Step 1: Understanding the Problem

We are given the following information about a class of 70 students:

• $n(U) = 70$ (total number of students)
• $n(M) = 45$ (students who offer Mathematics)
• $n(C) = 37$ (students who offer Chemistry)
• $n(P) = 43$ (students who offer Physics)
• $n(M \cap C \cap P) = 5$ (students who offer all three subjects)
• 20 students offer both Physics and Chemistry only.
• 3 students offer Physics only.
• Each student offers at least one of the three subjects.

Step 2: Set Up the Venn Diagram

Let's denote:

• $x_1$ = number of students who offer Mathematics only.
• $x_2$ = number of students who offer Chemistry only.
• $x_3$ = number of students who offer Physics only (given as 3).
• $x_4$ = number of students who offer both Mathematics and Chemistry, but not Physics.
• $x_5$ = number of students who offer both Mathematics and Physics, but not Chemistry.
• $x_6$ = number of students who offer both Chemistry and Physics, but not Mathematics (given as 20).
• $x_7$ = number of students who offer all three subjects (given as 5).

Step 3: Venn Diagram Representation

To calculate the unknowns:

1. Total for Mathematics:
   $x_1 + x_4 + x_5 + x_7 = 45$

2. Total for Chemistry:
   $x_2 + x_4 + x_6 + x_7 = 37$

3. Total for Physics:
   $x_3 + x_5 + x_6 + x_7 = 43$

Step 4: Substituting Given Values

• $x_3 = 3$ (Physics only)
• $x_6 = 20$ (Physics and Chemistry only)
• $x_7 = 5$ (All three subjects)

Using the equation for Physics: $3 + x_5 + 20 + 5 = 43 \implies x_5 = 15$

Using the equation for Mathematics: $x_1 + x_4 + 15 + 5 = 45 \implies x_1 + x_4 = 25$

Using the equation for Chemistry: $x_2 + x_4 + 20 + 5 = 37 \implies x_2 + x_4 = 12$

Step 5: Find $x_4$, $x_1$, and $x_2$

Subtract the equations for Chemistry and Mathematics to solve for $x_1$ and $x_2$: $(x_1 + x_4) - (x_2 + x_4) = 25 - 12 \implies x_1 - x_2 = 13$

Now, add $x_1 + x_2 = 25 + 12 = 37$, then solve the system: $x_1 + x_2 = 37 \quad \text{and} \quad x_1 - x_2 = 13$ Solving gives: $2x_1 = 50 \implies x_1 = 25, \quad x_2 = 12$ Thus, $x_4 = 0$.

Step 6: Summary of Students

• $x_1 = 25$ (Mathematics only)
• $x_2 = 12$ (Chemistry only)
• $x_3 = 3$ (Physics only)
• $x_4 = 0$ (Mathematics and Chemistry only)
• $x_5 = 15$ (Mathematics and Physics only)
• $x_6 = 20$ (Physics and Chemistry only)
• $x_7 = 5$ (All three subjects)

Step 7: Answer the Questions

(a) Chemistry only:
$n(C \text{ only}) = x_2 = 12$

(b) Only one subject:
$n(\text{only one subject}) = x_1 + x_2 + x_3 = 25 + 12 + 3 = 40$

(c) Only two subjects:
$n(\text{only two subjects}) = x_4 + x_5 + x_6 = 0 + 15 + 20 = 35$

Summary

• (a) 12 students offer Chemistry only.
• (b) 40 students offer only one subject.
• (c) 35 students offer only two subjects.

Would you like more details on any part? Here are some questions to further your understanding:

1. What would change if 10 students offered all three subjects instead of 5?
2. How would the numbers change if 10 students offered Physics only?
3. Can you explain how the Venn diagram visually represents the problem?
4. What if no student offered all three subjects? How would that affect the solution?
5. How do these calculations relate to the principle of inclusion-exclusion?

Tip: Always start by assigning variables to unknowns and use the given conditions step by step.