Operations, Logistics and Scheduling Data Previous Year Questions (PYQs) for CAT: 22+ Solved Questions with Step-by-Step Solutions

Solve 22+ Operations, Logistics and Scheduling Data previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Operations, Logistics and Scheduling Data

Chapter Journey

Transport Capacity and Transit Networks Focus: Segment-wise seat occupancy, transit times, and metro map routing. Current Topic

Delivery Routing and Sales Coverage Focus: Optimizing paths for multiple deliveries and calculating total distances.

Project Scheduling and Completion Tracking Focus: Gantt charts, critical paths, and tracking project completion percentages.

Equipment Operation and Process Control Focus: Machine efficiency, temperature control cycles, and operational modes.

By the end of this chapter, you will master the art of extracting logical constraints from operational data.

Transport Capacity and Transit Networks

Mastering the math of moving people and goods from point A to point B.

Segment-wise

Tracking Occupancy

Transit Time

Logic & Halts

Network

Routing & Maps

Chapter Context: Operations, Logistics and Scheduling Data

Operations, Logistics and Scheduling Data: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [30 - 34]
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.

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. 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?

A.
18%
B.
5.4%
C.
3.24%
D.
30%

Correct Answer:

Step-by-Step Solution

Key idea: Conditional probability with deterministic routing rules. Recognisable by network graphs with probabilistic demands and strict tie-breaking/routing algorithms.

Step 1: Analyze the condition "First location is Ahmednagar (A)".

Rule: Visit in decreasing order of demand. Tie-break: Closest to current location (Warehouse W).

For A to be first, $D_{A}$ must be the highest, or tied for highest with a favorable tie-break.

Max Demands: $A = 70, B = 60, C = 100, D = 50$ .

If $C = 100$ , C is first. So for A to be first, $C$ cannot be 100. Thus $C = 70$ .

If $C = 70$ , we compare A and C.

$D_{A}$ can be 50 or 70.

If $D_{A} = 50$ , $C (70) > A (50)$ , so C is first.

If $D_{A} = 70$ , $A (70) = C (70)$ . Tie-break: Distance from W.

$d (W, A) = 5$ , $d (W, C) = 12$ . A is closer. So A is first.

Condition E: $D_{A} = 70$ AND $D_{C} = 70$ .

$P (E) = P (D_{A} = 70) \times P (D_{C} = 70) = 0.6 \times 0.3 = 0.18$ .

Step 2: Analyze Route Distance given E.

Sequence starts W -> A. Remaining: B, C, D.

Current Loc: A. Next highest demand among B, C, D determines next stop.

We know $D_{C} = 70$ . $D_{B} \in {40, 60}$ . $D_{D} \in {30, 50}$ .

So $D_{C}$ is always the highest among remaining. Next stop is C.

Path: W -> A -> C.

Dist(W,A) = 5.

Dist(A,C): No direct road. Via W: $A \to W \to C = 5 + 12 = 17$ .

Total so far: $5 + 17 = 22$ .

Current Loc: C. Remaining: B, D.

Step 3: Determine rest of route to get Total 40 km.

Target Total = 40. Remaining Distance needed = $40 - 22 = 18$ .

Case 1: $D_{B} > D_{D}$ . Next is B, then D.

Path: $C \to B \to D$ .

$d (C, B) = 4$ (Direct).

$d (B, D)$ : No direct. Via W: $B \to W \to D = 10 + 2 = 12$ .

Leg Dist = $4 + 12 = 16$ . Total = $22 + 16 = 38$ . (Not 40).

Case 2: $D_{D} > D_{B}$ . Next is D, then B.

Path: $C \to D \to B$ .

$d (C, D) = 6$ (Direct).

$d (D, B)$ : No direct. Via W: $D \to W \to B = 2 + 10 = 12$ .

Leg Dist = $6 + 12 = 18$ . Total = $22 + 18 = 40$ . (Match!)

So we need $D_{D} > D_{B}$ .

Step 4: Calculate Probability of Case 2 given E.

We need $D_{D} > D_{B}$ .

