Quantitative Aptitude - Arithmetic Ability Questions

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.

Given 2000 ---- 54 days

The provisions for 2000 men for 39 days can be completed by 'm' men for only 20 days.

i.e, 2000 ----- 39 days == m ---- 20 days

=> m x 20 = 2000 x 39

m = 3900

So total men for 20 days is 3900

=> 2000 old and 1900 new reinforcement.

Hence, reinforcement = 1900.

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

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

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