Bank Clerk Questions

Days in 4 years => 

Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years.

4 years => 365 + 365 + 365 + 366(Leap year)

4 years => 730 + 731 = 1461

Therefore, Number of Days in 4 Years = 1461 Days.

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.

Let the number be n. Then,

50% of n - 35% of n = 24

50/100 n - 35/100 n = 24

n = (24 x 100)/15 = 160.

We have P = 8748, R = -10 and n = 3
Therefore, the purchase price of the machine
= P1+R100n= 87481-101003
= 8748 x (100/90) x (100/90) x (100/90)
= 12,000.

Therefore, the purchase price of the machine was Rs. 12,000.

Given X can do in 10 days 

=> 1 day work of X = 1/10

Y can do in 15 days 

=> 1 day work of Y = 1/15

1day work of (X + Y) = 1/10 + 1/15 = 1/6

Given they are hired for 5 days

 => 5 days work of (X + Y) = 5 x 1/6 = 5/6

 Therefore, Unfinished work = 1 - 5/6 = 1/6 

 It can be possible in only B.C. i.e, Before Christ. That was about 2000 years ago.

 Given, 

 In 1990 B.C = 2000 + 1990 = 3990 years ago, a person is 15 years Old  

=> In 1995 B.C = 2000 + 1995 = 3995 years ago, the same person will be 10 years Old 

Therefore, In 2000 B.C = 2000 + 2000 = 4000 years ago, the same person will be 5 years Old.

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.