Possible pairs $(D_{B}, D_{D})$ :

$(40, 30) \to 40 > 30$ (No)

$(40, 50) \to 50 > 40$ (Yes)

$(60, 30) \to 60 > 30$ (No)

$(60, 50) \to 60 > 50$ (No)

Only $(40, 50)$ works.

$P (D_{B} = 40, D_{D} = 50) = 0.3 \times 0.6 = 0.18$ .

Step 5: Final Calculation.

Question asks: Chance that distance is 40 GIVEN A is first.

$P (Dist = 40∣ A First) = \frac{P ( Dist = 40 \cap A First )}{P ( A First )}$ .

Numerator: $P (D_{A} = 70, D_{C} = 70, D_{B} = 40, D_{D} = 50) = 0.6 \times 0.3 \times 0.3 \times 0.6 = 0.0324$ .

Denominator: $P (D_{A} = 70, D_{C} = 70) = 0.18$ .

Result: $\frac{0.0324}{0.18} = 0.18 = 18%$ .

Answer: A

Question 2 · Data Interpretation and Logical Reasoning NAT

Common Description: A train travels from Station A to Station E, passing through stations B, C, and D, in that order. The train has a seating capacity of 200. A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station.
A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B - C, C - D, and D - E.
The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200.
The following information is known.
1. Segment C - D had an occupancy factor of 95%. Only segment B - C had a higher occupancy factor.
2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.
3. Among the seats reserved on segment D - E, exactly four-sevenths were from stations before C.
4. The number of tickets booked from A to C was equal to that booked from A to E, and it was higher than that from B to E.
5. No tickets were booked from A to B, from B to D and from D to E.
6. The number of tickets booked for any segment was a multiple of 10. How many tickets were booked from Station A to Station E?

Correct Answer:

Step-by-Step Solution

Key idea: This is a network flow problem with capacity constraints. We need to determine ticket counts between stations such that segment occupancies satisfy the given conditions.

Step 1: Define Variables.

Let $n_{X Y}$ be the number of tickets from station X to Y.

Given: $n_{B C} = 40$ , $n_{B E} = 30$ , $n_{A B} = 0$ , $n_{B D} = 0$ , $n_{D E} = 0$ .

Let $n_{A C} = n_{A E} = x$ . Let $n_{A D} = a$ , $n_{C D} = c$ , $n_{C E} = y$ .

Step 2: Analyze Segment Occupancies.

Capacity = 200. Segment C-D occupancy = 95% of 200 = 190.

Tickets using C-D: $A \to D$ , $A \to E$ , $B \to E$ , $C \to D$ , $C \to E$ .

Load( $C - D$ ) = $n_{A D} + n_{A E} + n_{B E} + n_{C D} + n_{C E} = a + x + 30 + c + y = 190$ .

So, $a + x + c + y = 160$ . (Eq 1)

Segment B-C is used by: $A \to C, A \to D, A \to E, B \to C, B \to E$ .

Load( $B - C$ ) = $n_{A C} + n_{A D} + n_{A E} + n_{B C} + n_{B E} = x + a + x + 40 + 30 = 2 x + a + 70$ .

Condition: Only B-C has higher occupancy than C-D (190). Max capacity is 200.

Also, load must be a multiple of 10. The only multiple of 10 strictly greater than 190 and $\leq 200$ is 200.

So, $2 x + a + 70 = 200 ⟹ 2 x + a = 130$ . (Eq 2)

Step 3: Analyze Segment D-E.

Tickets using D-E: $A \to E$ , $B \to E$ , $C \to E$ .

Load( $D - E$ ) = $x + 30 + y$ .

Condition: 4/7 of seats reserved on D-E are from stations before C (i.e., A and B).

Tickets from before C: $n_{A E} + n_{B E} = x + 30$ .

So, $\frac{x + 30}{x + 30 + y} = \frac{4}{7} ⟹ 7 (x + 30) = 4 (x + 30 + y) ⟹ 3 x + 90 = 4 y$ . (Eq 3)

Step 4: Solve for Integer Solutions.

From Eq 2: $a = 130 - 2 x$ . Since $a \geq 0$ , $x \leq 65$ .

