Bank Clerk Questions

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

 = 1-16C1× 14C1×12C1×10C116C4=1539

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.

An English word with five letters in which if you remove the last four letters, still pronounced the same is QUEUE.

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%

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.

Given 4men, 12 women and 20 children work for  2 days.

Workdone for 2 days by 4men, 12 women and 20 children = 4 x 26 x 12 + 12 x 28 x 18 + 20 x 218 x 10 = 12

Therefore, remaining work = 1 - 12 = 12

To complete the same work by only men in 1 day,

We know that M1 x D1 = M2 x D2

Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1

12 x 6 = M x 1

=> M = 12 x 6 = 72

=> But the remaining work = 1/2

Men required => 1/2 x 72 = 36

Only men required to Complete the remaining work in 1 day = 36.

Let us name the trains as A and B.

Then, (A's speed) : (B's speed)

= √b : √a = √16 : √9 = 4:3