Blackberry farm Interview Questions, Process, and Tips

8/15 left.

1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15

u c 5 numbers…..
we add the combination of digits…..
a.17->1+7=8(even)
b.23->2+3=5(odd)!!!!!
c.37->3+7=10(even)
d.13->1+3=4(even)
e.73->7+3=10(even)
all combinations are even xept 23
ans:23

4:3
A—–>B (first train)
B——>A(second train)
A/B=Srureroot((time to B)/(time to A))

128 is incorrect. There is a difference of multiples of 3 in the number pattern, which starts at 24 and reduces by 3 each time. Therefore the correct pattern would read 190, 166, 145, 127, 112, 100, 91.
this would leave the pattern with differences in the pattern 24, 21, 18, 15, 12, 9. The next number in the pattern would be 6 lower than 91. 91 – 6 = 85. Then 85 -3 – 82, followed by 82 – 0 = 82. After this, if -3 were taken the pattern would start to rise so 82 – – 3 = 85.

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

C

9

9/25