Last visit was: 07 Sep 2026, 06:29

It is currently 07 Sep 2026, 06:29

Toolkit

Practice Tests

Data Insights Questions

Two-part Analysis (TPA)

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Go to My Error Log Learn more

Request Expert Reply

Confirm Cancel

Sep 07
The Free 12-Week GMAT Study Plan (Full Walkthrough)
10:00 AM PDT
-
11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost.
Sep 02
Save on All Magoosh Study Plans
01:00 PM EDT
-
11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 04
BIG NEWS! We’ve been nominated, and we’d love your vote!
10:00 AM EDT
-
11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 07
All GMAT Club Tests are free and open for Labor Day on Monday, September 7th!
12:01 AM PDT
-
11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT
Sep 09
Masters' Spotlight Fair
07:00 AM PDT
-
10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway
Sep 09
How Aman Bounced Back from 3 MBA Rejections to Get Into Cambridge
10:00 AM PDT
-
11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application...
Sep 15
Live Online Class with Top GMAT Expert Rich Zwelling
07:00 PM EDT
-
09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 21
Live Online Bootcamp Class with Top GMAT Expert Chris Peckover
08:00 PM EDT
-
10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST

Back to Forum

Create Topic

GMAT Club World Cup 2026 (Day 6): A security architect is implementing

1 2 3 4

ajain31

Joined: 09 Nov 2024

Last visit: 07 Sep 2026

Posts: 70

Own Kudos:

[1]

Given Kudos: 68

Location: United Kingdom

Schools: Wharton '27 Stanford '27 HBS '27 CBS '27

GPA: 71%

Schools: Wharton '27 Stanford '27 HBS '27 CBS '27

Posts: 70

Kudos: 57

[1]

13 Jul 2026, 23:51

Kudos

Bookmarks

lets see, each measure foes to at least one zone,
M1 is in S and it given to us
M4 can be in S since M4 cant share a zone with M1
So M4 lives in R or P or both
M3 in R forces M1 into R

Finding cannot be true
Measure 2 and measure 4 are both applied to the public zone
suppose M2 in P, rule 2 then forces M3 into P as well. Add M4 ad the public zone now holds

M2, M3 and M4 = 3 policies

that breaks the max 2 policies per zone so cannot be true

finding x(must be true)
looking at the rules, M4 must be in P, unavoidable

Answer
X - if measure 3 is applied to the restricted zone measure must be applied
Y - measure 2 and measure 4 are both applied to the public zone

GamePine

Joined: 31 May 2026

Last visit: 04 Aug 2026

Posts: 61

Own Kudos:

[1]

Given Kudos: 1

GMAT Focus 1: 645 Q79 V85 DI82 Verified score

Posts: 61

Kudos: 55

[1]

13 Jul 2026, 23:54

Kudos

Bookmarks

First I tried to find Measure 1(M1)'s possible zones.

M1 must be in Secure (S), but could also be in Restricted (R), Public (P), or all three. Since M4 can't share a zone with M1, M1 cannot be in all three because there would be no zones left for M4. So M1 is either S, S&P, or S&R

If M3 is in R then M1 must be in R

If M3 is in R then M2 must be in S & R which would only leave P for M4. This shows that if M3 is in R then M3 must be applied to the Public Zone is always true.

Next I tested whether M2 and M4 are both applied to the public zone

If M4 is in P, M1 cannot be in P. Also if M2 is in P then M3 is in P based on rule 2. That puts M1,M2, and M3 in P which goes against the cap of 2 per zone so this cannot be true

SwethaReddyL

Joined: 28 Nov 2023

Last visit: 06 Sep 2026

Posts: 186

Own Kudos:

Given Kudos: 271

Location: India

Posts: 186

Kudos: 37

14 Jul 2026, 00:01

Kudos

Bookmarks

by the given conditions,
must be true is M4 is applied to R zone
cannot be true is M2 and M4 are both applied to P zone

because, if M2 is applied to P zone then M3 must also be applied to P zone along with M4 it makes 3 policies in a single zone, then by the given conditions no more than 2 policies should be applied to a single zone. So this must be false

If M1 is applied to S zone
M2 applied to P zone and M3 applied to P zone
M4 then must be applied to R zone - this doesn't break any of the constraints, this sits good.

thanks,
Swetha

Bunuel

A security architect is implementing four new policies—Measure 1, Measure 2, Measure 3, and Measure 4—across three defined network zones: Secure (S), Restricted (R), and Public (P). Each policy will be applied to at least one zone. Further, each zone must have at least one policy applied, and no more than two policies applied.

