Csep scpe Interview Questions, Process, and Tips

6 pairs of earrings. So 12 earrings.

The total has to be 26, so let’s say there’s 6 pairs of earrings and the number of pairs has to be 4 less than the number of rings. So that would make 10 rings. The 10 rings has to be 2 1/2 greater than the number of pins. So you would divide 10/2.5 = 4 Pins. That would equate to a total of 12 earrings, 10 rings, and 4 pins. A total of 26 pieces.

3800

7 Hr 30 Min

The work done by A in 8 days is = 8/ 12 = 2/3

Means A alone completes 2/3 part of work.

Remaining work which is (1–2/3) = 1/3 is completed by B in 8–2 = 6 days

So the complete work done by B in 6/(1/3) = 18 days.

B alone can complete the work in 18 days.

50

60*5/18*9=150

6.25

Let the revolutins made by bigger wheel is: x
then revolution by smaller wheel will be: x+10
Now, the distance covered by both wheels will be same.
So,
7(x+10)=9x
7x+70=9x
70=2x
35=x
so distance travelled=9*35=315
or 7*45=315

there will be no banana left

12

its a tricky question and very funny too.
the answer is 45% only which is gained by the minute hand.

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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