Question

In a town of $10,000$ families it was found that $40$ family buy newspaper $A$, $20$ buy newspaper $B$ and $10\%$ families buy newspaper $C$, $5\%$ families buy $A$ and $B$, $3\%$ buy $B$ and $C$ and $4\%$ buy (A\) and $C$. If $2\%$ families buy all the three newspapers, then number of families which buy $A$ only is

• A. $3100$
• B. $3300$
• C. $2900$
• D. $1400$

Answer 1

The correct option is B

$3300$

Let $A,B$ and $C$ denotes the types of newspapers.
From the given information,
$n\left(A\right)=40\%$ of $10,000=4,000$

$n\left(B\right)=20\%$ of $10,000=2,000$

$n\left(C\right)=10\%$ of $10,000=1,000$

$n(A\cap B)=5\%$ of $10,000=500$

$n(B\cap C)=3\%$ of $10,000=300$

$n(C\cap A)=4\%$ of $10,000=400$

$n(A\cap B\cap C)=2\%$ of $10,000=200$

As the number of families which buy only newapaper $A=n\left(A\right)-\left[n\right(A\cap B)+n(A\cap C)-n(A\cap B\cap C\left)\right]$

$=4000-[500+400-200]=4000-700=3300$