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

Given the information, our task is to deduce which patient has the disease based on the test results of the vials: vial C tests positive, and vials A, E, and H test negative. According to the data table, the strategy involves eliminating patients whose blood samples are in the negative-testing vials.

First, let's analyze the information for each vial:

• If a vial tests negative, none of the patients whose samples are in that vial can have the disease.
• Vials A, E, and H have tested negative, so we can eliminate all patients involved in these vials.

From the table, we have:

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

Negative result elimination:

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

Remaining candidate (after eliminating those who were in negative-tested vials and not in Vial C):

• Patient 6, who is in vials B, C, F, and G.

Since Vial C tested positive and Patient 6's blood is in Vial C, and all other patients in Vial C were eliminated either by the negative results of other vials, Patient 6 must have the disease.

Therefore, the correct answer is:

Patient 6

Answer 2

Vial Composition:

• Vial A: \(9, 10, 11, 12, 13, 14, 15, 16\)
• Vial B: \(1, 2, 3, 4, 5, 6, 7, 8\)
• Vial C: \(5, 6, 7, 8, 13, 14, 15, 16\)
• Vial D: \(1, 2, 3, 4, 9, 10, 11, 12\)
• Vial E: \(3, 4, 7, 8, 11, 12, 15, 16\)
• Vial F: \(1, 2, 5, 6, 9, 10, 13, 14\)
• Vial G: \(2, 4, 6, 8, 10, 12, 14, 16\)
• Vial H: \(1, 3, 5, 7, 9, 11, 13, 15\)

Test Result Summary:

• Vial C: Positive
• Vials A, E, and H: Negative

Deduction Logic:

Vial C contains: \[ \{5, 6, 7, 8, 13, 14, 15, 16\} \] Since Vial C is positive, **at least one of these patients is infected**.

Now check which of these patients are also in Vials A, E, or H:

• Vial A (negative): rules out 9, 10, 11, 12, 13, 14, 15, 16
• Vial E (negative): rules out 3, 4, 7, 8, 11, 12, 15, 16
• Vial H (negative): rules out 1, 3, 5, 7, 9, 11, 13, 15

That eliminates: \[ \boxed{5, 7, 8, 13, 14, 15, 16} \] from suspicion because all are present in one or more negative vials.

The **only remaining patient in Vial C not appearing in any of the negative vials is**: \[ \boxed{6} \]

Final Answer:

The diseased patient is: \[ \boxed{\text{Option (C): Patient 6}} \]

Answer 3

To solve this problem, we need to determine which vials should be tested next to identify the patient with the disease. We know the following based on the test results and patient distribution:

• Vial A tests positive: This means one of the patients in vial A (9, 10, 11, 12, 13, 14, 15, 16) could have the disease.
• Vials D and G test negative: Thus, none of the patients in vials D (1, 2, 3, 4, 9, 10, 11, 12) and G (2, 4, 6, 8, 10, 12, 14, 16) could have the disease.

By eliminating patients in vials D and G:

• From vial D: Remove patients 1, 2, 3, 4, 9, 10, 11, and 12.
• From vial G: Remove patients 2, 4, 6, 8, 10, 12, 14, and 16.

After removing these patients from consideration, we remain with patients 13, 14, 15, and 16 from the positive Vial A. However, patient 14 and 16 should also be removed since they appear in vial G, which tested negative. Thus, the possible patients are reduced to 13 and 15.

Reviewing the vial patient distribution:

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

In this scenario, to distinguish between patients 13 and 15, we should test Vial E. This is because:

• Vial E only contains patient 15 (from the remaining candidates) who hasn't been entirely ruled out by other negative vials. If Vial E tests negative, we will know patient 13 has the disease.

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

Answer 4

Vial Distribution:

• Vial A: \(9, 10, 11, 12, 13, 14, 15, 16\)
• Vial B: \(1, 2, 3, 4, 5, 6, 7, 8\)
• Vial C: \(5, 6, 7, 8, 13, 14, 15, 16\)
• Vial D: \(1, 2, 3, 4, 9, 10, 11, 12\)
• Vial E: \(3, 4, 7, 8, 11, 12, 15, 16\)
• Vial F: \(1, 2, 5, 6, 9, 10, 13, 14\)
• Vial G: \(2, 4, 6, 8, 10, 12, 14, 16\)
• Vial H: \(1, 3, 5, 7, 9, 11, 13, 15\)

Given:

• Vial A tests positive
• Vials D and G test negative

Analysis:

From the **negative vials D and G**, we eliminate the following patients: 
From Vial D (negative): \(1, 2, 3, 4, 9, 10, 11, 12\) → Not diseased 
From Vial G (negative): \(2, 4, 6, 8, 10, 12, 14, 16\) → Also not diseased

Combining both, the only remaining candidates for infection are: \[ \boxed{13 \text{ and } 15} \] because they appear in Vial A (positive) but not in D or G.

