Quantitative Aptitude - Arithmetic Ability Questions

Let the third number be 100.

Then, the first number = 100 - 20 = 80 and second number = 100 - 30 = 70.

Difference between the first and second number = 80-70 = 10

 Required percentage = 1080 x 100% = 100/8 = 12.5%

We already know that the calendar after a leap year repeats again after 28 years.

Here 1988 is a Leap year, then the same calendar will be in the year = 1988 + 28 = 2016.

?2 = 325 – 144 + 75 + 68
= 468 – 144 = 324
(?) = 18,-18

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3,4),(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(E) = 27.

P(E) = n(E)/n(S) = 27/36 = 3/4.

P(X) = 15, P(Y) =14 , P(Z) = 13

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

=14*13*45+34*13*15+23*14*15+14*13*15

= 460+360+260+160= 1060= 16

Given that, 

Number of windows = 50

Each window covering covers 15 windows

=> 50 windows requires 50/15 window coverings

= 50/15 = 3.333

Hence, more than 3 window coverings are required. In the options 4 is more than 3. 

Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.

speed = 12 km/h = 12×518=103m/s 

distance covered = 20×2×227×50=440007m 

time taken  = distance /speed =  440007×310sec=4400×37×160min=2207min