Question
In a class of 55 students 21 studies physics 24 study geography and 23 study economics is six study both physics and geography AIDS study both geography and economics and five study both economics and physics if x study all three subjects and 2x studies none of the three subjects find the value of x
Answers
Answered by
GPT-4o mini
To solve the problem, we will use the principle of inclusion-exclusion to find \( x \), the number of students studying all three subjects.
Let:
- \( P \) = number of students studying Physics = 21
- \( G \) = number of students studying Geography = 24
- \( E \) = number of students studying Economics = 23
- \( PG \) = students studying both Physics and Geography = 6
- \( GE \) = students studying both Geography and Economics = 4
- \( EP \) = students studying both Economics and Physics = 5
- \( x \) = students studying all three subjects
According to the principle of inclusion-exclusion, the total number of students studying at least one subject can be expressed as:
\[
|P \cup G \cup E| = |P| + |G| + |E| - |PG| - |GE| - |EP| + |P \cap G \cap E|
\]
Substituting the known values into the equation, we have:
\[
|P \cup G \cup E| = 21 + 24 + 23 - 6 - 4 - 5 + x
\]
Calculating the total:
\[
|P \cup G \cup E| = 21 + 24 + 23 - (6 + 4 + 5) + x
\]
\[
|P \cup G \cup E| = 68 - 15 + x
\]
\[
|P \cup G \cup E| = 53 + x
\]
According to the problem, there are a total of 55 students in the class, and \( 2x \) students study none of the three subjects. Therefore, the number of students studying at least one subject is given by:
\[
|P \cup G \cup E| = 55 - 2x
\]
Setting the two expressions for \( |P \cup G \cup E| \) equal gives:
\[
53 + x = 55 - 2x
\]
Now, we will solve for \( x \):
1. Add \( 2x \) to both sides:
\[
53 + x + 2x = 55
\]
\[
53 + 3x = 55
\]
2. Subtract 53 from both sides:
\[
3x = 2
\]
3. Divide by 3:
\[
x = \frac{2}{3}
\]
Thus, the value of \( x \) is \( \frac{2}{3} \).
This suggests that \( x \), which represents the number of students studying all three subjects, is a fraction, which is unusual in a context involving a whole number of students. Please check the problem statement for any potential errors or reassess the scenario, as typically we expect \( x \) to be a whole number.
Let:
- \( P \) = number of students studying Physics = 21
- \( G \) = number of students studying Geography = 24
- \( E \) = number of students studying Economics = 23
- \( PG \) = students studying both Physics and Geography = 6
- \( GE \) = students studying both Geography and Economics = 4
- \( EP \) = students studying both Economics and Physics = 5
- \( x \) = students studying all three subjects
According to the principle of inclusion-exclusion, the total number of students studying at least one subject can be expressed as:
\[
|P \cup G \cup E| = |P| + |G| + |E| - |PG| - |GE| - |EP| + |P \cap G \cap E|
\]
Substituting the known values into the equation, we have:
\[
|P \cup G \cup E| = 21 + 24 + 23 - 6 - 4 - 5 + x
\]
Calculating the total:
\[
|P \cup G \cup E| = 21 + 24 + 23 - (6 + 4 + 5) + x
\]
\[
|P \cup G \cup E| = 68 - 15 + x
\]
\[
|P \cup G \cup E| = 53 + x
\]
According to the problem, there are a total of 55 students in the class, and \( 2x \) students study none of the three subjects. Therefore, the number of students studying at least one subject is given by:
\[
|P \cup G \cup E| = 55 - 2x
\]
Setting the two expressions for \( |P \cup G \cup E| \) equal gives:
\[
53 + x = 55 - 2x
\]
Now, we will solve for \( x \):
1. Add \( 2x \) to both sides:
\[
53 + x + 2x = 55
\]
\[
53 + 3x = 55
\]
2. Subtract 53 from both sides:
\[
3x = 2
\]
3. Divide by 3:
\[
x = \frac{2}{3}
\]
Thus, the value of \( x \) is \( \frac{2}{3} \).
This suggests that \( x \), which represents the number of students studying all three subjects, is a fraction, which is unusual in a context involving a whole number of students. Please check the problem statement for any potential errors or reassess the scenario, as typically we expect \( x \) to be a whole number.
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