John burns real estate consulting Interview Questions, Process, and Tips

When I went to my native place the form is so good environment

e. 56 is 7*8 which is the product of 7 with even no.

When none of the digits are repeated:

The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.

Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240

1 ball – 4 runs (1st batter – 94+4=98)
2 ball – 3 runs and one run is short run during running between the wicket (1st batter- 98+3-1 =100)and strike changes for 2nd batter
3 ball – 6 runs ( 2nd batter 94+6=100)

Error is at 48. Correct once is 58

2:3

as i assume the time taken to complete the job is not to be
changed while increasing the employees as it does not
clearly states so……….
30 men wrk for 9 hrs
therefore for the work to be done if no of workers is
increases the no of hrs / day to work should decrease so
here s a inverse relation
solving further we get
30 = K /9 ==> K= 270
now
40 = 270/y ==> y= 6.75
therefore ans is d)none

Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? 28 m

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

Question is wrong i think

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6.25