Question : If $xy+yz+zx=1$ , then the value of $\frac{1\:+\:y^2}{(x\:+\:y)(y\:+\:z)}$ is:
Option 1: 2
Option 2: 3
Option 3: 4
Option 4: 1
Answer (1)
24th Jan, 2024
Correct Answer: 1
Solution :
Given:
$xy+yz+zx=1$
Now, $\frac{1\:+\:y^2}{(x\:+\:y)(y\:+\:z)}$
= $\frac{1\:+\:y^2}{xy\:+\:yz\:+\:zx\:+\:y^2}$
= $\frac{1\:+\:y^2}{1\:+\:y^2}$
= $1$
Hence, the correct answer is 1.
Comments (0)
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Get Updates Brochure
Thank You!
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Application Date:
23 Oct, 2024
- 25 Dec, 2024
Application Date:
9 Dec, 2024
- 30 Dec, 2024
Application Date:
2 Dec, 2024
- 31 Dec, 2024
Exam Date:
3 Jan, 2025
- 16 Jan, 2025
General Insurance Corporation Of India Assistant Manager Exam
- (GIC ASSISTANT MANAGER SCALE I EXAM)
Exam Date:
5 Jan, 2025
- 5 Jan, 2025
