Question

A dinner party is to be fixed for a group of $100$ persons. In this party, $50$ persons do not prefer fish, $60$ prefer chicken, and $10$do not prefer either chicken or fish. The number of persons who prefer both fish and chicken is.

• A. $20$
• B. $22$
• C. $25$
• D. $\mathrm{none}\mathrm{of}\mathrm{these}$

Answer 1

The correct option is A

$20$

Explanation for the correct option:

Step1. Define events.

The number of people who prefer fish $=\mathrm{n}\left(\mathrm{F}\right)$

The number of people who prefer chicken $=\mathrm{n}\left(\mathrm{C}\right)=60$

People who do not prefer fish $=\mathrm{n}\left({\mathrm{F}}^{\text{'}}\right)$

The number of people who prefer both fish and chicken $=\mathrm{n}(\mathrm{F}\cap \mathrm{C})$

Total number of people $=\mathrm{n}\left(\mathrm{U}\right)$

The number of people who prefer either fish or chicken $=\mathrm{n}(\mathrm{F}\cup \mathrm{C})$

The number of people who do not prefer either fish or chicken $=\mathrm{n}\left(\mathrm{FUC}\right)\text{'}=10$

Number of people who prefer neither fish nor chicken therefore$=10$

Therefore, The number of people who prefer either fish or chicken $=\mathrm{n}(\mathrm{F}\cup \mathrm{C})=100-10=90$

Step2. Calculate the number of people who prefer both fish and chicken:

Given,

$\begin{array}{rcl}n\left(U\right)& =& 100\\ n\left(C\right)& =& 60\\ \mathrm{n}(\mathrm{F}\cup \mathrm{C})\text{'}& =& 10\\ n\left(F^{\text{'}}\right)& =& 50\\ n\left(F\right)& =& 100-50;\mathbf{\left[}\mathbf{\because }\mathbf{n}\mathbf{\left(}\mathbf{F}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{n}\mathbf{\left(}\mathbf{U}\mathbf{\right)}\mathbf{-}\mathbf{n}\mathbf{(}\mathbf{F}\mathbf{\text{'}}\mathbf{)}\mathbf{\right]}\\ & =& 50\\ & & \\ n(F\cup C)& =& 100-10;\left[\mathbf{\because }\mathbf{n}\mathbf{(}\mathbf{F}\mathbf{\cup }\mathbf{C}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{n}\mathbf{\left(}\mathbf{U}\mathbf{\right)}\mathbf{-}\mathbf{n}\mathbf{(}\mathbf{F}\mathbf{\cup }\mathbf{C}\mathbf{)}\mathbf{\text{'}}\right]\\ & =& 90\\ & & \\ \mathrm{n}(\mathrm{F}\cup \mathrm{C})& =& \mathrm{n}\left(\mathrm{F}\right)+\mathrm{n}\left(\mathrm{C}\right)-\mathrm{n}(\mathrm{F}\cap \mathrm{C})\end{array}$

Substitute the value now,

$$\begin{gathered} 90=50+60-n(F\cap C) \\ \Rightarrow \mathrm{n}(\mathrm{F}\cap \mathrm{C})=110-90 \\ \Rightarrow \mathrm{n}(\mathrm{F}\cap \mathrm{C})=20 \end{gathered}$$

Hence, Option(A) is the correct answer.