Average Questions

 Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given.

So, the data provided is inadequate.

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]

Clearly, we  have : x = (3y + 3z ) / 6 or 2x = y + z.

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.

Numbers of students in section A = x
∴ Numbers of students in section B and C = (100 – x)
∴ x 70 + (100 – x) 87.5 = 84 100
=> 70x + 87.5 100 – 87.5x = 8400
=> 8750 – 17.5x = 8400
=> 17.5x = 8750 – 8400 => x = 20.

Since we know that the difference b/w the age of any two persons remains always constant, while the ratio of their ages gets changed as the time changes.

so, if the age of his child be x (presently)

Then the age of  wife be x + 26 (presently)

Thus the total age = x + ( x + 26) = 32  [∵252-220 =32]

=> x = 3

Therefore, The age of her child is 3 years and her self  is 29 years. Hence her age at the time of the birth of her child was 26 years.

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.

Total of results = 50 x 9 = 450

Total of first four results = 52 x 4 = 208

Total of Last four results = 49 x 4 = 196

Hence the fifth result = 450 - (208 + 196) = 46