The bitu-mill group Interview Questions, Process, and Tips

1(100-1)+ 2(100-2)+…..99(100-99)
N=100
solution: N(N-1)
100(99)= 9900

2:3

Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:

D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins

n+p

4400+(97 leap year )=4497

10 x 27=270

160600

A

21

but how?

2400

Gain=(profit/c.p)*100

Gain =selling price-cost price =900-675
=225.
Gain=(225/675)*100.
=33.3%

I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps

The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.

The time to travel 400m at 20mps is 20secs