Chd living Interview Questions, Process, and Tips

Second

4

n+p

28 hrs

let the 4 digit number be ABCD.
First Digit is A :
Therefore; according to the question
A=B/3
B=3A
C=A+B=A+3A=4A
D=3B=3(3A)=9A
Since the last Digit is D and it can neither be double-digit nor 0 ;
Therefore ;
A=1;
B=3A=3(1)=3
C=A+B=1+3=4
D=9A=9(1)=9
Therefore, the 4 digit number is 1349.

71

B. 437950

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

ans 810 i think

3

12

Take one fruit from box with label mixture. If we see
orange, because the basket lies (it cant have a mixture),
then it has only oranges. The other 2 are labeled apples and
oranges. The one labeled apples,
cannot have oranges inside, cos they are allready been
identified, and because it lies, it cannot have apples either.
So it has a mixture. And we are left with the one labeled
oranges that lies and has apples.

b

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