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Since 8 pairs will be formed in each tracking session, this means that:
  • Session 1: 8 dominant animals and 8 subordinate animals.
  • Similarly, session two will also have 8 dominant animals and 8 subordinate animals. Now here in the 8 subordinates, there are 5 old ones from the 1st session (mentioned in the question) and hence, 3 will be new ones.
    8 subordinates= 5 old subordinates from 1st tracking session + 3 new onesThis means that the 3 new ones in the subordinate moved from the dominant animals in session 1, leaving only (8-3) common dominants in both the first and second sessions.

Therefore, the answer is (D) 5.

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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D . 5

A is Dominant in both
B is dominant in first and Subordinate in second phase
C is sub in first and d in second
D is sub in both

A+b= 8
b+c=8
a+d=8
c+d= 8
a+c=8
A+B= 8
C+D= 8
d = 5

substitute





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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there are 16 animals and 8 pairs, so there are 8 and 8 for dominant and subordinate - given per stem; 5 of those 8 dominant remain dominant in the second session, and 3 become subordinate; therefore 5 remained dominant throughout both sessions -- similar to the count for the ones who remained subordinate throughout both sessions. there's probably someone to explain that algebraically as well given the whole "pairs" setup.

Going with answer choice D.
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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S1
D: 8
S: 8
S2
D: 8 (3 were S in S1)
S: 8 (5 were S and 3 were D in S1)
So, 8-3=5 were D in both sessions
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Option D

Let the pairings for test 1 be as follows
Dominant: A1, A2, A3, A4, A5. A6, A7, A8
Subordinate: A9, A10, A11, A12, A13, A14, A15, A16

Of these pairings, let A9-A13 (5 animals) be subordinates in the next test as well. => A14, A15 and A16 will be replaced as subordinates by 3 animals who were dominant in test 1. 5 animals who were dominant in test 1 will remain dominant in test 2 as well.
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16 - 2 groups of 8
D and subordinate
1st group - 8, dom and 8 sub
3 B
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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Dominant 1st session = 8
Sub 1st session = 8

Sub 2nd session = 5 (sub from 1st) and 3 dom from 1st
Dom 2nd session = 3 (remaining sub from 1st) and 5 dom (from 1st)

So the answer is there were 5 dominant in both sessions.
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This is how I am thinking about it:

16 animals:

  • Session 1: 8 sub, 8 dom
  • Session 2: 5 sub stays sub (meaning 3/8 subs --> dom).
    • However, this means that in order to fulfill the pairing of subs and doms, 3 doms --> subs.
      • This tells us that there were 5 animals that were doms in both sessions too.
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Given, there are 16 total animals. There are 2 tracking sessions wherein the animals were split evenly into 8 dominants and 8 submissives in each session. 5 animals were submissive in both sessions.

Asked to find how many animals were dominant in both sessions.

3 animals switched from being submissive to dominant in the second session. Because there are 8 dominant animals in the second session, 3 animals switched from dominant to submissive.

5 - animals that were submissive in both sessions
3 - animals that went from submissive to dominant
3 - animals that went from dominant to submissive to compensate for the 3 that switched
5 - therefore, there must be 5 animals who were dominant in both sessions.
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Let's denote the status of an animal in the first session as S1 or D1, and in the second session as S2 or D2.

So, for Session 1: * Number of S1 animals = 8 * Number of D1 animals = 8
And for Session 2: * Number of S2 animals = 8 * Number of D2 animals = 8


[color=#333333]Status in Session 1 \ Status in Session 2[/color][color=#333333]Subordinate (S2)[/color][color=#333333]Dominant (D2)[/color][color=#333333]Total in Session 1[/color]
[color=#333333]Subordinate (S1)[/color][color=#333333]5[/color] [color=#333333]8[/color]
[color=#333333]Dominant (D1)[/color] [color=#333333]8[/color]
[color=#333333]Total in Session 2[/color][color=#333333]8[/color][color=#333333]8[/color][color=#333333]16[/color]

From the row "Subordinate (S1)": If 5 S1 animals were also S2, then the remaining S1 animals must have become D2.
Number of (S1 & D2) = Total S1 - (S1 & S2) = 8 - 5 = 3


Status in Session 1 \ Status in Session 2Subordinate (S2)Dominant (D2)Total in Session 1
Subordinate (S1)538
Dominant (D1) 8
Total in Session 28816
Total S2 animals = 8. We know 5 of them were S1. So the remaining S2 animals must have been D1.
Number of (D1 & S2) = Total S2 - (S1 & S2) = 8 - 5 = 3


Status in Session 1 \ Status in Session 2Subordinate (S2)Dominant (D2)Total in Session 1
Subordinate (S1)538
Dominant (D1)3 8-3=58
Total in Session 28816
[color=#333333]Total D1 - (D1 & S2) = 8 - 3 = 5. Ans (D)[/color]

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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We have 16 Animals in this question.
These can be split in two groups: D and S. (50/50)

Knowing that and assuming, that none of the animals die during the tracking or is getting replaced for any other reasons, the relations of change in D or S, do equally impact S or D.

So if 7 D from the first round stay D, we know that one must be S now.
-> Vice versa: 7 from S are still S and one switched to D.

Therefore, the answer is D) 5
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The language is confusing but an easy problem to solve:
16 animals made into 8 pairs
each pair (1D (dominant), 1S (subordinate))

Pairs in first session
D1S1, D2S2, D3S3, D4S4, D5S5, D6S6, D7S7, D8S8

Now for the second session animals gets reshuffled and 5 subordinates from the first session are subordinates in the second session let say (S1, S2, S3, S4, S5) and S6, S7, and S8 are new dominants and 3 dominants from the last sessions are now new subordinates let say (D6, D7, D8)

Pairs in 2nd session
S6S1, S7S2, S8S3, D1S4, D2S5, D3D6, D4D7, D5D8

The count of dominants in second session that were dominant in the first session is 5 (D1, D2, D3, D4, and D5)
Option D - 5
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in first round we have 8 dominant and 8 subordinate. out of 8 subordinates, 5 are subordinates again. which is 3 became dominant and their corresponding animal in the pair is subordinate which leaves us with 5 animals which were dominant in both sessions

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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As per given conditions,
Total of 8 animals were dominant in Session 2.
3 animals from the Session 1 - Subordinate group became Session 2 - Dominant

Therefore, the remaining 8−3=5 animals that were dominant in Session 2 must have also been dominant in Session 1.
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Let S1 = subordinate in Session 1
S2 = subordinate in Session 2

dominant in both Session 1 and 2 would be neither S1 nor S2

Since
S1 = S2 = 8 and S1 ∩ S2 = 5,
S1 ∪ S2 = S1 + S2 - (S1 ∩ S2) = 8+8-5 = 11
And Neither S1 nor S2 = total 16 - S1 ∪ S2 = 5

Answer: D
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D. Since in first session we covers 8 animals and in second 5 were same which means we were left with 3 animals choosen from other 8 animals from first session. So we are left with 5
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Given that 5 subordinate from the first session are still tagged the same, it mean 5 are dominant in both session and remaining 3 are interchancing roles.
The answer here is (D)
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