Question

Which of the following CANNOT be obtained from the given information?

• A. Number of students choosing E1
• B. Number of students choosing either E1 or E2 or both, but not E3
• C. Number of students choosing both E1 and E2
• D. Number of students choosing E3
• E. Number of students choosing exactly one elective
• A. Number of students choosing only E2, and number of students choosing both E2 and E3
• B. Number of students choosing both E1 and E2, number of students choosing both E2 and E3, and number of students choosing both E3 and E1
• C. Number of students choosing only E1, and number of students choosing both E2 and E3
• D. No extra information is necessary
• J. Number of students choosing both E1 and E2

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the information provided and determine which piece of information cannot be directly obtained.
The given data is:

• Total students (let this be T = 190) must choose at least one elective and at most two: E1, E2, E3.
• The number of students choosing E1 (x) is half of those choosing E2 (y): \(x = \frac{y}{2}\).
• The number of students choosing E1 (x) is a third of those choosing E3 (z): \(x = \frac{z}{3}\).
• 50 students choose exactly two electives.

From this, directly calculating each described option:

1. Number of students choosing E1: Given by \(x = \frac{y}{2} = \frac{z}{3}\). This can be found by obtaining explicit values by further relations.
2. Number of students choosing either E1 or E2 or both, but not E3: Here, we focus on the union of E1 and E2 minus any common with E3, which can be derived easily from known students.
3. Number of students choosing both E1 and E2: This requires specific data about conjunction (intersection) which is not directly provided.
4. Number of students choosing E3: Given by \(z = 3x\), thus can be derived directly if x is known.
5. Number of students choosing exactly one elective: Total of T minus those choosing exactly two (50) will give a clue here. The split can aid the calculation.

Therefore, the option that cannot be directly determined from the given information is the Number of students choosing both E1 and E2. This requires explicit counts of intersections of E1 and E2, not provided.

Answer 2

To determine the necessary and sufficient information to compute the number of students choosing only E1, only E2, and only E3, we need to analyze the given data and establish equations for the problem. Given:
1. Total students = 190
2. Students choosing two electives = 50
3. Students choosing E1 is half of those choosing E2, and one-third of those choosing E3.

For students:
Let:

• E1 be students choosing E1
• E2 be students choosing E2
• E3 be students choosing E3
• x be students choosing only E1
• y be students choosing only E2
• z be students choosing only E3
• a be students choosing both E1 and E2
• b be students choosing both E2 and E3
• c be students choosing both E3 and E1

From the given:

• E1 = a + c + x
• E2 = a + b + y
• E3 = b + c + z

Since:
E1 = 0.5 * E2 and E1 = (1/3) * E3
a + c + x = 0.5 * (a + b + y)
a + c + x = (1/3) * (b + c + z)
Total students choosing 2 electives:
a + b + c = 50
Total students choosing at least 1 elective:
x + y + z + a + b + c = 190

We can conclude that:

The information required is: "Number of students choosing only E2, and number of students choosing both E2 and E3."