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14 Jul 2026, 06:42
Kudos
Bookmarks
X (Must be true)
If measure 3 is in the restricted zon then measure 1 also has to be Measure 1 & measure 4 cany be in the same zone so measure 4 has to be in the Public zone.
Y (Cannot be true)
Measure 2 & Measure 4 cannot both be in Public zone. If Measure 2 is in Public Measure 3 must also be.
That would make Measures 2,3 and 4 all in Public but a zone can have only 2 measures.
So this is impossiblee
If measure 3 is in the restricted zon then measure 1 also has to be Measure 1 & measure 4 cany be in the same zone so measure 4 has to be in the Public zone.
Y (Cannot be true)
Measure 2 & Measure 4 cannot both be in Public zone. If Measure 2 is in Public Measure 3 must also be.
That would make Measures 2,3 and 4 all in Public but a zone can have only 2 measures.
So this is impossiblee
Bunuel
14 Jul 2026, 07:10
Kudos
Bookmarks
Bunuel
It is given that each zone must have at least one policy and no more than two policies.
As M1 is in S, M4 cannot be in S. Therefore, M4 needs to be in either R or P.
If M2 is in P, then M3 is also in P. Consequently, M2 and M4 or M2 and M1 cannot be together in P.
If M3 is in R, then M1 is also in R. Consequently, M4 cannot be in R.
A. This cannot be true. Since if M2 is there, then M3 is also there in P automatically and each zone can have at most two policies. So M4 cannot be in P.
B. This may or may not be true. If M2 is in R, then M4 may or may not be in P. Multiple setups possible.
C. This must be true. If M3 is in R then it will also have M1. Also S must have M1. Therefore, both R and S cannot have M4, so M4 has to go in P. Consequently, M2 will go in S as each measure has to be used once.
D. This may or may not be true. If setup in option C is followed, M4 cannot be in R. On the other hand, if S has M1 and P has M2 and M3, then M4 can go in R.
E. This may or may not be true. Result of the inverse conditionality is not known.
14 Jul 2026, 07:13
Kudos
Bookmarks
Based on the conditions, let check the options
1) If Measure 2 is applied on Public zone then as per condition 2, measure 3 must also be applied to Public zone and one zone cannot have more than 2 policies. Hence M4 cannot be applied to Public zone- CANNOT BE TRUE
2) There is no relation between M2 and M4- can be true
3)M3 is applied to Restricted zone then as per condition 4, M1 is also applied to the same zone therefore M4 cannot be applied in the Restricted zone if M3 is applied there- MUST BE TRUE.
4) Can be true
5) Can be true
1) If Measure 2 is applied on Public zone then as per condition 2, measure 3 must also be applied to Public zone and one zone cannot have more than 2 policies. Hence M4 cannot be applied to Public zone- CANNOT BE TRUE
2) There is no relation between M2 and M4- can be true
3)M3 is applied to Restricted zone then as per condition 4, M1 is also applied to the same zone therefore M4 cannot be applied in the Restricted zone if M3 is applied there- MUST BE TRUE.
4) Can be true
5) Can be true
Bunuel
). If M1 is applied to restricted zone, it is also applied to Secure zone (as per orange color logic in photo), then M4 cannot be applied to either of these (as per the green color logic in the photo). Hence it has to be applied to public zone and is a MUST BE TRUE statement.
