Question

Suppose vial C tests positive and vials A, E and H test negative. Which patient has the disease?

• A. Patient 14
• B. Patient 2
• C. Patient 6
• D. Patient 8
• A. Vial E
• B. Vial H
• C. Vial C
• D. Vial B
• A. Vials A and G positive, vials D and E negative
• B. Vials B and D positive, vials F and H negative
• C. Vial B positive, vials C, F and H negative
• D. Vials A and E positive, vials C and D negative
• A. {4,5}
• B. {5,6,7,8}
• C. {4,5,6,7,8}
• D. {4,5,6,7}

Answer 1

The Correct Option is C

The objective is to identify the patient with the disease using test results from vials. Vial C tested positive, while vials A, E, and H tested negative. The methodology involves excluding patients whose blood samples are present in vials that tested negative.

Let's first analyze the implications of each vial's test result:

• A negative test result for a vial signifies that no patient whose sample is in that vial has the disease.
• As vials A, E, and H yielded negative results, all patients associated with these vials are eliminated as potential carriers.

Consulting the data table reveals the following patient assignments to these vials:

• Vial A: Patients 9, 10, 11, 12, 13, 14, 15, 16
• Vial E: Patients 3, 4, 7, 8, 11, 12, 15, 16
• Vial H: Patients 1, 3, 5, 7, 9, 11, 13, 15

Elimination based on negative test results:

• From Vial A: Exclude patients 9, 10, 11, 12, 13, 14, 15, 16.
• From Vial E: Exclude patients 3, 4, 7, 8, 11, 12, 15, 16.
• From Vial H: Exclude patients 1, 3, 5, 7, 9, 11, 13, 15.

The sole remaining candidate, after eliminating individuals from negatively tested vials and who were not in Vial C:

• Patient 6, whose sample is included in vials B, C, F, and G.

Given that Vial C produced a positive test result, and Patient 6 is the only remaining individual in Vial C after all other patients in that vial were excluded due to negative test results in other vials, Patient 6 must be the one with the disease.

Consequently, the confirmed patient is:

Patient 6

Answer 2

To resolve this, we must identify the next vials to test to pinpoint the patient with the disease. The current findings are:

• Vial A is positive: This indicates a potential disease carrier among patients 9, 10, 11, 12, 13, 14, 15, 16.
• Vials D and G are negative: Consequently, patients 1, 2, 3, 4, 9, 10, 11, 12 (from Vial D) and 2, 4, 6, 8, 10, 12, 14, 16 (from Vial G) are excluded.

Excluding patients from vials D and G yields:

• From Vial D exclusion: Patients 1, 2, 3, 4, 9, 10, 11, 12 are eliminated.
• From Vial G exclusion: Patients 2, 4, 6, 8, 10, 12, 14, 16 are eliminated.

After these exclusions, the potential candidates from Vial A are reduced to patients 13, 14, 15, and 16. However, patients 14 and 16 are also eliminated as they were part of the negative Vial G. This leaves patients 13 and 15 as the remaining possibilities.

Consulting the vial patient distribution:

Patient	Vials
13	A,C,F,H
15	A,C,E,H

To differentiate between patients 13 and 15, the next test should be Vial E. This is because:

• Vial E contains patient 15 (from the remaining candidates) and has not been ruled out by other negative tests. If Vial E tests negative, patient 13 is confirmed to have the disease.

Therefore, the subsequent vial to test is: Vial E.

Answer 3

To identify an impossible test result combination, we must examine the distribution of patient blood samples across vials. A patient diagnosed with the disease will yield positive results in all vials containing their blood. Conversely, a negative test result from a vial indicates that none of the patients whose blood is in that vial have the disease.

From the table:

Patient	Vials	Patient	Vials
1	B,D,F,H	9	A,D,F,H
2	B,D,F,G	10	A,D,F,G
3	B,D,E,H	11	A,D,E,H
4	B,D,E,G	12	A,D,E,G
5	B,C,F,H	13	A,C,F,H
6	B,C,F,G	14	A,C,F,G
7	B,C,E,H	15	A,C,E,H
8	B,C,E,G	16	A,C,E,G

Evaluate each scenario:

• Option 1: Vials A and G positive; vials D and E negative.
  Patients associated with vial A: 9, 10, 11, 12, 13, 14, 15, 16. Patients associated with vial G: 2, 4, 6, 8, 10, 12, 14, 16. For both A and G to be positive, patients such as 10, 12, 14, or 16 could be diseased. If vials D and E are negative, this scenario is feasible.
• Option 2: Vials B and D positive; vials F and H negative.
  Patients associated with vial B: 1, 2, 3, 4, 5, 6, 7, 8. Patients associated with vial D: 1, 2, 3, 4, 9, 10, 11, 12. Patient 2, for example, could be the diseased patient due to the overlap. Negative results for F and H do not conflict, as positive B and D can coexist with these negative results.
• Option 3: Vial B positive; vials C, F, and H negative.
  Patients associated with vial B: 1, 2, 3, 4, 5, 6, 7, 8. With only B testing positive and C, F, and H testing negative, there is no conflict. A diseased patient could be among those contributing to B's positive result without their blood being in vials C, F, or H.
• Option 4: Vials A and E positive; vials C and D negative.
  Patients causing vial A to be positive: 9, 10, 11, 12, 13, 14, 15, 16. Patients causing vial E to be positive: 3, 4, 7, 8, 11, 12, 15, 16. If C is negative, patients 5, 6, 7, 8, 13, 14, 15, 16 are excluded. If D is negative, patients 1, 2, 3, 4, 9, 10, 11, 12 are excluded. No remaining patient can satisfy the conditions of A and E being positive while simultaneously ensuring C and D are negative.

Consequently, the combination "Vials A and E positive, vials C and D negative" is not possible due to a contradiction in assigning patients to vials based on the test results.

Answer 4

To address this issue, the distribution of blood samples and the testing methodology must be analyzed. With 16 patient samples distributed across 8 vials, we need to determine the number of vials that will yield a positive result if one of these samples belongs to an infected patient.

The crucial factor is that a single infected patient's sample will cause all vials containing it to test positive. Each patient's sample is present in 4 distinct vials, as indicated in the table. Given the mixed nature of the vials, we will examine the combined test outcomes for two potential scenarios where one of two patients is infected.

Patient	Vials	Patient	Vials
1	B,D,F,H	9	A,D,F,H
2	B,D,F,G	10	A,D,F,G
3	B,D,E,H	11	A,D,E,H
4	B,D,E,G	12	A,D,E,G
5	B,C,F,H	13	A,C,F,H
6	B,C,F,G	14	A,C,F,G
7	B,CE,H	15	A,C,E,H
8	B,C,E,G	16	A,C,E,G

If a single patient is infected, exactly 4 vials will test positive. In cases of sample mix-ups where either of two patients could be infected, all tested vials may still turn positive. Consequently, the number of positive results can fluctuate.

Consider this: if one patient's samples are distributed across 4 vials and these vials overlap with another patient's distribution, the number of positive vials can increase to 5 or more when the affected samples have further overlaps. Through systematic evaluation:

• If the two patients involved have entirely separate vial sets, there will be 4 positive vials.
• If the two patients share some common vials, the number of potential positive tests increases to include these overlapping vials, resulting in outcomes of 5 to 8 positive vials in specific configurations.

Therefore, the set of possible positive test outcome counts includes all integers from 4 to 8, corresponding to the set: {4,5,6,7,8}.