Aptitude
November 20, 2023Question 7130 – NVS PGT CS 2019 Part-A
November 20, 2023Operating-Systems
| Question 48 |
A counting semaphore was initialized to 10. Then 6P (wait) operations and 4V (signal) operations were completed on this semaphore. The resulting value of the semaphore is
| 0 | |
| 8 | |
| 10 | |
| 12 |
Question 48 Explanation:
Let the semaphore be S which is initially 10.
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 – 6 + 4 = 8
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 – 6 + 4 = 8
Correct Answer: B
Question 48 Explanation:
Let the semaphore be S which is initially 10.
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 – 6 + 4 = 8
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 – 6 + 4 = 8
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