Question

Atal Academy focuses on the performing arts. In the current batch, 40 students can sing, 60 students can dance and 24 students can act. No student is a triple threat (some one who can sing, dance and act). However, 16 students can sing as well as dance, 12 students can dance and act and 8 students can sing and act. How many students have only one performing talent?

• A. 44
• B. 52
• C. 60
• D. 66

Hint:

When the triple intersection is zero, the number of people having exactly one attribute is simply: $\sum(\text{Total}) - 2 \times \sum(\text{Intersections of two sets})$.
Calculation: $(40 + 60 + 24) - 2 \times (16 + 12 + 8) = 124 - 2 \times 36 = 124 - 72 = 52$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves set theory and Venn diagrams for three overlapping sets (Singing, Dancing, and Acting). We need to calculate the number of students in the "only one" category for each set.
Step 2: Key Formula or Approach:
Let $S$, $D$, and $A$ represent the sets of students who can sing, dance, and act respectively.
$n(S) = 40, n(D) = 60, n(A) = 24$.
Intersections: $n(S \cap D) = 16, n(D \cap A) = 12, n(S \cap A) = 8$.
Triple intersection: $n(S \cap D \cap A) = 0$ (No student is a triple threat).
Step 3: Detailed Explanation:
To find students with only one talent, we subtract the overlapping categories from the total of each set. Since the triple intersection is zero, the calculation is simplified:
1. Singing only:
\[ \text{Only } S = n(S) - [n(S \cap D) + n(S \cap A)] = 40 - (16 + 8) = 40 - 24 = 16 \]
2. Dancing only:
\[ \text{Only } D = n(D) - [n(S \cap D) + n(D \cap A)] = 60 - (16 + 12) = 60 - 28 = 32 \]
3. Acting only:
\[ \text{Only } A = n(A) - [n(S \cap A) + n(D \cap A)] = 24 - (8 + 12) = 24 - 20 = 4 \]
Total students with only one talent = $16 + 32 + 4 = 52$.
Step 4: Final Answer:
There are 52 students who have only one performing talent.