Question

Let A = {x : x is positive integral multiple of 3 less than 31} and B = { x : x is prime number less than 30}, Then n(AUB) + n(AՈB) is equal to: *​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$n(AUB) = n(A) + n(B) - n(AՈB)$

GIVEN

A = {x : x is positive integral multiple of 3 less than 31}

B = { x : x is prime number less than 30}

TO DETERMINE

n(AUB) + n(AՈB)

CALCULATION

A = {x : x is positive integral multiple of 3 less than 31} = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

B = { x : x is prime number less than 30}

= { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

So

$n(A) = 10 \: \: and \: \: \: n(B) = 10$

Hence

$n(AUB) + n(AՈB)$

$= n(A) + n(B) - n(AՈB) + n(AՈB)$

$= n(A) + n(B) = 10 + 10 = 20$

Answer 2

Step-by-step explanation:

FORMULA TO BE IMPLEMENTED

n(AUB) = n(A) + n(B) - n(AՈB)n(AUB)=n(A)+n(B)−n(AՈB)

GIVEN

A = {x : x is positive integral multiple of 3 less than 31}

B = { x : x is prime number less than 30}

TO DETERMINE

n(AUB) + n(AՈB)

CALCULATION

A = {x : x is positive integral multiple of 3 less than 31} = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

B = { x : x is prime number less than 30}

= { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

So

n(A) = 10 \: \: and \: \: \: n(B) = 10n(A)=10andn(B)=10

Hence

n(AUB) + n(AՈB)n(AUB)+n(AՈB)

= n(A) + n(B) - n(AՈB) + n(AՈB)=n(A)+n(B)−n(AՈB)+n(AՈB)

= n(A) + n(B) = 10 + 10 = 20=n(A)+n(B)=10+10=20

Hope it helps!