HKS Interview Questions, Process, and Tips

1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500,
Shyam = 4/13 * 3250 = 1000,
Mohan = 3/13 * 3250 = 750

Camel is the answer
Camel can store water for many days or months in there intestines

say the work is w and let no of days taken by b is ‘x’ which
we have to calculate
so work done by a in one day is w/6
work done by b in one day is w/x
a and b together can do work in 4 days ie=(w/6)+(w/x)=(w/4)
solving equation x=12
so no of days taken by b=12

Answer:

24

Step-by-step explanation:

A + B = 40.

And at Rs 7 a kg for 40 kg, you want a total of Rs.280.

So the second equation is 9A + 4B = 280

From the first equation: B = 40 – A

and sub into the second equation:

9A + 4(40-A) = 280

9A + 160 – 4A = 280

5A = 120

A = 24.

And you should check: B should equal 40-24 = 16. Check with the final equation: 9*24 + 4*16 = 216 + 64 = 280. So it works.

Your answer, of course, is A = 24

C

135 is wrong because from starting difference between 7 after that 9 after 11 ,13,17 and so on

20 minutes

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