Question

There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics?

Answer 1

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:

$A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)$

In which a is the number of students that take only calculus, $A \cap B$ is the number of students that take both calculus and physics, $A \cap C$ is the number of students that take both calculus and chemistry and $A \cap B \cap C$ is the number of students that take calculus, physics and chemistry.

By the same logic, we have:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

This diagram has the following subsets:

$a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C), D$

There are 360 people in my school. This means that:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360$

The problem states that:

15 take calculus, physics, and chemistry, so:

$A \cap B \cap C = 15$

15 don't take any of them, so:

$D = 15$

75 take both calculus and chemistry, so:

$A \cap C = 75$

75 take both physics and chemistry, so:

$B \cap C = 75$

30 take both physics and calculus, so:

$A \cap B = 30$

Solution:

The problem states that 180 take calculus. So

$a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 180$

$a + 30 + 75 + 15 = 180$

$a = 180 - 120$

$a = 60$

Twice as many students take chemistry as take physics:

It means that: $C = 2B$

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$B = b + 75 + 30 + 15$

$B = b + 120$

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$C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)$

$C = c + 75 + 75 + 15$

$C = c + 165$

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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.

$C = 2B$

$c + 165 = 2(b+120)$

$c = 2b + 240 - 165$

$c = 2b + 75$

The equation that sums all the subsets is:

$a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360$

$60 + b + 2b + 75 + 30 + 75 + 15 + 15 = 360$

$3b + 270 = 360$

$3b = 90$

$b = \frac{90}{3}$

$b = 30$

30 students take only physics.

The number of student that take physics is:

$B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)$

$B = b + 75 + 30 + 15$

$B = 30 + 120$

$B = 150$

150 students take physics.