If $X$ and $Y$are two sets such that $(X\cup Y)$has $60$ elements, $X$has $38$ elements and $Y$has $42$ elements, how many elements does $(X\cap Y)$ have?
A. \[11\]
B. \[20\]
C. \[13\]
D. None of these.
Answer
302.7k+ views
Hint: To solve this question we will use the formula of $(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$. We will substitute the given values of $X$, $Y$ and $(X\cup Y)$ in the formula of $(X\cap Y)$ and then simplify the equation. After simplification of equation we will get the number of elements set $(X\cap Y)$ have.
Formula Used:The union of two sets can be calculated with the help of the formula.
$(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$
Complete step by step solution:We are given two sets $X$ and $Y$such that $(X\cup Y)$has $60$ elements, $X$has $38$ elements and $Y$has $42$ elements and we have to find the number of elements $(X\cap Y)$ have.
We will substitute the given values in the formula of $(X\cup Y)$to calculate the number of elements $(X\cap Y)$ have.
$\begin{align}
& (X\cup Y)=n(X)+n(Y)-n(X\cap Y) \\
& 60=38+42-X\cap Y
\end{align}$
We will now simplify the equation.
$\begin{align}
& 60=80-X\cap Y \\
& X\cap Y=20
\end{align}$
Option ‘B’ is correct
Note: The union of two or more sets is the collection of all the elements which are present in the individual sets. $Y$. The symbol of union is read as “or”. It can be written as $X\cup Y=\{x:x\in A\,\,or\,\,x\in B\}$. In the formula of union of two sets $(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$, $n(X)$ is the number of elements in set $X$, $n(Y)$ is the number of elements in set $Y$and $(X\cap Y)$ is the number of elements which are common in both the sets $X$ and The cardinal number of the union of two set can be also calculated with the formula $(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$.
The union of the two or more sets follows commutative law, associative law, identity law, idempotent law, and domination law.
Formula Used:The union of two sets can be calculated with the help of the formula.
$(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$
Complete step by step solution:We are given two sets $X$ and $Y$such that $(X\cup Y)$has $60$ elements, $X$has $38$ elements and $Y$has $42$ elements and we have to find the number of elements $(X\cap Y)$ have.
We will substitute the given values in the formula of $(X\cup Y)$to calculate the number of elements $(X\cap Y)$ have.
$\begin{align}
& (X\cup Y)=n(X)+n(Y)-n(X\cap Y) \\
& 60=38+42-X\cap Y
\end{align}$
We will now simplify the equation.
$\begin{align}
& 60=80-X\cap Y \\
& X\cap Y=20
\end{align}$
Option ‘B’ is correct
Note: The union of two or more sets is the collection of all the elements which are present in the individual sets. $Y$. The symbol of union is read as “or”. It can be written as $X\cup Y=\{x:x\in A\,\,or\,\,x\in B\}$. In the formula of union of two sets $(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$, $n(X)$ is the number of elements in set $X$, $n(Y)$ is the number of elements in set $Y$and $(X\cap Y)$ is the number of elements which are common in both the sets $X$ and The cardinal number of the union of two set can be also calculated with the formula $(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$.
The union of the two or more sets follows commutative law, associative law, identity law, idempotent law, and domination law.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Electron Gain Enthalpy and Electron Affinity Explained

Other Pages
NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

Hybridisation in Chemistry – Concept, Types & Applications

