CAT 2024 Slot 2 DILR Set 1 Q1–Q4

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2024 Slot 2, with 4 questions (Routes & Networks). Answers are CAT's official key.

The set

This set relies on a figure in the original paper, which is not reproduced here. The questions and answers are below.

[Figure in the original paper]
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.

Question 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
    2700
  3. C
    3000
  4. D
    3400

Answer: (A) 3200 Routes & Networks MCQ

Solution

  1. Set coordinates from the given lengths: columns at A/H/I/P = 0, B/G/J/O = 150, C/F/K/N = 450, D/E/L/M = 750; rows: top line 1000, H-E line 600, I-L line 400, bottom 0. So horizontal pieces are 150, 300, 300; vertical pieces are 400, 200, 400; diagonals IG = 250 and OK = 500.
  2. Shortest L to P: L-K-O-P \(= 300 + 500 + 150 = 950\) (other routes are 1150).
  3. Shortest P to B: P-I-G-B \(= 400 + 250 + 400 = 1050\).
  4. Shortest B to L: e.g. B-G-J-K-L \(= 400 + 200 + 300 + 300 = 1200\) (every route is at least 1200).
  5. Round trip L, P, B, L \(= 950 + 1050 + 1200 = 3200\) m (the reverse order gives the same).

Question 2

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
    3800
  2. B
    2800
  3. C
    3000
  4. D
    3200

Answer: (A) 3800 Routes & Networks MCQ

Solution

  1. Set coordinates from the given lengths: columns at A/H/I/P = 0, B/G/J/O = 150, C/F/K/N = 450, D/E/L/M = 750; rows: top line 1000, H-E line 600, I-L line 400, bottom 0. So horizontal pieces are 150, 300, 300; vertical pieces are 400, 200, 400; diagonals IG = 250 and OK = 500.
  2. Paths next to a lake are the four sides of each lake: lake CDEF \(= 300+400+300+400 = 1400\), lake GFKJ \(= 300+200+300+200 = 1000\), lake KLMN \(= 300+400+300+400 = 1400\).
  3. In this sub-network F and K have degree 4 and every other point degree 2, so all degrees are even and it is connected: he can walk every such path exactly once and return to C.
  4. Distance \(= 1400 + 1000 + 1400 = 3800\) m.

Question 3

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?

TITA: type in the answer. There are no options and no negative marking.

Answer: 5100 Routes & Networks TITA

Solution

  1. Set coordinates from the given lengths: columns at A/H/I/P = 0, B/G/J/O = 150, C/F/K/N = 450, D/E/L/M = 750; rows: top line 1000, H-E line 600, I-L line 400, bottom 0. So horizontal pieces are 150, 300, 300; vertical pieces are 400, 200, 400; diagonals IG = 250 and OK = 500.
  2. A has only two paths (A-B 150, A-H 400), so the loop uses both. To be long, the loop should visit all 16 points and prefer the 400 m verticals.
  3. Loop A-B-G-F-C-D-E-L-M-N-K-J-O-P-I-H-A visits every point once.
  4. Length \(= 150+400+300+400+300+400+200+400+300+400+300+400+150+400+200+400 = 5100\) m, which is the maximum (it uses all eight 400 m verticals; any loop using diagonal OK must drop longer pieces).

Question 4

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?

TITA: type in the answer. There are no options and no negative marking.

Answer: 3500 Routes & Networks TITA

Solution

  1. Set coordinates from the given lengths: columns at A/H/I/P = 0, B/G/J/O = 150, C/F/K/N = 450, D/E/L/M = 750; rows: top line 1000, H-E line 600, I-L line 400, bottom 0. So horizontal pieces are 150, 300, 300; vertical pieces are 400, 200, 400; diagonals IG = 250 and OK = 500.
  2. Residences lie on AH and ML, so points A, H, M, L and those paths are off-limits. The visitor must start and end at C (the only gate) without repeating any other point: a simple loop through C.
  3. Loop C-B-G-J-I-P-O-N-K-F-E-D-C uses every remaining point once.
  4. Length \(= 300+400+200+150+400+150+300+400+200+300+400+300 = 3500\) m, the maximum possible.

Same concept in other years

See every question on Network of roads/lines with travel times or distances; routes and schedules.

More Routes & Networks questions from CAT

All 41 Routes & Networks questions asked since 2017.