If n(A)=50,n(B)=20 and n(A∩B)=10 then n(AΔB)=k2+1 then k=______

n(AΔB)=n(A∪B)−n(A∩B) =nA+nB-2nA∩B =50+20-20 =50Hence, k2+1=50k2=49k=±7Therefore, k=±7

If n(A)=50,n(B)=20 and n(A∩B)=10 then n(AΔB)=k2+1 then k=______

n(AΔB)=n(A∪B)−n(A∩B) =nA+nB-2nA∩B =50+20-20 =50Hence, k2+1=50k2=49k=±7Therefore, k=±7

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If $n (A) = 50, n (B) = 20$ and $n (A \cap B) = 10$ then $n (A Δ B) = k^{2} + 1$ then $k =______$

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$\pm 5$

$\pm 6$

$\pm 8$

$\pm 7$

answer is B.

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Detailed Solution

$\begin{array}{rcl} n (A Δ B) & = & n (A \cup B) - n (A \cap B) \\ = & n (A) + n (B) - 2 n (A \cap B) \\ = & 50 + 20 - 20 \\ = & 50 \end{array}$

Hence,

$\begin{array}{rcl} k^{2} + 1 & = & 50 \\ k^{2} & = & 49 \\ k & = & \pm 7 \end{array}$

Therefore, $k = \pm 7$

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If n(A)=50,n(B)=20 and n(A∩B)=10 then n(AΔB)=k2+1 then k=______

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If $n (A) = 50, n (B) = 20$ and $n (A \cap B) = 10$ then $n (A Δ B) = k^{2} + 1$ then $k =______$