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.

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.

$$\begin{gathered} ?2=400+16650x100 \\ ?= 8 \end{gathered}$$

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.

Let the total population be 'p'

Given ratio of male and female in a city is 7 : 8

In that percentage of children among male and female is 25% and 20%

=> Adults male and female % = 75% & 80%

But given adult females is = 156800

=> 80%(8p/15) = 156800

=> 80 x 8p/15 x 100 = 156800

=> p = 156800 x 15 x 100/80 x 8

=> p = 367500

Therefore, the total population of the city = p = 367500

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.