Aptitude and Reasoning Questions

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

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

Father's father ---->Grandfather;  Grandfather's granddaughter -----> Sister; Sister's husband ---->Brother-in-law. So, Y is X's Brother-in-law.

A red ball can be drawn in two mutually exclusive ways

 (i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore 

P(E1) = 1/2  and  P(E2) = 1/2

Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7

  PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6

 Using the law of total probability, we have 

 P(red ball) = P(A) = PE1×PAE1+PE2×PAE2 

                          = 12×47+12×26=1942