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

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Dominant. Subordinate Total
Session 1: 5 3 8
Session 2: 3 5 8
Total 8 8 16

Ans Option D

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Bunuel

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- Total animals: 16
- Session 1: 8 D, 8 S
- Session 2: 8 D, 8 S
Number of animals subordinate in Session 1 and dominant in Session 2 = 8 - 5 = 3.
Number of animals dominant in Session 1 and dominant in Session 2 = 8 - 3 = 5 .

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Bunuel

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Let us assume that T1,T2 are the sessions:

Session T1: Lets consider the dominant 8 to be A,B,C,D,E,F,G,H
Subordinate would be 1,2,3,4,5,6,7,8.

So irrespective of the grouping, the alphabets (A...H) will always be dominant in comparison with the (1,2,..8).

Session T2: It is said that 5 of the subordinates are common. Irrespective of the 5, we can assume 1,...5 are subordinate.
So the other 3 cannot be subordinate, which means these 3 will be dominant. Hence, there will be only 5 that are dominant in both groups.

Hence IMO D

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	1st dominant	1st subordinate	Total
2nd dominant	5	3	8
2nd subordinate	3	5	8
Total	8	8	16

So the number of animals dominant in both the animals = 5. Ans (D)

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There were 8 subordinate animals in Session 1. The problem says 5 of them stayed subordinate in Session 2.

This means the other 3 animals from that group (8 - 5 = 3) must have switched to become dominant in Session 2.

We know there are 8 dominant animals in total in Session 2. Since 3 of them came from the subordinate group, the rest must have been dominant in both sessions.

So, 8 (total dominant in S2) - 3 (who switched) = 5.

5 animals were dominant in both sessions.

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8 get the dominant tag
8 get the subordinate tag
Of those 8 subordinate, 5 stay subordinate in Session 2.
That means 3 of them switch and become dominant next time.

So 5 animals keep the dominant tag in both sessions.

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In first session by combining both the pairs 8 Dominant and 8 Subordinates animals wer there. In the second session again there will be a 8 Dominant and 8 Subordinates , and out of these 8 Subordinates, 5 were Subordinates in the earlier session, which means 3 of the earlier Subordinates are now Dominant. Therefore, animals who were Dominant in both sessions are 5.

Bunuel

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Since each animal appears in at most one morning and one afternoon session, and each session has a maximum of 2 animals, the best-case overlap happens when each animal is placed in one morning and one afternoon session without duplication. So the max number of animals that could have been in both is 5.

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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?

In session one, 8 Dominant and 8 Subordinate
Of the animals that were subordinate in the first session, 5 were also subordinate in the second session. Means 3 animals that were subordinate in first session became dominant in second session. And out of 8 animals that were dominant in first session, 3 became subordinate.
So, 5 animals were dominant in both sessions

Answer: 5

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We’re told: There are 16 tagged animals.

In each session, animals are randomly grouped into pairs of 8, with one dominant and one subordinate per pair.

So in each session, there are: 8 dominant animals and 8 subordinate animals

We’re told: Of the 8 animals that were subordinate in the first session, 5 were also subordinate in the second session. So the remaining 3 of those 8 became dominant in session 2. So a total of 5 have continued to be dominant and hence the answer is 5 as they are dominant in both the sessions.

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Since, out of 8 subordinates, 5 were same in the 2nd session, 3 dominants from 1st session would have turned into subordinates in 2nd session. Which leaves us with 5 dominants which were carried forward into 2nd session. and Thus these 5 dominants are common for both group.

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D. 5

During the second session, since 5 subordinate animals were from the 8 subordinate animals from the first session, remaining 3 have to be from the 8 animals which were dominant during the first session. So the answer will be 8 dominant from the 2nd session - 3 which we deteremined to be subordinate in the first session = 5 animals which were dominant in both the sessions.

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Answer: 5 (choice D)

Each session classifies the 16 animals into
• 8 dominants and 8 subordinates.
Let’s label the animals by their status in session 1:
• 8 subordinates
• 8 dominants.
From the problem: of the 8 subordinates in session 1, 5 stayed subordinate in session 2.
⇒ The other 3 of those subordinates became dominant in session 2.
Session 2 still has 8 dominants in total. They consist of
• 3 animals that switched from subordinate to dominant (step 3)
• x animals that were already dominant in session 1 and stayed dominant.
So 3 + x = 8 ⇒ x = 5.

Therefore, 5 animals were dominant in both tracking sessions.

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In the first session we know that 8 would have been subordinate and 8 would have been dominate. If in the second session, 5 of the 8 initial subordinate animals continued to be subordinate... then that means we would have to make up the difference of 3 to get to 8 total subordinate animals in the second session.

If we need 3 NEW subordinate animals to complete our 8, that means we have to remove 3 dominant animals from our initial 8 in the first session. This leaves 5 animals that were dominant in BOTH sessions.

The answer is D

Bunuel

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In the first session, 8 Dominant, 8 Subordinate
In the 2nd session, 8 Dominant, 8 Subordinate
8 Subordinate here includes 5 which were subordinate earlier, then remaining 3 would have been dominant earlier.
8 will include 5 which were also dominant earlier, hence ans 5

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It's given that 5 animals were Subordinate in 2nd Session, that means, 3 Animals that were Dominant in Session 1 are Subordinate in Session 2.
Similarly, when you create a 2/2 Table, you will see that animals eligible to be Dominant in Session two are:-
- 3 Subordinate Animals from Session 1 & 5 Dominant from Session 1

Which clearly tells us that 5 animals are in dominant in both the sessions.

Session 1 / Session 2	Dominant	Subordinate	Total
Dominant	5	3	8
Subordinate	3	5	8
Total	8	8	16

5 - Given (Subordinate Animals in both Sessions)

Bunuel

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S1 = Subordinate animals in session 1
S2 = Subordinate animals in session 2
D1 = Dominant animals in session 1
D2 = Dominant animals in session 2

S1 = S2 = D1 = D2 = 8 because there were 8 pairs and in each pair, one animal was observed as dominant and the other as subordinate.

	S1	D1	Total
S2	5 (given)	3	8
D2	3	5 (to be calculated)	8
Total	8	8	16

Answer: (D) 5

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30 Jun 2025, 09:41

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Subordinate (S), Dominant (D)
Session 1 (ss1), Session 2 (ss2)

Given information

	S ss1	D ss1	Total
S ss2	5	3	8
D ss2	3	5	8
Total	8	8	16

Answer: D

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30 Jun 2025, 09:42

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since 5 remain subordinates, then 3 of the dominants will be added to complete 8 pairs. Thus, 5 remain as dominants in both sessions.

thus (D) 5 is the right answer in my opinion!

Bunuel

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30 Jun 2025, 09:43

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IMO D. 5
Since the 5 animals which were subordinate in session 2 also, the remaining 3 subordinate will be dominant in the session 2. Hence the remaining 5 from session 2 will be the same dominant animals from session 1.