Quantitative Aptitude - Arithmetic Ability Questions

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.

Let the edge = a cm   

So increase by 50 % = a + a/2 = 3a/2  

Total surface Area of original cube = 6a2  

TSA of new cube = 63a22 =69a24=  13.5a2   

Increase in area = 13.5a2-6a2 =7.5a2  

 7.5a2 Increase % =7.5a26a2×100 = 125%

Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50

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