Society of Petroleum Engineers SPE Interview Questions, Process, and Tips

a:37
b:22
c:50
d:11
e:25
f:27

Total game played= 60
%won =30%
Total won= 60*30/100 i.e. 18
now team plays x games and win all of those to increase the
average to 50%.
So,
(60+x)*50/100=18+x
(60+x)/2=18+x
60+x=36+2x
24=x
So the final answer is 24.

5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000

d=5
g=1
a=4, b=6,c=2,d=5,e=8,f=3,g=1,h=9,i=7

Pen can’t be sharpened by sharpener
Pencil is only sharpened by sharpner
Likewise rigth man can only fill the rigth position so we have to select it. This is human resource recruiting.
In school teachers manage and motivate you to do your work.
The same is done to employees this is hrms

cost price(c.p) of one fruit = 24/16
(c.p) = Rs 1.50
selling price(s.p) of one fruit = 18/8
(s.p) =Rs 2.25
profit for one fruit = 2.25 – 1.50 = 0.75
profit
profit % = ———– * 100
cost price
0.75
= ——- * 100
1.50
= 50%

let a = 9 * 9 * 9;
let b = 12 * 12 * 12;
let c = 15 * 15 * 15;

sum = a + b + c;

edge = Math.cbrt(sum);

console.log(edge);

20×20+22×22+24×24= 1460
Numbers is less then 30 because 30 square is 900 we have to take three consecutive number if we take numbers greater than 30 the sum of that numbers wil be greater than 1460.

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