
If X and Y are two sets such that \[n(X) = 17,n(Y) = 23\] and \[n(X \cup Y) = 38\], find \[n(X \cap Y)\].
Answer
584.4k+ views
Hint: Two sets $A$ and $B$ are said to be disjoint, if $A \cap B = \phi $. If $A \cap B \ne \phi $, then $A$ and $B$ are called overlapping or intersecting sets.
$n$, the number of elements in a set is called the cardinal number of the set, denoted by $n(A)$.
Cardinal Property of Sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ and if $A$ and $B$ are disjoint, then $n(A \cup B) = n(A) + n(B)$.
Complete step by step answer:
We know that any well-defined collection of objects, which are different from each other, and which we can see or think of is called a set. The objects which belong to a set are called its members or elements.
Two sets are said to be disjoint if they have no element in common.
We are given two sets \[X\] and \[Y\] such that \[n(X) = 17,n(Y) = 23\]and\[n(X \cup Y) = 38\].
We have to find the value of \[n(X \cap Y)\].
According to the Cardinal property of sets we have,
If \[X\] and \[Y\] are two joint sets then $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.
By substituting the values given in the question, we get
$\Rightarrow 38 = 17 + 23 - n(X \cap Y)$
By transposing $n(X \cap Y)$ to the L.H.S and $38$ to the R.H.S we get,
$\Rightarrow n(X \cap Y) = 17 + 23 - 38$
On solving the above equation
$ \Rightarrow n(X \cap Y) = 40 - 38$
$ \Rightarrow n(X \cap Y) = 2$
$\therefore $ The value of $n(X \cap Y) = 2$.
Note:
The union of two sets $A$ and $B$, denoted by $A \cup B$ (read as $A$ union $B$), is the set of all those elements which are either in sets $A$ or in sets $B$ or in both.
Symbolically, $A \cup B = \{ x:x \in A \, or \, x \in B\}.$
The intersection of two sets $A$ and $B$, denoted by $A \cap B$ (read as $A$ intersection $B$), is the set of all those elements which are common to both sets $A$ and $B$.
Symbolically, $A \cap B = \{ x:x \in A\, and\, x \in B\}.$
Union and Intersection are two main set operations.
$n$, the number of elements in a set is called the cardinal number of the set, denoted by $n(A)$.
Cardinal Property of Sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ and if $A$ and $B$ are disjoint, then $n(A \cup B) = n(A) + n(B)$.
Complete step by step answer:
We know that any well-defined collection of objects, which are different from each other, and which we can see or think of is called a set. The objects which belong to a set are called its members or elements.
Two sets are said to be disjoint if they have no element in common.
We are given two sets \[X\] and \[Y\] such that \[n(X) = 17,n(Y) = 23\]and\[n(X \cup Y) = 38\].
We have to find the value of \[n(X \cap Y)\].
According to the Cardinal property of sets we have,
If \[X\] and \[Y\] are two joint sets then $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.
By substituting the values given in the question, we get
$\Rightarrow 38 = 17 + 23 - n(X \cap Y)$
By transposing $n(X \cap Y)$ to the L.H.S and $38$ to the R.H.S we get,
$\Rightarrow n(X \cap Y) = 17 + 23 - 38$
On solving the above equation
$ \Rightarrow n(X \cap Y) = 40 - 38$
$ \Rightarrow n(X \cap Y) = 2$
$\therefore $ The value of $n(X \cap Y) = 2$.
Note:
The union of two sets $A$ and $B$, denoted by $A \cup B$ (read as $A$ union $B$), is the set of all those elements which are either in sets $A$ or in sets $B$ or in both.
Symbolically, $A \cup B = \{ x:x \in A \, or \, x \in B\}.$
The intersection of two sets $A$ and $B$, denoted by $A \cap B$ (read as $A$ intersection $B$), is the set of all those elements which are common to both sets $A$ and $B$.
Symbolically, $A \cap B = \{ x:x \in A\, and\, x \in B\}.$
Union and Intersection are two main set operations.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