● Measure 1 must be applied to the Secure zone.
● If Measure 2 is applied to the Public zone, then Measure 3 must also be applied to the Public zone.
● Measure 4 cannot be applied to any zone to which Measure 1 is applied.
● If Measure 3 is applied to the Restricted zone, then Measure 1 must also be applied to the Restricted zone.

For the deployment plan to comply with the conditions, select the statement that must be true in the first column (X) and the statement that cannot be true in the second column (Y).

Reon

Joined: 16 Sep 2025

Last visit: 06 Sep 2026

Posts: 296

Own Kudos:

200

[1]

Given Kudos: 19

GMAT Focus 1: 575 Q78 V78 DI79 Verified score

Posts: 296

Kudos: 200

[1]

14 Jul 2026, 00:30

Kudos

Bookmarks

S———————-R————————P
M1 ————- M3, M1 ————- M2, M3
.....—————- M4 —————— M4

1. M2 and M4 are both applied to P. If M2 is applied to P, then according to conditions given in question M3 must be applied to P. Atmost only 2 measures can be applied to a zone. Therefore, if P has M2, then M4 cannot be applied to P. IT MUST BE FALSE (Y)

3. If M3 is applied to R, M4 must be applied to P. If M3 is applied to R, then M1 must also be applied to R and one zone can atmost have two measures only. M4 cannot be applied to S because it has M1. And M1 and M4 cannot be applied to same zone. So M4 can only be applied to P. IT MUST BE TRUE (X)

MakB

Joined: 05 May 2025

Last visit: 03 Sep 2026

Posts: 73

Own Kudos:

Given Kudos: 146

Posts: 73

Kudos: 29

14 Jul 2026, 01:00

Kudos

Bookmarks

We know that measure 1 is for Secure zone only. Keep that aside.
Now as first option says meaaure 2 and 4 applied to public zone, but in question it is clearly stated than 2 has to be with 3, hence this is absolutely not true.

For must be true,
We know if 2+3 are together in one zone, then 4 will be in the next zone automatically.
Statement have word play.
Statement 2 says if Measure 2 is in restrict zone then 4 WILL BE APPLIED to Public
and statement 3 says if measure 3 is in restrict then 4 MUST BE APPLIED to public.
For statement to to MUST BE TRUE, we chose statement 3.

Saurav1408

Joined: 26 Oct 2023

Last visit: 02 Sep 2026

Posts: 81

Own Kudos:

[1]

Given Kudos: 287

Status:Active

Location: India

Products:

Posts: 81

Kudos: 61

[1]

14 Jul 2026, 01:20

Kudos

Bookmarks

Before starting remember the 4 rules given to us. And
Constraints: Each zone must contain 1 or 2 policies but not more than 2
Every policy must be used at least once

For MUST BE TRUE -
If M3 is in R, then Rule 4 forces forces M1 to be in R
IF M1 is in R then Rule 3 dictates that M4 cannot be in R

We already know M4 cannot be in S(because M1 is in S)

Since M4 must be used at least once(as given) so it cannot go into S or 4 so M4 must be placed in P
Therefore, If M3 is applied to R zone, M4 must be applied to P zone MUST BE TRUE

For Cannot be True:
Let us see if both M2 and M4 are assigned to P zone
According to given rule 2, M3 must be assigned to P too.
And P would contain M2, M3 and M4 then which does not obey the constraint
hence it cannot be true.
Therefore,
M2 and M4 are both applied to the P zone CANNOT BE TRUE

twinkle2311

Joined: 05 Nov 2021

Last visit: 07 Sep 2026

Posts: 211

Own Kudos:

247

[1]

Given Kudos: 54

Location: India

Concentration: Finance, Real Estate

Schools: ISB '27 Wharton '27

GPA: 9.041

Products:

Schools: ISB '27 Wharton '27

Posts: 211

Kudos: 247

[1]

14 Jul 2026, 01:59

Kudos

Bookmarks

M1 is always applied to Secure.
Since M4 cannot be applied wherever M1 is applied, M4 can never be in Secure.

