CAT 2022 Slot 2 DILR Set 2 Q6–Q10

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2022 Slot 2, with 5 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.

Every day a widget supplier supplies widgets from the warehouse (W) to four locations – Ahmednagar (A), Bikrampore (B), Chitrachak (C), and Deccan Park (D). The daily demand for widgets in each location is uncertain and independent of each other. Demands and corresponding probability values (in parenthesis) are given against each location (A, B, C, and D) in the figure below. For example, there is a 40% chance that the demand in Ahmednagar will be 50 units and a 60% chance that the demand will be 70 units. The lines in the figure connecting the locations and warehouse represent two-way roads connecting those places with the distances (in km) shown beside the line. The distances in both the directions along a road are equal. For example, the road from Ahmednagar to Bikrampore and the road from Bikrampore to Ahmednagar are both 6 km long.

[Figure in the original paper]

Every day the supplier gets the information about the demand values of the four locations and creates the travel route that starts from the warehouse and ends at a location after visiting all the locations exactly once. While making the route plan, the supplier goes to the locations in decreasing order of demand. If there is a tie for the choice of the next location, the supplier will go to the location closest to the current location. Also, while creating the route, the supplier can either follow the direct path (if available) from one location to another or can take the path via the warehouse. If both paths are available (direct and via warehouse), the supplier will choose the path with minimum distance.

Question 6

If the last location visited is Ahmednagar, then what is the total distance covered in the route (in km)?

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

Answer: 35 Routes & Networks TITA

Solution

  1. Shortest legs (direct, or via W if shorter): WA 5, WB 10, WC 12, WD 2, AB 6, AD 7 (via W beats the direct 8), BC 4, CD 6, AC 17 (via W), BD 12 (via W).
  2. For A (at least 50) to be visited last, every other location needs demand at least as high: B = 60, D = 50 (tied with A at 50), and C is 70 or 100.
  3. So the order is C, B, then the 50-unit pair A and D. The question tells us A is last, so the route must be W-C-B-D-A (this is the only route that can end at A).
  4. Distance \(=12+4+12+7=35\) km.

Question 7

If the total number of widgets delivered in a day is 250 units, then what is the total distance covered in the route (in km)?

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

Answer: 38 Routes & Networks TITA

Solution

  1. Shortest legs (direct, or via W if shorter): WA 5, WB 10, WC 12, WD 2, AB 6, AD 7 (via W beats the direct 8), BC 4, CD 6, AC 17 (via W), BD 12 (via W).
  2. The minimum total is \(50+40+70+30=190\). An extra 60 can only be \(20+20+20\) (A, B, D high, C low): A 70, B 60, C 70, D 50.
  3. A and C tie at 70; from W, A (5) is closer than C (12), so the order is A, C, B (60), D (50).
  4. Distance \(=5+17+4+12=38\) km.

Question 8

What is the chance that the total number of widgets delivered in a day is 260 units and the route ends at Bikrampore?
  1. A
    10.80%
  2. B
    33.33%
  3. C
    7.56%
  4. D
    17.64%

Answer: (C) 7.56% Routes & Networks MCQ

Solution

  1. Shortest legs (direct, or via W if shorter): WA 5, WB 10, WC 12, WD 2, AB 6, AD 7 (via W beats the direct 8), BC 4, CD 6, AC 17 (via W), BD 12 (via W).
  2. 260 = 190 + 70, and 70 can only be \(30+20+20\): C = 100 plus two of A, B, D at their higher values.
  3. To end at B, B must have the lowest demand, so B = 40, with A = 70 and D = 50 (order C, A, D, B). With B = 60 the route ends at D.
  4. Probability \(=0.7\times0.6\times0.6\times0.3=0.0756=7.56\%\).

Question 9

If the first location visited from the warehouse is Ahmednagar, then what is the chance that the total distance covered in the route is 40 km?
  1. A
    3.24%
  2. B
    30%
  3. C
    5.4%
  4. D
    18%

Answer: (D) 18% Routes & Networks MCQ

Solution

  1. Shortest legs (direct, or via W if shorter): WA 5, WB 10, WC 12, WD 2, AB 6, AD 7 (via W beats the direct 8), BC 4, CD 6, AC 17 (via W), BD 12 (via W).
  2. A is first only if A = 70 and C = 70 (a tie that W breaks in A's favour, 5 vs 12). The route starts W-A-C: \(5+17=22\).
  3. From C: if B = 60, then C-B-D \(=4+12\), total 38. If B = 40 and D = 30, then C-B-D, total 38. If B = 40 and D = 50, then C-D-B \(=6+12\), total 40.
  4. Given that A is first, a 40 km route needs B = 40 and D = 50: chance \(0.3\times0.6=0.18=18\%\).

Question 10

If Ahmednagar is not the first location to be visited in a route and the total route distance is 29 km, then which of the following is a possible number of widgets delivered on that day?
  1. A
    210
  2. B
    200
  3. C
    250
  4. D
    220

Answer: (A) 210 Routes & Networks MCQ

Solution

  1. Shortest legs (direct, or via W if shorter): WA 5, WB 10, WC 12, WD 2, AB 6, AD 7 (via W beats the direct 8), BC 4, CD 6, AC 17 (via W), BD 12 (via W).
  2. If A is not first, C is first: W-C = 12, leaving 17 km for three legs.
  3. The only 17 km order is C-B-A-D \(=4+6+7\), which needs B = 60, A = 50 and D at most 50 (a 50-50 tie goes to A, which is closer to B).
  4. Totals: \(C+60+50+D\) gives 210, 230, 240 or 260. Among the options only 210 works.

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.