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Question: A newspaper agent sells \[TOI\], \[HT\] and \[NBT\] in equal numbers to \[302\] persons. Seven got \...

A newspaper agent sells TOITOI, HTHT and NBTNBT in equal numbers to 302302 persons. Seven got HTHT and NBTNBT, twelve got TOITOI and NBTNBT, nine got TOITOI and HTHT & three got all the three newspapers. The details are given in the Venn diagram.

What percentage gets TOITOI or HTHT but not NBTNBT?
A). More than 65%65\%
B). Less than60%60\%
C). 64%\cong 64\%
D). None of these

Explanation

Solution

Here, in the question, we have been given a Venn diagram which represents the data of the newspaper readers of three different brands. We are asked to find the percentage of people who read TOITOI or HTHT but not NBTNBT. To find that, we will first find the number of persons who read only TOITOI, only HTHT and only NBTNBT and then reach the desired result.
Formula used:
n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(AC)+n(ABC)n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {A \cap C} \right) + n\left( {A \cap B \cap C} \right)

Complete step-by-step solution:
Given, Total number of persons who read newspaper, n(TOIHTNBT)n\left( {TOI \cup HT \cup NBT} \right)=302302

n\left( {HT \cap NBT} \right) = 7 \\\ n\left( {TOI \cap NBT} \right) = 12 \\\ n\left( {TOI \cap HT} \right) = 9 \\\ n\left( {TOI \cap HT \cap NBT} \right) = 3 $$ Now, using identity, $$n\left( {A \cup B \cup C} \right) = n\left( A \right) + n\left( B \right) + n\left( C \right) - n\left( {A \cap B} \right) - n\left( {B \cap C} \right) - n\left( {A \cap C} \right) + n\left( {A \cap B \cap C} \right)$$ $$n\left( {TOI \cup HT \cup NBT} \right) = n\left( {TOI} \right) + n\left( {HT} \right) + n\left( {NBT} \right) - n\left( {TOI \cap HT} \right) - n\left( {HT \cap NBT} \right) - n\left( {TOI \cap NBT} \right) + n\left( {TOI \cap HT \cap NBT} \right)$$ $$ \Rightarrow 302 = n\left( {TOI} \right) + n\left( {HT} \right) + n\left( {NBT} \right) - 9 - 7 - 12 + 3$$ Now, we are given that $$n\left( {TOI} \right) = n\left( {HT} \right) = n\left( {NBT} \right)$$ Therefore, $$ \Rightarrow 302 = 3 \times n\left( {TOI} \right) - 25$$ Simplifying it, we get,

\Rightarrow n\left( {TOI} \right) = \dfrac{{327}}{3} \\
\Rightarrow n\left( {TOI} \right) = 109 Hence, Hence,n\left( {TOI} \right) = n\left( {HT} \right) = n\left( {NBT} \right) = 109Now,wewillcalculatethenumberofpersonswhoreadonly Now, we will calculate the number of persons who read onlyTOI,only, only HTandonlyand onlyNBT$$ using Venn diagram

n\left( {TOI\;only} \right) = 109 - \left( {6 + 9 + 3} \right) \\\ \Rightarrow n\left( {TOI\;only} \right) = 91 $$ Similarly, we get,

n\left( {HT;only} \right) = 109 - \left( {6 + 4 + 3} \right) \\
\Rightarrow n\left( {HT;only} \right) = 96 $$,
and

n\left( {NBT\;only} \right) = 109 - \left( {9 + 4 + 3} \right) \\\ \Rightarrow n\left( {NBT\;only} \right) = 93 $$ Let us draw a fresh Venn diagram again and shade the required region, ![](https://www.vedantu.com/question-sets/ed085433-7ddb-4671-83a4-04684317b7a22534421229413622648.png) Now, using Venn diagram, No. of persons reading $$TOI \text{ or }HT\text{ but not }NBT$$ is calculated as:

n\left( {TOI \text{ or }HT\text{ but not }NBT} \right) = 91 + 96 + 6 \\
\Rightarrow n\left( {TOI \text{ or }HT\text{ but not }NBT} \right) = 193 \\

Percentage of persons reading $$TOI \text{ or }HT\text{ but not }NBT$$=$$\dfrac{{193}}{{302}} \times 100\% $$ $$ = 63.907$$ Hence the correct option is C, $$ \cong 64\% $$ is the correct option. **Note:** Whenever we face such types of questions, we should try to solve the question using Venn diagram to a possible extent. Venn diagrams are easy to read and understand. We can understand the data just by looking at the Venn diagram. And it is equally important to remember and understand the basic formula related to Venn diagram problems.