Wog Interview Questions, Process, and Tips

23. 15s squre is 225and 8 squre is 64 addition is 289 product also 120 addition 15+8=23

1, 2, 4, 13, 31, 112,?
This series is written in base 5 numbers for decimal nos of 1,2,4,8,16,32,64

so next number is 224 .
224 in base 5 = 64 in base 10.

so 224 is next number.

The ratio flat vs hill is unknown so you can’t calculate this normally. But since the average for uphill/downhill is also 4 kmph ((1.5h)/6km)), the calculation is: 6 hours * 4kmph = 24km.

4 years

B. 279

360

Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74

ANS = 74

the answer is E

Explanation:
They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
6000 : 8000 => 3:4

Rs.265.80

4 boy 3 sis

a

17 secs

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

1. 1g
2. 3g with 1g counter
3. 3g
4. 3g plus 1g
5. 3g plus 1 g plus 1g weighed medicine
6. 9g with 3g counter
7. 9g with 1g of counter and 1g weighed medicine
8. 9g with 1g counter
9. 9g
10. 9g plus 1g
11. 9g plus 3g with 1g on counter
12. 9g plus 3g
13. All 3 weights on one side.