Quantitative Aptitude - Arithmetic Ability Questions

n(S) = C220 = 190 

n(E) = C215 = 105 

Therefore, P(E) = 105/190 = 21/38

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

The number can be 4 digited but greater than 4000 or 5 digited.

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x P34 = 3 x 24 = 72. 

5 digited numbers = P55 = 5! = 120 

So the total numbers greater than 4000 = 72 + 120 = 192

We know that perimeter of rectangle = 2(l+b) = 2(8+7)= 30cm
Given perimeter of square is twice the perimeter of rectangle = 2(30) = 60cm
Therefore, side of the square is = 1/4 x 60 = 15cm
Circumference of the required semicircle = πr + 2r = 227× 152 + 2×152 = 23.57 + 15 = 38.57cm.

Number of shares = 44558.25= 540.

Face value = Rs. (540 x 10) = Rs. 5400.

Annual income = Rs.12100*5400 = Rs. 648.

Let the length of the train be L metres

Speed of first train = 360018xL m/hour

Speed of secxond train = 360012xL m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken = 2L500Lx3600 sec

                 =14.4 sec

Let total population = p
number of females = 5p/9
number of males =(p-5p/9) = 4p/9
married females = 30% of 5p/9 = 30x5p/100x9 = p/6
married males = p/6
unmarried males =(4p/9-p/6) = 5p/18
Percentage of unmarried males in the total males = {(5p/18)/4p/9}x100 = (5p/18)x(9/4p) = 125/2 % = 62.5%

If Rahul age is x, then Sachin age is x - 7,

so, (x-7)/x =7/9

 9x - 63 = 7x
 2x = 63
 x = 31.5

So Sachin age is 31.5 - 7 = 24.5

Number of gaps between 41 poles = 40
So total distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute train is moving 2 km/minute.
Speed in hour = 2*60 = 120 km/hour