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GMAT Club Olympics 2025 (Day 1): At a wildlife reserve, two tracking

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lvillalon

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30 Jun 2025, 09:46

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(D) 5

Of the 8 dominants of the first session, 8-5=3 became subordinate in the second one, hence, leaving the other 5 again as dominant.

Bunuel

At a wildlife reserve, two tracking sessions were conducted to study 16 tagged animals. In each session, the animals were randomly grouped into 8 pairs, and in each pair, one animal was observed as dominant and the other as subordinate. Of the animals that were subordinate in the first session, 5 were also subordinate in the second session. How many animals were dominant in both sessions?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

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30 Jun 2025, 09:48

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A
Tried working backwards but got confused with all the numbers. Only a few animals could stay dominat right? But didn't account for all the switching properly.

B
Mixed up the numbers here. Saw that 3 animals switched from subordinate to dominant, so thought maybe 3 also stayed dominant. That's not how it works out tho.

C
Was close but made an error somwhere. Didn't keep track of all the animals correctly when they were switching between roles.

D (Correct)
Worked it out step by step. If 5 stayed subordinate and 3 switched from sub to dominant, then 3 others had to switch the other way to keep numbers balanced. Leaves 5 animals that stayed dominant both times.

E
I Thnk there was overcounting somewhere. Maybe forgot that the total has to add up to 16, or mixed up which direction animals were switching.

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30 Jun 2025, 09:48

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No of sessions : 2

First session, Dominant animals (D1) = 8, Subordinate animals (S1) = 8
Second session, Dominant animals (D2) = 8, Subordinate animals (S2) = 8

Now, 5 out of S1 are part of S2, meaning 3 are part of D2.
By similar logic, if S2 must have 3 from D1.

=> D2 must have 5 animals from D1.

Answer is 5

Bunuel

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30 Jun 2025, 09:53

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There were 8 animals that were subordinate in the first session.

The problem states that 5 of these animals were also subordinate in the second session.

This means the other animals from this group must have been dominant in the second session.

Number of animals that were subordinate in Session 1 but dominant in Session 2 = 8−5=3 animals.

In Session 2, there were also 8 dominant animals in total.

We already identified that 3 of these dominant animals were subordinate in Session 1.

Therefore, the remaining animals in this dominant group must have also been dominant in Session 1.

Number of animals dominant in both sessions = 8−3=5 animals.

nevaan9

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30 Jun 2025, 09:54

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Both sessions is a clue that this problem can be solved with a 2x2 matrix

If you map session 1 in the rows and session 2 in the columns, and and 8 to the totals (at the end), we can fill in the blanks to solve this problem.

Session1(Dominant)
Session1(Sub)
Total

Session	Session2(Dominant)	Session2(Sub)
Session1(Dominant)	5	3	8
Session1(Sub)	3	5	8
	8	8

Answer is 5 [where Session1(Dominant) and Session2(Dominant) overlap ]

Bunuel

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30 Jun 2025, 09:55

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Since there are 16 tagged animals and while grouping into 8 pairs one animal is dominant and other is subordinate. This imply that there are atleast 8 dominant animals and 8 subordinate
also in second session there were 5 who were also subordinate
S1 and S2 = 5
We need to find D1 and D2

Since D1, D2 and S1 has 8 members each

So x+(8-5) = 8
x=5

Answer is D

Bunuel

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30 Jun 2025, 09:56

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Two tracking session to study 16 animals.

For each study session, 8 pairs were studied with one dominant animal another subordinate.

From information we know, 5 subordinate in session were subordinate in session 2 as well.

Total subordinates in session 1 and 2 are 8. 3 of them we dont know if they are dominant or subordinate in session 2.

So lets take scenario 1: All Subordinates were also subordinates in session 2 as well

So All 8 will be dominant in session 2

Scenario 2: 5 Subordinate were Subordinates in session 2 as well but remaining 3 were dominant

Here 3 dominant lets assume is pared with 3 of the previous subodinates. So 2 previous dominants were paired with previous subordinates. and 3 previous dominants were paird with previous dominants but session 2 subordinate.

So total 5 of previous dominants have to be paired as dominants to make it work.

There 5 must be total animals who were dominant in both sessions.

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Bunuel

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Subordinate animals in both sessions = 5
So, 3 were subordinate in S1 but dominant in S2. Likewise, 3 were dominant in S1 but subordinate in S2.
Total of these combinations = 5 (sub-sub) + 3 (sub-dom) + 3 (dom-sub) = 11
animals dominant in both sessions = total animals - (subordinate in both sessions + dominant in S1 but subordinate in S2) = 16 - 11 = 5

Answer: (D)

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Bunuel

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Session 1: 1d-1s, 2d-2s, 3d-3s, 4d-4s, 5d-5s, 6d-6s, 7d-7s, 8d-8s (16 animals - 8 pairs - 1 dominant & 1 subordinate)
Session 2: 1s to 5s were still subordinate, so only from 6s to 8s were switched to dominant, meaning that only 3 dominant animals in session one could switch to subordinate.
Therefore, 5 animals were dominant in both sessions.
Answer (D)

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30 Jun 2025, 10:07

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Out of 8, 5 were subordinates.
now this 5 remains constant as subordinates in new group.
Rest 3 subordinates come from those who were dominant. Select any 3.
Thus, 5 dominants continue as dominants.

