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Easy

Question

In a town of $10000$ families $blank$ it was found that 40% families buy newspaper $A comma 20 percent sign$ families buy newspaper $B$ and 10% families buy newspaper $C comma blank 5 percent sign$ families buy $A$ and $B comma$ 3% buy $B$ and $C$ and 4% buy $A$ and $C.$ If 2% families buy all the three newspapers, then the number of families which buy $A$ only is

3100
3300
2900
1400

The correct answer is: 3300

We have,
$N equals 10000 comma n open parentheses A close parentheses equals 40 percent sign blank o f blank 10000 equals 4000 comma$
$n open parentheses B close parentheses equals 2000 comma n open parentheses C close parentheses equals 1000 comma n open parentheses A intersection B close parentheses equals 500 comma$
$n open parentheses B intersection C close parentheses equals 300 comma n open parentheses C intersection A close parentheses equals 400 comma n open parentheses A intersection B intersection C close parentheses equals 200$
Now,
Required number of families =
$n open parentheses A intersection stack B with minus on top intersection stack C with minus on top close parentheses equals n left parenthesis A intersection open parentheses B union C close parentheses to the power of ´ end exponent right parenthesis$
$equals n open parentheses A close parentheses minus n left parenthesis A intersection left parenthesis B union C right parenthesis right parenthesis$
$equals n open parentheses A close parentheses minus n open parentheses open parentheses A intersection B close parentheses union open parentheses A intersection C close parentheses close parentheses$
$equals n open parentheses A close parentheses minus open curly brackets n open parentheses A intersection B close parentheses plus n open parentheses A intersection C close parentheses minus n open parentheses A intersection B intersection C close parentheses close curly brackets$
$equals 4000 minus open parentheses 500 plus 400 minus 200 close parentheses equals 3300$

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