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CAT 2020 Slot 3 DILR · Set 1 of 5Logical ReasoningEasy DifficultyQ1–Q4 · 4 MCQ

CAT 2020 Slot 3 DILR Set 1: Grids (4 Questions) with Solutions

The Grids set from the official CAT 2020 Slot 3 DILR paper: the caselet, all 4 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2020 Slot 3 DILR · Questions 1–4
Sixteen patients in a hospital must undergo a blood test for a disease. It is known that exactly one of them has the disease. The hospital has only eight testing kits and has decided to pool blood samples of patients into eight vials for the tests. The patients are numbered 11 through 1616, and the vials are labelled A, B, C, D, E, F, G, and H. The following table shows the vials into which each patient's blood sample is distributed.
PatientVialsPatientVials
1B, D, F, H9A, D, F, H
2B, D, F, G10A, D, F, G
3B, D, E, H11A, D, E, H
4B, D, E, G12A, D, E, G
5B, C, F, H13A, C, F, H
6B, C, F, G14A, C, F, G
7B, C, E, H15A, C, E, H
8B, C, E, G16A, C, E, G
If a patient has the disease, then each vial containing his/her blood sample will test positive. If a vial tests positive, one of the patients whose blood samples were mixed in the vial has the disease. If a vial tests negative, then none of the patients whose blood samples were mixed in the vial has the disease.

In the exam this set competed with 4 others for 40 minutes. Budget about 8 minutes if you attempt it.

Q1. Grids

MCQ · +3 / −1
Suppose vial C tests positive and vials A, E and H test negative. Which patient has the disease?
  1. A) Patient 2
  2. B) Patient 6
  3. C) Patient 14
  4. D) Patient 8
Correct Answer
B) Patient 6
Step-by-step solution
Clue Breakdown:
1
Vial C is positive.
2
Vials A, E, and H are negative.
Step-by-Step Deduction Grid:
1
Eliminate patients in vial A (negative):
Patients 9, 10, 11, 12, 13, 14, 15, 16 are eliminated.
2
Eliminate patients in vial E (negative):
Remaining patients in E: 3, 4, 7, 8.
These patients are eliminated.
3
Eliminate patients in vial H (negative):
Remaining patients in H: 1, 5, 7.
These patients are eliminated.
4
Remaining patients: 2 and 6.
5
Vial C is positive:
Patients in C: 5, 6, 7, 8, 13, 14, 15, 16.
After elimination, only patient 6 remains.
Question Resolution: Since vial C is positive and only patient 6 is left in vial C, patient 6 has the disease.
Final Answer Confirmation: Correct Answer: B

Q2. Grids

MCQ · +3 / −1
Suppose vial A tests positive and vials D and G test negative. Which of the following vials should we test next to identify the patient with the disease?
  1. A) Vial C
  2. B) Vial H
  3. C) Vial B
  4. D) Vial E
Correct Answer
D) Vial E
Step-by-step solution
📌 Core Concept
The vials come in complementary pairs (A/B, C/D, E/F, G/H). Every patient is in exactly one vial of each pair, so each test result splits the 16 patients into halves.
🔢 Step-by-Step Solution
1
Vial A holds patients 9 to 16 and vial B holds patients 1 to 8. A positive A therefore limits the disease to patients 9 to 16.
2
Vial D holds patients 1 to 4 and 9 to 12. A negative D removes 9 to 12, leaving 13, 14, 15, 16.
3
Vial G holds the even-numbered patients. A negative G removes 14 and 16, leaving patients 13 and 15.
4
Patient 13 is in A, C, F, H and patient 15 is in A, C, E, H. They share C, H and A, and differ only in F versus E.
5
A test that separates them must be E or F. E is the option given: it is positive only if patient 15 is infected and negative if patient 13 is.
Why Not the Others
Vial C: both patients 13 and 15 are in C, so it is positive either way.
Vial H: both are in H, so it gives no information.
Vial B: neither is in B, since B is already known to be negative. It cannot separate them.
Correct Answer: D

Q3. Grids

MCQ · +3 / −1
Which of the following combinations of test results is NOT possible?
  1. A) Vial B positive, vials C, F and H negative
  2. B) Vials B and D positive, vials F and H negative
  3. C) Vials A and E positive, vials C and D negative
  4. D) Vials A and G positive, vials D and E negative
Correct Answer
C) Vials A and E positive, vials C and D negative
Step-by-step solution
📌 Core Concept
Each patient is in exactly one vial of each pair A/B, C/D, E/F, G/H. With one infected patient, exactly one vial in every pair is positive and the other is negative.
🔢 Step-by-Step Solution
1
Check each option against the rule that one vial of each pair is positive and the other negative.
2
Option A: B positive means A is negative. C, F, H negative means D, E, G are positive. The pattern B, D, E, G is patient 4. Possible.
3
Option B: B and D positive, F and H negative. Then A and C are negative and E and G are positive. The pattern B, D, E, G is again patient 4. Possible.
4
Option C: C and D are both negative. They are a complementary pair, so at least one of them must be positive for any patient. This cannot happen.
5
Option D: A and G positive, D and E negative. Then C and F are positive, which gives A, C, F, G, patient 14. Possible.
Quick Check
Only option C breaks the pairing rule, because it has both vials of one pair negative.
Correct Answer: C

Q4. Grids

MCQ · +3 / −1
Suppose one of the lab assistants accidentally mixed two patients' blood samples before they were distributed to the vials. Which of the following correctly represents the set of all possible numbers of positive test results out of the eight vials?
  1. A) {5,6,7,8}
  2. B) {4,5}
  3. C) {4,5,6,7,8}
  4. D) {4,5,6,7}
Correct Answer
C) {4,5,6,7,8}
Step-by-step solution
📌 Core Concept
A mixed sample goes into every vial that either of the two patients would have used. The positive vials are therefore the union of the two patients' vial sets, or just one patient's four vials if the infected patient was not mixed.
🔢 Step-by-Step Solution
1
Each patient is in exactly one vial of each of the four pairs, so each uses 4 vials.
2
If the infected patient is not one of the two mixed patients, only that patient's 4 vials are positive. This gives 4.
3
If the infected patient is one of the mixed pair, the positive vials are the union of the two patients' sets. Two patients differ in dd of the four pairs, where dd is 1, 2, 3 or 4, and in each differing pair both vials become positive.
4
The union therefore has 4+d4 + d vials. For example, patients 1 and 2 differ only in the E/F pair... (B, D, F, H versus B, D, F, G differ in the G/H pair), giving 5 vials. Patients 1 and 16 differ in all four pairs, giving 8.
5
So the number of positive vials can be 4, 5, 6, 7 or 8.
Why Not the Others
{5,6,7,8} leaves out 4, which occurs when the infected patient is not in the mixed pair.
{4,5} and {4,5,6,7} leave out larger unions that occur when the two patients differ in more pairs.
Correct Answer: C
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CAT 2020 Slot 3 DILR at a Glance

Sets in this section5
Questions in this section24
Section time40 minutes
Convenor IIMIIM Indore

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