Given $n_{A C} > n_{B E} ⟹ x > 30$ . So $30 < x \leq 65$ .

From Eq 3: $y = \frac{3 x + 90}{4}$ . For $y$ to be an integer, $3 x + 90$ must be divisible by 4.

$3 x + 90 \equiv 0 (mod 4) ⟹ 3 x + 2 \equiv 0 (mod 4) ⟹ 3 x \equiv 2 (mod 4) ⟹ x \equiv 2 (mod 4)$ .

Candidates for $x$ : 34, 38, 42, 46, 50, 54, 58, 62.

Step 5: Apply "Multiple of 10" Constraint to Segment Loads.

Load( $D - E$ ) = $x + 30 + y = x + 30 + \frac{3 x + 90}{4} = \frac{7 x + 210}{4}$ .

This load must be a multiple of 10. So $7 x + 210$ must be divisible by 40.

$7 x + 210 \equiv 0 (mod 40) ⟹ 7 x \equiv - 210 \equiv 30 (mod 40)$ .

Multiplying by 23 (inverse of 7 mod 40): $x \equiv 30 \times 23 = 690 \equiv 10 (mod 40)$ .

The only candidate in $(30, 65]$ satisfying $x \equiv 10 (mod 40)$ is $x = 50$ .

Check: If $x = 50$ , $a = 30$ , $y = 60$ . Load( $D - E$ ) = $50 + 30 + 60 = 140$ (Multiple of 10). Valid.

The question asks for tickets from A to E, which is $x = 50$ .

Answer: 50

Question 3 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [40 - 44]
Given above is the schematic map of the metro lines in a city with rectangles denoting terminal stations (e.g. A), diamonds denoting junction stations (e.g. R) and small filled-up circles denoting other stations. Each train runs either in east-west or north-south direction, but not both. All trains stop for 2 minutes at each of the junction stations on the way and for 1 minute at each of the other stations. It takes 2 minutes to reach the next station for trains going in east-west direction and 3 minutes to reach the next station for trains going in northsouth direction. From each terminal station, the first train starts at 6 am; the last trains leave the terminal stations at midnight. Otherwise, during the service hours, there are metro service every 15 minutes in the north-south lines and every 10 minutes in the east-west lines. A train must rest for at least 15 minutes after completing a trip at the terminal station, before it can undertake the next trip in the reverse direction. (All questions are related to this metro service only. Assume that if someone reaches a station exactly at the time a train is supposed to leave, (s)he can catch that train.)

Terminal stations: A, B, C, D, M, N, P, Q
Junction stations: R, S, T, V

A -- ● -- ● -- R -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- S -- ● -- ● -- ● -- ● -- ● -- N
                |                                            |
                ●                                            ●
                |                                            |
                ●                                            ●
                |                                            |
M -- ● -- ● -- ● -- R                                            S -- ● -- ● -- ● -- ● -- ● -- N
                |                                            |
                ●                                            ●
                |                                            |
                ●                                            ●
                |                                            |
                T -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- V
                |                                            |
                ●                                            ●
                |                                            |
                ●                                            ●
                |                                            |
                B                                            D

P -- ● -- ● -- ● -- T -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- V -- ● -- ● -- ● -- ● -- ● -- Q

What is the minimum number of trains that are required to provide the service in this city?

Correct Answer:

Step-by-Step Solution

Key idea: This is a fleet sizing problem. We need to calculate the minimum number of trains required for each independent line to maintain the specified headway (frequency), considering the round trip time including terminal rest.

Step 1: Understand the Formula.

Minimum Trains on a Line = $⌈ \frac{Round Trip Time (RTT)}{Headway} ⌉$ .

RTT = One-Way Transit Time + Terminal Rest + Return Transit Time + Terminal Rest.

Since One-Way Transit = Return Transit, RTT = $2 \times (One-Way Transit + Terminal Rest)$ .

Step 2: Analyze Line 1 (A-N, East-West).

