The rainfall in City M during May of last year was 3 times the average (arithmetic mean) monthly rainfall for...

1. Translate the problem requirements: We need to find May's rainfall in City M. We know that May's rainfall was 3 times the average of the other 11 months, and the overall yearly average was 3.5 inches per month.
2. Set up variables for the two different averages: Define May's rainfall and the average for the other 11 months as separate quantities, since they have a specific relationship.
3. Apply the overall average constraint: Use the fact that the yearly average of 3.5 inches must account for all 12 months, including May's unusually high rainfall.
4. Solve using the 3:1 relationship: Substitute the given relationship (May = \(3 \times \mathrm{other\,months'\,average}\)) to find May's actual rainfall.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

We're looking for how much rain fell in City M during May. We have two key pieces of information:

• May's rainfall was 3 times higher than the typical month
• The average rainfall for the entire year was 3.5 inches per month

Think of it this way: imagine 11 months had "normal" rainfall, but May was unusually rainy - exactly 3 times what a normal month would be. We need to figure out what May's rainfall was.

Process Skill: TRANSLATE - Converting the problem's language into clear mathematical relationships

2. Set up variables for the two different averages

Let's define our unknowns using everyday language first:

Let's call the average rainfall for the 11 "normal" months = \(\mathrm{A}\) inches

Then May's rainfall = \(3\mathrm{A}\) inches (since it's 3 times the normal average)

So we have:

• 11 months with average rainfall of \(\mathrm{A}\) inches each
• 1 month (May) with rainfall of \(3\mathrm{A}\) inches

This gives us a clear picture: most months were typical, but May was exceptional.

3. Apply the overall average constraint

Now here's the key insight: when we average all 12 months together, we get 3.5 inches per month.

In plain English: if we add up all the rainfall from all 12 months and divide by 12, we get 3.5 inches.

Total rainfall for the year = (11 months × \(\mathrm{A}\)) + (1 month × \(3\mathrm{A}\)) = \(11\mathrm{A} + 3\mathrm{A} = 14\mathrm{A}\)

Since the average for all 12 months is 3.5 inches:

Total rainfall ÷ 12 months = 3.5 inches

\(14\mathrm{A} ÷ 12 = 3.5\)

Process Skill: APPLY CONSTRAINTS - Using the yearly average to create our equation

4. Solve using the 3:1 relationship

Now we can solve for \(\mathrm{A}\) (the average of the normal 11 months):

\(14\mathrm{A} ÷ 12 = 3.5\)

\(14\mathrm{A} = 3.5 \times 12\)

\(14\mathrm{A} = 42\)

\(\mathrm{A} = 42 ÷ 14\)

\(\mathrm{A} = 3\) inches

So the average rainfall for the 11 normal months was 3 inches.

Since May's rainfall was 3 times this amount:

May's rainfall = \(3 \times \mathrm{A} = 3 \times 3 = 9\) inches

Let's verify: (11 months × 3 inches) + (1 month × 9 inches) = \(33 + 9 = 42\) inches total

Average = \(42 ÷ 12 = 3.5\) inches ✓

Final Answer

The rainfall in City M during May was 9 inches.

Looking at our answer choices, this corresponds to choice E.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the 3:1 relationship

Students often misread "May's rainfall was 3 times the average monthly rainfall for the other 11 months" as "May's rainfall was 3 times the overall yearly average." This leads them to incorrectly calculate May's rainfall as \(3 \times 3.5 = 10.5\) inches, which isn't even among the answer choices.

2. Confusing which average to use as the baseline

Students may incorrectly think May's rainfall is 3 times the overall yearly average (3.5 inches) rather than 3 times the average of just the other 11 months. This fundamental misunderstanding derails the entire solution approach.

Errors while executing the approach

1. Setting up the wrong equation

Even when students understand the relationship correctly, they often struggle to set up the equation properly. They might write something like "\(11\mathrm{A} + 3(3.5) = 12(3.5)\)" instead of recognizing that both \(\mathrm{A}\) and May's rainfall (\(3\mathrm{A}\)) are unknowns that need to be expressed in terms of the same variable.

2. Arithmetic errors in fraction calculations

When solving \(14\mathrm{A}/12 = 3.5\), students frequently make calculation errors. They might incorrectly compute \(3.5 \times 12 = 36\) (instead of 42) or make errors when dividing 42 by 14, especially if they don't recognize that \(42/14 = 3\) immediately.

Errors while selecting the answer

1. Reporting the wrong value

Students who correctly calculate \(\mathrm{A} = 3\) inches sometimes forget the final step and select an answer choice closest to 3 (like choice B: 2) instead of calculating May's actual rainfall of \(3\mathrm{A} = 9\) inches. They report the average of the 11 normal months instead of May's exceptional rainfall.