If n- 700- nA - 200- nB - 300 and nA- B - 100- then nA- B- - -

From de Morgan's law of complementation, we have A' ∩ B' = (A∪ B)'. ⇒ n(A' ∩ B') = n((A∪ B)') But, nbsp;n((A∪ B)' nbsp;) nbsp;= nbsp;n(U) n(A ∪ B) by d ...

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Standard VIII

Mathematics

Intersection of Sets

If n∪= 700, n...

Question

$If n (\cup) = 700, n (A) = 200, n (B) = 300$ and $n (A \cap B) = 100,$ then $n (A^{^{'}} \cap B^{^{'}}) =$ ___.

400

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240

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300

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500

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Solution

The correct option is C
300

From de-Morgan's law of complementation, we have $A^{'} \cap B^{'} = (A \cup B)^{'} .$
$\Rightarrow n (A^{'} \cap B^{'}) = n ((A \cup B)^{'})$
But, $n ((A \cup B)^{'}) = n (U) - n (A \cup B)$ by definition of complement of a set.
$∴ n (A^{'} \cap B^{'}) = n (U) - n (A \cup B)$
$= n (U) - [n (A) + n (B) - n (A \cap B)]$
$= 700 - (200 + 300 - 100)$
$= 300$

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Similar questions

If $n (U) = 700, n (A) = 200, n (B) = 300$ and $n (A \cap B) = 100,$ then $n (A^{'} \cap B^{'}) =$ ___.

Q. Let

n (u) = 700, n (A) = 200, n (B) = 300, n (A \cap B) = 100

, then

n (A^{'} \cap B^{'}) =

Q. If

n (U) = 700

n (A) = 200

n (B) = 300

n (A \cap B) = 100

, then

n (A^{^{'}} \cap B^{^{'}})

is equal to

Q. Let

n (u) = 700, n (A) = 200, n (B) = 300

n (A \cap B) = 100, n (A^{'} \cap B^{'}) =

If $n (U) = 700, n (A) = 200, n (B) = 300$ and $n (A \cap B) = 100,$ then $n (A^{'} \cap B^{'}) =$ ___.

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If n(∪)=700, n(A)=200, n(B)=300 and n(A∩B)=100, then n(A′∩B′)= ___.

The correct option is C 300 From de-Morgan's law of complementation, we have A′∩B′=(A∪B)′. ⇒n(A′∩B′)=n((A∪B)′) But, n((A∪B)′)=n(U)−n(A∪B) by definition of complement of a set. ∴n(A′∩B′)=n(U)−n(A∪B) =n(U)−[n(A)+n(B)−n(A∩B)] =700−(200+300−100) =300

$If n (\cup) = 700, n (A) = 200, n (B) = 300$ and $n (A \cap B) = 100,$ then $n (A^{^{'}} \cap B^{^{'}}) =$ ___.