Question

What is the maximum number of patients that the clinic can cater to on any single day?

• A. 30
• B. 12
• C. 31
• D. 15
• A. Dr. Wayne
• B. Dr. Kane
• C. Both Dr. Wayne and Dr. Kane
• D. Dr. Ben
• A. Monday
• B. Friday
• C. Wednesday
• D. Same duration on all three days
• A. 15
• B. 30
• C. 10
• D. 0

Answer 1

The Correct Option is C

The problem requires determining the maximum number of patients the clinic can handle in a single day from 9 a.m. to 11:30 a.m. This is equivalent to 150 minutes, as calculated below: 

• Duration from 9 a.m. to 11:30 a.m.: 2 hours 30 minutes = 150 minutes

Since three doctors are sharing the available rooms, it is essential to compute the number of patients each doctor can see individually, assuming optimal room allocation:

• Dr. Ben: Sees each patient for 10 minutes.
  • Maximum patients: \( \frac{150 \text{ minutes}}{10 \text{ minutes per patient}} = 15 \text{ patients} \)
• Dr. Kane: Sees each patient for 15 minutes.
  • Maximum patients: \( \frac{150 \text{ minutes}}{15 \text{ minutes per patient}} = 10 \text{ patients} \)
• Dr. Wayne: Sees each patient for 25 minutes.
  • Maximum patients: \( \frac{150 \text{ minutes}}{25 \text{ minutes per patient}} = 6 \text{ patients} \)

Considering simultaneous use of rooms and continuous patient handling, let's allocate time to maximize output:

1. Allocate the doctors into rooms according to the shortest available consultation time for maximal overlap.
2. In 150 minutes, manage room turnover so that the next available doctor gets the room as soon as one becomes free.

The combination that provides the largest number of patients under room constraints and time is:

• Dr. Ben (15 patients) manages to see patients in rotation due to quick time per patient.
• Dr. Kane (10 patients) complements Dr. Ben by also using quick turnovers compared to Dr. Wayne.
• Interleave Dr. Wayne (6 patients) in places to utilize room gaps effectively, but this doctor contributes the least due to longer times.

By optimized allocation using room freedoms and patients ranking upon room availability, we calculate effects:

• Maximum patients seen by overlapping treatments: 15 (Dr. Ben) + 10 (Dr. Kane) + 6 (Dr. Wayne) = 31 patients.

Doctor	Number of Patients
Dr. Ben	15
Dr. Kane	10
Dr. Wayne	6

Thus, the correct answer is that the clinic can cater to 31 patients on any single day based on the given situation and optimal use of room and time resources.

Answer 2

To determine which doctor earns the maximum consultation charges on a particular Saturday when the queue is never empty, we need to analyze their patient handling capabilities and earnings. The clinic operates for 2.5 hours (150 minutes) from 9 a.m. to 11.30 a.m.

• Dr. Ben: Consultation Time: 10 minutes per patient; Fee: Rs. 100/- per patient.
  Maximum Patients: 150/10 = 15 patients;
  Total Earnings: 15 x 100 = Rs. 1500/-.
• Dr. Kane: Consultation Time: 15 minutes per patient; Fee: Rs. 200/- per patient.
  Maximum Patients: 150/15 = 10 patients;
  Total Earnings: 10 x 200 = Rs. 2000/-.
• Dr. Wayne: Consultation Time: 25 minutes per patient; Fee: Rs. 300/- per patient.
  Maximum Patients: 150/25 = 6 patients;
  Total Earnings: 6 x 300 = Rs. 1800/-.

Comparing the total earnings of each doctor, Dr. Kane earns the most at Rs. 2000/- on that particular Saturday. Thus, the answer is: Dr. Kane.

Answer 3

To determine the day Mr. Singh was at the clinic for the maximum duration, we need to calculate the time he spent on each day based on the doctors' schedules. Mr. Singh, with token number 13, arrived at 8:50 a.m. When the clinic opens at 9 a.m., his wait time depends on the doctors' consultation times per patient.

Doctor	Consultation Time	Charges	Room
Dr. Ben	10 min	Rs. 100	1
Dr. Kane	15 min	Rs. 200	2
Dr. Wayne	25 min	Rs. 300	3

From 9:00 a.m., doctors start consulting patients. The sequence in which patients are attended is based on token numbers and room availability.

• Dr. Ben (Room 1): Each patient takes 10 minutes. Patients 1, 4, 7, 10 are attended by 9:40 a.m..
• Dr. Kane (Room 2): Each patient takes 15 minutes. Patients 2, 5, 8, 11 are attended by 10:00 a.m.
• Dr. Wayne (Room 3): Each patient takes 25 minutes. Patients 3, 6, 9, 12 are attended by 10:30 a.m..

Mr. Singh, with token number 13, will be attended next. As it is calculated:

• Monday: Dr. Ben will see the 13th patient in room 1 at 10:00 a.m., completing the consultation by 10:10 a.m..
• Wednesday: Dr. Kane would be free around 10:15 to take Mr. Singh if another followed order in practitioner or assistant's shift exists.
• Friday: Filling back to Dr. Wayne for Mr. Singh in place of any medical assistant and current shifted time circumstance may slightly vary toward 10:25 a.m.

Thus, Mr. Singh was at the clinic for the maximum duration on Monday, as there was the earliest start without necessity for significant wait or patient lineup adjustment could facilitate earlier on-site consultation, conjoined with every other noted factor in practice.

Answer 4

In this problem, we need to determine the time when all three doctors are simultaneously free. Let's analyze the given information:

1. Three doctors with different consultation times:

• Dr. Ben: 10 minutes per patient. 
• Dr. Kane: 15 minutes per patient.
• Dr. Wayne: 25 minutes per patient.

2. The clinic's operating hours are from 9 a.m. to 11:30 a.m.

3. Patients start arriving at 9 a.m., with two patients waiting. More patients arrive at 15-minute intervals starting from 9:15 a.m.

4. Patients go to the available room with the lowest number.

We need to calculate when all three doctors will be free simultaneously.

Initially, at 9 a.m., two patients enter rooms 1 and 2 to see Dr. Ben and Dr. Kane, respectively.

At 9:10 a.m., Dr. Ben finishes with the first patient.

At 9:15 a.m., Dr. Kane finishes with the second patient, and two more patients arrive. Room 1 becomes free, the next patient enters Dr. Ben's room.

At 9:20 a.m., Dr. Ben finishes with the next patient and another patient enters.

At 9:25 a.m., Dr. Wayne finishes and another patient enters.

When does the cycle repeat?

• Dr. Ben finishes every 10 minutes.
• Dr. Kane finishes every 15 minutes.
• Dr. Wayne finishes every 25 minutes.

The Least Common Multiple (LCM) of these times is the earliest time when they all will be free. Calculating the LCM: The prime factors are 2¹ * 5² = 50 minutes.

Since the calculation starts at 9 a.m., after 50 minutes, the time will be 9:50 a.m.

Let's verify if they are free at 9:50 a.m.:

• Dr. Ben's cycle: 9:00, 9:10, ..., 9:50.
• Dr. Kane's cycle: 9:00, 9:15, ..., 9:45, 10:00.
• Dr. Wayne's cycle: 9:00, 9:25, 9:50.

Dr. Wayne finishes his patient at 9:50 a.m., but Dr. Kane's next simultaneous finish is at 10:00 a.m. Thus, all doctors aren't free simultaneously. Repeating this for subsequent intervals, they do not align simultaneously by the clinic closing time.

Conclusively, the total time during which all doctors are simultaneously free is 0 minutes.