Quantitative Aptitude - Arithmetic Ability Questions

The pattern followed is *1.5 , * 2 , *2.5 , *3 , *3.5

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.

Double Difference pattern is followed in the above series. +2, +5, + 10, +17, +26

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.

T= 6 months = 1/2 year

R = 6%

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

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

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