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CAT 2024 Slot 2 DILR · Set 2 of 5Logical ReasoningHard DifficultyQ6–Q9 · 2 MCQ + 2 TITA

CAT 2024 Slot 2 DILR Set 2: Grids (4 Questions) with Solutions

The Grids set from the official CAT 2024 Slot 2 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 2024 Slot 2 DILR · Questions 6–9
Official Reference Chart - CAT 2024 Slot 2 Official Reference Chart - CAT 2024 Slot 2
The above is a schematic diagram of walkways (indicated by all the straight lines) and lakes (3 of them, each in the shape of rectangles - shaded in the diagram) of a gated area. Different points on the walkway are indicated by letters (A through P) with distances being OP = 150 m, ON = MN = 300 m, ML = 400 m, EL = 200 m, DE = 400 m.
The following additional information about the facilities in the area is known.
1
The only entry/exit point is at C.
2
There are many residences within the gated area; all of them are located on the path AH and ML with four of them being at A, H, M, and L.
3
The post office is located at P and the bank is located at B.

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

Q6. Grids

MCQ · +3 / −1
One resident whose house is located at L, needs to visit the post office as well as the bank. What is the minimum distance (in m) he has to walk starting from his residence and returning to his residence after visiting both the post office and the bank?
  1. A) 3200
  2. B) 3000
  3. C) 2800
  4. D) 2700
Correct Answer
A) 3200
Step-by-step solution
📌 Core Concept
The walk is a closed loop L → P → B → L. Its minimum length is the sum of the three shortest legs between these points.
🔢 Step-by-Step Solution
1
Set up the grid distances. The columns are at 0, 150, 450 and 750 m (from OP=150OP=150, ON=300ON=300, MN=300MN=300). The row gaps are DE=400DE=400, EL=200EL=200 and ML=400ML=400. The diagonal OKOK is the hypotenuse of a 300-400 triangle, so OK=500OK=500. The diagonal IGIG is 1502+2002=250\sqrt{150^2+200^2}=250.
2
L to P. The route L-K-O-P gives 300+500+150=950300+500+150=950. L-M-N-O-P gives 400+300+300+150=1150400+300+300+150=1150, and L-K-J-I-P also gives 300+300+150+400=1150300+300+150+400=1150. The shortest is 950.
3
P to B. The route P-I-G-B uses the diagonal IGIG and gives 400+250+400=1050400+250+400=1050. P-I-H-A-B gives 400+200+400+150=1150400+200+400+150=1150, and P-O-J-G-B gives 150+400+200+400=1150150+400+200+400=1150. The shortest is 1050.
4
B to L. Every route has to cover 400 m down and 750 m across, and none of them can use a diagonal. For example, B-G-F-K-L gives 400+300+200+300=1200400+300+200+300=1200, and B-C-D-E-L gives 300+300+400+200=1200300+300+400+200=1200. The shortest is 1200.
5
Total: 950+1050+1200=3200950+1050+1200=3200. Visiting B before P gives the same loop in reverse, so the total is the same.
Quick Check
The loop L-K-O-P-I-G-B-G-F-K-L is 3200 m. Its legs are 950, 1050 and 1200.
Correct Answer: A

Q7. Grids

MCQ · +3 / −1
One person enters the gated area and decides to walk as much as possible before leaving the area without walking along any path more than once and always walking next to one of the lakes. Note that he may cross a point multiple times. How much distance (in m) will he walk within the gated area?
  1. A) 3200
  2. B) 2800
  3. C) 3000
  4. D) 3800
Correct Answer
D) 3800
Step-by-step solution
📌 Core Concept
Walking only next to a lake restricts the walker to the lake-boundary segments. Using each segment at most once, the longest walk is the full set of these segments if an Euler circuit through C exists.
🔢 Step-by-Step Solution
1
Lake CDEF has boundary CD = 300, DE = 400, EF = 300 and FC = 400, a total of 1400.
2
Lake GFKJ has boundary GF = 300, FK = 200, KJ = 300 and JG = 200, a total of 1000.
3
Lake KLMN has boundary KL = 300, LM = 400, MN = 300 and NK = 400, a total of 1400.
4
The segment EL is not next to any lake, because it lies in the gap between CDEF and KLMN. So it is excluded. The three boundaries share no segment, so they add up without overlap.
5
The three boundaries touch at F and K. Every point on them has an even number of boundary segments: C, D, E, G, J, L, M and N have 2, and F and K have 4. So a closed walk from C that uses every boundary segment exactly once exists, for example C→D→E→F→G→J→K→L→M→N→K→F→C.
6
The total distance is 1400+1000+1400=38001400+1000+1400 = 3800 m.
Why Not the Others
3200, 2800 and 3000 each leave out some lake-side segments that can be added without repeating a segment.
Correct Answer: D

