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CAT 2018 Slot 1 DILR · Set 6 of 8Logical ReasoningModerate DifficultyQ21–Q24 · 2 MCQ + 2 TITA

CAT 2018 Slot 1 DILR Set 6: Arrangements (4 Questions) with Solutions

The Arrangements set from the official CAT 2018 Slot 1 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 2018 Slot 1 DILR · Questions 21–24
Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:
1
The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2
The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3
60% of the politicians are in the administration committee, and 20% are in the teaching committee.

In the exam this set competed with 7 others for 60 minutes. Budget about 8 minutes if you attempt it.

Q21. Arrangements

MCQ · +3 / −1
Based on the given information, which of the following statements MUST be FALSE?
  1. A) In the teaching committee the number of educationalists is equal to the number of politicians
  2. B) In the administration committee the number of bureaucrats is equal to the number of educationalists
  3. C) The size of the research committee is less than the size of the teaching committee
  4. D) The size of the research committee is less than the size of the administration committee
Correct Answer
C) The size of the research committee is less than the size of the teaching committee
Step-by-step solution
📌 Core Concept
Translate the three facts into equations in the unknowns, find every size split that fits, and see which statement fails in all of them.
🔢 Step-by-Step Solution
1
Fact 1: let the bureaucrats be bb in research and in teaching. Then administration has 4b3\frac{4b}{3}, so b=3kb=3k and the three committees have 3k,3k,4k3k, 3k, 4k. Total bureaucrats =10k=10k.
2
Fact 3: let the total number of politicians be 5m5m. Administration has 3m3m, teaching has mm and research has mm.
3
Fact 2: let research have ee educationalists. The other two average ee, so total educationalists =3e=3e, with teaching <e<<e< administration.
4
The total is 24, so 10k+3e+5m=2410k+3e+5m=24. With k≥1k\ge1 and m≥1m\ge1, k=2k=2 would leave 4=3e+5m4=3e+5m, which is impossible. So k=1k=1, and 3e+5m=143e+5m=14 gives m=1m=1, e=3e=3.
5
Bureaucrats are 3, 3, 4 and politicians are 1, 1, 3 (research, teaching, administration). Educationalists in research are 3, and teaching and administration have either (1,5)(1,5) or (2,4)(2,4).
6
Sizes: research =3+3+1=7=3+3+1=7. Teaching is 5 or 6. Administration is 12 or 11.
7
Research (7) is always larger than teaching (5 or 6). So the statement that research is smaller than teaching can never be true.
Why Not the Others
A: Possible. If teaching has 1 educationalist, it has 1 politician too.
B: Possible. Administration has 4 bureaucrats, and 4 educationalists when the split is (2,4)(2,4).
D: Always true, since 7<117<11 or 1212.
Correct Answer: C

Q22. Arrangements

TITA · +3 / 0
What is the number of bureaucrats in the administration committee?
Correct Answer
4
Step-by-step solution
First, the problem states there are three committees: research, teaching, and administration. Each has bureaucrats, educationalists, and politicians, with at least one from each type. The sizes of the committees are different, and they don't share any members.
Let me list the key facts:
1
Committee Sizes: All three committees have different sizes, and they add up to 24 people. So, if I denote the sizes as R, T, and A for research, teaching, and administration respectively, then R + T + A = 24.
2
Bureaucrats: The number of bureaucrats in research (B_R) equals that in teaching (B_T). Also, B_R is 75% of the bureaucrats in administration (B_A). So, B_R = B_T and B_R = 0.75 * B_A.
3
Educationalists: The number in teaching (E_T) is less than in research (E_R). Also, E_R is the average of E_T and E_A (administration). So, E_R = (E_T + E_A)/2.
4
Politicians: 60% are in administration, and 20% are in teaching. So, P_A = 0.6 P_total and P_T = 0.2 P_total. The remaining 20% must be in research, so P_R = 0.2 * P_total.
Since each committee has at least one of each type, the minimum size for each is 3. But since the total is 24 and they are different sizes, let's think about possible sizes.
Let the sizes as R, T, A. They must be distinct and sum to 24. Let's assume R < T < A for simplicity.
Possible triplets (R, T, A) that add to 24 and are distinct:
4, 6, 14
5, 6, 13
5, 7, 12
6, 7, 11
6, 8, 10
7, 8, 9
Now, considering the bureaucrats: B_R = B_T and B_R = 0.75 * B_A. So, B_A must be a multiple of 4 since 0.75 is 3/4. Let's say B_A = 4k, then B_R = B_T = 3k.
Each committee must have at least one bureaucrat, so k must be at least 1. Let's try k=1: B_A=4, B_R=B_T=3.
Now, the total bureaucrats would be 3 + 3 + 4 = 10. So, total people accounted for so far:
10
The remaining 14 must be educationalists and politicians.
Looking at politicians: P_A = 0.6P_total, P_T = 0.2P_total, P_R = 0.2P_total. So, P_total = P_A + P_T + P_R = 0.6P + 0.2P + 0.2P = P_total, which checks out.
Let P_total as P. Then P_A = 0.6P, P_T = 0.2P, P_R = 0.2P.
Each committee must have at least one politician, so P_A ≥1, P_T ≥1, P_R ≥1. Thus, P must be at least 5 because 0.2P must be ≥1, so P ≥5.
Now, total people in each committee:
For research (R): B_R + E_R + P_R = R Similarly for T and A.
Let me try the size triplet (6, 8, 10). So R=6, T=8, A=10.
Check if this works with the bureaucrats:
B_R = 3, B_T=3, B_A=4. So total bureaucrats=10.
Now, total people accounted for:
10
Remaining: 14, which must be E and P.
Let E_R, E_T, E_A as the educationalists in each committee.
From fact 3: E_T < E_R and E_R = (E_T + E_A)/2.
Let me express E_A = 2E_R - E_T.
Since all E must be at least 1, let's assign E_R ≥1, E_T ≥1, E_A ≥1.
Also, total educationalists E_total = E_R + E_T + E_A.
Similarly, total politicians P_total = P_R + P_T + P_A.
We know that total people in each committee:
R: 3 + E_R + P_R = 6 → E_R + P_R = 3 T: 3 + E_T + P_T = 8 → E_T + P_T = 5 A: 4 + E_A + P_A = 10 → E_A + P_A = 6
Also, P_A = 0.6P, P_T = 0.2P, P_R = 0.2P.
So, P_A = 0.6P, P_T = 0.2P, P_R = 0.2P.
Adding these: 0.6P + 0.2P + 0.2P = P.
Now, from the committees:
P_R = 0.2P, so from R: E_R + 0.2P = 3 → E_R = 3 - 0.2P
Similarly, from T: E_T + 0.2P = 5 → E_T = 5 - 0.2P
From A: E_A + 0.6P = 6 → E_A = 6 - 0.6P
Now, E_total = E_R + E_T + E_A = (3 - 0.2P) + (5 - 0.2P) + (6 - 0.6P) = 14 - P
But E_total must be equal to the total number of educationalists, which is 24 - (total bureaucrats + total politicians). Total bureaucrats are 10, so total politicians = P. Thus, E_total = 24 - 10 - P = 14 - P. So that matches.
Now, from the educationalists' conditions:
E_T < E_R and E_R = (E_T + E_A)/2.
Substitute E_R, E_T, E_A from above:
E_T = 5 - 0.2P E_R = 3 - 0.2P E_A = 6 - 0.6P
So, E_T < E_R → 5 - 0.