Path: A to N.
Segments: 18 segments. Travel Time: $18 \times 2 = 36$ mins.
Intermediate Stations: 15 regular stops (15 mins) + 2 junction stops (R, S, 4 mins).
One-Way Transit = $36 + 15 + 4 = 55$ mins.
Terminal Rest = 15 mins.
RTT = $2 \times (55 + 15) = 140$ mins.
Headway = 10 mins.
Trains = $140/10 = 14$ .

Step 3: Analyze Line 2 (M-B, North-South).

One-way transit = 45 mins (derived from map segments and stops).
Terminal Rest = 15 mins.
RTT = $2 \times (45 + 15) = 120$ mins.
Headway = 15 mins.
Trains = $120/15 = 8$ .

Step 4: Analyze Line 3 (P-Q, East-West).

One-way transit = 43 mins.
Terminal Rest = 15 mins.
RTT = $2 \times (43 + 15) = 116$ mins.
Headway = 10 mins.
Trains = $⌈ 116/10 ⌉ = ⌈ 11.6 ⌉ = 12$ .

Step 5: Analyze Line 4 (C-D, North-South).

One-way transit = 49 mins.
Terminal Rest = 15 mins.
RTT = $2 \times (49 + 15) = 128$ mins.
Headway = 15 mins.
Trains = $⌈ 128/15 ⌉ = ⌈ 8.53 ⌉ = 9$ .

Step 6: Sum the trains.

Total minimum trains = $14 + 8 + 12 + 9 = 43$ .

Answer: 43

Question 4 · Data Interpretation and Logical Reasoning NAT

Correct Answer:

Step-by-Step Solution

Key idea: This is a network flow capacity problem. We model ticket bookings as flows on edges (segments) and use occupancy constraints to form a system of Diophantine equations. The question asks for the total tickets booked FROM Station C, which corresponds to the sum of tickets starting at C ( $n_{C D} + n_{C E}$ ).

Step 1: Define variables for unknown ticket counts.

Let $n_{A C} = x$ , $n_{A E} = x$ (given equal). Let $n_{A D} = a$ .

Let $n_{C D} = c$ and $n_{C E} = y$ .

Knowns: $n_{B C} = 40, n_{B E} = 30$ . Others are 0 ( $n_{A B}, n_{B D}, n_{D E}$ ).

Step 2: Formulate segment occupancy equations.

Segment B-C Flow: $F_{B C} = n_{A C} + n_{A D} + n_{A E} + n_{B C} + n_{B E} = x + a + x + 40 + 30 = 2 x + a + 70$ .

Given: C-D is 95% of 200 = 190. Only B-C had a higher occupancy factor. Max capacity is 200. So $190 < F_{B C} \leq 200$ . Also $F_{B C}$ is a multiple of 10. The only multiple of 10 in $(190, 200]$ is 200.

Eq 1: $2 x + a + 70 = 200 ⟹ 2 x + a = 130$ .

Step 3: Analyze Segment D-E.

Flow $F_{D E} = n_{A E} + n_{B E} + n_{C E} = x + 30 + y$ .

Given: 4/7 of $F_{D E}$ comes from stations before C (i.e., A and B origins). Contribution = $n_{A E} + n_{B E} = x + 30$ .

Eq 2: $\frac{x + 30}{x + 30 + y} = \frac{4}{7} ⟹ 7 (x + 30) = 4 (x + 30 + y) ⟹ 3 (x + 30) = 4 y ⟹ 3 x + 90 = 4 y$ .

Step 4: Solve integer constraints.

From Eq 2: $y = \frac{3 x + 90}{4}$ . For $y$ to be an integer, $3 x + 90$ must be divisible by 4. Since $90 \equiv 2 (mod 4)$ , $3 x \equiv 2 (mod 4) ⟹ x \equiv 2 (mod 4)$ .

From Eq 1: $a = 130 - 2 x$ . Since $a \geq 0$ , $x \leq 65$ .

Also given $x > n_{B E} = 30$ . So $30 < x \leq 65$ .

Candidates for $x$ ( $x \equiv 2 (mod 4)$ ): 34, 38, 42, 46, 50, 54, 58, 62.

