Bank PO Questions

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.

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%

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

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

 = 1-16C1× 14C1×12C1×10C116C4=1539

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

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.

Given lb =96

Area of pathway = [(l-4)(b-4)-lb] = 16-4(l+b)

Which cannot be determined. so, data is inadequate.

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51).

Therefore, HCF of 923, 1207 and 1349 is 71.