Quantitative Aptitude - Arithmetic Ability Questions

Total number of elementary events = 50C5
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = 29C2 ways and remaining two in 20C2 ways.

Therfore,favourable number of events = 29C2*20C2

Hence,required probability = 29C2*20C2/50C5 =551 / 15134

T= 6 months = 1/2 year

R = 6%

TD= BD×100100+TR=18.54×100100+12×6=Rs.18

Let inner radius be r metres. Then, 2πr = 440 ; r = 440×744= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

Let the areas of the two parts be x and (700-x) hectares

therefore, x-700-x=15x+700-x2

 ⇒2x-700=70

 ⇒x=385

So, the two parts are 385 and 315.

 Hence, Area of the smaller = 315 hectares

Given the cost price of the articles = Rs. 450

To get overall 20% gain,

Total Selling Price = (20/100) x 450 = 540

One third of the CP = 1/3 x 450 = Rs. 150

But given 1/3 of articles are sold at 10% loss

S.P of 1/3 of articles = 90% of 150

= 90 x 150/100 = 135

Then, S.P on remaining 2/3 goods must be

= 450 - 135 = 405 ...........(1)

CP on remaining goods

= 2/3 x 450 = 300 ............(2)

Profit = SP - CP = 405 - 300 = 105

Profit %  = (105/300) x 100

= 35%.

Let A's 1 day's work=x and B's 1 day's work=y

Then x+y = 1/40 and 20x+60y=1 

Solving these two equations , we get : x= 1/80 and y= 1/80

Therefore B's  1 day work = 1/80

Hence,B alone shall finish the whole work in 80 days