In a certain town,25% of families own a phone | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $P$ be the set of families owning a phone and $C$ be the set of families owning a car.Given: $n(P) = 25\%$,$n(C) = 15\%$,and $n(P^c \cap C^c)

Vedclass · Accepted Answer

$2$ and $3$

Vedclass · Answer

(C) Let $P$ be the set of families owning a phone and $C$ be the set of families owning a car.Given: $n(P) = 25\%$,$n(C) = 15\%$,and $n(P^c \cap C^c) = 65\%$.We know that $n(P^c \cap C^c) = n((P \cup C)^c) = 100\% - n(P \cup C)$.Therefore,$n(P \cup C) = 100\% - 65\% = 35\%$. This confirms statement $2$ is correct.Using the formula $n(P \cup C) = n(P) + n(C) - n(P \cap C)$:$35\% = 25\% + 15\% - n(P \cap C)$.$n(P \cap C) = 40\% - 35\% = 5\%$.Since $5\%$ of the total families $= 2000$,the total number of families $= \frac{2000 	imes 100}{5} = 40,000$. This confirms statement $3$ is correct.Statement $1$ claims $10\%$ own both,which is incorrect as we calculated $5\%$.Thus,statements $2$ and $3$ are correct.

Similar Questions

$A$ survey shows that $63 \%$ of the people in a city read newspaper $A$ whereas $76 \%$ read newspaper $B$. If $x \%$ of the people read both the newspapers,then a possible value of $x$ can be

Let $X$ be a set of $5$ elements. The number $d$ of ordered pairs $(A, B)$ of subsets of $X$ such that $A \neq \phi, B \neq \phi, A \cap B \neq \phi$ satisfies:

If sets $A$ and $B$ have $3$ and $6$ elements respectively,what is the minimum number of elements in $A \cup B$?

If a set $A$ is such that $A \cup \{1, 2\} = \{1, 2, 3, 5, 9\}$,then the set $A$ could be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module