Step 5: Apply the 'multiple of 10' rule to Segment D-E.

Flow $F_{D E} = x + 30 + y$ . This must be a multiple of 10.

Substitute $y$ : $F_{D E} = x + 30 + \frac{3 x + 90}{4} = \frac{4 x + 120 + 3 x + 90}{4} = \frac{7 x + 210}{4}$ .

For $F_{D E}$ to be an integer multiple of 10, $7 x + 210$ must be divisible by 40.

$7 x + 210 \equiv 0 (mod 40) ⟹ 7 x \equiv - 210 \equiv 30 (mod 40)$ .

Multiplying by 23 (inverse of 7 mod 40): $x \equiv 30 \times 23 = 690 \equiv 10 (mod 40)$ .

The only candidate in range satisfying this is $x = 50$ .

Step 6: Calculate remaining variables.

If $x = 50$ , then $a = 130 - 100 = 30$ .

$y = \frac{150 + 90}{4} = 60$ .

Now find $c$ using Segment C-D flow.

$F_{C D} = n_{A D} + n_{A E} + n_{B E} + n_{C D} + n_{C E} = a + x + 30 + c + y = 190$ .

$30 + 50 + 30 + c + 60 = 190 ⟹ 170 + c = 190 ⟹ c = 20$ .

Step 7: Answer the specific question.

"How many tickets were booked from Station C?"

Tickets from C are $n_{C D} + n_{C E} = c + y = 20 + 60 = 80$ .

Answer: 80

Question 5 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [30 - 34]
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.

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. 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?

A.
210
B.
220
C.
200
D.
250

Correct Answer:

Step-by-Step Solution

Key idea: This is a reverse-engineering routing problem combining network distances with probabilistic demand sorting. The trigger is a fixed total distance and a negative constraint ("A not first"), requiring deduction of the specific route and subsequently the demand scenario.

Step 1: Establish the distance matrix.

Nodes: A(TL), B(TR), C(BR), D(BL). Center W.

Direct Links: A-B (6), B-C (4), C-D (6), A-D (8).

Radials: W-A (5), W-B (10), W-C (12), W-D (2).

Effective Distances (Min of Direct vs Via W):

A-D: Direct 8 vs Via W $5 + 2 = 7$ . Min = 7.

A-C: Via W = $5 + 12 = 17$ (no direct).

B-D: Via W = $10 + 2 = 12$ (no direct).

Step 2: Identify valid routes with Total Distance = 29 km.

Constraint: A is NOT first.

Possible starts: B, C, D.

Let's test C-start: W $\to$ C $\to$ B $\to$ A $\to$ D.

Dist: W-C (12) + C-B (4) + B-A (6) + A-D (7).

Total: $12 + 4 + 6 + 7 = 29$ . Matches perfectly.

Let's check if other routes work.

W $\to$ B $\to$ C $\to$ D $\to$ A: $10 + 4 + 6 + 7 = 27$ . (No)

W $\to$ D $\to$ C $\to$ B $\to$ A: $2 + 6 + 4 + 6 = 18$ . (No)

W $\to$ C $\to$ D $\to$ A $\to$ B: $12 + 6 + 7 + 6 = 31$ . (No)

So the route MUST be W $\to$ C $\to$ B $\to$ A $\to$ D.

Step 3: Determine Demand Scenario for W-C-B-A-D.

Sorting Rule: Decreasing Demand. Tie-break: Closest.

C is first $⟹ D e man d (C) \geq D e man d s (B, A, D)$ . C $\in {70, 100}$ . Always max or tied. OK.
B is second $⟹ D e man d (B) \geq D e man d s (A, D)$ . B $\in {40, 60}$ .

For B to be chosen over A, need $B \geq A$ .

If B=40: Cannot be $\geq A$ (min A=50). So B MUST be 60.

If B=60: A must be $\leq 60$ . A $\in {50, 70}$ . So A MUST be 50.

Crucial Deduction: $D e man d (B) = 60$ and $D e man d (A) = 50$ .

A is third $⟹ D e man d (A) \geq D e man d s (D)$ . A=50. D $\in {30, 50}$ . Always $\geq$ . OK.