If M3 is applied to restricted, M1 must also be applied to restricted. That fills Restricted (max 2 policies), so M4 can't be there either.

Since M4 must be applied somewhere, it has to be in Public

So, if M3 is applied to restricted, M4 must be applied to Public

For the 2nd part, M2 and M4 cant both be in public. If M2 is in Public, M3 must also be in Public. That would put M2, M3 and M4 all in public, violating the max of 2 policies per zone.

X : If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone
Y: Measure 2 and Measure 4 are both applied to the Public Zone

Kinshook

Major Poster

Joined: 03 Jun 2019

Last visit: 07 Sep 2026

Posts: 6,129

Own Kudos:

6,090

[1]

Given Kudos: 164

Location: India

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

WE:Engineering (Transportation)

Products:

Schools: HBS '26 CBS '26 Wharton '26

GMAT 1: 690 Q50 V34 Verified score

Posts: 6,129

Kudos: 6,090

[1]

14 Jul 2026, 02:40

Kudos

Bookmarks

A security architect is implementing four new policies - Measure 1 (M1), Measure 2 (M2), Measure 3 (M3) and Measure 4 (M4) - across three defined network zones: Secure (S), Restricted (R) and Public (P).

Each policy will be applied to at least one zone.
Further, each zone must have at lease one policy applied, and no more than 2 policies applied.

C1. Measure 1 (M1) must be applied to Secure (S) zone; Secure (S) zone may have one more measure; M1 may be applied to other zones (R & P).
C2. If Measure 2 (M2) is applied to Public (P) zone, then Measure 3 (M3) must also be applied to Public (P) zone. Public (P) zone will have either none of M2 & M3, M3 alone or M2 & M3 both applied.
C3. Measure 4 (M4) can not be applied to any zone to which Measure 1 (M1) is applied. Measure 4 (M4) is not applied to Secure (S) zone and any other zone to which Measure 1 (M1) is applied.
C4. If Measure 3 is applied to the Restricted zone, then Measure 1 must also be applied to the Restricted zone. Restricted (R) zone will either none of M3 & M1, M1 alone or M1 & M3 both.

Statements: -
1. Measure 2 & Measure 4 are both applied to the Public (P) zone, then no other policy can be applied to Public zone since max 2 policy limit is reached. But as per C2 above, If Measure 2 is applied to Public (P) zone, then Measure 3 must also be applied to Public zone. This condition is violated since Public zone can not have 3 policies.
CANNOT BE TRUE

2. If Measure 2 is applied to the Restricted zone, Measure 4 must be applied to Public zone.

Case1 :
Measure 1 - Secure
Measure 2 - Restricted
Measure 3 - Public
Measure 4 - Restricted

Case 2:
Measure 1 - Secure
Measure 2 - Restricted
Measure 3 - Public
Measure 4 - Public

There are multiple possibilities

3. As per C4, If Measure 3 is applied to the Restricted zone then Measure 1 must also be applied to Restricted zone, no other policy can be applied to Restricted zone since max 2 policies limit is reached. Measure 1 is applied to both Secure and Restricted zone, then Measure 1 can not be applied to Public zone. Measure 4 can not be applied to Secure zone as per C1 & C4 above. Measure 4 can not be applied to Restricted and Secure zones, then it must be applied to Public zone . Therefore, if Measure 3 is applied to Restricted zone, Measure 4 must be applied to Public zone.
MUST BE TRUE

There is no need to check other statement.

CAN NOT BE TRUE	MUST BE TRUE
Measure 2 & Measure 4 are both applied to the Public zone	If Measure 3 is applied to Restricted zone then Measure 4 must be applied to the Public zone

remdelectus

Joined: 01 Sep 2025

Last visit: 07 Sep 2026

Posts: 120

Own Kudos:

112

[1]

Given Kudos: 6

Posts: 120

Kudos: 112

[1]

14 Jul 2026, 03:20

Kudos

Bookmarks

Policies 1,2,3,4 must be used at least once each.
zones S,R,P must have 1 or 2 policies each.
Rule 1: 1 applied to S
Rule 2: If 2 applied to P, then 3 applied P
Rule 3: 4 and 1 never in the same zone.
Rule 4: If 3 applied to R then 1 applied to R
Must Be True (X)
If measure 3 is applied to the restricted zone, measure 4 must be applied to the public zone.
if 3 is in R, by rule 4, 1 must also be in R.
as per rule 1, 1 is already in S, so 1 is in both S and R.
4 cannot be with 1 as per rule 3, so 4 cannot be in S or in R.
since every policy must be applied at least once and 4 cannot be in S or R, it must be applied to P.
This statement must be true.

