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
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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 $\sqrt{\frac{831}{23}}$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Question : If $x^2+\frac{1}{x^2}=29$, then find the value of $x-\frac{1}{x}$.
Question : If $\left(x+\frac{1}{x}\right)=5 \sqrt{2}$, and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right) ?$
Question : If $\sqrt x-\sqrt y=1$ , $\sqrt x+\sqrt y=17$, then $\sqrt {xy}=?$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile