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13 Jul 2026, 11:10
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Measure 3 is applied to the restricted zone.
than measure 1 must also be applied to the restricted zone. Measure 1 is always applied to secure zone. Therefore measure 1 occupies both secure and restricted.
Since measure 4 cannot be applied to any zone containing measure 1, measure 4 cannot be in secure or resticted. Measure 4 must be applied to the public zone.
Measure 3 must be restricted zone and measure 4 to public zone.
Can not be true.
If measure 2 and measure 4 are both applied to public zone. If measure 2 is public then measure 3 must also be public. That would mean measure 2,3 and 4 are all public. but each zone can not have more than two policies. Therefore measure 2 and 4 can not be public zones.
than measure 1 must also be applied to the restricted zone. Measure 1 is always applied to secure zone. Therefore measure 1 occupies both secure and restricted.
Since measure 4 cannot be applied to any zone containing measure 1, measure 4 cannot be in secure or resticted. Measure 4 must be applied to the public zone.
Measure 3 must be restricted zone and measure 4 to public zone.
Can not be true.
If measure 2 and measure 4 are both applied to public zone. If measure 2 is public then measure 3 must also be public. That would mean measure 2,3 and 4 are all public. but each zone can not have more than two policies. Therefore measure 2 and 4 can not be public zones.
13 Jul 2026, 11:18
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For Y (cannot be true): Measure 2 and Measure 4 are both applied to the Public zone.
This is because, the second condition mentions that if measure 2 is applied to the Public Zone, then Measure 3 must be too. And hence, if the above statement was true, Public zone would have Measures 2, 3 AND 4; and we know that a zone cannot have more than 2 policies applied to it. Hence, this statement cannot be true.
For X (Must be Tue): If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone.
Explanation: Assume Measure 3 is applied to the Restricted zone. According to the fourth condition, if measure 3 is applied to the restricted zone, then Measure 1 must be too. Therefore, the restricted zone will already have 2 policies applied to it i.e., 1 and 3. Also, the first condition states that measure 1 must be applied to the secure zone. Hence, we have the restricted zone with measures 1 and 3 and the secure zone with measure 1. Now, the third condition states that Measure 4 cannot be applied to any zone which has Measure 1 applied to it. This leaves only the Public zone where measure 4 can be applied. And hence, it must be true that if Measure 3 is applied to the Restricted Zone, then Measure 4 must be applied to the Public Zone.
This is because, the second condition mentions that if measure 2 is applied to the Public Zone, then Measure 3 must be too. And hence, if the above statement was true, Public zone would have Measures 2, 3 AND 4; and we know that a zone cannot have more than 2 policies applied to it. Hence, this statement cannot be true.
For X (Must be Tue): If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone.
Explanation: Assume Measure 3 is applied to the Restricted zone. According to the fourth condition, if measure 3 is applied to the restricted zone, then Measure 1 must be too. Therefore, the restricted zone will already have 2 policies applied to it i.e., 1 and 3. Also, the first condition states that measure 1 must be applied to the secure zone. Hence, we have the restricted zone with measures 1 and 3 and the secure zone with measure 1. Now, the third condition states that Measure 4 cannot be applied to any zone which has Measure 1 applied to it. This leaves only the Public zone where measure 4 can be applied. And hence, it must be true that if Measure 3 is applied to the Restricted Zone, then Measure 4 must be applied to the Public Zone.
13 Jul 2026, 11:31
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I went for X: If M3 applied to R then M4 applied to P, Y: M2 and M4 both applied to P
The rules given to us are as follows,
a. There are 4 policies - M1, M2, M3, M4
b. Three Zones - S, R, P
c. Every zone must have atleast one and at most two policies applied to it
d. Every policy must be applied to atleast one zone
also,
1. M1 must be applied to S
2. If M2 is applied to P, M3 must also be applied to P
3. M4 cannot be applied to any zone where M1 is applied
4. If M3 is applied to R, M1 must also be applied to R
We have to find X (which must be true) and Y (which cannot be true)
Lets look at the option statements
- M2 and M4 are both applied to P: If we look at rule 2 above, we know that if M2 is applied M3 must also be applied to P. Also, if you look at rule c above, at most only 2 policies can apply. Hence, if M2 is being applied to P, only M3 can be the second policy to be applied. We found our Y (Cannot be true statement)
- If M2 is applied to R than M4 must be applied to P: Let's try and see how the policies can apply.
We know that M1 is applied to S. The option states that M2 is applied to R
Thus,
S: M1 __
R: M2 __
P: __ __
There is no constraint stopping us from applying M4 to R and M3 to P. This arrangement does not violate any rules. Hence, this does not HAVE to be true. We can eliminate this. Additionally, we can check for cannot be true as well. We can apply M3 to S and M4 to either R or P.
- If M3 is applied to R then M4 must be applied to P: We know from Rule 4 above, that if M3 is applied to R, the second policy to be applied to R is M1. Also, Rule 3 states that M4 and M1 cannot be applied together. Since rule 1 says that M1 must apply to S, we know M4 does not apply to it. Thus, M4 must be applied to P. This is our MUST BE TRUE statement (X).
- M4 must be applied to R: A quick look at the rules tells us that M4 can or can not be applied to R. Eg. configurations as below:
S: M1
R: M4 (M4 applied to R)
P: M2 M3
OR
S: M1
R: M2 M3 (M4 not applied to R)
P: M4
Thus, we can eliminate this option.
- If M1 applies to R, M3 is also applied to R: This is the trap option. We are told under rule 4 that if M3 is applied to R then M1 must also be applied to it. Also, under rule 2, we are told that if M2 is applied to P then M3 must be applied to P. However, the reverse is not compulsory in either of the cases. The configuration can be as follows:
S: M1
R: M1 M2
P: M3 M4
Thus, we know that this option can or cannot be true. Eliminate.
The rules given to us are as follows,
a. There are 4 policies - M1, M2, M3, M4
b. Three Zones - S, R, P
c. Every zone must have atleast one and at most two policies applied to it
d. Every policy must be applied to atleast one zone
also,
1. M1 must be applied to S
2. If M2 is applied to P, M3 must also be applied to P
3. M4 cannot be applied to any zone where M1 is applied
4. If M3 is applied to R, M1 must also be applied to R
We have to find X (which must be true) and Y (which cannot be true)
Lets look at the option statements
- M2 and M4 are both applied to P: If we look at rule 2 above, we know that if M2 is applied M3 must also be applied to P. Also, if you look at rule c above, at most only 2 policies can apply. Hence, if M2 is being applied to P, only M3 can be the second policy to be applied. We found our Y (Cannot be true statement)
- If M2 is applied to R than M4 must be applied to P: Let's try and see how the policies can apply.
We know that M1 is applied to S. The option states that M2 is applied to R
Thus,
S: M1 __
R: M2 __
P: __ __
There is no constraint stopping us from applying M4 to R and M3 to P. This arrangement does not violate any rules. Hence, this does not HAVE to be true. We can eliminate this. Additionally, we can check for cannot be true as well. We can apply M3 to S and M4 to either R or P.
- If M3 is applied to R then M4 must be applied to P: We know from Rule 4 above, that if M3 is applied to R, the second policy to be applied to R is M1. Also, Rule 3 states that M4 and M1 cannot be applied together. Since rule 1 says that M1 must apply to S, we know M4 does not apply to it. Thus, M4 must be applied to P. This is our MUST BE TRUE statement (X).
- M4 must be applied to R: A quick look at the rules tells us that M4 can or can not be applied to R. Eg. configurations as below:
S: M1
R: M4 (M4 applied to R)
P: M2 M3
OR
S: M1
R: M2 M3 (M4 not applied to R)
P: M4
Thus, we can eliminate this option.
- If M1 applies to R, M3 is also applied to R: This is the trap option. We are told under rule 4 that if M3 is applied to R then M1 must also be applied to it. Also, under rule 2, we are told that if M2 is applied to P then M3 must be applied to P. However, the reverse is not compulsory in either of the cases. The configuration can be as follows:
S: M1
R: M1 M2
P: M3 M4
Thus, we know that this option can or cannot be true. Eliminate.
13 Jul 2026, 11:34
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Rules:
1. S = M1
2. if P= M2 then P= M3 as well.
3. S= XM4( No M4 For sure)
4. if R=M3 then R=M1
So=> S=M1, R= if M3 Then M1, P= if M2 then M3
Option,
1st=> M2 and M4 both to P. If P=M2 then P=M3 as well so M4 can't be in P. So this is can not be true for sure(Y)
2nd=> R=M2, then P=M4, this can be and cannot be. So ignore.
3rd=> if R=M3, Then R=M1 for sure, so M4 definitely in P, which is exactly the option stating So this is must be true (X)
X= 3st option, Y= 1st option.
1. S = M1
2. if P= M2 then P= M3 as well.
3. S= XM4( No M4 For sure)
4. if R=M3 then R=M1
So=> S=M1, R= if M3 Then M1, P= if M2 then M3
Option,
1st=> M2 and M4 both to P. If P=M2 then P=M3 as well so M4 can't be in P. So this is can not be true for sure(Y)
2nd=> R=M2, then P=M4, this can be and cannot be. So ignore.
3rd=> if R=M3, Then R=M1 for sure, so M4 definitely in P, which is exactly the option stating So this is must be true (X)
X= 3st option, Y= 1st option.
Bunuel
