In a certain town,25% of the families own a p | 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' \cap C') =

Vedclass · Accepted Answer

All $(A), (B)$ and $(C)$ are correct.

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 $n(P' \cap C') = 65\%$.Since $n(P' \cap C') = 1 - n(P \cup C)$,we have $n(P \cup C) = 100\% - 65\% = 35\%$.Statement $(B)$ is correct because $35\%$ of families own either a phone or a car.Using the formula $n(P \cup C) = n(P) + n(C) - n(P \cap C)$,we get $35\% = 25\% + 15\% - n(P \cap C)$,which implies $n(P \cap C) = 5\%$.Statement $(A)$ is correct because $5\%$ of families own both.Given $n(P \cap C) = 2,000$,let $x$ be the total number of families. Then $5\% 	ext{ of } x = 2,000$,so $0.05x = 2,000$,which gives $x = 40,000$.Statement $(C)$ is correct because there are $40,000$ families in the town.Thus,all statements $(A), (B),$ and $(C)$ are correct.

Similar Questions

If $A = \{2, 3, 4, 8, 10\}, B = \{3, 4, 5, 10, 12\}, C = \{4, 5, 6, 12, 14\}$,then $(A \cap B) \cup (A \cap C)$ is equal to

If $A = \{x : x = \frac{n-1}{n+1}, n \in W\}$ and $\{n \leq 10\}$,point out the correct statement from the following:

Let $A$ and $B$ be two non-empty subsets of a set $X$ such that $A$ is not a subset of $B$, then

The domain of the function $f(x) = \sqrt{\log \frac{1}{|\sin x|}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module