CAT 2023 Slot 1 · DILR · Set 1: Visa Processing Office Schedule

The full set and all 5 questions (Q1–Q5) from CAT 2023 Slot 1, with answers and solutions.

A visa processing office (VPO) accepts visa applications in four categories - US, UK, Schengen, and Others. The applications are scheduled for processing in twenty 15-minute slots starting at 9:00 am and ending at 2:00 pm. Ten applications are scheduled in each slot.

There are ten counters in the office, four dedicated to US applications, and two each for UK applications, Schengen applications and Others applications. Applicants are called in for processing sequentially on a first-come-first-served basis whenever a counter gets freed for their category. The processing time for an application is the same within each category. But it may vary across the categories. Each US and UK application requires 10 minutes of processing time. Depending on the number of applications in a category and time required to process an application for that category, it is possible that an applicant for a slot may be processed later.

On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15 am slot but entered the VPO at 9:20 am. When they entered the office, exactly six out of the ten counters were either processing applications, or had finished processing one and ready to start processing the next.

Mahira and Osman were scheduled in the 9:30 am slot on that day for visa processing in the Others category.

The following additional information is known about that day.

1. All slots were full.

2. The number of US applications was the same in all the slots. The same was true for the other three categories.

3. 50% of the applications were US applications.

4. All applicants except Ira, Vijay and Nandini arrived on time.

5. Vijay was called to a counter at 9:25 am.

Question 1 · Caselet DI

How many UK applications were scheduled on that day?

Answer & solution

Answer: 0

Solving the set

Step 1 – Applications per slot. 20 slots × 10 = 200 applications. 50% are US, so 100 US, i.e., 5 per slot (every slot has the same mix).

The other 5 per slot: Ira, Vijay and Nandini are all Schengen in the 9:15 slot, so at least 3 Schengen per slot; Mahira and Osman are both Others in the 9:30 slot, so at least 2 Others per slot. 3 + 2 = 5 leaves nothing for UK.

Category	Per slot	Reason	Total for the day (20 slots)
US	5	50% of 10	100
Schengen	3	Ira, Vijay, Nandini share a slot (at least 3)	60
Others	2	Mahira, Osman share a slot (at least 2)	40
UK	0	10 − 5 − 3 − 2	0
Total	10		200

Step 2 – Schengen processing time (t). In the 9:00 slot, two Schengen applicants start at 9:00 and finish at 9:00 + t; the third starts at 9:00 + t and finishes at 9:00 + 2t. Ira arrives at 9:20.

• If t ≤ 20, a counter is free by 9:20 and Ira is called at 9:20. Vijay is called when the other counter frees, at 9:00 + 2t = 9:25, so t = 12.5 minutes.

• If t > 20, nobody is free at 9:20; Ira and the third 9:00 applicant would both start at 9:00 + t, and Vijay could not start before 9:00 + 2t > 9:40. Not possible.

So each Schengen application takes 12.5 minutes:

Schengen counter	Applicant	Start	End
Counter 1	9:00 slot – 1st	9:00	9:12:30
Counter 2	9:00 slot – 2nd	9:00	9:12:30
Counter 1	9:00 slot – 3rd	9:12:30	9:25
Counter 2	Ira	9:20	9:32:30
Counter 1	Vijay	9:25	9:37:30
Counter 2	Nandini	9:32:30	9:45

Step 3 – Others processing time. At 9:20 exactly six counters were active:

Counters	Status at 9:20 am	Counted
4 US	Busy (US queue never empties)	4
2 UK	No UK applications at all	0
2 Schengen	One processing the 3rd 9:00 applicant, one starting Ira	2
2 Others	Must be idle	0
Total		6

So the two Others counters were idle at 9:20: the two 9:15 Others applicants (started at 9:15) had finished by 9:20. Each Others application therefore takes at most 5 minutes.

Others applicant	Counter	Start	End (processing time t ≤ 5 min)
9:15 slot – 1st	1	9:15	by 9:20
9:15 slot – 2nd	2	9:15	by 9:20
Mahira	1	9:30	by 9:35
Osman	2	9:30	by 9:35

Step 4 – US counters. 5 US applicants arrive every 15 minutes for 4 counters of 10 minutes each.

Slot	Start times of its 5 US applicants	Last one ends
9:00	9:00, 9:00, 9:00, 9:00, 9:10	9:20
9:15	9:15, 9:15, 9:15, 9:20, 9:25	9:35
9:30	9:30, 9:30, 9:30, 9:35, 9:40	9:50
…	same pattern every slot	slot time + 20 min
1:45 pm (last)	1:45, 1:45, 1:45, 1:50, 1:55	2:05 pm

This question

From Step 1, the 5 non-US places in each slot are taken by 3 Schengen and 2 Others applications, leaving 0 for UK. So no UK applications were scheduled.

Hence, the answer is 0.

Question 2 · Caselet DI

What is the maximum possible value of the total time (in minutes, nearest to its integer value) required to process all applications in the Others category on that day?

Answer & solution

Answer: 200

Solving the set

Step 1 – Applications per slot. 20 slots × 10 = 200 applications. 50% are US, so 100 US, i.e., 5 per slot (every slot has the same mix).

Category	Per slot	Reason	Total for the day (20 slots)
US	5	50% of 10	100
Schengen	3	Ira, Vijay, Nandini share a slot (at least 3)	60
Others	2	Mahira, Osman share a slot (at least 2)	40
UK	0	10 − 5 − 3 − 2	0
Total	10		200

Step 2 – Schengen processing time (t). In the 9:00 slot, two Schengen applicants start at 9:00 and finish at 9:00 + t; the third starts at 9:00 + t and finishes at 9:00 + 2t. Ira arrives at 9:20.

• If t ≤ 20, a counter is free by 9:20 and Ira is called at 9:20. Vijay is called when the other counter frees, at 9:00 + 2t = 9:25, so t = 12.5 minutes.

• If t > 20, nobody is free at 9:20; Ira and the third 9:00 applicant would both start at 9:00 + t, and Vijay could not start before 9:00 + 2t > 9:40. Not possible.

So each Schengen application takes 12.5 minutes:

Schengen counter	Applicant	Start	End
Counter 1	9:00 slot – 1st	9:00	9:12:30
Counter 2	9:00 slot – 2nd	9:00	9:12:30
Counter 1	9:00 slot – 3rd	9:12:30	9:25
Counter 2	Ira	9:20	9:32:30
Counter 1	Vijay	9:25	9:37:30
Counter 2	Nandini	9:32:30	9:45

Step 3 – Others processing time. At 9:20 exactly six counters were active:

Counters	Status at 9:20 am	Counted
4 US	Busy (US queue never empties)	4
2 UK	No UK applications at all	0
2 Schengen	One processing the 3rd 9:00 applicant, one starting Ira	2
2 Others	Must be idle	0
Total		6

So the two Others counters were idle at 9:20: the two 9:15 Others applicants (started at 9:15) had finished by 9:20. Each Others application therefore takes at most 5 minutes.

Others applicant	Counter	Start	End (processing time t ≤ 5 min)
9:15 slot – 1st	1	9:15	by 9:20
9:15 slot – 2nd	2	9:15	by 9:20
Mahira	1	9:30	by 9:35
Osman	2	9:30	by 9:35

Step 4 – US counters. 5 US applicants arrive every 15 minutes for 4 counters of 10 minutes each.

Slot	Start times of its 5 US applicants	Last one ends
9:00	9:00, 9:00, 9:00, 9:00, 9:10	9:20
9:15	9:15, 9:15, 9:15, 9:20, 9:25	9:35
9:30	9:30, 9:30, 9:30, 9:35, 9:40	9:50
…	same pattern every slot	slot time + 20 min
1:45 pm (last)	1:45, 1:45, 1:45, 1:50, 1:55	2:05 pm

This question

From Step 3, each Others application takes at most 5 minutes. There are 2 × 20 = 40 Others applications, so the total processing time is at most 40 × 5 = 200 minutes.

Hence, the answer is 200.

Question 3 · Caselet DI

Which of the following is the closest to the time when Nandini’s application process got over?

A
9:50 am
B
9:37 am
C
9:35 am
D
9:45 am

Pick an option, then check.

Answer & solution

Answer: Option D

Solving the set

Step 1 – Applications per slot. 20 slots × 10 = 200 applications. 50% are US, so 100 US, i.e., 5 per slot (every slot has the same mix).

Category	Per slot	Reason	Total for the day (20 slots)
US	5	50% of 10	100
Schengen	3	Ira, Vijay, Nandini share a slot (at least 3)	60
Others	2	Mahira, Osman share a slot (at least 2)	40
UK	0	10 − 5 − 3 − 2	0
Total	10		200

Step 2 – Schengen processing time (t). In the 9:00 slot, two Schengen applicants start at 9:00 and finish at 9:00 + t; the third starts at 9:00 + t and finishes at 9:00 + 2t. Ira arrives at 9:20.

• If t ≤ 20, a counter is free by 9:20 and Ira is called at 9:20. Vijay is called when the other counter frees, at 9:00 + 2t = 9:25, so t = 12.5 minutes.