Ans 5 ..option D

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D. as the subordinate animals in first session 5 were subordinate but in second session the remaining 3 were dominant so 5 mist be dominant in both sessions

Bunuel

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30 Jun 2025, 10:12

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There are 16 total animals.
They are divided into pairs across two sessions.
That means there are 8 pairs in each session and every category (dominant/subordinate) has 8 animals in it per session.
If 5 animals were subordinate in both sessions, that would mean 3 animals switched from subordinate to dominant and vice-versa across sessions.
Out of the 8 animals that were subordinates in session 1, 5 of them were also subordinate in session 2. Therefore 3 animals that were subordinate in session 1 became dominant in session 2.
Session 2 also had 8 animals in the dominant category. If 3 of those animals were subordinates in session 1, that would mean 5 of the animals that were dominant in session 1 are also dominant in session 2.
Answer - 5

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30 Jun 2025, 10:13

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Bunuel

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First Session -> 8D 8S

Second session -> 3 S of the first session are now dominant thus the remaining 5 dominants of the second session were also dominant on the first sessions

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30 Jun 2025, 10:15

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I picked D)5 and arrived at the answer by creating a matrix.

Column 1: Group 1 Dominant, Group 1 Sub (or not dominant), Total
Row 1: Group 2 Dominant, Group 2 Sub (or not dominant), Total

The cell that will answer the question/what we are solving for is Group 1 Dominant + Group 2 Dominant

Since the stimulus says there were 8 pairs/session and one animal in each pair was dominant and the other submissive, there is an even split of 8 and 8 for the totals:

Group 1 Dominant + Total: 8
Group 1 Sub + Total: 8
Group 2 Dominant + Total: 8
Group 2 Sub + Total: 8
Total + Total: 16

The stimulus also states that of subordinate animals in the first session (Group 1 Sub), 5 were also subordinate in the second session (Group 2 Sub). This Then fills in the cells:

Group 1 Subordinate + Group 2 Dominant: 3
Group 1 Subordinate + Group 2 Subordinate: 5

Since we now have only one cell missing for Group 1 Dominant and Group 2 Dominant (Group 1 Dominant + Group 2 Dominant), we can solve:

(Group 1 Dominant + Group 2 Dominant) + 3 = 8
Therefore, Group 1 Dominant + Group 2 Dominant = 8 - 3 = 5

In case this helps anyone else just getting their footing with overlapping sets and matrices, the mistake I made that slowed me down at first was choosing incorrect variables:
Column 1: Group 1, Group 2, Total
Row 1: Dominant, Subordinate, Total

This doesn't help because it doesn't create the overlap in session sets needed to solve (i.e., you need a cell where Group 1 and Group 2 overlap, which this matrix didn't provide; and you need a specific cell where Group 1 Dominant and Group 2 Dominant overlap to account for the 5 animals presented in the stimulus).

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30 Jun 2025, 10:23

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ANS D

First group

Dominant = 8
Subordinate = 8
5 from this 8 subordinate kept subordinated and 3 of them became dominant in the second study , that means the we need 3 subordinate to pair with them, so from the first group of dominant 3 of 8 became subordinate and 5 them kept dominant in the second study also. Leaving 5 always dominant.

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30 Jun 2025, 10:24

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As per the Question Stem

Total Subordinate in Session 1 (S1) = 8
Total Dominant in Session 1 (D1) = 8
Total Subordinate in Session 2 (S2) = 8
Total Dominant in Session 2 (D2) = 8
Animals subordinate in Session 1 & subordinate in Session 2 (S1∩S2) = 5

Subordinate in Session 1(S1)

[table][tr][td]Subordinate in Session 2 (S2)

[/td]
Dominant in Session 2 (D2) Total in Session 1[/tr]
538Dominant in Session 1 (D1)358Total in Session 28816[/table]

If 8 animals were subordinate in Session 1, and 5 of them remained subordinate in Session 2, then the remaining 8−5=3 must have become dominant in Session 2.

We know there are 8 animals dominant in Session 2 (D2) in total.
Above, We just found that 3 of these (D2) came from the group that was subordinate in Session 1 (S1).
Therefore, the remaining 8−3=5 animals that were dominant in Session 2 (D2) must have come from the group that was dominant in Session 1 (D1

Therefore, the remaining 8−5=3 animals that were dominant in Session 1 (D1). must have become subordinate in Session 2 (S2).

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30 Jun 2025, 10:28

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The answer is 5, since only 3 subordinates became dominant, while 5 animals remained dominant in both sessions.

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30 Jun 2025, 10:28

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4 cases -
A = DD, B = SS = 5 (given), C = DS, D = SD

(given)
A+B+C+D = 16
A+C+D = 16-5 = 11

Also:

In Session 1, 8 animals were dominant,
A+C = 8
C = 8-A
Similarly, in Session 2, 8 animals were dominant,
A+D= 8
D = 8-A

Thus, A+(8−A)+(8−A)=11
8+8−A=11
16−A=11

A=5

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30 Jun 2025, 10:32

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30 Jun 2025, 10:35

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This Q is quite simple the since 5 remained sub in the second time so 3 became dom. So from the first trial 5 remain dom and 3 become sub in the second trial.