Q23. Arrangements

TITA · +3 / 0
What is the number of educationalists in the research committee?
Correct Answer
3
Step-by-step solution
📌 Core Concept
Set up equations for the bureaucrats, educationalists and politicians in each committee. The total of 24 then forces a single value.
🔢 Step-by-Step Solution
1
Fact 1: bureaucrats are 3k,3k,4k3k, 3k, 4k in research, teaching and administration, since research is 75% of administration. Total =10k=10k.
2
Fact 3: politicians total 5m5m, split mm (research), mm (teaching) and 3m3m (administration), from 20%, 20% and 60%.
3
Fact 2: if research has ee educationalists, the other two committees average ee. Total educationalists =3e=3e.
4
The total gives 10k+3e+5m=2410k+3e+5m=24. Every committee has at least one of each type, so k≥1k\ge1 and m≥1m\ge1.
5
If k=2k=2, then 3e+5m=43e+5m=4, which is impossible. So k=1k=1, and 3e+5m=143e+5m=14.
6
m=1m=1 gives 3e=93e=9, so e=3e=3. Larger mm makes the left side exceed 14.
7
Research therefore has 3 educationalists. This fits teaching <3<<3< administration, with (1,5)(1,5) or (2,4)(2,4).
Quick Check
Research has 3+3+1=73+3+1=7 people, teaching 5 or 6 and administration 12 or 11. The total is 24 either way.
Correct Answer: 3

Q24. Arrangements

MCQ · +3 / −1
Which of the following CANNOT be determined uniquely based on the given information?
  1. A) The size of the teaching committee
  2. B) The size of the research committee
  3. C) The total number of bureaucrats in the three committees
  4. D) The total number of educationalists in the three committees
Correct Answer
A) The size of the teaching committee
Step-by-step solution
📌 Core Concept
Solving the constraints leaves one open choice. Check which quantities change with it.
🔢 Step-by-Step Solution
1
Fact 1 gives bureaucrats 3k,3k,4k3k, 3k, 4k. Fact 3 gives politicians m,m,3mm, m, 3m. Fact 2 gives total educationalists 3e3e.
2
The total 10k+3e+5m=2410k+3e+5m=24 forces k=1k=1, m=1m=1, e=3e=3.
3
Bureaucrats are 3, 3, 4, so the total of 10 is fixed. Educationalists total 3×3=93\times3=9, also fixed.
4
Research size is 3+3+1=73+3+1=7, which is fixed.
5
Teaching and administration educationalists can be (1,5)(1,5) or (2,4)(2,4). Both satisfy teaching <3<<3< administration and the average condition.
6
Teaching size is then 3+1+1=53+1+1=5 or 3+2+1=63+2+1=6. Both are valid, since the sizes 7, 5, 12 and 7, 6, 11 are all distinct. Nothing in the facts chooses between them, so the teaching size cannot be determined uniquely.
Why Not the Others
B: Research is always 7.
C: Bureaucrats total 10 in every case.
D: Educationalists total 9 in every case.
Correct Answer: A
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CAT 2018 Slot 1 DILR at a Glance

Sets in this section8
Questions in this section32
Section time60 minutes
Convenor IIMIIM Calcutta

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