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CAT 2018 Slot 1 DILR · Set 8 of 8Logical ReasoningHard DifficultyQ29–Q32 · 2 MCQ + 2 TITA

CAT 2018 Slot 1 DILR Set 8: 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 29–32
A company administers a written test comprising of three sections of 20 marks each - Data Interpretation (DI), Written English (WE) and General Awareness (GA), for recruitment. A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections. Candidates who score less than 70% marks in two or more sections are disqualified. From among the rest, the four with the highest composite scores are recruited. If four or less candidates qualify, all who qualify are recruited.Ten candidates appeared for the written test. Their marks in the test are given in the table below. Some marks in the table are missing, but the following facts are known:
1
No two candidates had the same composite score.
2
Ajay was the unique highest scorer in WE.
3
Among the four recruited, Geeta had the lowest composite score.
4
Indu was recruited.
5
Danish, Harini, and Indu had scored the same marks the in GA.
6
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
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In the exam this set competed with 7 others for 60 minutes. Budget about 8 minutes if you attempt it.

Q29. Arrangements

MCQ · +3 / −1
Which of the following statements MUST be true?
1
Jatin's composite score was more than that of Danish.
2
Indu scored less than Chetna in DI.
3
Jatin scored more than Indu in GA.
  1. A) Both 2 and 3
  2. B) Only 1
  3. C) Only 2
  4. D) Both 1 and 2
Correct Answer
D) Both 1 and 2
Step-by-step solution
📌 Core Concept
The 70% rule means a score below 14 out of 20 counts as a failed section. Combining it with the composite-score facts fixes the missing marks of Jatin and Indu.
🔢 Step-by-Step Solution
1
Jatin has 16 in WE and 14 in GA, and neither is 100%. Fact 6 says he scored 100% in exactly one section, so his DI is 20. His composite is 2(20)+16+14=702(20)+16+14=70.
2
Indu's composite is 70−10=6070-10=60. Her WE is 8, which is below 14. She was recruited (fact 4), so she cannot fail another section, and her DI and GA must both be at least 14.
3
Indu scored 100% in exactly one section, so either DI or GA is 20. If DI were 20, then 40+8+GA=6040+8+\text{GA}=60 gives GA = 12, which is below 14 and not allowed. So GA = 20 and 2 DI+8+20=602\,\text{DI}+8+20=60 gives DI = 16.
4
Danish has the same GA as Indu (fact 5), so his GA is 20. His composite is 16+15+20=5116+15+20=51.
5.
Statement 1
Jatin's 70 is greater than Danish's 51. This must be true.
6.
Statement 2
Indu's DI is 16 and Chetna's is 19. This must be true.
7.
Statement 3
Jatin's GA is 14 and Indu's is 20. This is false.
Statements 1 and 2 must be true.
Why Not the Others
A and C include statement 3, which is false.
B leaves out statement 2, which is also true.
Correct Answer: D

Q30. Arrangements

MCQ · +3 / −1
Which of the following statements MUST be FALSE?
  1. A) Bala scored same as Jatin in DI
  2. B) Harini’s composite score was less than that of Falak
  3. C) Bala’s composite score was less than that of Ester
  4. D) Chetna scored more than Bala in DI
Correct Answer
A) Bala scored same as Jatin in DI
Step-by-step solution
📌 Core Concept
Composite scores must all be different (fact 1). A statement must be false if it forces two candidates onto the same composite.
🔢 Step-by-Step Solution
1
Fact 6 gives Jatin's DI as 20, because he has 16 in WE and 14 in GA and scored 100% in exactly one section. His composite is 70 and Indu's is 60. Indu's marks are DI 16, WE 8 and GA 20.
2
Option A: if Bala's DI were 20, his composite would be 40+9+11=6040+9+11=60. That equals Indu's 60, which fact 1 forbids. So this must be false.
3
Option B: Falak's composite is 30+7+10=4730+7+10=47. Harini has DI 5 and GA 20, so her composite is 30+WE30+\text{WE}. A WE below 17 gives a score below 47, so this can be true.
4
Option C: Ester's composite is 24+18+16=5824+18+16=58. Bala's is 2 DI+202\,\text{DI}+20, which is below 58 for any DI up to 18. This can be true.
5
Option D: Chetna's DI is 19. Bala's DI can be any value below 19 that avoids a tie, so this can be true.
Why Not the Others
B is possible when Harini's WE is below 17.
C is possible for Bala's DI of 18 or less.
D is possible for Bala's DI below 19.
Correct Answer: A

