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

By de Morgan's law of complementation, n(A' nbsp;∩ B') = n((A∪ B)') But, by the definition of a complement of a set, n((A∪ B)')=n(U) n(A ∪ B). = n(U) [n( ...

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Byju's Answer

Standard VIII

Mathematics

Complement of a Set

If nU = 700, ...

Question

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

400

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600

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300

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200

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Solution

The correct option is C
300

By de-Morgan's law of complementation,
$n (A^{'} \cap B^{'}) = n ((A \cup B)^{'})$
But, by the definition of a complement of a set,
$n ((A \cup B)^{'}) = n (U) - n (A \cup B) .$

$= n (U) - [n (A) + n (B) - n (A \cap B)]$

$= 700 - [200 + 300 - 100]$

$= 300$ .

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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. 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

, then

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

Q. If

n (U) = 700, n (A) = 200, n (B) = 240, n (A \cap B) = 100,

then

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

is equal to

Q. Let

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

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

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

The correct option is C 300 By de-Morgan's law of complementation, n(A′∩B′)=n((A∪B)′) But, by the definition of a 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 $n (U) = 700, n (A) = 200, n (B) = 300$ and $n (A \cap B) = 100,$ then $n (A^{'} \cap B^{'}) =$ ___.

The correct option is C
300

By de-Morgan's law of complementation,
$n (A^{'} \cap B^{'}) = n ((A \cup B)^{'})$
But, by the definition of a complement of a set,
$n ((A \cup B)^{'}) = n (U) - n (A \cup B) .$

$= n (U) - [n (A) + n (B) - n (A \cap B)]$

$= 700 - [200 + 300 - 100]$

$= 300$ .