Exam SK0-005 topic 1 question 321 discussion

For RAID 1 we need 2 disks as it works on mirrored volumes To Calculate the Number of Disk for RAID 5, we need to calculate like this Useable Capacity = (Number of disks -1) x Disk Size Or Useabel Capaciity= (N-1) x Disk Size Here Useable Capacity = 20GB Disk Size = 10GB N=? Place the values in the formula 20GB = (N-1) x 10 20GB = 10N-10 10N =20+10 10N = 30 N = 3 Same calculations will be performed for the 2nd RAID 5 array Final calculation will be = 2 + 3 + 3=8

RAID 1 for Log Files: 2 disks RAID 5 for Data Files: 3 disks RAID 5 for Operating System: 3 disks

C is not correct, it does not specify that the 2x 20gb of Raid5 need to be on separate arrays so it is 2 disk for the raid1 = 10gb and 5 disk of raid5 = 40gb.

C. 8 Disks (per AzadOB's explanation)

To calculate the minimum number of disks needed for the given RAID configurations, let's break down each requirement: RAID 1 for log files: This requires mirroring, so each disk contributes half of its capacity. Since we need a total of 10GB usable space, we'll need 20GB raw capacity (10GB mirrored). Thus, we need 20GB / 10GB = 2 disks. RAID 5 for data files (20GB): RAID 5 requires at least three disks, and one disk worth of capacity is used for parity. So, for 20GB of usable space, we'll need 3 disks. RAID 5 for the operating system (20GB): Same as the data files, we'll need 3 disks for 20GB of usable space. Adding these up: RAID 1 (log files): 2 disks RAID 5 (data files): 3 disks RAID 5 (operating system): 3 disks Total disks needed: 2 + 3 + 3 = 8 disks. So, the correct answer is: C. 8