Flexicode India Interview Questions, Process, and Tips

It is assumed to be constant. Now, to cross past the pole, the train should cover a distance of x meters. Now, the time taken by the train to cross a platform of length 100 m is 25 seconds. Hence, the length of the train is 150 m.

There should be 5(2+2+1) decimal places in the answer and the decimal place should end with 2(3*2*2=12) so the correct answer is a, no need for calculator. You can rule out all the other options because it does not satisfy this condition.

The answer cannot be determined as there is a particular formula where the consecutive numbers start.

First nos series is 7,9,11,?
ie odd number siries ie 7,9,11,13
Second number series is 16,15,14
ie 1 less the previous number 16,15,14,13
Ans —-series is 7,16,9,15,11,14,13,13

Its 2 times faster than the other train
v1*t=v2
v2*t=4*v1
solving these two,we get
v2/v1=2

Explanation:

The number of ways of selecting three men, two women and three children is:

= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 ways.

30

maternal grand mother

Let the person ate x banana on first day, so below is the equation for five days
X+x+6+x+12+x+18+x+24=100
So, he ate 8 banana first day

72

132

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

C

E