Quantitative Aptitude - Arithmetic Ability Questions

Circumference = No.of revolutions * Distance covered

Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls =5C2 = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21.

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