Question

$U=\{x:x\text{is a natural number \&}x\le 10\}$
$A=\{a:a\text{is an even number \&}a\le 10\}$
$B=\{b:b\text{is a prime number \&}b\le 9\}$

Then, $A^c\cap B^c$ is

• A. $\{2,4,6,8\}$
• B. $\{3,5,7\}$
• C. $\{1,9,10\}$
• D. $\{2,3,5,7\}$

Answer 1

The correct option is C

$\{1,9,10\}$

$U=\{1,2,3,4,5,6,7,8,9,10\}$

$A=\{2,4,6,8\}$

$B=\{2,3,5,7\}$

$A^c=U-A$ ≡ All the elements in $U$ which does not belong to $A$

$=\{1,3,5,7,9,10\}$

$B^c=U-B$ ≡ Elements which belongs to $U$ and does not belong to $B$

$=\{1,4,6,8,9,10\}$

$A^c\cap B^c=\{1,9,10\}$

Alternative solution :
Using De Morgan's Law

$A^c\cap B^c=(A\cup B{)}^c$
$A\cup B=\{2,3,4,5,6,7,8\}$
$(A\cup B{)}^c=\{1,9,10\}$