BITSAT 2024
Exam BITSAT
Year 2024
Subject Mathematics
Chapter Sets and Relations BITSAT 2024 · Mathematics · Sets and Relations
In a statistical investigation of 1003 families of Calcutta, it was...
In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a TV, 794 families has a radio and 187 has TV. The number of families in that group having both a radio and a TV is
Options A 36 B 41 C 32 D None of these
Step-by-step solution
Given, n ( R ) = 794 n ( T ) = 187 n ( R ∪ T ) ′ = 63 n ( T o t a l ) = n ( R ∪ T ) + n ( R ∪ T ) ′ ⇒ 1003 = n ( R ∪ T ) + 63 ⇒ n ( R ∪ T ) = 940 \begin{aligned}
\text{ Given, } n(R) & =794 \\
n(T) & =187 \\
n(R \cup T)^{\prime} & =63 \\
n(\mathrm{Total}) & =n(R \cup T)+n(R \cup T)^{\prime} \\
\Rightarrow \quad 1003 & =n(R \cup T)+63 \\
\Rightarrow \quad n(R \cup T) & =940
\end{aligned} Given, n ( R ) n ( T ) n ( R ∪ T ) ′ n ( Total ) ⇒ 1003 ⇒ n ( R ∪ T ) = 794 = 187 = 63 = n ( R ∪ T ) + n ( R ∪ T ) ′ = n ( R ∪ T ) + 63 = 940
By set theory
⇒ n ( R ∪ T ) = n ( R ) + n ( T ) − n ( R ∩ T ) ⇒ 940 = 794 + 187 − n ( R ∩ T ) ⇒ n ( R ∩ T ) = 981 − 940 = 41 \begin{array}{lc}
\Rightarrow & n(R \cup T)=n(R)+n(T)-n(R \cap T) \\
\Rightarrow & 940=794+187-n(R \cap T) \\
\Rightarrow & n(R \cap T)=981-940=41
\end{array} ⇒ ⇒ ⇒ n ( R ∪ T ) = n ( R ) + n ( T ) − n ( R ∩ T ) 940 = 794 + 187 − n ( R ∩ T ) n ( R ∩ T ) = 981 − 940 = 41
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