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 average expenditure was Rs. R

13 x 79 + 6x(R + 4) = 19R

=> R = Rs. 80.84

Total money = 19 x 80.84 = Rs. 1536.07.

Let the number of boys = x

From the given data,

=> [21x + 16(18)]/(x+16) = 19

=> 21x - 19x = 19(16) - 16(18)

=> 2x = 16

=> x = 8

Therefore, the number of boys in the class = 8.

Given,

mean of 100 observations is 40

=> Total of 100 observations = 40 x 100 = 4000

84 is misread as 48

=> Difference = 84 - 48 = 36

=> Now, new total of 100 observations = 4000 + 36 = 4036

Correct Mean = 4036/100 = 40.36

Marks in English  = 85×7 – 83×6

= 595 – 498

= 97

Assume Hebah has Rs. M

Since 25% more money at Poonam

=> Money at Poonam = M + (25% of M)

=> Money at Poonam = M + 0.25M = Rs. 1.25M

Money with Navaneet is thrice the money with Poonam,

=> Money at Navaneet = 3(1.25M) = Rs. 3.75M

Now, sum of all Money = (M + 1.25M + 3.75M) = Rs. 6M

But given the average of the money is Rs. 350

=> 6M/3 = 350

=> 2M = 350

=> M = 350/2 = Rs. 175

=> Amount of money Navaneet has = Rs. 3.75M = 3.75 x 175 = Rs. 656.25.

Using Trial and error method,

we get the number of girls in the hall = 78

Let the avg age of 7 boys be 'p' years
Let the age of the girl be 'q' years
From given data,
The age of 7 boys = 7p years
Now the new average = (p + 1) when 22 yrs is replaced by q
Now the age of all 7 will become = 7(p + 1) yrs
Hence, 7p - 22 + q = 7(p + 1) yrs
7p - 22 + q = 7p + 7
q = 22 + 7 = 29

Therefore, the age of girl = q = 29 years.

Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x.
Let k be the quantity of third variety of oil in the final mixture.
Let the ratio of initial quantity of oils be 'y'
From given details,
4x + 5 = 5y ..... (1)
5x + 10 = 7y .....(2)
8x + k = 9y ......(3)

By solving (1) & (2), we get
x = 5 & y = 5

From (3) => k = 5

Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.