• If t > 20, nobody is free at 9:20; Ira and the third 9:00 applicant would both start at 9:00 + t, and Vijay could not start before 9:00 + 2t > 9:40. Not possible.

So each Schengen application takes 12.5 minutes:

Schengen counter	Applicant	Start	End
Counter 1	9:00 slot – 1st	9:00	9:12:30
Counter 2	9:00 slot – 2nd	9:00	9:12:30
Counter 1	9:00 slot – 3rd	9:12:30	9:25
Counter 2	Ira	9:20	9:32:30
Counter 1	Vijay	9:25	9:37:30
Counter 2	Nandini	9:32:30	9:45

Step 3 – Others processing time. At 9:20 exactly six counters were active:

Counters	Status at 9:20 am	Counted
4 US	Busy (US queue never empties)	4
2 UK	No UK applications at all	0
2 Schengen	One processing the 3rd 9:00 applicant, one starting Ira	2
2 Others	Must be idle	0
Total		6

So the two Others counters were idle at 9:20: the two 9:15 Others applicants (started at 9:15) had finished by 9:20. Each Others application therefore takes at most 5 minutes.

Others applicant	Counter	Start	End (processing time t ≤ 5 min)
9:15 slot – 1st	1	9:15	by 9:20
9:15 slot – 2nd	2	9:15	by 9:20
Mahira	1	9:30	by 9:35
Osman	2	9:30	by 9:35

Step 4 – US counters. 5 US applicants arrive every 15 minutes for 4 counters of 10 minutes each.

Slot	Start times of its 5 US applicants	Last one ends
9:00	9:00, 9:00, 9:00, 9:00, 9:10	9:20
9:15	9:15, 9:15, 9:15, 9:20, 9:25	9:35
9:30	9:30, 9:30, 9:30, 9:35, 9:40	9:50
…	same pattern every slot	slot time + 20 min
1:45 pm (last)	1:45, 1:45, 1:45, 1:50, 1:55	2:05 pm

This question

From the Schengen timeline in Step 2, Nandini is called at 9:32:30 when Ira finishes and is done at 9:32:30 + 12.5 min = 9:45 am.

Hence, the answer is option 4.

Question 4 · Caselet DI

Which of the following statements is false?

A
The application process of Osman was completed before 9:45 am.
B
The application process of Mahira started after Nandini’s.
C
The application process of Osman was completed before Vijay’s.
D
The application process of Mahira was completed before Nandini’s.

Pick an option, then check.

Answer & solution

Answer: Option B

Solving the set

Step 1 – Applications per slot. 20 slots × 10 = 200 applications. 50% are US, so 100 US, i.e., 5 per slot (every slot has the same mix).

Category	Per slot	Reason	Total for the day (20 slots)
US	5	50% of 10	100
Schengen	3	Ira, Vijay, Nandini share a slot (at least 3)	60
Others	2	Mahira, Osman share a slot (at least 2)	40
UK	0	10 − 5 − 3 − 2	0
Total	10		200

Step 2 – Schengen processing time (t). In the 9:00 slot, two Schengen applicants start at 9:00 and finish at 9:00 + t; the third starts at 9:00 + t and finishes at 9:00 + 2t. Ira arrives at 9:20.

• If t ≤ 20, a counter is free by 9:20 and Ira is called at 9:20. Vijay is called when the other counter frees, at 9:00 + 2t = 9:25, so t = 12.5 minutes.

• If t > 20, nobody is free at 9:20; Ira and the third 9:00 applicant would both start at 9:00 + t, and Vijay could not start before 9:00 + 2t > 9:40. Not possible.

So each Schengen application takes 12.5 minutes:

Schengen counter	Applicant	Start	End
Counter 1	9:00 slot – 1st	9:00	9:12:30
Counter 2	9:00 slot – 2nd	9:00	9:12:30
Counter 1	9:00 slot – 3rd	9:12:30	9:25
Counter 2	Ira	9:20	9:32:30
Counter 1	Vijay	9:25	9:37:30
Counter 2	Nandini	9:32:30	9:45

Step 3 – Others processing time. At 9:20 exactly six counters were active:

Counters	Status at 9:20 am	Counted
4 US	Busy (US queue never empties)	4
2 UK	No UK applications at all	0
2 Schengen	One processing the 3rd 9:00 applicant, one starting Ira	2
2 Others	Must be idle	0
Total		6

So the two Others counters were idle at 9:20: the two 9:15 Others applicants (started at 9:15) had finished by 9:20. Each Others application therefore takes at most 5 minutes.

