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

In a class of $125$ students,$70$ passed in Mathematics,$55$ in Statistics,and $30$ in both. The probability that a student selected at random from the class has passed in only one subject is

If $A$ and $B$ are two sets such that $n(A) = 70$,$n(B) = 60$,and $n(A \cup B) = 110$,then $n(A \cap B)$ is equal to:

If $X$ and $Y$ are two sets such that $X$ has $40$ elements,$X \cup Y$ has $60$ elements and $X \cap Y$ has $10$ elements,how many elements does $Y$ have?

In a college of $300$ students,every student reads $5$ newspapers and every newspaper is read by $60$ students. The number of newspapers is

Vedclass Test Series

Exam Paper Generator

Online Exam Module