The hoover institution, stanford university Interview Questions, Process, and Tips

FOR 15 = RS.20
FOR 20 = RS.18
FOR 25 = RS.16
FOR 30 = RS.14
FOR 35 = RS.12
FOR 40 = RS.10

Ans : – 215
Explanation :-
1^3 = 1-1=0
2^3 = 8-1 = 7
3^3 = 27-1=26
4^3 = 64 – 1 = 63
5^3 = 125-1 = 124
Hence, Ans is
6^3 = 216-1=215

The 3-digit number can be written as the sequence [n, 2n, 3n]
n = 1 → [1, 2, 3] → valid
n = 2 → [2, 4, 6] → valid
n = 3 → [3, 6, 9] → valid

n = 4 → [4, 8, 12] is not valid because this would lead to a 4-digit number
any value of n > 4 would also produce invalid answer

Answer: Three numbers: {123, 246, 369}

A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m

8-3 = 5
15-8 = 7
24-15 = 9
34-24 = 10 this one should be 35-24 = 11
48 – 34 = 14 based on the correction will be 48-35 = 13
63-48 = 15
the difference will create a series 5,7,9,11,13,15…..etc

The phrase is usually ‘like looking for a needle in a haystack’. It describes a task that is virtually impossible because you would have to search a huge area

Assume Q can do the work alone in Q days:

work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.

it is based on the right shift operation on the perticular digit

Answer:270
Explanation: Formula is SI= P*R*T/100
Where P is principal amount
R is Rate
T is Time
So, SI= 500*6*9/100
i.e SI= 27000/100
SI= 270

1:2:3 or 1:2:4

suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3

time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20

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