Q8. Grids

TITA · +3 / 0
One resident takes a walk within the gated area starting from A and returning to A without going through any point (other than A) more than once. What is the maximum distance (in m) she can walk in this way? □\square
Correct Answer
5100
Step-by-step solution
📌 Core Concept
The walk is a cycle through A with no point repeated, so the aim is a cycle that uses as many long segments as possible.
🔢 Step-by-Step Solution
1
Read the segment lengths. The 400 m segments are AHAH, BGBG, CFCF, DEDE, LMLM, KNKN, JOJO and PIPI. KOKO is 500 m, and every other segment is 150 to 300 m.
2
Points A, D, M and P each have only two walkways, so a cycle through them must use both. This forces A-BA\text{-}B, A-HA\text{-}H, C-DC\text{-}D, D-ED\text{-}E, L-ML\text{-}M, M-NM\text{-}N, O-PO\text{-}P and P-IP\text{-}I.
3
Build the cycle: A-B-G-F-C-D-E-L-M-N-K-J-O-P-I-H-A. It uses all eight 400 m segments, and it visits all 16 points exactly once.
4
Add up the segments:
150+400+300+400+300+400+200+400+300+400+300+400+150+400+200+400=5100150+400+300+400+300+400+200+400+300+400+300+400+150+400+200+400=5100
5
The diagonal KOKO (500 m) does not help. Taking K-O instead of K-J-O replaces 300+400=700300+400=700 with 500 and loses length.
Quick Check
The route has 16 segments. Eight are 400 m (3200 m in total), and the other eight add 1900 m. That gives 3200+1900=51003200+1900=5100.
Correct Answer: 5100

Q9. Grids

TITA · +3 / 0
Visitors coming for morning walks are allowed to enter as long as they do not pass by any of the residences and do not cross any point (except C) more than once. What is the maximum distance (in m) that such a visitor can walk within the gated area?
Correct Answer
3500
Step-by-step solution
📌 Core Concept
The visitor enters and leaves at C, so the walk is a cycle through C. It must avoid every residence, which means the segments AHAH and MLML and the points A, H, M and L. The longest such cycle is wanted.
🔢 Step-by-Step Solution
1
Remove A, H, M and L. Twelve points remain: B, C, D, E, F, G, I, J, K, N, O, P.
2
Without L, point E connects only to D and F, and D connects only to C and E. So the stretch C-D-E-F is forced if the walk uses D. It is worth 300+400+300=1000300+400+300=1000.
3
Start with C-D-E-F, then take the longest route from F to B that covers the other points: F-K (200200), K-N (400400), N-O (300300), O-P (150150), P-I (400400), I-J (150150), J-G (200200), G-B (400400).
4
The closing segment is B-C (300300). The cycle is C-D-E-F-K-N-O-P-I-J-G-B-C, and it visits all 12 permitted points.
5
Total:
300+400+300+200+400+300+150+400+150+200+400+300=3500300+400+300+200+400+300+150+400+150+200+400+300=3500
Quick Check
The cycle covers all 12 permitted points once each, and no residence lies on it. So it cannot be extended, and the maximum is 3500 m.
Correct Answer: 3500
Timed Practice

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CAT 2024 Slot 2 DILR at a Glance

Sets in this section5
Questions in this section22
Section time40 minutes
Convenor IIMIIM Calcutta

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