Step 4: Calculate possible total widgets.

Total = $C + B + A + D$ .

We know $B = 60, A = 50$ .

C can be 70 or 100.

D can be 30 or 50.

Possible Totals:

$C = 70, D = 30 ⟹ 70 + 60 + 50 + 30 = 210$ .
$C = 70, D = 50 ⟹ 70 + 60 + 50 + 50 = 230$ .
$C = 100, D = 30 ⟹ 100 + 60 + 50 + 30 = 240$ .
$C = 100, D = 50 ⟹ 100 + 60 + 50 + 50 = 260$ .

Check Options:

A) 210 -> Possible.

B) 220 -> Not possible.

C) 200 -> Not possible.

D) 250 -> Not possible.

Answer: A

A question from this chapter

Question 1

Common Description: Instructions [30 - 34] 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. A B C D W 6846510212 [50 (40%), 70 (60%)] [40 (30%), 60 (70%)] [70 (30%), 100 (70%)] [30 (40%), 50 (60%)] 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. 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?

Question 2

Common Description: A train travels from Station A to Station E, passing through stations B, C, and D, in that order. The train has a seating capacity of 200. A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station. A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B - C, C - D, and D - E. The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200. The following information is known. 1. Segment C - D had an occupancy factor of 95%. Only segment B - C had a higher occupancy factor. 2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E. 3. Among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. 4. The number of tickets booked from A to C was equal to that booked from A to E, and it was higher than that from B to E. 5. No tickets were booked from A to B, from B to D and from D to E. 6. The number of tickets booked for any segment was a multiple of 10. How many tickets were booked from Station A to Station E?

Question 3

Common Description: Instructions [40 - 44] Given above is the schematic map of the metro lines in a city with rectangles denoting terminal stations (e.g. A), diamonds denoting junction stations (e.g. R) and small filled-up circles denoting other stations. Each train runs either in east-west or north-south direction, but not both. All trains stop for 2 minutes at each of the junction stations on the way and for 1 minute at each of the other stations. It takes 2 minutes to reach the next station for trains going in east-west direction and 3 minutes to reach the next station for trains going in northsouth direction. From each terminal station, the first train starts at 6 am; the last trains leave the terminal stations at midnight. Otherwise, during the service hours, there are metro service every 15 minutes in the north-south lines and every 10 minutes in the east-west lines. A train must rest for at least 15 minutes after completing a trip at the terminal station, before it can undertake the next trip in the reverse direction. (All questions are related to this metro service only. Assume that if someone reaches a station exactly at the time a train is supposed to leave, (s)he can catch that train.) Terminal stations: A, B, C, D, M, N, P, Q Junction stations: R, S, T, V A -- ● -- ● -- R -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- S -- ● -- ● -- ● -- ● -- ● -- N | | ● ● | | ● ● | | M -- ● -- ● -- ● -- R S -- ● -- ● -- ● -- ● -- ● -- N | | ● ● | | ● ● | | T -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- V | | ● ● | | ● ● | | B D P -- ● -- ● -- ● -- T -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- ● -- V -- ● -- ● -- ● -- ● -- ● -- Q What is the minimum number of trains that are required to provide the service in this city?

Question 4

Common Description: A train travels from Station A to Station E, passing through stations B, C, and D, in that order. The train has a seating capacity of 200. A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station. A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B - C, C - D, and D - E. The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200. The following information is known. 1. Segment C - D had an occupancy factor of 95%. Only segment B - C had a higher occupancy factor. 2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E. 3. Among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. 4. The number of tickets booked from A to C was equal to that booked from A to E, and it was higher than that from B to E. 5. No tickets were booked from A to B, from B to D and from D to E. 6. The number of tickets booked for any segment was a multiple of 10. How many tickets were booked from Station C?

Question 5

Common Description: Instructions [30 - 34] 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. A B C D W 6846510212 [50 (40%), 70 (60%)] [40 (30%), 60 (70%)] [70 (30%), 100 (70%)] [30 (40%), 50 (60%)] 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. 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?

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