Reaching the place of meeting on Tuesday 15 min before 8:30 h, Anuj found himself half an hour earlier than the man who was 40 min late. What was the scheduled time of the meeting?

4th observation i.e., 8 is the median.

27 is in the middle, so is the median and its the mean too.

Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4, and 5) with either blue or red balls 2⁵ = 32.
Two adjacent boxes with blue can be obtained in 4 ways, i.e., (12), (23), (34) and (45).
Three adjacent boxes with blue can be obtained in 3 ways, i.e., (123), (234)and (345). Four boxes with blue can be obtained in 2 ways, i.e., (1234) and (2345). And five boxes with blue can be got in 1 way. Hence, the number of ways of filling the boxes such that adjacent boxes have blue
= (4 + 3 + 2 + 1) = 10.
Hence, the number or ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22.

The digit in the unit's place should be greater than that in the tens' place.
Hence, if digit 5 occupies the unit place, then remaining four digits need not to follow any order,hence required number = 4!
However, if digit 4 occupies the unit place then 5 cannot occupy the ten;s position. Hence, digit at the ten's place and it will be filled by the digit 1, 2 or 3. This can happen in 3 ways. The remaining 3 digit can be filled in the remaining three place in 3! ways.
Hence, in all, we have (3 x 3!) numbers ending in 4. Similarly, if we have 3 in the unit's place and it will be either 1 or 2. this can happen in 2 ways. Hence, we will have (2 x 3! ) number ending in 3 . Similarly, we can find that there will be 3! numbers ending in 2 and no number with 1. Hence, total number of numbers
= 4! + (3) x 3! + (2 x 3!) + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60

The available digits are 0, 1, 2,...., 9. The first digit can be chosen in 9 ways (0 not acceptable), the second digit can be accepted in 9 ways (digit repetition not allowed). Thus, the code can be made in 9 x 9 = 81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways
Total number of ways confusion can arise
= 4 x 3 = 12
Thus, the ways in which no such confusion arise = 81-12 =69

Number of buzzes in a day
       = 12(12 + 1)/2 x 2 = 156

In 12 h, they are at a right angles, 22 times.
So, in 24 h, they are at right angles, 44 times.

From the properties of the clock, we know that hands of a clock coincide once in every hour but between 11 O'clock and 1 O'clock, they coincide only once. Therefore, the hands of a clock coincide 11 time in every 12 h.
Hence, they will coincide (11 x 2) = 22 times in 24 h.
89 h of this clock = 90 h of correct clock
Therefore, it is clear that in 89 h this clock losses 1 h and hence, the correct time is 11 : 00 pm when this clock shown 10 : 00 pm.

The hands of a clock point towards each other 11 times in every 12 h (because between 5 and 7, they point towards each other only once at 6 O'clock).
Therefore, in a day, the hands points 22 times in all, towards each other.

Here, 363/7 = 51 weeks + 6 days