Logical Reasoning

Optimization

Medium

Three doctors, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges Rs. 100/-. Dr. Kane sees each patient for 15 minutes and charges Rs. 200/-, while Dr. Wayne sees each patient for 25 minutes and charges Rs. 300/-.

The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.

Room No.	Monday & Tuesday	Wednesday & Thursday	Friday & Saturday
1	Ben	Wayne	Kane
2	Kane	Ben	Wayne
3	Wayne	Kane	Ben

The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.

On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.

On a slow Thursday, only two patients are waiting at $9$ a.m. After that two patients keep arriving at exact $15$ minute intervals starting at $9:15$ a.m. -- i.e. at $9:15$ a.m., $9:30$ a.m., $9:45$ a.m. etc. Then the total duration in minutes when all three doctors are simultaneously free is

✅ Correct Option: 1

Slide 1/2

Understanding the set :-

This set is a logical reasoning and scheduling problem

Here we are given information about a clinic -

a) Open: Monday to Saturday

b) Time: 9:00 AM – 11:30 AM → 150 minutes total per day

Also, we are given details of each doctor as -

Doctor	Time per Patient	Charge per Patient
Ben	10 minutes	₹100
Kane	15 minutes	₹200
Wayne	25 minutes	₹300

=> So, in 150 minutes:

a) Ben can see 15 patients

b) Kane can see 10 patients

c) Wayne can see 6 patients

There is a room schedule as -

Room No.	Mon & Tue	Wed & Thu	Fri & Sat
1	Ben	Wayne	Kane
2	Kane	Ben	Wayne
3	Wayne	Kane	Ben

So, the doctors rotate across rooms based on days.

Also, there is a token system -

i) Patients enter the clinic and receive tokens (1, 2, 3, ...).

ii) When any doctor becomes free, the next patient in the queue goes in.

iii) If multiple rooms are free, patient with lowest token chooses the smallest numbered room.

Example:

If token 7 is next and rooms 1 and 3 are free, token 7 → Room 1

and, Token 8 → Room 3

Now, lets try to solve the set using these clues.

movement on Thursday as per condition

Wayne	Ben	Kane
Token no	Time	Token no	Time	Token no	Time
1	9:00-9:25	2	9:00-9:10	3	9:10-9:30
5	9:30-9:55	6	9:30-9:40	4	9:15-9:30
9	10:00-10:25	10	10:00-10:10	8	9:45-10:00

As shown above token number $11,12$ will have same movement as of token number $3$ and $4$

and the same sequence will follow between $10:11$ and between $11:00-11:30$ .

Hence there is no time duration in which all the three doctors are simultaneously free.

Question 17

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

Question 18

The queue is never empty on one particular Saturday. Which of the three doctors would earn the maximum amount in consultation charges on that day?

Question 19

Mr. Singh visited the clinic on Monday, Wednesday, and Friday of a particular week, arriving at 8:50 a.m. on each of the three days. His token number was 13 on all three days. On which day was he at the clinic for the maximum duration?

Solution

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