Question : If $x$ and $y$ are positive numbers such that $x - y = 5$ and $xy = 150$, the value of $(x + y)$ is:
Option 1: 45
Option 2: 25
Option 3: 35
Option 4: 15
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 25
Solution : Given: $x - y = 5$ and $xy = 150$ Squaring both sides of the given equation $x - y = 5$, we get, $(x - y)^2 = 5^2$ ⇒ $x^2 + y^2 - 2xy = 25$ ⇒ $x^2 + y^2 - 2 × 150 = 25$ ⇒ $x^2 + y^2 = 325$ So, the value of $(x + y)$ is calculated as, $(x + y)^2 = x^2 + y^2 + 2xy$ ⇒ $(x + y)^2 = 325 + 2 × 150$ ⇒ $(x + y)^2 = 625$ ⇒ $(x + y) =\sqrt{625}$ ⇒ $(x+ y) = 25$ Hence, the correct answer is $25$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Question : If $x^2-xy+y^2=2$ and $x^4+x^2y^2+y^4=6$, then the value of $(x^2+xy+y^2)$ is:
Question : If $(4 x-7 y)=11$ and $x y=8$, what is the value of $16 x^2+49 y^2$, given that $x$ and $y$ are positive numbers?
Question : If $x+ \sqrt{5} = 5+\sqrt{y}$ are positive integers, then the value of $\frac{\sqrt{x}+y}{x+ \sqrt{y}}$ is:
Question : If $x+y=5$, $x^{3}+ y^{3}=35$, what is the positive difference between $x$ and $y$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile