Quantitative Aptitude - Arithmetic Ability Questions

n(S) = C220 = 190 

n(E) = C215 = 105 

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

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%

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) 

 = 1-16C1× 14C1×12C1×10C116C4=1539

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

speed of the man = 4 x (5/18) m/sec = 10/9 m/sec

time taken = 3 x 60 sec = 180 sec

length of diagonal = speed x  time = (10/9) x 180 = 200m

Area of the field = (1/2) x (dioagonal)²

= (1/2) x 200 x 200 sq.m = 20000sq.m

The day of the week repeats every 7 days.

Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.

Hence, Friday is on the 63rd day, as 63 is multiple of 7.

Now, the required day of the week on the 65th day is Sunday.

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

Total ages of 80 boys = 15 x 80 = 1200 yrs.
Total age of 16 boys = 15 x 16 = 240 yrs
Total age of 25 boys = 14 x 25 = 350 yrs.
Average age of remaining boys = 1200 - (240+350) / 80 - (25+15) = 610/41 = 15.25 yrs.