The Smith Family (Australia) Interview Questions, Process, and Tips

Suppose that total 100 employees are in company….. out of that 35 are man and remaining 65 are women.
20% of man 35 = 20*35/100
40% of women = 40*65/100
total employees = 7+26 = 33 out of 100
so, Ans = 33 %

When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.

Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.

So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.

So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.

cant say

If c is 5 times faster than a then he can do it in 2 days alone.

17 secs

C. 50

Speed = (45×518)m/sec=252m/sec

Total distance covered = (360+140) m = 500 m.

∴Required time= 500×225sec = 40 sec.

D

1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0

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

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