The old mill hotel and restaurant Interview Questions, Process, and Tips

NON

x and y can be equal to 1..
If so,then he gets 1/1 of Rs.10 which is equal to 10 and
again 1/1 of Rs.10=Rs.10..so,he gets total of 20..and
returns 20..so,no gain and no loss..
If x=1,y=2,he gets, 5+20=25..and returns 20..so he may not
lose..
So,whatever be the values of x and y,only these two answers
are possible..
so its a)He never loses..

C

answer can be 1,2,3,4 can’t be determined exact number
just explaining a case:
there are 10 people in the party.
name of people no. of people with they made handshake list of those people(this can vary but showing the possibility)
1 1 9
2 2 8, 9
3 3 7, 8, 9
4 4 6, 7, 8, 9
5 5 6, 7, 8, 9. 10
6 6 4, 5, 7, 8, 9, 10
7 7 3, 4, 5, 6, 8, 9, 10
8 8 2, 3, 4, 5, 6, 7, 9, 10
9 9 1, 2, 3, 4, 5, 6, 7, 8, 10
10 4 5, 6, 7, 8, 9
jack got 9 different answer so jack can be either 4th number or 10th number and jack’s wife know jack very well so she can’t have handshake with jack so if 4th is jack then she can’t be handshake with 6,7,8,9, in this case she can be 1,2,3, 5, 10 and now depending upon which no is jack’s wife she can have hand shake with- 1- 4 people, and if jack is number 10 then she can’t be 5,6,7,8,9 so again depending upon her number she can handshake with people in range of 1-4

let the speed limit be x.
speed by first rider S1 and 2nd rider be S2.
S1=x+10;
ans S2=x+2*10; and given S2=35.
solving we get x=15mph.

I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps

The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.

The time to travel 400m at 20mps is 20secs

I have been teaching some students for 1 year.

225

Ratio: 3/4

GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC

21

13

In 30 mins- Thief – 20km, Owner – 0km
In 1hr – Thief – 40km, Owner – 25km
In 1.5hrs – Thief – 60km, Owner – 50km
In 2hrs – Thief – 80km, Owner – 75km
In 2.5hrs – Thief – 100km, Owner – 100km

Answer: In 2.5 hours the owner would have overtaken the Theif

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