In a class of 80 students, 40 study Physics, 48 study Mathematics and 44 study Chemistry. 20 study Physics and Mathematics, 24 study Physics and Chemistry and 32 study only two of... In a class of 80 students, 40 study Physics, 48 study Mathematics and 44 study Chemistry. 20 study Physics and Mathematics, 24 study Physics and Chemistry and 32 study only two of the three subjects. If every student studies at least one of the three subjects, find: (a) the number of students who study all three subjects, (b) the number of students who study only Mathematics and Chemistry. Read more

Steps to Solve

1. Define the Variables

Let:

• $n(P)$ = number of students studying Physics = 40
• $n(M)$ = number of students studying Mathematics = 48
• $n(C)$ = number of students studying Chemistry = 44
• $n(P \cap M)$ = number studying both Physics and Mathematics = 20
• $n(P \cap C)$ = number studying both Physics and Chemistry = 24
• $n(M \cap C)$ = number studying both Mathematics and Chemistry (unknown)
• $n(P \cap M \cap C)$ = number studying all three subjects (unknown)
• Total students = 80

2. Use the Inclusion-Exclusion Principle

According to the principle, the total number of students can be expressed as:

$$ n(P \cup M \cup C) = n(P) + n(M) + n(C) - n(P \cap M) - n(P \cap C) - n(M \cap C) + n(P \cap M \cap C) $$

Substituting known values:

$$ 80 = 40 + 48 + 44 - 20 - 24 - n(M \cap C) + n(P \cap M \cap C) $$

3. Rearrange to Extract Unknowns

Combine the known quantities:

$$ 80 = 88 - n(M \cap C) + n(P \cap M \cap C) $$

This simplifies to:

$$ n(M \cap C) - n(P \cap M \cap C) = 8 \quad \text{(1)} $$

4. Calculate the Number of Students Studying Exactly Two Subjects

We know that the total number of students studying exactly two subjects is 32, which can be defined as:

$$ n(P \cap M) + n(P \cap C) + n(M \cap C) - 3n(P \cap M \cap C) = 32 $$

Substituting known values:

$$ 20 + 24 + n(M \cap C) - 3n(P \cap M \cap C) = 32 $$

Simplifying gives:

$$ n(M \cap C) - 3n(P \cap M \cap C) = -12 \quad \text{(2)} $$

5. Solve the System of Equations

Now we have a system of equations:

From (1):

$$ n(M \cap C) = n(P \cap M \cap C) + 8 $$

Substituting into (2):

$$ (n(P \cap M \cap C) + 8) - 3n(P \cap M \cap C) = -12 $$

Combining gives:

$$ -2n(P \cap M \cap C) + 8 = -12 $$

From which we find:

$$ -2n(P \cap M \cap C) = -20 $$

Thus:

$$ n(P \cap M \cap C) = 10 $$

6. Calculate Students Studying Only Mathematics and Chemistry

Now substituting $n(P \cap M \cap C)$ back into equation (1):

$$ n(M \cap C) = 10 + 8 = 18 $$

To find the number of students studying only Mathematics and Chemistry:

Let $n(M \cap C \text{ only}) = x$.

Using:

$$ x + n(P \cap M \cap C) = n(M \cap C) $$

We have:

$$ x + 10 = 18 $$

Thus:

$$ x = 8 $$