Respuesta :
Answer:
150 students take physics.
Step-by-step explanation:
To solve this problem, we must build the Venn's Diagram of this set.
I am going to say that:
-The set A represents the students that take calculus.
-The set B represents the students that take physics
-The set C represents the students that take chemistry.
-The set D represents the students that do not take any of them.
We have that:
[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]
In which a is the number of students that take only calculus, [tex]A \cap B[/tex] is the number of students that take both calculus and physics, [tex]A \cap C[/tex] is the number of students that take both calculus and chemistry and [tex]A \cap B \cap C[/tex] is the number of students that take calculus, physics and chemistry.
By the same logic, we have:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
This diagram has the following subsets:
[tex]a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C), D[/tex]
There are 360 people in my school. This means that:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360[/tex]
The problem states that:
15 take calculus, physics, and chemistry, so:
[tex]A \cap B \cap C = 15[/tex]
15 don't take any of them, so:
[tex]D = 15[/tex]
75 take both calculus and chemistry, so:
[tex]A \cap C = 75[/tex]
75 take both physics and chemistry, so:
[tex]B \cap C = 75[/tex]
30 take both physics and calculus, so:
[tex]A \cap B = 30[/tex]
Solution:
The problem states that 180 take calculus. So
[tex]a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 180[/tex]
[tex]a + 30 + 75 + 15 = 180[/tex]
[tex]a = 180 - 120[/tex]
[tex]a = 60[/tex]
Twice as many students take chemistry as take physics:
It means that: [tex]C = 2B[/tex]
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]B = b + 75 + 30 + 15[/tex]
[tex]B = b + 120[/tex]
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[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]
[tex]C = c + 75 + 75 + 15[/tex]
[tex]C = c + 165[/tex]
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Our interest is the number of student that take physics. We have to find B. For this we need to find b. We can write c as a function o b, and then replacing it in the equations that sums all the subsets.
[tex]C = 2B[/tex]
[tex]c + 165 = 2(b+120)[/tex]
[tex]c = 2b + 240 - 165[/tex]
[tex]c = 2b + 75[/tex]
The equation that sums all the subsets is:
[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360[/tex]
[tex]60 + b + 2b + 75 + 30 + 75 + 15 + 15 = 360[/tex]
[tex]3b + 270 = 360[/tex]
[tex]3b = 90[/tex]
[tex]b = \frac{90}{3}[/tex]
[tex]b = 30[/tex]
30 students take only physics.
The number of student that take physics is:
[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]
[tex]B = b + 75 + 30 + 15[/tex]
[tex]B = 30 + 120[/tex]
[tex]B = 150[/tex]
150 students take physics.
