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CAT 2018 Slot 1 DILR · Set 4 of 8Logical ReasoningModerate DifficultyQ13–Q16 · 4 MCQ

CAT 2018 Slot 1 DILR Set 4: 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 13–16
Fuel contamination levels at each of 20 petrol pumps P1,P2,…,P20P1, P2, \ldots, P20 were recorded as either high, medium, or low.
1
Contamination levels at three pumps among P1−P5P1 - P5 were recorded as high.
2
P6P6 was the only pump among P1−P10P1 - P10 where the contamination level was recorded as low.
3
P7P7 and P8P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4
High contamination levels were not recorded at any of the pumps P16−P20P16 - P20.
5
The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

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

Q13. Arrangements

MCQ · +3 / −1
Which of the following MUST be true?
  1. A) The contamination level at P20 was recorded as medium.
  2. B) The contamination level at P13 was recorded as low.
  3. C) The contamination level at P12 was recorded as high.
  4. D) The contamination level at P10 was recorded as high.
Correct Answer
D) The contamination level at P10 was recorded as high.
Step-by-step solution
📌 Core Concept
Clue 3 means every adjacent pair differs except P7–P8, so each stretch is forced to alternate. Fix P1–P10 first, then use the count condition to test the two possible cases for P10.
🔢 Step-by-Step Solution
1
Clue 2 makes P6 the only low pump in P1–P10, so every other pump there is H or M. Clue 3 says only P7–P8 match, so P1–P5 must alternate between H and M.
2
Clue 1 needs exactly 3 H among P1–P5, so P1–P5 = H M H M H. P1, P3 and P5 are high.
3
P6 = L, so P7 ≠ L. Let P7 = P8 = xx, with xx being H or M. P9 differs from P8 and cannot be L, so P9 is the other level. P10 differs from P9, so P10 = xx.
4
Clue 4 allows no H in P16-P20, and these five alternate. So they are M L M L M (2 lows) or L M L M L (3 lows).
5
Let P11–P15 hold hh highs and ll lows. Clue 5 says the number of H equals 2×2 \times the number of L.
6
Case x=Mx = M: P1-P10 then has 4 H (P1, P3, P5, P9) and 1 L. With mm lows in P16-P20 (mm = 2 or 3), the clue 5 equation is 4+h=2(1+l+m)4 + h = 2(1 + l + m), so h=2l+2m−2h = 2l + 2m - 2.
7
Since h≤3h \le 3 (no two adjacent pumps match), l=0l = 0 is forced. Then m=2m = 2 and h=2h = 2.
8
With l=0l = 0, P11-P15 uses only H and M, so it alternates. P10 = M forces P11 = H, giving H M H M H with h=3h = 3. This contradicts h=2h = 2, so x=Mx = M is impossible.
9
Therefore x=Hx = H, and P10 = H in every valid arrangement.
10
The other options can fail. With P11–P15 = M H M H M and P16–P20 = L M L M L, all clues hold and P20 = L, which refutes A. With P11–P15 = M H M H L and P16–P20 = M L M L M, all clues hold and P13 = M, which refutes B. With P11–P15 = M L H M H and P16–P20 = M L M L M, all clues hold and P12 = L, which refutes C.
Why Not the Others
A: P20 can be L, as in the first arrangement in step 10.
B: P13 can be M, as in the second arrangement in step 10.
C: P12 can be L, as in the third arrangement in step 10.
Quick Check
In the first arrangement there are 8 H (P1, P3, P5, P7, P8, P10, P12, P14) and 4 L (P6, P16, P18, P20), so H = 2L.
Correct Answer: D

