Aptitude and Reasoning Questions

Daughter of your father — Your sister. So, the person's mother is woman's sister or the woman is person's aunt. 

Let their initial investments be x, 3x and 5x respectively. Then,

A:B:C = (x*4+2x*8) : (3x*4+(3x/2)*8) : (5x*4+(5x/2)*8)

20x : 24x : 40x = 5 : 6 : 10          

In given Series, all the numbers except 54 are multiples of 5.

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days

(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 dayss = (21 weeks + 1 day) = 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.

Given day is Sunday

Average of 20 numbers = 0. 

Sum of 20 numbers = (0 * 20) = 0. 

It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (- a).

In this type of problems the formuae is
(5*x+ or - t)*12/11
Here x is replaced by the first interval of given time. Here x is 5.
t is spaces apart

  Case 1 : (5*x + t) * 12/11

 (5*5 + 3) * 12/11

 28 * 12/11 = 336/11= 31511 min

therefore the hands will be 3 min apart at 31 5/11 min past 5.

Case 2 : (5*x - t) * 12/11

(5*5 -3 ) * 12/11

 22 *12/11 = 24 min 

 therefore the hands will be 3 min apart at 24 min past 5

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 

Jan   Feb   Mar   Apr   May   Jun   Jul  

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 

Total number of odd days = (0 + 2) = 2.  

Required day was 'Tuesday'.