Cannot be True (Y)
2 and 4 are both applied to P.
assume measure2,4 applied to p
rule 2 says if 2 is applied to P, 3 must be in P.
so P =2,4,3
rule 2 states that each zone can have no more than two policies.
since this results in 3 policies in one zone, it cannot be true
ANS: X (Must be True) - If Measure 3 is applied to restricted zone, Measure 4 must be applied to the public zone.
Y (Cannot be True) - Measure 2 and Measure 4 are both applied to the public zone.

jkkamau

Joined: 25 May 2020

Last visit: 07 Sep 2026

Posts: 285

Own Kudos:

244

[1]

Given Kudos: 147

Location: Kenya

Schools: Jones '26 (A$$)

GMAT 1: 730 Q50 V46 Verified score

GPA: 3.5

Products:

Schools: Jones '26 (A$$)

GMAT 1: 730 Q50 V46 Verified score

Posts: 285

Kudos: 244

[1]

14 Jul 2026, 04:14

Kudos

Bookmarks

Must be true: C (IF M3 is put in R then based on 4th rule M1 is forced to R but according rule 3 M4 cannot share a zone with M1 meaning M4 is excluded from both S and R an hence can only take P
Must be false: A
Suppose M2 and M4 are placed in P. This forces M3 to be also included in P which breaks the more than 2 policies in a single zone.
Ans CA

Bunuel

MANASH94

Joined: 25 Jun 2025

Last visit: 05 Sep 2026

Posts: 165

Own Kudos:

121

[1]

Given Kudos: 27

Location: India

Schools: IIM IIM ISB

GMAT Focus 1: 525 Q80 V76 DI72 Verified score

GPA: 2.9

Products:

Schools: IIM IIM ISB

GMAT Focus 1: 525 Q80 V76 DI72 Verified score

Posts: 165

Kudos: 121

[1]

14 Jul 2026, 05:26

Kudos

Bookmarks

Some Conditions that we can understand from the question stem:
Measure 1 (M1) must be applied to secure zone (S).
Measure 4 (M4) cannot be with Measure 1 so 4 cannot be secure.
P can have max of 2 measures.
Calling Measure 2 as M2 and Measure 3 as M3 and Restricted as R and Public as P:

M3 is in R, M1 is in R.
M1 is already in S.
M4 cannot be where M1 is.
So M4 cannot be in S or R.
If Measure 3 is in R then M4 must be in P (Must be True).
Which is option 3.

M2 in P ⇒ M3 in P. then as per second statement P has M2, M3, M4 = 3 measures.
But a zone can have at-most 2 measures which is not possible (Cannot be true)
Which is option 1.
IMO X: If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone.
Y: Measure 2 and Measure 4 are both applied to the Public Zone.

Bunuel

redandme21

Joined: 14 Dec 2025

Last visit: 01 Aug 2026

Posts: 157

Own Kudos:

144

[1]

Posts: 157

Kudos: 144

[1]

14 Jul 2026, 05:45

Kudos

Bookmarks

S: M1, no M4
R: if M3 then M1
P: if M2 then M3

X (Must Be True):
If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone.

Using "R: if M3 then M1", M4 no in R because R cannot have a third policy (only two at most).
As M4 no in S too, M4 must be in P (each policy will be applied to at least one zone).

Y (Cannot Be True):
Measure 2 and Measure 4 are both applied to the Public Zone

"P: if M2 then M3", P has M2 and M3, so cannot have a third policy (only two at most).

