Question

Consider three sets $A,B,C$ such that set A contains all three digit numbers that are multiples of $4,$ set $B$ contains all three digit even numbers that are multiples of $3$ and set $C$ contains all three digit numbers that are multiples of $5.$ Then the value of $\frac{n\left[\right(A\cap B)\times (A\cap C\left)\right]}{n\left[\right(B\cap C)\times (A\cap B\cap C\left)\right]}$ is

• A. $\frac{15}{2}$
• B. $\frac{15}{4}$
• C. $15$
• D. $30$

Answer 1

The correct option is A $\frac{15}{2}$

$A=\{100,104,108,...,996\}$
$B=\{102,108,114,...,996\}$
$C=\{100,105,110,...,995\}$

$A\cap B=$ All $3$ digit multiples of $12$
$\Rightarrow \{108,120,132,...,996\}$
$\Rightarrow \frac{996-108}{12}+1=75$ elements.

$B\cap C=$ All $3$ digit multiples of $30$
$\Rightarrow \{120,150,180,...,990\}$
$\Rightarrow \frac{990-120}{30}+1=30$ elements.

$C\cap A=$ All $3$ digit multiples of $20$
$\Rightarrow \{100,120,140,160,...,980\}$
$\Rightarrow \frac{980-100}{20}+1=45$ elements.

$A\cap B\cap C=$ All $3$ digit multiples of $60$
$\Rightarrow \{120,180,...,960\}$
$\Rightarrow \frac{960-120}{60}+1=15$ elements.

$\frac{n\left[\right(A\cap B)\times (A\cap C\left)\right]}{n\left[\right(B\cap C)\times (A\cap B\cap C\left)\right]}$
$=\frac{n(A\cap B)\times n(A\cap C)}{n(B\cap C)\times n(A\cap B\cap C)}$
$=\frac{75\times 45}{30\times 15}$
$=\frac{15}{2}$