Peter alexander Interview Questions, Process, and Tips

to reach land
first time it travels 8 mtrs
next time (half=4) 4+4 mtrs (up&down)
next time (half=4) 2+2 mtrs (up&down)
next time (half=4) 1+1 mtrs (up&down)
and so on.
Altogether,close to 24 mtrs.

54

40

3

Let x and y be the two numbers. And x > y.
Then,
x + y = 80—–(1)
y=80-y
Also,
x = 4y – 5
Solving the two equations,
80 – y = 4y – 5
85 = 5y
Therefore, y = 17
put the value of y in equation(1)
then
x+y=80
x+17=80
x=63
the numbers are 17&63

8/15 left.

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

Let\: the\: total\: price \: be\: Rs. X\: then,

B= 2x/7 & A= \left ( x-2 \right )/7

So, A:B = 5x/7 : 2x/7 = 5 : 2

Let \: B’s \: capital\; be\: Rs.Y\: then\: \left ( 16000\ast 8 \right )/4y= 5/2

=> (16000\ast 8\ast 2)/(4\ast 5) = y

=> y = Rs. 12800

Use 3pt. Formula
( (x-x1)/(x2-x1) )=( (y-y1)/(y2-y1) )
We get 150

94

The answer is depend on the distance between Mysore &
Bangalore which is not given in the question.Bird
travel=Distance/35*25km

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

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

D=5, G=1