Question

there are 190 students taking band, chorus, or both. if there are 180 students taking band and 60 students in both band and chorus, how many students are only in chorus? 60 70 120 10

Explanation:

Answer:

To solve for the number of students only in chorus, we use the principle of inclusion - exclusion for sets. Let \( B \) be the set of students in band and \( C \) be the set of students in chorus. We know that \( n(B\cup C)=190 \), \( n(B) = 180 \), and \( n(B\cap C)=60 \).

First, we find the number of students in chorus (\( n(C) \)) using the formula \( n(B\cup C)=n(B)+n(C)-n(B\cap C) \).

Substituting the known values: \( 190=180 + n(C)-60 \)

Simplify the right - hand side: \( 190=n(C)+120 \)

Then, solve for \( n(C) \): \( n(C)=190 - 120=70 \)

The number of students only in chorus is \( n(C)-n(B\cap C) \).

Substitute \( n(C) = 70 \) and \( n(B\cap C)=60 \): \( 70 - 60 = 10 \)

Wait, there is a mistake above. Let's re - do it.

We know that \( n(B\cup C)=n(\text{only }B)+n(\text{only }C)+n(B\cap C) \)

We know that \( n(B)=n(\text{only }B)+n(B\cap C) \), so \( n(\text{only }B)=n(B)-n(B\cap C)=180 - 60 = 120 \)

And \( n(B\cup C)=n(\text{only }B)+n(\text{only }C)+n(B\cap C) \)

Substitute \( n(B\cup C) = 190 \), \( n(\text{only }B)=120 \), and \( n(B\cap C)=60 \) into the formula:

\( 190=120 + n(\text{only }C)+60 \)

\( 190=n(\text{only }C)+180 \)

Then \( n(\text{only }C)=190 - 180=10 \)

Wait, but let's check again. The correct formula for two sets \( A \) and \( B \) is \( n(A\cup B)=n(A)+n(B)-n(A\cap B) \)

We know \( n(B\cup C) = 190 \), \( n(B)=180 \), \( n(B\cap C)=60 \)

So \( 190=180 + n(C)-60 \)

\( 190=120 + n(C) \)

\( n(C)=70 \)

The number of students only in chorus is \( n(C)-n(B\cap C)=70 - 60 = 10 \)

But the options are 60, 70, 120, 10. So the answer is 10.