In a survey of brand preference for toothpastes, $82$ of the population (number of people covered for the survey is $100$ ) liked at least one of the brands: I, II and III. $40$ of those liked brand I, $25$ liked brand II and $35$ liked brand III. If $8$ of those asked, showed liking for all the three brands, then what percentage of those liked more than one of the three brands?

13

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10

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8

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Solution

The correct option is A $10$

By set theory

$n (I \cup I I \cup I I I) = n (I) + n (I I) + n (I I I) - n (I \cap I I) - n (I I \cap I I I) - n (I \cap I I I) + n (I \cap I I \cap I I I)$

$82 = 40 + 25 + 35 - x + 8$

$\Rightarrow x = 26$

But

$n (I \cap I I) + n (I I \cap I I I) + n (I \cap I I I)$ counts $n (I \cap I I \cap I I I)$ thrice whereas it needs to be counted only once.

Hence answer $= 26 - 2 \times 8 = 10$

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The correct option is A 10By set theoryn(I∪II∪III)=n(I)+n(II)+n(III)−n(I∩II)−n(II∩III)−n(I∩III)+n(I∩II∩III)82=40+25+35−x+8⇒x=26Butn(I∩II)+n(II∩III)+n(I∩III) counts n(I∩II∩III) thrice whereas it needs to be counted only once.Hence answer =26−2×8=10