(C) Let the total population be $100$.$n(A) = 25$,$n(B) = 20$,and $n(A \cap B) = 8$.Number of people who read $A$ but not $B$ is $n(A \setminus B) = n

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(C) Let the total population be $100$.$n(A) = 25$,$n(B) = 20$,and $n(A \cap B) = 8$.Number of people who read $A$ but not $B$ is $n(A \setminus B) = n(A) - n(A \cap B) = 25 - 8 = 17$.Number of people who read $B$ but not $A$ is $n(B \setminus A) = n(B) - n(A \cap B) = 20 - 8 = 12$.Number of people who read both $A$ and $B$ is $n(A \cap B) = 8$.Percentage of population who look into advertisements:$= (30\% \text{ of } 17) + (40\% \text{ of } 12) + (50\% \text{ of } 8)$$= (0.30 \times 17) + (0.40 \times 12) + (0.50 \times 8)$$= 5.1 + 4.8 + 4.0 = 13.9$.

Class 11 Mathematics Set Theory Word Problem - Set Theory

Difficult

Two newspapers $A$ and $B$ are published in a city. It is known that $25\%$ of the city population reads $A$ and $20\%$ reads $B$,while $8\%$ reads both $A$ and $B$. Further,$30\%$ of those who read $A$ but not $B$ look into advertisements,$40\%$ of those who read $B$ but not $A$ look into advertisements,and $50\%$ of those who read both $A$ and $B$ look into advertisements. The percentage of the population who look into advertisements is:

JEE MAIN 2019 All JEE MAIN

A
$12.8$
B
$13.5$
C
$13.9$
D
$13$

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Let $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$. If $A = \{a_{1} + a_{2} + \ldots + a_{k} : k \in N, a_{1}, a_{2}, \ldots, a_{k} \in S\}$,then the sum of all the elements in the set $T - A$ is equal to:

DifficultJEE MAIN 2022

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In a classroom,one-fifth of the boys leave the class and the ratio of the remaining boys to girls is $2:3$. If further $44$ girls leave the class,the ratio of boys to girls becomes $5:2$. How many more boys should leave the class so that the number of boys equals that of girls?

DifficultKVPY 2017

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The number of $3$-digit numbers that are divisible by either $3$ or $4$ but not divisible by $48$ is:

DifficultJEE MAIN 2023

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Let $S = \{1, 2, 3, \ldots, 2022\}$. The probability that a randomly chosen number $n$ from the set $S$ satisfies $\operatorname{HCF}(n, 2022) = 1$ is:

DifficultJEE MAIN 2022

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In a class of $55$ students,the number of students studying different subjects are $23$ in Mathematics,$24$ in Physics,$19$ in Chemistry,$12$ in Mathematics and Physics,$9$ in Mathematics and Chemistry,$7$ in Physics and Chemistry,and $4$ in all the three subjects. The total number of students who have taken exactly one subject is

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Let $S = \{4, 6, 9\}$ and $T = \{9, 10, 11, \ldots, 1000\}$. If $A = \{a_{1} + a_{2} + \ldots + a_{k} : k \in N, a_{1}, a_{2}, \ldots, a_{k} \in S\}$,then the sum of all the elements in the set $T - A$ is equal to:

In a classroom,one-fifth of the boys leave the class and the ratio of the remaining boys to girls is $2:3$. If further $44$ girls leave the class,the ratio of boys to girls becomes $5:2$. How many more boys should leave the class so that the number of boys equals that of girls?

The number of $3$-digit numbers that are divisible by either $3$ or $4$ but not divisible by $48$ is:

Let $S = \{1, 2, 3, \ldots, 2022\}$. The probability that a randomly chosen number $n$ from the set $S$ satisfies $\operatorname{HCF}(n, 2022) = 1$ is:

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