
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
232.8k+ 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
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

