Consider the following frequency distribution:

x     f

8     6

5     4

6     5

10    8

9     9

4     6

7     4

What is the median for the distribution?

The 5 consecutive odd numbers whose average is k are  (k-4), (k-2), k, (k+2), (k+4)

Again the average of (k-4), (k-2), k, (k+2), (k+4), (k+6), (k+8) is (k+2)

M+A+S3=11111 

also S+M2=11111   (∵ M - 11111 = 11111 - S) 

=>A = 11111

Amithab's total expenditure for Jan - June  = 4200 x 6 = 25200

Expenditure for Feb - June = 25200-1200 = 24000

Expenditure for the months of Feb - July = 24000 + 1500 =25500

The average expenditure = 25500/6 = 4250

It is the same as a person with 20 years more age  replaces an existing person of the group ( or village)

Since the total age of the village having n persons, is being increased by 20 years and the average age of village is being increased by 1 year, hence there are total 20 people in the village.

Alternatively :  ( n x 42 ) + 20 = ( n x 43 )

                          => n=20

Use unitary method:

In  hours, 123 typists can type 984 paper

In 1 minute, 123 typists can type   = 246 papers

In 1 minute, 1 typist can type  =2 papers      [ Hint  hours = 4 minutes]

Four tenths = 0.4

Five thousandths = 0.005

The average is (0.4 + 0.005)/2 = 0.2025

You could solve this by drawing a Venn diagram. A simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking French to find the number taking only French. Likewise find those taking only German. Then we have:Total = only French + only German + both + neither

78 = (41-9) + (22-9) + 9 + neither.

Not enrolled students = 24

One way to deal with fractions is to convert them all to decimals. 

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7.