Q31. Arrangements

TITA · +3 / 0
If all the candidates except Ajay and Danish had different marks in DI, and Bala's composite score was less than Chetna's composite score, then what is the maximum marks that Bala could have scored in DI?
Correct Answer
13
Step-by-step solution
📌 Core Concept
Bala's composite must stay below Chetna's. The distinct-DI condition then rules out the DI values already used by other candidates.
🔢 Step-by-Step Solution
1
Chetna's composite is 2(19)+4+12=542(19)+4+12=54.
2
Bala's composite is 2 DI+9+11=2 DI+202\,\text{DI}+9+11=2\,\text{DI}+20. The condition 2 DI+20<542\,\text{DI}+20<54 gives DI<17\text{DI}<17, so DI≤16\text{DI}\le 16.
3
Only Ajay and Danish may share a DI value. Everyone else has a different DI, so Bala's DI cannot equal that of any other candidate.
4
The DI values already taken are 20 (Jatin, from fact 6), 19 (Chetna), 16 (Indu, derived from facts 4, 5 and 6), 15 (Falak), 14 (Geeta), 12 (Ester) and 5 (Harini).
5
So 16, 15 and 14 are all unavailable for Bala. The largest free value is 13.
6
Bala's composite is then 26+20=4626+20=46. This is below 54 and differs from the other composites, for example Falak's 47.
Quick Check
A DI of 13 gives 46 and stays below Chetna's 54. DI values of 14 to 16 are blocked, and 17 or more breaks the composite condition.
Correct Answer: 13

Q32. Arrangements

TITA · +3 / 0
If all the candidates scored different marks in WE then what is the maximum marks that Harini could have scored in WE?
Correct Answer
14
Step-by-step solution
📌 Core Concept
Harini's composite score must be unique, and each WE mark must be different. Together these rule out several high WE values for her.
🔢 Step-by-Step Solution
1
Jatin has DI 20 (fact 6) and a composite of 70. Indu has a composite of 60, with DI 16 and GA 20. Danish and Harini share Indu's GA of 20 (fact 5), so Harini's composite is 2(5)+WE+20=30+WE2(5)+\text{WE}+20=30+\text{WE}.
2
Ester has the highest known WE of 18 and Ajay is the unique WE topper (fact 2). So Ajay's WE is 19 or 20. His composite is 32+WE32+\text{WE}, and Danish's is 16+15+20=5116+15+20=51. Ajay must be 52 to avoid a tie, so his WE is 20.
3
Geeta must be the lowest of the four recruited (fact 3). Jatin (70), Indu (60) and Ester (58) are above her. Her composite 34+WE34+\text{WE} must therefore beat Ajay's 52, so her WE is 19.
4
WE values already used are 20, 19, 18, 16, 15, 9, 8, 7 and 4. Harini's WE must be below 20 and different from all of these, so 17 is the highest candidate.
5
WE of 17 gives Harini a composite of 47. Falak's composite is 30+7+10=4730+7+10=47, which is a tie that fact 1 forbids.
6
The next values, 16 and 15, are already used by Jatin and Danish. The next free value is 14, giving a composite of 44. This is unique.
Quick Check
WE of 14 gives 44, which matches no other composite. WE of 17 clashes with Falak at 47.
Correct Answer: 14
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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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