How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

+91 99359 65550 contact@maarula.in

Mathem SolvexGet the solution of every question

Home

Previous Year Questions

NIMCET CUET PG JAMIA MAH-CET AMU VITMEE

PYQ Paper Downloads Practice Paper

Latest Update

Resources

PYQ PDF Downloads Our Courses Test Series About Us

Home

Question Archive

Explore All PYQs NIMCET CUET PG JAMIA MAH-CET AMU VITMEE

Downloads & Practice

PDF Material Downloads Solve Practice Paper

Quick Links

Latest Updates About Us Our Premium Courses Contact Support

Preparing for Excellence...

MAARULA CLASSES

MCA Entrance Coaching

Quick Links

Home
Blog
About
Contact Us
Faculty

Exams

NIMCET
CUET PG
VITMEE
JAMIA
MAH-CET

Free Resources

Previous Year Papers
Free Mock Tests
Strategy Videos
Syllabus

Our Courses

Live Classes
Test Series
Comprehensive Notes

Download Maarula Mathem App

For Seamless learning experience

Google Play

Designed & Developed with ❤️ by Vivek Kumar | LinkedIn

Home PYQs

Updates Resources About

JEE 2023 Mathematics PYQ — In a group of 100 persons 75 speak English and 40 speak Hindi. Ea… | Mathem Solvex | Mathem Solvex

Home
Questions
JEE
Mathematics
Q. 2659

Tip:A–D to answerE for explanationV for videoS to reveal answer

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons, who speak only English is $α$ and the number of persons who speak only Hindi is $β$ , then the eccentricity of the ellipse $25 (β^{2} x^{2} + α^{2} y^{2}) = α^{2} β^{2}$ is :

Choose the correct answer:

A.
$\frac{129}{12}$
B.
$\frac{117}{12}$

Correct Answer:

$\frac{119}{12}$

Explanation

$n (A \cup B) = n (A) + n (B) - n (A \cap B)$

$n (A \cap B) = 75 + 40 - 100$

$n (A \cap B) = 15$

Only English = 60, $α = 60$

Only Hindi = 25, $β = 25$

$\Rightarrow 25 (β^{2} x^{2} + α^{2} y^{2}) = α^{2} β^{2}$

$\Rightarrow \frac{25 x ^{2}}{α ^{2}} + \frac{25 y ^{2}}{β ^{2}} = 1$

$\Rightarrow \frac{25 x ^{2}}{( 60 ) ^{2}} + \frac{25 y ^{2}}{( 25 ) ^{2}} = 1$

$\Rightarrow e^{2} = 1 - [\frac{25 \times 25}{( 60 ) ^{2}}]$

$\Rightarrow e^{2} = \frac{( 60 ) ^{2} - ( 25 ) ^{2}}{( 60 ) ^{2}}$

$\Rightarrow e = \frac{119}{12}$

Explanation

$n (A \cup B) = n (A) + n (B) - n (A \cap B)$

$n (A \cap B) = 75 + 40 - 100$

$n (A \cap B) = 15$

Only English = 60, $α = 60$

Only Hindi = 25, $β = 25$

$\Rightarrow 25 (β^{2} x^{2} + α^{2} y^{2}) = α^{2} β^{2}$

$\Rightarrow \frac{25 x ^{2}}{α ^{2}} + \frac{25 y ^{2}}{β ^{2}} = 1$

$\Rightarrow \frac{25 x ^{2}}{( 60 ) ^{2}} + \frac{25 y ^{2}}{( 25 ) ^{2}} = 1$

$\Rightarrow e^{2} = 1 - [\frac{25 \times 25}{( 60 ) ^{2}}]$

$\Rightarrow e^{2} = \frac{( 60 ) ^{2} - ( 25 ) ^{2}}{( 60 ) ^{2}}$

$\Rightarrow e = \frac{119}{12}$

Same Topic Questions

JEE

If the tangent at a point

Explore More

TopicAll "ellipse" PYQs →SubjectMore Mathematics Questions →YearJEE 2023 Paper →All JEE Questions →📄 Download JEE PYQ Papers →📝 Preparation Tips & Articles →

C.
$\frac{119}{12}$
(Correct Answer)

D.
$\frac{3 15}{12}$