Others applicant	Counter	Start	End (processing time t ≤ 5 min)
9:15 slot – 1st	1	9:15	by 9:20
9:15 slot – 2nd	2	9:15	by 9:20
Mahira	1	9:30	by 9:35
Osman	2	9:30	by 9:35

Step 4 – US counters. 5 US applicants arrive every 15 minutes for 4 counters of 10 minutes each.

Slot	Start times of its 5 US applicants	Last one ends
9:00	9:00, 9:00, 9:00, 9:00, 9:10	9:20
9:15	9:15, 9:15, 9:15, 9:20, 9:25	9:35
9:30	9:30, 9:30, 9:30, 9:35, 9:40	9:50
…	same pattern every slot	slot time + 20 min
1:45 pm (last)	1:45, 1:45, 1:45, 1:50, 1:55	2:05 pm

This question

Check each statement using Steps 2 and 3:

• Osman finished by 9:35, before 9:45 – true.

• Mahira started at 9:30, Nandini at 9:32:30 – so 'Mahira started after Nandini' is false.

• Osman (by 9:35) finished before Vijay (9:37:30) – true.

• Mahira (by 9:35) finished before Nandini (9:45) – true.

Hence, the answer is option 2.

Question 5 · Caselet DI

When did the application processing for all US applicants get over on that day?

A
2:05 pm
B
2:25 pm
C
2:00 pm
D
3:40 pm

Pick an option, then check.

Answer & solution

Answer: Option A

Solving the set

Step 1 – Applications per slot. 20 slots × 10 = 200 applications. 50% are US, so 100 US, i.e., 5 per slot (every slot has the same mix).

Category	Per slot	Reason	Total for the day (20 slots)
US	5	50% of 10	100
Schengen	3	Ira, Vijay, Nandini share a slot (at least 3)	60
Others	2	Mahira, Osman share a slot (at least 2)	40
UK	0	10 − 5 − 3 − 2	0
Total	10		200

Step 2 – Schengen processing time (t). In the 9:00 slot, two Schengen applicants start at 9:00 and finish at 9:00 + t; the third starts at 9:00 + t and finishes at 9:00 + 2t. Ira arrives at 9:20.

• If t ≤ 20, a counter is free by 9:20 and Ira is called at 9:20. Vijay is called when the other counter frees, at 9:00 + 2t = 9:25, so t = 12.5 minutes.

• If t > 20, nobody is free at 9:20; Ira and the third 9:00 applicant would both start at 9:00 + t, and Vijay could not start before 9:00 + 2t > 9:40. Not possible.

So each Schengen application takes 12.5 minutes:

Schengen counter	Applicant	Start	End
Counter 1	9:00 slot – 1st	9:00	9:12:30
Counter 2	9:00 slot – 2nd	9:00	9:12:30
Counter 1	9:00 slot – 3rd	9:12:30	9:25
Counter 2	Ira	9:20	9:32:30
Counter 1	Vijay	9:25	9:37:30
Counter 2	Nandini	9:32:30	9:45

Step 3 – Others processing time. At 9:20 exactly six counters were active:

Counters	Status at 9:20 am	Counted
4 US	Busy (US queue never empties)	4
2 UK	No UK applications at all	0
2 Schengen	One processing the 3rd 9:00 applicant, one starting Ira	2
2 Others	Must be idle	0
Total		6

So the two Others counters were idle at 9:20: the two 9:15 Others applicants (started at 9:15) had finished by 9:20. Each Others application therefore takes at most 5 minutes.

Others applicant	Counter	Start	End (processing time t ≤ 5 min)
9:15 slot – 1st	1	9:15	by 9:20
9:15 slot – 2nd	2	9:15	by 9:20
Mahira	1	9:30	by 9:35
Osman	2	9:30	by 9:35

Step 4 – US counters. 5 US applicants arrive every 15 minutes for 4 counters of 10 minutes each.

Slot	Start times of its 5 US applicants	Last one ends
9:00	9:00, 9:00, 9:00, 9:00, 9:10	9:20
9:15	9:15, 9:15, 9:15, 9:20, 9:25	9:35
9:30	9:30, 9:30, 9:30, 9:35, 9:40	9:50
…	same pattern every slot	slot time + 20 min
1:45 pm (last)	1:45, 1:45, 1:45, 1:50, 1:55	2:05 pm

This question

From Step 4, the last US applicant of every slot (from 9:15 on) finishes 20 minutes after the slot time. The last slot is 1:45 pm, so all US processing ends at 2:05 pm.

Hence, the answer is option 1.

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