X (Must Be True) = If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone.
Y (Cannot Be True) = Measure 2 and Measure 4 are both applied to the Public Zone.

tanvi0915

Joined: 01 Oct 2025

Last visit: 07 Sep 2026

Posts: 32

Own Kudos:

Given Kudos: 7

Posts: 32

Kudos: 21

14 Jul 2026, 05:49

Kudos

Bookmarks

For X I choose If measure 2 is applied to R Z measure 4 will be applied to P Z because we know measure 1 is applied to SZ so measure 4 cannot be applied there and if measure 2 is applied to R Z the measure 3 is also applied to RZ and since atmost 2 measure can be applied to a zone that leaves measure 4 to be applied to PZ and for the same reason For Y I choose Measure 2 and 4 both applied to public zone as not true

gaurikhanna17

Joined: 16 Jan 2025

Last visit: 07 Sep 2026

Posts: 73

Own Kudos:

[1]

Products:

Posts: 73

Kudos: 46

[1]

14 Jul 2026, 05:58

Kudos

Bookmarks

IMO :
Must be true : If M3 applied to R ----> M4 must be applied to P
Cannot be True : M2 and M4 both applied to R

Reasoning : ATQ: if M2=P --> M3 =P
M4 is not equal to M1
If M3=R ---> M1 =R
therefore when M3 applied to R then M1 will be applied to R
and that means M4 will neither have R not S since S has been assigned in the question itself
therefore atleast 1 --- therefore M4 = P

2nd:
M2 =P then M3 =P and in option it says M4 also =P but three are not possible as per question

AngelUPENN

Joined: 17 Jun 2026

Last visit: 25 Jul 2026

Posts: 36

Own Kudos:

[1]

Given Kudos: 1

Products:

Posts: 36

Kudos: 29

[1]

14 Jul 2026, 05:59

Kudos

Bookmarks

The answer is X must be: if measure three is applied to the restricted zone, measure four must be applied to the public zone.
Why: measure three in restricted forces measure one into restricted too, filling restricted with two measures. Measure four cannot go to secure measure one is there, and restricted is now full and holds measure one anyway. So measure four has to land in public.
Y cannot be true: measure two and measure four are both applied to the public zone.
Why: measure two in public drags measure three into public. Adding measure four makes three measures in public, which breaks the two per zone cap.

Bunuel

geocircle

Joined: 14 Dec 2025

Last visit: 27 Jul 2026

Posts: 150

Own Kudos:

143

[1]

Posts: 150

Kudos: 143

[1]

14 Jul 2026, 06:10

Kudos

Bookmarks

From the rule: "If Measure 2 is applied to the Public zone, then Measure 3 must also be applied to the Public zone" then "Measure 2 and Measure 4 are both applied to the Public Zone" cannot be true because there cannot be three Measures applied (2,3,4) to the Public Zone. Each zone can have no more than two Measures.

From the rule: "Measure 1 must be applied to the Secure zone" and "Measure 4 cannot be applied to any zone to which Measure 1 is applied", then Measure 4 cannot be applied to the Secure Zone.
"If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone" must be true because, as we have seen, Measure 4 cannot be applied to the Secure Zone, and it cannot be applied either to Restricted Zona because using the rule "If Measure 3 is applied to the Restricted zone, then Measure 1 must also be applied to the Restricted zone", the Restricted Zone would already have two Measures applied (Measure 1 and Measure 3) and, as all Measures must be applied to at least one Zone, Measure 4 should be applied to the Public Zone.

X (Must Be True): "If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone."
Y (Cannot Be True): "Measure 2 and Measure 4 are both applied to the Public Zone."

Nsanghavi

Joined: 08 Jul 2025

Last visit: 29 Aug 2026

Posts: 59

Own Kudos:

Given Kudos: 20

Products:

Posts: 59

Kudos: 46

14 Jul 2026, 06:29

Kudos

Bookmarks

This is a little confusing question - It mentions the following facts: 3 Network Zones are there and each zone can have a minimum of 1 policy and maximum of 2 policies
a) Measure 1 = Secure Zone
b) If Measure 2 = Public Zone, then Measure 3 = Public zone
c) Measure 4 and Measure 1 can not be together
d) If Measure 3 = Restricted Zone, then Measure 1 = Restricted Zone

Option I - Measure 2 and Measure 1 are both applied to Public Zone - But this contradicts the option b and the initial statement above hence this statement is false

Option II - If measure 2 is applied to restricted zone , Measure 4 will be applied to Public zone - This also has 2 possibilities - Measure 4 can be both Public Zone or restricted zone. hence not always true

Option III - If measure 3 is applied to restricted zone, Measure 4 must be applied to Public zone - This seems to be true since it fulfills all the conditions

Option IV - Measure 4 is applied to restricted zone - There can be other possibilities as well hence not always true

Option V - If Measure 1 is applied to restricted zonce then Measure 3 is also applied to restricted zone - Over here as well Measure 3 can be in any zone and not only restricted zone.

