Bristol airport Interview Questions, Process, and Tips

I think the answer is: 16
the woman faster *1.5 from the man, every day.
because that, the exersize is:
24/1+1.5+x=6
x=1.5
24/1.5=16!

1+0^3=1
1+1^3=2
2+2^3=10
10+3^3=37
37+4^3=101
101+5^3=226
1,2,10,37,101,226……..

3:7

8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here

40

x – 30 = 1/4 x || *4
4x – 120 = x || -x + 120
3x = 120 || /3
x = 40

Urinary Tract Infection

let’s say the Cost Price is 1000x. (CP)
The selling price is also the same as the Cost price. So, here SP=CP (But he sells 950gm instead of 1000gm)

Instead of 1000gm, he is selling 950 grams at the CP. So he sells 050 gms @1000x price. So his net profit is 50gm.
Now 1000gm is 1000x Rs
So, 50gms is 50x rs. [Apllied Unitary Method]

So his profit percentage is:
Profit Percentage Formula: {(Profit/CP)*100%}
So, Here profit is 50x;
CP is 1000x;
so putting the value in the formula we get, Profit Percentage is: (50x/1000x)*100% =(5000x/1000x)%=5%.

Tuesday

17

B. 27

( Number × 2 ) + 1
3 × 2 + 1 = 7
7 × 2 + 1 = 15
15 × 2 + 1 = 31 ( not 27 )
31 × 2 + 1 = 63
63 × 2 + 1 = 127
127 × 2 + 1 = 255

5/9

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