Question

Let the sets $A=\{2,4,6,8,...\}$ and $B=\{3,6,9,12,...\}$, and $n\left(A\right)=200,n\left(B\right)=250$. Then

• A. $n(A\cap B)=67$
• B. $n(A\cup B)=450$
• C. $n(A\cap B)=66$
• D. $n(A\cup B)=384$

Answer 1

Answer: (C), (D)

The correct options are
C $n(A\cup B)=384$
D $n(A\cap B)=66$

In A, last term will be $400$.

In B, the terms are also in A.P having a common difference of $3$.

Hence

$a_n=a_1+(n-1)d$.

Now $n=250$ for the last term.

Hence

$a_{250}=3+(250-1).3$
$=3(1+250-1)$
$=750.$

Now $A\cap B$ will have elements which are multiples of $6$.

Last term will be $400-4=396$.

Hence
$a_n=a+(n-1).d$
$d=6,n=?,a=6$ and $a_n=396$

Hence
$396=6+(n-1).6$
Or

$66=n$.

Hence
$n(A\cap B)=66$.

Now
$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$
$=200+250-66$
$=384$.