Now check which vial can distinguish between Patient 13 and 15:

• Vial B: Contains neither 13 nor 15 → Not helpful
• Vial C: Contains both 13 and 15 → Can't distinguish
• Vial H: Contains both 13 and 15 → Also not helpful
• Vial E: Contains only 15 (not 13)

Therefore:

• If Vial E tests positive → Patient 15 is diseased
• If Vial E tests negative → Patient 13 is diseased

Final Answer:

The correct vial that will help determine the diseased patient is: \[ \boxed{\text{Option (A): Vial E}} \]

Answer 5

To determine which combination of test results is not possible, we need to analyze the patient blood sample distribution in the vials. If a patient has the disease, all vials containing their blood will test positive, and if a vial tests negative, then none of the patients whose blood samples are 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

Check each option:

• Option 1: Vials A and G positive, vials D and E negative
  Patients that affect vial A are 9, 10, 11, 12, 13, 14, 15, 16. Patients that affect vial G are 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 have the disease. However, with vials D and E negative, this combination is possible.
• Option 2: Vials B and D positive, vials F and H negative
  Patients that affect vial B are 1, 2, 3, 4, 5, 6, 7, 8. Patients that affect vial D are 1, 2, 3, 4, 9, 10, 11, 12. There is an overlap such as patient 2 who could be the diseased patient. No conflict arises with F and H negative, as the positive B and D can exist with these negatives.
• Option 3: Vial B positive, vials C, F, and H negative
  Patients for vial B are 1, 2, 3, 4, 5, 6, 7, 8. By having only B positive while C, F, and H are negative, there is no conflict because a diseased patient could still exist from B but not be shared with C, F, and H.
• Option 4: Vials A and E positive, vials C and D negative
  Patients that cause vial A to be positive: 9, 10, 11, 12, 13, 14, 15, 16. For positive E: 3, 4, 7, 8, 11, 12, 15, 16. However, if C is negative: 5, 6, 7, 8, 13, 14, 15, 16 are negative; and if D is negative: 1, 2, 3, 4, 9, 10, 11, 12 are negative. There's no patient left that can satisfy A and E being positive without conflicting with C and D being negative.

Therefore, the combination "Vials A and E positive, vials C and D negative" is not possible as it leads to a contradiction in patient allocation based on vial positivity.

Answer 6

Vial Composition:

• Vial A: \(9, 10, 11, 12, 13, 14, 15, 16\)
• Vial B: \(1, 2, 3, 4, 5, 6, 7, 8\)
• Vial C: \(5, 6, 7, 8, 13, 14, 15, 16\)
• Vial D: \(1, 2, 3, 4, 9, 10, 11, 12\)
• Vial E: \(3, 4, 7, 8, 11, 12, 15, 16\)
• Vial F: \(1, 2, 5, 6, 9, 10, 13, 14\)
• Vial G: \(2, 4, 6, 8, 10, 12, 14, 16\)
• Vial H: \(1, 3, 5, 7, 9, 11, 13, 15\)

Given Condition:

If vials C and D are negative, then the patients contained in them can be confirmed as disease-free.

From the vial compositions:

• Vial C (negative): Rules out patients \(5,6,7,8,13,14,15,16\)
• Vial D (negative): Rules out patients \(1,2,3,4,9,10,11,12\)

Together, vials C and D cover **all 16 patients**. Hence: \[ \text{If both C and D are negative} \Rightarrow \text{no patient has the disease} \] Which contradicts the assumption that at least one patient is infected.

Therefore, **both vials C and D cannot be negative together unless no one is infected**, which is not the case.

Valid Test Result Pattern:

• Vial A: positive
• Vial E: positive
• Vial C: negative
• Vial D: negative

This combination is consistent with at least one patient having the disease and is logically possible.

Final Answer:

The correct option is (D): \[ \boxed{\text{Vials A and E positive, vials C and D negative}} \]

Answer 7

To solve this problem, we must consider the distribution of blood samples and the testing method employed. Given that the blood samples from 16 patients are distributed into 8 vials, we need to understand how many vials will test positive if one of the samples mixed in the vials belongs to the patient with the disease.

The key point is that one patient has the disease, and their sample will cause all vials containing it to test positive. Each patient contributes to 4 different vials as per the table. Since the vials are mixed, we'll consider the merging of the test scenarios of two possible patients having the disease.

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 one patient has the disease, exactly 4 vials will test positive. If a mix-up occurs, and either of two possible patients has the disease, all tested vials may still return positive. Therefore, the number of positive results can vary.

Consider the scenario: if one patient contributes to 4 vials and has distributions such that there is an overlap with another patient's set, it increases to 5 or more positive vials when the affected samples overlap further. Through systematic evaluation:

• If both patients involved have unique vial sets per their sharing, there will be 4 positive vials.
• If the two patients share some common vials, the number of possible positive tests expands to other overlapped vials, yielding numbers 5 to 8 in particular configurations.

Therefore, the possible sets of positive test outcomes include all numbers from 4 to 8, matching the option: {4,5,6,7,8}.

Answer 8

Consider a scenario where patients' blood samples are pooled and distributed across 8 vials, each containing a subset of patients. A vial tests positive if at least one diseased patient's blood is included in it.

Case Analysis:

• Case 1: Patient 1 or Patient 16 has the disease
  Since their blood appears in all 8 vials, if either is infected: \[ \Rightarrow \text{All 8 vials will test positive} \] So, 8 must be included.
• Case 2: Patient 2 or Patient 16 has the disease
  These patients are in 7 out of 8 vials. If one of them is infected: \[ \Rightarrow 7 vials will test positive \] So, 7 must also be included.
• Case 3: Patient 1 is diseased and mixed with Patient 7
  The combination covers only 6 vials: \[ \Rightarrow 6 vials will test positive \]
• Case 4: Patient 1 is diseased and mixed with Patient 9
  Their combination appears in only 5 vials: \[ \Rightarrow 5 vials will test positive \]
• Case 5: Patient 1 is diseased and tested individually (not mixed)
  His sample appears in only 4 vials: \[ \Rightarrow 4 vials will test positive \]

Conclusion:

Depending on how the infected blood is distributed among the vials, the number of positive tests can be: \[ \boxed{4, 5, 6, 7, \text{or } 8} \]

Final Answer:

The correct option is (C): \[ \boxed{\{4, 5, 6, 7, 8\}} \]