Each of 90 students participated in at least one of three track tryouts: high jump, long jump, and 100-meter dash....

1. Translate the problem requirements: We need to find students who participated in exactly two tryouts (not one, not three). Given: 90 total students, 20 in high jump, 40 in long jump, 60 in 100-meter dash, and 5 in all three tryouts.
2. Apply the inclusion-exclusion principle to find total overlaps: Calculate how many students participated in at least two tryouts by comparing individual participation totals with the overall total.
3. Isolate the 'exactly two' group: Subtract the 'all three' group from the 'at least two' group to find students in exactly two tryouts.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English first. We have 90 students total, and each student tried out for at least one of three events. Think of this like three overlapping circles - some students are in just one circle, some are where two circles overlap, and some are in the center where all three circles meet.

Here's what we know:

• 20 students tried high jump
• 40 students tried long jump
• 60 students tried 100-meter dash
• 5 students tried all three events
• 90 students total participated

We need to find how many students participated in exactly two tryouts - not one, not three, but exactly two.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding of what we're looking for

2. Apply the inclusion-exclusion principle to find total overlaps

Now let's think about this step by step. If we simply add up all the individual tryout numbers (\(20 + 40 + 60 = 120\)), we get more than our total of 90 students. This means we're counting some students multiple times - specifically, we're double-counting students who participated in two events, and triple-counting students who participated in all three.

The difference between our sum and the actual total tells us about overlaps:
\(120 - 90 = 30\)

This difference of 30 represents the "extra counting" we did. Every student who participated in exactly two tryouts gets counted one extra time (they appear in two of our totals instead of one). Every student who participated in all three tryouts gets counted two extra times (they appear in all three totals instead of just one).

So if we call the number of students in exactly two tryouts "T", and we know 5 students are in all three tryouts:
\(\mathrm{T} \times 1 + 5 \times 2 = 30\)
\(\mathrm{T} + 10 = 30\)
\(\mathrm{T} = 20\)

Process Skill: INFER - Drawing the non-obvious conclusion about how overcounting relates to overlaps

3. Isolate the 'exactly two' group

Let's verify this makes sense. We found that 20 students participated in exactly two tryouts.

Checking our work:

• Students in exactly one tryout: \(90 - 20 - 5 = 65\) students
• Students in exactly two tryouts: 20 students
• Students in all three tryouts: 5 students
• Total: \(65 + 20 + 5 = 90\) ✓

Also checking the overcounting:

• Students in exactly two tryouts contribute 1 extra count each: \(20 \times 1 = 20\)
• Students in all three tryouts contribute 2 extra counts each: \(5 \times 2 = 10\)
• Total extra counting: \(20 + 10 = 30\) ✓

This matches our calculation of \(120 - 90 = 30\).

4. Final Answer

The number of students who participated in exactly two tryouts is 20.

This corresponds to answer choice B.

Common Faltering Points

Errors while devising the approach

Students often confuse what the question is asking for. The problem asks for students who participated in "only two" tryouts, which means exactly two. However, students might think this includes students who participated in all three tryouts, since those students also participated in two of the events (plus a third one). This fundamental misunderstanding would lead them to include the 5 students who tried all three events in their final answer.

Many students try to solve this problem by guessing or using basic addition/subtraction without recognizing this as a classic overlapping sets problem. They might attempt to work backwards from the total or try to create equations without understanding the systematic approach needed. This leads to confusion and incorrect setup of the problem.

Students often fail to understand why adding individual totals (\(20 + 40 + 60 = 120\)) gives more than the actual total (90). They don't realize that this difference represents systematic overcounting of students who participated in multiple events. Without this key insight, they cannot set up the correct equation to solve for students in exactly two tryouts.

Errors while executing the approach

Even when students understand the overcounting concept, they often make errors in calculating how much each group contributes to the overcounting. They might think students in exactly two tryouts contribute 2 extra counts instead of 1, or that students in all three tryouts contribute 3 extra counts instead of 2. This leads to setting up the wrong equation: for example, writing \(\mathrm{T} \times 2 + 5 \times 3 = 30\) instead of \(\mathrm{T} \times 1 + 5 \times 2 = 30\).

Students may correctly set up \(\mathrm{T} + 10 = 30\) but then make basic arithmetic mistakes when solving for T, such as getting \(\mathrm{T} = 40\) instead of \(\mathrm{T} = 20\), or forgetting to subtract 10 from both sides properly.

Errors while selecting the answer

Students often calculate multiple values during their solution process (like the 30 representing total overcounting, the 65 students in exactly one tryout, etc.) and then select the wrong number as their final answer. They might choose 30, 65, or 10 instead of the correct answer of 20 students who participated in exactly two tryouts.