Q14. Arrangements

MCQ · +3 / −1
What best can be said about the number of pumps at which the contamination levels were recorded as medium?
  1. A) At least 8
  2. B) More than 4
  3. C) Exactly 8
  4. D) At most 9
Correct Answer
C) Exactly 8
Step-by-step solution
📌 Core Concept
Work out the fixed part (P1–P10), then use H = 2L and the total of 20 pumps to fix the number of mediums.
🔢 Step-by-Step Solution
1
From the clues, P1–P5 = H M H M H and P6 = L. P7–P10 must be H H M H. The alternative M M H M fails the count condition (it would give H = 4 and L = 1, and then P11–P15 cannot be completed).
2
So P1–P10 has 6 H, 1 L and 3 M.
3
P16–P20 has no H (clue 4) and no equal neighbours, so it is M L M L M (2 L) or L M L M L (3 L).
4
Let the total be L lows and H = 2L highs. P11–P15 can hold at most 2 H, because P11 ≠ H, and no two neighbours are equal.
5
If L = 3, then H = 6, so P11–P15 would have no H and 0 L, which is impossible. If L ≥ 5, then H ≥ 10, which needs at least 4 H in P11–P15, also impossible. So L = 4 and H = 8.
6
Mediums = 20 − 8 − 4 = 8. Two valid grids confirm this: M H M H M with L M L M L, and M H M H L with M L M L M.
Quick Check
8 H + 4 L + 8 M = 20, and H = 2 × L.
Correct Answer: C

Q15. Arrangements

MCQ · +3 / −1
If the contamination level at P11 was recorded as low, then which of the following MUST be true?
  1. A) The contamination level at P12 was recorded as high.
  2. B) The contamination level at P15 was recorded as medium.
  3. C) The contamination level at P18 was recorded as low.
  4. D) The contamination level at P14 was recorded as medium.
Correct Answer
D) The contamination level at P14 was recorded as medium.
Step-by-step solution
📌 Core Concept
Once P11 is fixed as low, the count condition forces the whole of P11–P20.
🔢 Step-by-Step Solution
1
From the clues, P1–P10 = H M H M H L H H M H. Highs must equal twice the lows, which gives 8 H and 4 L in total (so 8 M).
2
P16–P20 has no H and alternates, so it is either M L M L M or L M L M L.
3
P11 = L is one low. P1–P10 holds one low, and P16–P20 holds at least 2, so the total is already at least 4. It must be exactly 4. So P16–P20 is M L M L M, and P11 is the only low in P11–P15.
4
Highs: 6 so far in P1–P10, so P11–P15 needs 2 H. P11–P15 therefore has 1 L, 2 H and 2 M.
5
P16 = M, so P15 ≠ M. The only low is P11, so P15 = H.
6
P12 to P14 hold 1 H and 2 M with no equal neighbours. P14 ≠ H because P15 = H, and P12 ≠ L. The only fit is M H M.
7
The sequence is P11–P15 = L M H M H, so P14 = M.
Why Not the Others
A: P12 = M, not H.
B: P15 = H, not M.
C: P18 = M, not L.
Correct Answer: D

Q16. Arrangements

MCQ · +3 / −1
If contamination level at P15 was recorded as medium, then which of the following MUST be FALSE?
  1. A) Contamination levels at P13 and P17 were recorded as the same.
  2. B) Contamination levels at P11 and P16 were recorded as the same.
  3. C) Contamination level at P14 was recorded to be higher than that at P15.
  4. D) Contamination levels at P10 and P14 were recorded as the same.
Correct Answer
B) Contamination levels at P11 and P16 were recorded as the same.
Step-by-step solution
📌 Core Concept
Setting P15 = M forces the pattern in P16–P20, and the count condition then fixes P11–P15.
🔢 Step-by-Step Solution
1
From the clues, P1–P10 = H M H M H L H H M H. This has 6 H and 1 L.
2
P16–P20 has no H and alternates M and L. Since P15 = M, P16 ≠ M, so P16–P20 = L M L M L, which has 3 L.
3
Total lows are now at least 1 + 3 = 4. Highs must be at least 8, and the count condition then needs exactly 4 L and 8 H. So P11–P15 has no L and exactly 2 H.
4
P11–P15 uses only H and M, with no equal neighbours and P11 ≠ H (as P10 = H). The only pattern with 2 H ending in M is M H M H M.
5
The full tail is P11–P20 = M H M H M L M L M L.
6
Check each option:
A: P13 = M and P17 = M, so true.
B: P11 = M and P16 = L, so they differ. The statement is false.
C: P14 = H is higher than P15 = M, so true.
D: P10 = H and P14 = H, so true.
7
Only B is impossible.
Correct Answer: B
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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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