Ahs hospitality Interview Questions, Process, and Tips

d

d

1min = 60 seconds
6 min=6*60=360seconds
4 sec=10
360sec=360*10 /4=900

X+x+2+x+3=42
3x+6=42
3x=42-6=36
X=12
X+2=14
X+3=15
Middle number 14

Answer:50.24
Explanation:
Area of a circle =π*r*r
616=π*r*r
r*r= 616÷3.14
r*r =196
r =13
Circumference of a circle= 2*π*r
=2*π*13
=16*3.14=50.24

we can draw 12 tangents, 4 for 2 circles and 12 for 3
circles

Suppose that x glassses are supplied safely , therefore
(100-x) glases are damage. Then equation is :
3x – 3(100-x) = 270
on solving this equation value of x comes out to be 95.

required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)

$57.30

Assume his salary was Rs. X
He earns 5% raise. So his salary is (105*X)/100
A year later he receives 2.5% cut.
So his salary is ((105*X)/100)*(97.5/100)
which is Rs. 22702.68
So Ans is Rs.22176

only if the results are greater than my input

20(85+0)/2=850

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

20&130

(N * 1.1) * 0.9 = 7920
N=8000

2:3:4