Hif australia Interview Questions, Process, and Tips

Second

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

The number of ways of selecting a group of eight is
5 men and 3 women=5C5​×6C3​ =20
4 men and 4 women=5C4​×6C4​ =75
3 men and 5 women=5C3​×6C5​=60
2 men and 6 women=5C2​×6C6​=10
Thus the total possible cases is 20+75+60+10=165.

let the for digit number be = pqrs
p=q/3 ==>q = 3p
r=p+q=p+3p =4p
s=3q = 3(3P)=9p
number:
p 3p 4p 9p
let p=1 answer is 1349
if it 2 answer 2 6 8 18
so it becomes five digit number so correct answer is 1349

Either 4 or (-)4

Day 1: 30-3ft = 27+2ft = 29
Day 8: 23 jumps 3ft = 20 slips 2ft = 22
Day 16: 13 jumps 3ft = 10 slips 2ft = 12
Day 24: 3 jumps 3ft = Out?

21 days

17

C. Every real number is a rational number.

3

C)1440
2*2=4
4*3=12
12*4=48
48*5=240
240*6=1440

let x be sum.
SI of 18% for 1 year= x*18/100=0.18x;
SI of 18% for 2 years is 0.36x;

SI of 12% for 1 year= x*12/100=0.12x;
SI of 12% for 2 years is 0.24x;

Given, 0.36x-0.24x=840
0.12x=840
x=840/0.12=7000.

NCPQJG