# 898

Imaging Specialist interview questions shared by candidates

## Top Interview Questions

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Junior Seismic Imager was asked...21 February 2014

### Again, there are lots of logic questions. Expect some statistics questions and other math questions. Another example would be, "If you have 100 perfectly square cubes, how big of a shell (I.E. cubes makeup the outside but the inside is hollow) can you make?"

a^3-(a-2)^3&lt;=100 a=5

My bad, should be a^3-(a-2)^3 &lt;= 100.

4x4x4 if the shell has to be a cube

### about imaging quality ,patient care and safety ,daily operation

through phone if possible

Please fix the interview and intimate in the mail about date and place

### Given 1000 small cubics, what is the length of the edge of the biggest cubic you can make.

you dont have to create the cube with unit cubes in the volume...you can have all of the cubes on the edges alone. The answer is the solution of the inequality. X^3-(X-2)^3 &lt;1000 Less

13

shouldn't be 10? how do you get the value 13/83, can you explain a bit?

### three consecutive odd or even number, a, b and c. a^3 + b^3 = 3 * c^2, what are they

Problem maybe a^3+b^3=c^2 or =3c then answer is 1 2 3

So you know that C has to be 9 or less. If you go 3*c^2, with c=9, you get 243. Assuming a=1, b can't be higher than 6 (7^3 = 49*7=343). This tells you 1 or 2 don't work, as their cubes aren't high enough to make the difference between 243-216=27. Ironically, 27 is obviously 3^3, so it shows you the answer itself. The answer is 369. It may seem lucky, but knowing 5^3 is only 125 tells you that b has to be 6 to get the sum to work. Less

s

### Mainly two math problem: -When can you start to work? -Q1: 3-dimension rectangular solid, total surface 64 cm^2, total length of edges 40cm, find the length of longest diagonal. -Q2: A coin, probability to get head is less than 0.5, the probability to flip it four times and get two heads and two tiles is 1/6. Find the probability to get a head in one flip.

u don't need a,b,c 2ab+2bc+2ac=64 a+b+c=10 a^2 + b^2 + c^2 +2ab + 2ac+2bc = 100 d^2 = a^2 + b^2 +d^2 = 100 - 64 =36 d = 6 Less

Q1=36 Q2=(3-sqrt(3))/6

You have 2 equations and 3 unknowns. If you assume b=c, then b and c are not 4. 2*2*4+2*4*4+2*4*4=80 Less

### 1. Given a known parabola equation (y=1/2*x^2 or similar), find the point on it which is closest to the point (0, 4). 2. Randomly pick up 3 points on the circumference of a circle, what is the probability that the distance between all 3 points are all smaller than the radius?

Thanks for posting. How long does it take for them to issue the offer to you?

pick any point in parabola and calculate distance from that point to target point, then minimize it to find closest point Q1 x^2 + (y-4)^2 = d^2 x^2 + (0.5x^2 - 4)^2 = d^2 = f(x) minimize f(x), df/dx = 0 gives : x^3 - 6x = 0 x1 = 0 x2,3 = sqrt(6), -sqrt(6) then, y2,3 = 0.5x^2 = 3 both due to symmetry now, distance between (0,4) and (sqrt(6),3) is sqrt(15), which is less than 4. so, the answer is (sqrt(6),3) Less

when largest distance is equal to rarius , this two points and center point make isoscale triangle, while 3rd point reside in between this two points. Using this, p = (pi*r/3)/(2pi*r) = 1/6 Less

### (x-2)^2=y+4, (y-2)^2 =x+4, find x^2+y^2

0 or 15 or 50

50&amp;0

0 or 18 if x = -y (missed the 0)

### Not really difficult but at the time I was asked I looked like the biggest idiot: There are N boxes and N keys. Each key opens exactly one box. Suppose we lock each key inside a random box. We pick a box at random and break it in order to get the key. What is the probability that with that key we will be able to open the rest of the boxes without having to break any of the remaining ones?

Isn't it (n-1)/n, because if 1/n is the probabilty that the key can belong to that box itself, so in this way we won't be able to open other boxes. But if it had different key (i.e. N-1 possible keys,) it could open all the boxes. Less

It is only (n-1)/n if you were to successfully open the next box, and the question asks the probability of opening all the boxes, so each time you have to multiply the probability: (n-1)/n * (n-2)/(n-1) * ... * 1/2, thus the answer is 1/n. Less

No box must contain its correct key - for if it did, it would be impossible to open the box in the first place. The probability then would be given by the number of derangements !n of n. The probability of this would be !n/n! Less

for the first question, should n be 9?

Thanks for sharing this information~Does this matter if I run out time for solving only 2 problems? Less

1- it means the proportion of area and volume is 2:3, so 2:3 = 6n^2 : n^3 which leads to n=9. 2- The probability of getting 1 is 1/6 and the probability that it’s been removed is 1/6, same for the other numbers, so the final probability is 6(1/6*1/6)=1/6. Less

### how many 1s are there from 1-100

20

21 1s including 100

100-9*9+1+1=21