
Which of the following is correct?
I. $n(S\cup T)$ is maximum when $n(S\cap T)$ is least.
II. If $n(U)=1000$, $n(S)=720$, $n(T)=450$, then least value of $n(S\cap T)=170$.
A. Only I is true
B. Only II is true
C. Both I and II are true
D. Both I and II are false
Answer
163.2k+ views
Hint: In this question, we are to find the correct statement. This is obtained by using the formula. By substituting the given values into the formula, we can prove the given statements.
Formula Used: Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the natural number’s set - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
If two sets $A,B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.
Some of the important set operations:
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\
\end{align}$
Complete step by step solution: From the formula we have
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
On substituting given values, we get
\[\begin{align}
& n(S\cup T)=n(S)+n(T)-n(S\cap T) \\
& \Rightarrow n(S\cup T)=720+450-n(S\cap T) \\
\end{align}\]
From this, we can write
\[\begin{align}
& 1170-n(S\cap T)\le n(U) \\
& \Rightarrow n(S\cap T)\ge 1170-n(U) \\
& \Rightarrow n(S\cap T)\ge 1170-1000 \\
& \Rightarrow n(S\cap T)\ge 170 \\
\end{align}\]
Thus, if $n(S\cup T)$ is maximum when $n(S\cap T)$ is least. Therefore, both statements are true.
Option ‘C’ is correct
Note: By the definition, the value of $n(A\cup B)$ is maximum for the least value of $n(A\cap B)$. So, by this principle, the above given two statements are true.
Formula Used: Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the natural number’s set - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
If two sets $A,B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.
Some of the important set operations:
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\
\end{align}$
Complete step by step solution: From the formula we have
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
On substituting given values, we get
\[\begin{align}
& n(S\cup T)=n(S)+n(T)-n(S\cap T) \\
& \Rightarrow n(S\cup T)=720+450-n(S\cap T) \\
\end{align}\]
From this, we can write
\[\begin{align}
& 1170-n(S\cap T)\le n(U) \\
& \Rightarrow n(S\cap T)\ge 1170-n(U) \\
& \Rightarrow n(S\cap T)\ge 1170-1000 \\
& \Rightarrow n(S\cap T)\ge 170 \\
\end{align}\]
Thus, if $n(S\cup T)$ is maximum when $n(S\cap T)$ is least. Therefore, both statements are true.
Option ‘C’ is correct
Note: By the definition, the value of $n(A\cup B)$ is maximum for the least value of $n(A\cap B)$. So, by this principle, the above given two statements are true.
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