The median for the data 2, 4, 6, 8, 10, 12, 14 is

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

Any of the 4 colour can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The third stripe can again be coloured in 3 ways (we can repeat the colour of the first stripe but not use the colours of the second stripe).
Similarly, There are 3 ways to colour each of the remaining stripes.
? The number of ways the flag can be coloured is 4(3)⁵ = (12) (3)⁴ = 12 x 81

Anuj reached the place of meeting at 8:15 h, he reached 30 min earlier than the man who was 40 min late, Hence, the scheduled time of the meeting was 8:05 h.

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.