Rhomberg rail australia pty ltd Interview Questions, Process, and Tips

3600

B

Kerala

Since the total train length passed the tunnel so the distance would be the length of train added to the tunnel legth
D=150+300=450 mt.
speed = distance/time
speed = 450/(40.5/3600)
speed =40,000 mt/sec

39/50

( a ) 11

Let:
1. speed of boat in still water = x
2. speed of stream = y;

Then net speed of boat upstream = x – y = 16km / 4 h = 4kmph
Net speed of boat downstream = x + y = 16 / 2 = 8kmph

Now solve for x to get x = 6kmph in still water.

First we need to find out LCM of 2,3,5
that is 30,,,
then add 30 to 6 we get 36…
then divide it by 2 we get 18..
so 18 would be written interms of binay as 10010
means..Answer is
$**$*

56.25%

speed of the train respect to man
= (63 – 3) km/hr
= 60 km/hr
= 60 * 1000 / 3600 m/sec
= 50/3 m/sec

time
= distance/speed
= 500 * 3/ 50
= 30 sec

sunday

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

a/g

9000

1/3rd of it

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