Forty students watched films A, B and C over a week. Each student watched either only one film or all three. Thirteen students watched film A, sixteen students watched film B and nineteen students watched film C. How many students watched all three films?
Options:
(A) 0
(B) 2
(C) 4
(D) 8
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2018 Q.6)**
**1. Intuitive Physical Setup & Given Parameters**:
- Total students: $N = 40$.
- Each student watched **either only one film or all three films**.
- This means the number of students watching exactly two films is zero:
n(\text{exactly 2}) = 0
- Total film viewership:
- $n(A) = 13$, $n(B) = 16$, $n(C) = 19$.
**2. Step-by-Step Mathematical Derivation**:
1. **Let the unknown variables be**:
- $x$: number of students who watched all three films ($A \cap B \cap C$).
- $a, b, c$: number of students who watched only film $A$, only film $B$, and only film $C$ respectively.
2. **Formulate the system of equations**:
- Viewership for each film:
n(A) = a + x = 13
n(B) = b + x = 16
n(C) = c + x = 19
- Total students (since everyone watched either only one or all three):
(a + b + c) + x = 40 \implies a + b + c = 40 - x
3. **Summing the individual film equations**:
(a + x) + (b + x) + (c + x) = 13 + 16 + 19 = 48
(a + b + c) + 3x = 48
4. **Substitute $a + b + c = 40 - x$**:
(40 - x) + 3x = 48
40 + 2x = 48 \implies 2x = 8 \implies x = 4
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** 0: ❌ Incorrect. If $x=0$, total students would be $13+16+19 = 48 > 40$.
- **(B)** 2: ❌ Incorrect. Yields total students $40+2(2) = 44 \neq 48$.
- **(C)** 4: ✅ **CORRECT**. $40 + 2(4) = 48$, perfectly consistent with given numbers.
- **(D)** 8: ❌ Incorrect. Yields $40 + 2(8) = 56 \neq 48$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **C**
- **Exam Rule**: In Venn diagram problems where two-set intersections are zero: $\sum n_i = N_{total} + 2x_{all\text{ }3}$.