Question

9. If} = {natural numbers between 10 and 40}, A = {mutiples of 5) and B =
{multiples of 6}, verify that n(AUB) = n(A) + n(B) - n(ANB)
(5)​

Answer 1

SOLUTION

GIVEN

• U = { Natural numbers between 10 and 40 }

• A = { Multiples of 5 }

• B = { Multiples of 6 }

TO VERIFY

$\sf{n(A \cup B) = n(A) + n(B) - n(A \cap B) \: }$

EVALUATION

Here it is given that

U = { Natural numbers between 10 and 40 }

A = { Multiples of 5 }

= { 15, 20, 25, 30, 35 }

$\therefore \: \: \sf{n(A) = 5 \: }$

B = { Multiples of 6 }

= { 12, 18, 24, 30, 36 }

$\therefore \: \: \sf{n(B ) = 5 \: }$

Now

$\sf{A \cap B = \{30 \} \: }$

$\therefore \: \: \sf{n(A \cap B ) = 1 \: }$

Again

A U B = { 12, 15, 18, 20, 24, 25, 30, 35, 36 }

$\therefore \: \: \sf{n(A \cup B ) = 9 \: }$

LHS

$= \sf{n(A \cup B ) \: }$

$= \sf{ 9}$

RHS

$\sf{= n(A) + n(B) - n(A \cap B) \: }$

$\sf{ = 5 + 5 - 1}$

$= \sf{ 9}$

Therefore LHS = RHS

Hence verified

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