine what the total cost and number of fruits should be after removing oranges to achieve a 52-cent average
4. Find the difference: Compare the initial and target scenarios to determine how many oranges must be removed

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're working with in everyday terms. Mary is buying fruit, and we need to understand what "average price" really means.

When we say the average price is 56 cents, we mean: if Mary paid for all 10 pieces of fruit and then divided that total cost equally among the 10 pieces, each piece would "cost" 56 cents on average.

So if the average price is 56 cents for 10 pieces, the total cost = \(56 \times 10 = 560\) cents.

What we need to find: Mary wants to put back some oranges so that the remaining fruit has an average price of 52 cents. We need to figure out how many oranges she should return.

Process Skill: TRANSLATE - Converting the average price concept into total cost makes the problem much more manageable

2. Determine the initial fruit composition

Now let's figure out how many apples and oranges Mary initially picked.

We know:

• Apples cost 40 cents each
• Oranges cost 60 cents each
• She picked 10 pieces total
• The total cost is 560 cents

Let's say Mary picked A apples and O oranges.
Then: \(\mathrm{A} + \mathrm{O} = 10\) (total pieces)
And: \(40\mathrm{A} + 60\mathrm{O} = 560\) (total cost)

From the first equation: \(\mathrm{A} = 10 - \mathrm{O}\)
Substituting into the second equation:
\(40(10 - \mathrm{O}) + 60\mathrm{O} = 560\)
\(400 - 40\mathrm{O} + 60\mathrm{O} = 560\)
\(400 + 20\mathrm{O} = 560\)
\(20\mathrm{O} = 160\)
\(\mathrm{O} = 8\)

So \(\mathrm{A} = 10 - 8 = 2\)

Mary initially selected 2 apples and 8 oranges.

Let's verify: \(2 \times 40 + 8 \times 60 = 80 + 480 = 560\) cents ✓

3. Calculate the target scenario

Now we need to figure out what happens when Mary puts back some oranges to get an average price of 52 cents.

Important insight: Mary will only put back oranges (the more expensive fruit), not apples. This makes sense because removing expensive items will lower the average price.

So Mary will keep her 2 apples and some number of oranges. Let's say she keeps R oranges (where \(\mathrm{R} < 8\)).

After putting back oranges:

• Apples kept: 2 (costing \(2 \times 40 = 80\) cents)
• Oranges kept: R (costing \(\mathrm{R} \times 60 = 60\mathrm{R}\) cents)
• Total pieces kept: \(2 + \mathrm{R}\)
• Total cost of kept fruit: \(80 + 60\mathrm{R}\) cents
• Desired average: 52 cents

For the average to be 52 cents:
\(\frac{80 + 60\mathrm{R}}{2 + \mathrm{R}} = 52\)

\(80 + 60\mathrm{R} = 52(2 + \mathrm{R})\)
\(80 + 60\mathrm{R} = 104 + 52\mathrm{R}\)
\(60\mathrm{R} - 52\mathrm{R} = 104 - 80\)
\(8\mathrm{R} = 24\)
\(\mathrm{R} = 3\)

So Mary should keep 3 oranges.

4. Find the difference

Mary initially had 8 oranges and should keep 3 oranges.

Therefore, she must put back: \(8 - 3 = 5\) oranges.

Let's verify our answer:

• Fruit kept: 2 apples + 3 oranges = 5 pieces total
• Cost of kept fruit: \(2 \times 40 + 3 \times 60 = 80 + 180 = 260\) cents
• Average price: \(260 \div 5 = 52\) cents ✓

Final Answer

Mary must put back 5 oranges to achieve an average price of 52 cents for the remaining fruit.

The answer is (E) 5.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what "average price" means in terms of total cost. Students might try to work directly with averages throughout the problem instead of recognizing that average price × number of items = total cost, which makes the calculations much more straightforward.

Faltering Point 2: Assuming Mary puts back both apples and oranges randomly, rather than recognizing that to lower the average price from 56 cents to 52 cents, she must specifically put back the more expensive items (oranges only).

Faltering Point 3: Setting up the wrong relationship for the final scenario. Students might incorrectly think they need to find how many total pieces Mary keeps rather than specifically how many oranges she puts back.

Errors while executing the approach

Faltering Point 1: Making algebraic errors when solving the system of equations to find the initial composition. For example, incorrectly substituting A = 10 - O into the cost equation, or making arithmetic mistakes when solving 20O = 160.

Faltering Point 2: Setting up the final average equation incorrectly. Students might write (80 + 60R) ÷ (2 + R) = 52 but then make distribution or cross-multiplication errors, such as incorrectly expanding 52(2 + R) or making sign errors when collecting like terms.

Faltering Point 3: Calculation errors when solving for R, particularly in the step 8R = 24, where students might get R = 4 instead of R = 3, leading to an incorrect final answer.

Errors while selecting the answer

Faltering Point 1: Confusing what the question is asking for. Students might select the number of oranges Mary keeps (3) instead of the number of oranges she puts back (5), or select the total number of pieces she keeps (5) thinking that's the answer.

Faltering Point 2: Failing to verify the answer and catching computational mistakes. Students might get an incorrect value for R but not check whether their final answer actually produces an average price of 52 cents, missing the opportunity to catch their error.