Question

Let \( A, B, C \) be any three finite sets. If \( n(A \times B) = 160\), \( n(B \times C) = 80 \) and \( n(C \times A) = 200\), then \( n(A) = \)

• A. \(10 \)
• B. \(18 \)
• C. \(16 \)
• D. \(12 \)
• E. \(20 \)

Hint:

For Cartesian products, always convert into multiplication equations and eliminate variables step-by-step.

Answer 1

Answer: (E)

Concept: For finite sets, the number of elements in the Cartesian product is: \[ n(A \times B) = n(A)\cdot n(B) \]

Step 1: Form equations using given data.
\[ n(A)n(B) = 160 \quad ...(1) \] \[ n(B)n(C) = 80 \quad ...(2) \] \[ n(C)n(A) = 200 \quad ...(3) \]

Step 2: Multiply all three equations.
\[ [n(A)n(B)] [n(B)n(C)] [n(C)n(A)] = 160 \times 80 \times 200 \] \[ [n(A)]^2 [n(B)]^2 [n(C)]^2 = 160 \cdot 80 \cdot 200 \]

Step 3: Take square root.
\[ n(A)n(B)n(C) = \sqrt{160 \cdot 80 \cdot 200} \] \[ = \sqrt{(16 \cdot 10)(16 \cdot 5)(20 \cdot 10)} = \sqrt{256000} = 160\sqrt{10} \]

Step 4: Divide (1) by (2).
\[ \frac{n(A)n(B)}{n(B)n(C)} = \frac{160}{80} \Rightarrow \frac{n(A)}{n(C)} = 2 \Rightarrow n(A) = 2n(C) \]

Step 5: Substitute into (3).
\[ n(C)\cdot 2n(C) = 200 \Rightarrow 2n(C)^2 = 200 \Rightarrow n(C)^2 = 100 \Rightarrow n(C)=10 \] \[ n(A)=2n(C)=20 \]