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?

Let the three numbers be x, y, z.

From the gien data,

x = 2y ....(1)

x = z/2 => z = 2(2y) = 4y .....(From 1)  ...........(2)

Given average of three numbers = 56

Then,

$$\begin{gathered} x + y + z3 = 56 \\ 2y + y + 4y3 = 56 (From 1 & 2) \\ 7y = 56 x 3 \\ y = 24 \end{gathered}$$

Now,

x = 2y => x = 2 x 24 = 48

z = 4y = 4 x 24 = 96

Now, the highest number is z = 96 & smallest number is y = 24

Hence, required sum of highest number and smallest number

= z + y

= 96 + 24

= 120.

Given that, 

Number of windows = 50

Each window covering covers 15 windows

=> 50 windows requires 50/15 window coverings

= 50/15 = 3.333

Hence, more than 3 window coverings are required. In the options 4 is more than 3. 

Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.

Given number of boxes = 14

Number of workers = 4

Now, number of whole boxes per worker = 14/4 = 3.5 

Hence, number of whole boxes per each coworker = 3

Given Five boxes of bananas sell for Rs. 30.

=> 1 Box of Bananas for = 30/5 = Rs. 6

Then, for Rs. 9

=> 9/6 = 3/2 = 1.5

Hence, for Rs. 9, 1.5 box of bananas can buy.

The third proportional of two numbers p and q is defined to be that number r such that

p : q = q : r.

Here, required third proportional of 10 & 20, and let it be 'a'

=> 10 : 20 = 20 : a

10a = 20 x 20

=> a = 40

Hence, third proportional of 10 & 20 is 40.

Given total number of passengers in the bus = 45

First average weight of 45 passengers = 52 kgs

Average weight of 5 passengers who leave bus = 48

Average weight of 5 passengers who joined the bus = 54

Therefore, the net average weight of the bus is given by

$$\begin{gathered} = 45 x 52 - 5 x 48 + 5 x 5445 \\ = 237045 \\ = 1583 \\ = 52.66 kgs. \end{gathered}$$

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers.

Given that their average is 52

Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6

6x + 30 = 312

x = 47

So Product = 47 × 57 = 2679