
Which of the following is correct?
1. $n\left( {S \cup T} \right)$ is maximum when $n\left( {S \cap T} \right)$ is minimum.
2. If $n\left( U \right) = 1000$, $n\left( S \right) = 720$, $n\left( T \right) = 450$, then the last value of $n\left( {S \cap T} \right) = 170$.
A. Only 1 is true
B. Only 2 is true
C. Both 1 and 2 are true
D. Both 1 and 2 are false
Answer
162.6k+ views
Hint: We will use the formula cardinal numbers of the union of sets to check statement 1. As we know, $U$ denotes the universal set and $S \cup T$ is a subset of $U$. So, the cardinal of the union of two sets is less than the cardinal of the universal set. Then by using the formula cardinal numbers of the union of sets, we will calculate $n\left( {S \cap T} \right)$ and check the second statement.
Formula Used:
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
Complete step by step solution:
We know that, $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$.
For $S$ and $T$ we can say, $n\left( {S \cup T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cap T} \right)$.
Let $n\left( {S \cap T} \right) \ge x$.
Then $n\left( {S \cap T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cup T} \right)$.
$n\left( {S \cap T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cup T} \right) \ge x$
$ - n\left( {S \cup T} \right) \ge x - n\left( S \right) - n\left( T \right)$
$n\left( {S \cup T} \right) \le - x + n\left( S \right) + n\left( T \right)$
So, the least value of $n\left( {S \cap T} \right)$ we get the maximum value of $n\left( {S \cup T} \right)$.
Hence statement 1 is correct.
Now we will put the values of $n\left( S \right) = 720$, $n\left( T \right) = 450$ in the $n\left( {S \cup T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cap T} \right)$
$n\left( {S \cup T} \right) = 720 + 450 - n\left( {S \cap T} \right)$
Since $S \cup T$ is a subset of $U$. So, $n\left( {S \cup T} \right) \le n\left( U \right)$.
Putting $720 + 450 - n\left( {S \cap T} \right)$ in place $n\left( {S \cup T} \right)$ and $n\left( U \right) = 1000$ in the inequality $n\left( {S \cup T} \right) \le n\left( U \right)$.
$720 + 450 - n\left( {S \cap T} \right) \le 1000$
Solve the above inequality
$1170 - n\left( {S \cap T} \right) \le 1000$
Subtract $1170$ from both sides
$ \Rightarrow 1170 - n\left( {S \cap T} \right) - 1170 \le 1000 - 1170$
$ \Rightarrow - n\left( {S \cap T} \right) \le - 170$
Multiply both sides by -1 and change the direction of the inequality
$ \Rightarrow n\left( {S \cap T} \right) \ge 170$
So, the least value of $n\left( {S \cap T} \right)$ is 170.
Thus statement 2 is correct.
Option ‘C’ is correct
Note: The formula of the cardinal of the union of two sets will help to check statement 1. By using the formula we get the maximum value of $n\left( {S \cup T} \right)$ .
Remember the cardinal of the union of two sets must be less than or equal to the cardinal of the universal set.
Using the formula $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$ and $n\left( {S \cup T} \right) \le n\left( U \right)$ , we check statement 2.
Formula Used:
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
Complete step by step solution:
We know that, $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$.
For $S$ and $T$ we can say, $n\left( {S \cup T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cap T} \right)$.
Let $n\left( {S \cap T} \right) \ge x$.
Then $n\left( {S \cap T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cup T} \right)$.
$n\left( {S \cap T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cup T} \right) \ge x$
$ - n\left( {S \cup T} \right) \ge x - n\left( S \right) - n\left( T \right)$
$n\left( {S \cup T} \right) \le - x + n\left( S \right) + n\left( T \right)$
So, the least value of $n\left( {S \cap T} \right)$ we get the maximum value of $n\left( {S \cup T} \right)$.
Hence statement 1 is correct.
Now we will put the values of $n\left( S \right) = 720$, $n\left( T \right) = 450$ in the $n\left( {S \cup T} \right) = n\left( S \right) + n\left( T \right) - n\left( {S \cap T} \right)$
$n\left( {S \cup T} \right) = 720 + 450 - n\left( {S \cap T} \right)$
Since $S \cup T$ is a subset of $U$. So, $n\left( {S \cup T} \right) \le n\left( U \right)$.
Putting $720 + 450 - n\left( {S \cap T} \right)$ in place $n\left( {S \cup T} \right)$ and $n\left( U \right) = 1000$ in the inequality $n\left( {S \cup T} \right) \le n\left( U \right)$.
$720 + 450 - n\left( {S \cap T} \right) \le 1000$
Solve the above inequality
$1170 - n\left( {S \cap T} \right) \le 1000$
Subtract $1170$ from both sides
$ \Rightarrow 1170 - n\left( {S \cap T} \right) - 1170 \le 1000 - 1170$
$ \Rightarrow - n\left( {S \cap T} \right) \le - 170$
Multiply both sides by -1 and change the direction of the inequality
$ \Rightarrow n\left( {S \cap T} \right) \ge 170$
So, the least value of $n\left( {S \cap T} \right)$ is 170.
Thus statement 2 is correct.
Option ‘C’ is correct
Note: The formula of the cardinal of the union of two sets will help to check statement 1. By using the formula we get the maximum value of $n\left( {S \cup T} \right)$ .
Remember the cardinal of the union of two sets must be less than or equal to the cardinal of the universal set.
Using the formula $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$ and $n\left( {S \cup T} \right) \le n\left( U \right)$ , we check statement 2.
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