Quantitative Aptitude - Arithmetic Ability Questions

Let x = the youngest child. Each of the other four children will then be x+3, x+6, x+9, x+12. 

We know that the sum of their ages is 50 .

so, x+(x+3)+(x+6)+(x+9)+(x+12) = 50

=>  x= 4

 The youngest child is 4 years old.

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 year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

The day on 5th March, 2005 will be 1 day beyond the day on 5th March, 2004.

Given that, 5th March, 2005 is Monday.

5th March, 2004 is Sunday (1 day before to 5th March, 2005).

Ram age when he was born = 0 years
=> His brother's age = 6 year
=> His father's age = brother age + 32years = 6+32 = 38
=> His mother's age = father's age - 3 = 35
So sister's age = 35-25 = 10years.

R =OA = OB = Radius of hemisphere.
= Radius of cylinder ABCD.
Height of the cylinder = BC = radius

Volume of cylinder/Volume of hemisphere = πr323πr3