Quantitative Aptitude - Arithmetic Ability Questions

Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,

part filled by tap X in 1 minute = 1/30

part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,

part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute = [1/30 +1/40 - 1/60] = 5/120 = 1/24

i.e., the tank can be filled in 24 minutes.

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%

Ans. 
5 = 3 + (13 + 1)
15 = 5 + (23 + 2)
45 = 15 + (33 + 3)
113 = 45 + (43 + 4)
? = 113 + (53 + 5), i.e. ? = 243

Relative speed = 60 + 90 = 150 km/hr.

= 150 x 5/18 = 125/3 m/sec.

Distance covered = 1.10 + 0.9 = 2 km = 2000 m.

Required time = 2000 x 3/125 = 48 sec.

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.

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.

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