So - Must be True is option III - but I am not sure which to select for cannot be true - I have randomly guessed at Option IV since it is the only option which directly mentions about 1 measure without its linkage to any other measures.

LibrarianTurtle

Joined: 12 Nov 2024

Last visit: 06 Sep 2026

Posts: 36

Own Kudos:

[1]

Given Kudos: 8

GPA: 3.9

Products:

Posts: 36

Kudos: 24

[1]

14 Jul 2026, 06:34

Kudos

Bookmarks

We have 4 different measures (policy) that will be applied to 3 different network zones.
First we need to understand what the criteria are.

Each policy will be applied to at least one zone (Can't leave any policy outside)
Each zone must have at least one policy applied (No zone without policy)
No more than two policies applied to a single zone. (Max 2 Policies - Min 1 policy)

Measure 1 to Secure Zone is a must
If measure 2 to Public, then also measure 3 to Public (But not vice versa, If measure 3 to Public that does not mean measure 2 to public)
If the zone has measure 1, then measure 4 cannot be added to that specific zone.
If measure 3 to Restricted, then also measure 1 to Restricted (Attention! Not vice versa, we cannot reverse the relation)

Lets create a table according to these criteria

Security	Public	Restriction
1	1	1
1, 2	1,3	2
1, 3	2,3	3,1
	4	4
	4,3	1,2
	3	2,4

Now lets go through each option

M2 and M4 both to Public (Since if we add 2, we also must add 3 and maximum policy limit is 2 so we surely can't have both M2 and M4 at the same time in Public zone) (Cannot be True)
If M2 to Restricted, then M4 (will be) to Public (It may or it may not be nothing certain)
If M3 to Restricted, then M4 (must be) to Public (If we put M3 to restricted, then slots at Restricted is full and since we cannot add M4 to any zone with M1 only zone that must be added to is Public (No policy left out) (Must be True)

IMO

topgmat25

Joined: 15 Dec 2025

Last visit: 27 Jul 2026

Posts: 150

Own Kudos:

139

Posts: 150

Kudos: 139

14 Jul 2026, 06:34

Kudos

Bookmarks

Secure Zone: M1
Restricted Zone: M3->M1
Public Zone: M2->M3

X (Must Be True)='If Measure 3 is applied to the Restricted zone, Measure 4 must be applied to the Public Zone.'
Restricted Zone: M3->M1, M3 and M1, no more measures allowed (2 at most)
Secure Zone: M1, so no M4
M4 must be in Public Zone (M4 must be in at least 1 zone).

Y (Cannot Be True)='Measure 2 and Measure 4 are both applied to the Public Zone.'
Public Zone: M2->M3, M2 and M3, no more measures allowed (2 at most)

mehtyas

Joined: 27 May 2025

Last visit: 06 Sep 2026

Posts: 74

Own Kudos:

[1]

Given Kudos: 832

Location: India

Concentration: Finance, Strategy

Products:

Posts: 74

Kudos: 61

[1]

14 Jul 2026, 06:40

Kudos

Bookmarks

Key deductions:
-> M1 is always in S.
-> M4 can never be in S (it cannot be in the same zone as M1).
-> Since each zone can have at most 2 measures, the remaining placements are tightly constrained.

Evaluating the statements:
1. M2 and M4 are both in P - Cannot be true
--> If M2 is in P, them M3 must also be in P.
--> That would put M2, M3 and M4 in P, exceeding the limit of 2 measures

2. If M2 is in R, then M4 is in P - Not always true
3. If M3 is in R, then M4 is in P. - Must be true
--> M3 in R = M1 is also in R
--> M4 cannot be with M1, so it cannot be in R.
--> M4 also cannot be in S (since M1 is always in S).
--> Therefore, M4 must be in P.

4. M4 is in R - Can be true but not always
5. If M1 is in R, then M3 is also in R - Not always true.

Answer:
Must be true - Statement 3
Cannot be true - Statement 1