A survey shows that 73 % of the persons worki | vedclass

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

Question

(D) Let $C$ be the set of persons who like coffee and $T$ be the set of persons who like tea.Given: $n(C) = 73$,$n(T) = 65$.Let $x$ be the percentage

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) Let $C$ be the set of persons who like coffee and $T$ be the set of persons who like tea.Given: $n(C) = 73$,$n(T) = 65$.Let $x$ be the percentage of persons who like both,i.e.,$n(C \cap T) = x$.We know that the total percentage cannot exceed $100 \%$,so $n(C \cup T) \leq 100$.Using the formula $n(C \cup T) = n(C) + n(T) - n(C \cap T)$:$73 + 65 - x \leq 100$$138 - x \leq 100$$x \geq 38$.Also,the number of people who like only coffee or only tea cannot be negative:$n(C) - x \geq 0 \Rightarrow 73 - x \geq 0 \Rightarrow x \leq 73$$n(T) - x \geq 0 \Rightarrow 65 - x \geq 0 \Rightarrow x \leq 65$Combining these,we get $38 \leq x \leq 65$.Therefore,$x$ must be in the range $[38, 65]$.Among the given options,$36$ is outside this range. Thus,$x$ cannot be $36$.

$A$ survey shows that $73 \%$ of the persons working in an office like coffee,whereas $65 \%$ like tea. If $x$ denotes the percentage of them who like both coffee and tea,then $x$ cannot be

Similar Questions

In a certain office,$72 \%$ of the workers prefer cold drink and $44 \%$ prefer tea. If each of them prefers cold drink or tea and $40$ like both,then the total number of workers in the office is

The function $f:R \to R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$,then $f(R) \in $

Let $A = \{1, 2, 3\}$. The total number of distinct relations that can be defined over $A$ is

Domain of the function $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$ is

The equivalent function of $\log {x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module