Question

If A and B be two sets such that n(A) = 15, n(B) =25, then number of possible values of $n\left(A\Delta B\right)$(symmetric difference of A and B) is

• A. 30
• B. 16
• C. 26
• D. 40

Answer 1

The correct option is B 16

$n\left(A△B\right)=n(A\cup B)-n(A\cap B)$

for $n$ (A \triangle B)$tobemax.$n (A \cap B)=0$

We know, that

$n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)=15+25-0=40$

$\Rightarrow n(A△B{)}_{max}=40-0=40$

For minimum value of $n\left(A△B\right)$

$n(A\cup B)$ should be min, $n(A\cap B)$ should be max.

$n\left(A△B\right)$ min $=25-15=10$

So. value of

$n\left(A△B\right)=n(A\cup B)-n(A\cap B)$ lies om the set

$10,11,12,......,3,9,40$

Now, when $n\left(A△B\right)$ is max. i.e. when

$n(A\cup B)=40$ & $n(A\cap B)=0$

If we decrease $n(A\cup B)$ by $1$ then $n(A\cap B)$

Will increase by $1$
$n\left(A△B\right)=39-1=38$

Similarly on for the decrease of $1$ you will get in $\left(A△B\right)$ as $36$ and $30$ so on.

Hence

Range of $n\left(A△B\right)=10,12,14,16,18,20,......,38,40$

$=16$ values