If P and Q are two sets such that P- Q has 40 elements- P has 22 elements and Q has 28 elements- How many does P- Q have-

The correct option is C 10 Given, P∪ Q=40, n(P)=22, n(Q)=28 To find: n(P∩ Q) Therefore, n (P∪ Q) = n(P) + n(Q) n (P∩ Q) ⇒ n(P∩ Q)=n( ...

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Standard VII

Mathematics

Cardinality of Sets

If P and ...

Question

If $P$ and $Q$ are two sets such that $P \cup Q$ has $40$ elements, $P$ has $22$ elements and $Q$ has $28$ elements, How many does $P \cap Q$ have?

20

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28

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10

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Solution

The correct option is C $10$
Given, $P \cup Q = 40, n (P) = 22, n (Q) = 28$
To find: $n (P \cap Q)$
Therefore, $n (P \cup Q) = n (P) + n (Q) - n (P \cap Q)$
$\Rightarrow n (P \cap Q) = n (P) + n (Q) - n (P \cup Q)$
$= 22 + 28 - 40$
$= 10$

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If P and Q are two sets such that P∪Q has 40 elements, P has 22 elements and Q has 28 elements, How many does P∩Q have?

The correct option is C 10Given, P∪Q=40,n(P)=22,n(Q)=28To find: n(P∩Q)Therefore, n(P∪Q)=n(P)+n(Q)−n(P∩Q)⇒n(P∩Q)=n(P)+n(Q)−n(P∪Q)=22+28−40=10

If $P$ and $Q$ are two sets such that $P \cup Q$ has $40$ elements, $P$ has $22$ elements and $Q$ has $28$ elements, How many does $P \cap Q$ have?

The correct option is C $10$
Given, $P \cup Q = 40, n (P) = 22, n (Q) = 28$
To find: $n (P \cap Q)$
Therefore, $n (P \cup Q) = n (P) + n (Q) - n (P \cap Q)$
$\Rightarrow n (P \cap Q) = n (P) + n (Q) - n (P \cup Q)$
$= 22 + 28 - 40$
$= 10$