Question:hard

The regular mathematics faculty could not teach because of being sick. As a stop-gap arrangement, different visiting faculty taught different topics on 4 different days in a week. The scheduled time for class was 7:00 am, with a maximum permissible delay of 20 minutes. The monsoon made the city bus schedules erratic, so the classes started at different times on different days.

Mr. Singh did not teach on Thursday. Calculus was taught in the class that started at 7:20 am. Mr. Chatterjee took the class on Wednesday, but he did not teach probability. The class on Monday started at 7:00 am, but Mr. Singh did not teach it. Mr. Dutta did not teach ratio and proportion. Mr. Banerjee, who did not teach set theory, taught a class that started five minutes later than the class in which probability was taught. The teacher in Friday's class taught set theory. Wednesday's class did not start at 7:10 am. No two classes started at the same time.

77. The class on Wednesday started at:

Show Hint

Fix Mr. Singh to Friday (set theory) and Mr. Chatterjee to Wednesday first, then use the five-minutes-after-the-probability-class clue to place Mr. Dutta on Monday and Mr. Banerjee on Thursday.
Updated On: Jul 10, 2026
  • 7:05 am and the topic was ratio and proportion.
  • 7:20 am and the topic was calculus.
  • 7:00 am and the topic was calculus.
  • 7:20 am and the topic was calculus.
Show Solution

The Correct Option is B

Solution and Explanation

This is a 4-day, 4-teacher, 4-topic, 4-timing puzzle. Instead of building the schedule from scratch, this method checks each of the four listed timings against the clues directly and eliminates the ones that break a rule.

  • Mr. Singh must teach on Friday: he does not teach Monday or Thursday, and Wednesday belongs to Mr. Chatterjee, so Friday is the only day left for him, and Friday's topic is set theory.
  • Mr. Dutta must teach probability on Monday at 7:00 am: Mr. Chatterjee, Mr. Singh and Mr. Banerjee are all ruled out of teaching probability (Chatterjee by direct statement, Singh because he has set theory, Banerjee because his class starts five minutes after the probability class), leaving only Dutta; putting him on Thursday would force the probability class to start at 6:55 am, which breaks the 7:00 to 7:20 am window, so he must be on Monday.
  • Mr. Banerjee then teaches Thursday at 7:00 am + 5 minutes = 7:05 am, and since calculus starts at 7:20 am while Banerjee's class starts at 7:05 am, Banerjee's topic must be ratio and proportion, not calculus.

By elimination, the only slot left is Mr. Chatterjee's Wednesday class, and the only topic left is calculus, so it must start at 7:20 am, matching the requirement that calculus starts at 7:20 am, and not clashing with the rule that Wednesday did not start at 7:10 am.

  1. 7:05 am, ratio and proportion: wrong topic and wrong time for Wednesday, since Wednesday's topic is calculus.
  2. 7:20 am, calculus: matches every clue worked out above.
  3. 7:00 am, calculus: 7:00 am is already taken by Monday's class, and no two classes share a start time, so this is impossible.
  4. 7:20 am, calculus: identical in content to the option above; this repeated option is exactly why the question could not be scored with one unique correct choice and was officially withdrawn.

Let's summarize:

  • Monday, Wednesday, Thursday and Friday are the four visiting-faculty days, taken by Dutta, Chatterjee, Banerjee and Singh respectively.
  • The topics in that same order are probability, calculus, ratio and proportion, and set theory.

So the Wednesday class started at 7:20 am with calculus as the topic, even though the paper's own duplicated option made this question a dropped one.

Was this answer helpful?
0

Top Questions on Complex Arrangement

Questions Asked in XAT exam

Read the scenario and answer the question:
The upper hinge of a dataset is the median of all the values to the right of the median of the dataset in an ascending arrangement, while the lower hinge of the dataset is the median of all the values to the left of the median of the dataset in the same arrangement.
For example, consider the dataset 4, 3, 2, 6, 4, 2, 7. When arranged in the ascending order, it becomes 2, 2, 3, 4, 4, 6, 7. The median is 4 (the bold value), and hence the upper hinge is the median of 4, 6, 7, i.e., 6. Similarly, the lower hinge is 2. 
A student has surveyed thirteen of her teachers, and recorded their work experience (in integer years). Two of the values recorded by the student got smudged, and she cannot recall those values. All she remembers is that those two values were unequal, so let us write them as A and B, where A < B. The remaining eleven values, as recorded, are: 5, 6, 7, 8, 12, 16, 19, 21, 21, 27, 29. Moreover, the student also remembers the following summary measures, calculated based on all the thirteen values: 
Minimum: 2
Lower Hinge: 6.5 
Median: 12 
Upper Hinge: 21 
Maximum: 29
View Solution
Read the following scenario and answer the questions that follow.
A T20 cricket match consists of two teams playing twenty overs each, numbered 1 to 20. The runs scored in any over is a non-negative integer. The run rate at the end of any over is the average runs scored up to and including that over, i.e., the run rate at the end of the kth over is the average number of runs scored in overs numbered \(1, 2, …, k,\) where \(1 \leq k \leq 20, k\) a positive integer.
The following table indicates the run rate of a team at the end of some of the overs during a T20 cricket match (correct up to 2 decimal places), where \(1 \leq N – 2 < N + 6 \leq 20\), N a positive integer. It is also known that the team did not score less than 6 runs and more than 15 runs in any over.
Over NumberRun Rate
N-28.00
N7.43
N+28.11
N+48.45
N+68.08
View Solution