Quantitative Aptitude - Arithmetic Ability Questions

(150 x 460)/100 + (24 x 650)/100

= 690+156

= 846

Let the numbers be x, x + 1, x + 2,  x + 3, x + 4, x + 5 and x + 6,

Then   (x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)) / 7 = 20.

or  7x + 21 = 140 or 7x = 119 or x =17.

Latest number = x + 6 = 23.

Here the given series is 123 112 133 112 143

It follows a pattern that,

123

123 - 11 = 112

112 + 21 = 133

133 - 31 = 102 (not 122)

102 + 41 = 143

Hence, the odd man in the given number sequence is 122 instead of 102.

In each term second two numbers are multiples of the first number. But in 4-15-24 15 is not multiple of 4.

= (69x85/3x100) + 10.7

= 30.25.

A : B = 100 : 90.

A : C = 100 : 87.

B/C = (B/A*A/C) = (90/100*100/87) = 30/29

When B runs 30 m, C runs 29 m.

When B runs 180 m, C runs (29/30*180)m =174m

 B beats C by (180 - 174) m = 6 m.

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'.