Question : If $x^2+y^2=427$ and $xy=202$, then find the value of $\frac{x+y}{x-y}$.
Option 1: $\sqrt{\frac{835}{23}}$
Option 2: $\sqrt{\frac{830}{29}}$
Option 3: $\sqrt{\frac{831}{23}}$
Option 4: $\sqrt{\frac{830}{23}}$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $\sqrt{\frac{831}{23}}$
Solution :
Given: $x^2+y^2=427$ and $xy=202$.
Now, $(x+y)^2=x^2+y^2+2xy$
⇒ $(x+y)^2=427+2×202$
⇒ $(x+y)=\sqrt{831}$ ---------------------------------------(1)
Also, $(x-y)^2=x^2+y^2-2xy$
⇒ $(x-y)^2=427-2×202$
⇒ $(x-y)= \sqrt{23}$ ------------------------------------(2)
So, $\frac{x+y}{x-y}$
$= \frac{\sqrt{831}}{\sqrt{23}}$
$= \sqrt{\frac{831}{23}}$.
Hence, the correct answer is $\sqrt{\frac{831}{23}}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
