# Preguntas de entrevista de Actuarial intern

# 4 mil

Preguntas de entrevista para Actuarial Intern compartidas por los candidatos## Principales preguntas de entrevista

### How do you calculate the number of red cars in a city?

15 respuestas↳

First, I choose the city and estimate the number of cars in it. Second, I estimate the number of typical car colours. Third, I deduct 10-20% percent from the estimated number of cars to exclude cars with atypical colours. Fourth, I divide the resulting number of cars by the estimated number of typical car colours. Voila. Menos

↳

look up how many lawyers in the city.

↳

Polk Data.

### (Analytical test in excel) At the end of 2006 we had 800,000 Policies in Force (PIF). 75% of those policies were Preferred (in one of our preferred companies). What annualized growth rate would be needed to increase non-preferred PIF by 40,000 policies over two years?

7 respuestas### IQ question. There are 25 horses and you want to find the fastest 3. Given that you can only put 5 of them in each race, what is the minimum number of races that you need?

6 respuestas↳

Answer is 5. Unless the timings of the horses were not known, then the answer is 7. Menos

↳

6 race 5 each group, pick the best of each group and one fore the final 5

↳

Its 8. 7 races would ignore the fact that the top 3 could be in the same first race. Menos

### you have a 9x9x9 sided cube made up of 1 squaree inch squares. If all sides are painted how many cubes have no paint on any side?

6 respuestas↳

Since cubes on all sides are painted, I believe that you would need to eliminate a cube from each side, so I think that the answer is 7x7x7 = 343 Menos

↳

To add in general for nXnXn cube the answer is (n-2)X(n-2)X(n-2)

↳

I agree with John's answer, but the wording is terrible. You have to assume that the cube is built from 9^3 one inch cubes. Then the problem becomes simple to visualize and solve. I hope the interview explained the problem clearer than this poster. Menos

### I have a deck of 52 cards. I take one and throw it away. I pull the next one. What is the probability it is a king?

5 respuestas↳

The 1st is a king and the 2nd a king or not a king and the 2nd a king. (4/62)×(3/61)+(58/62)×(4/61)= Menos

↳

52 rather than 62

↳

And 51 rather than 61

### What degree is the clock at 2.15?

4 respuestas↳

The minute hand will be at 3 (perfectly vertical), and the hour hand will be a quarter of the way from the 2, so 30*(3/4)=22.5 degrees. Menos

↳

I agree with the methods mentioned above. But, an approximation is probably a lot easier to do, and will be a lot simpler. Imagine both the hands of the clock are at twelve. The minute hand in theory would have to travel 15 minutes, and the hour hand would have to travel IN THEORY, 10 minutes. Now, we know if the minute hand were to travel 15 minutes from minute 0, it would form a 90 degree angle. Therefore, it travels 6 degree/minute. The hour hand will definitely not be travelling 6 degree in a minute. But, for simplicity sake, lets ignore that for now. Consequently, for the hour hand to be at two (which realistically it won't be), it has to travel 10minute*6deg/minute = 60 degree. The difference between the degrees traveled by the minute hand and the hour hand should give you an approximate answer, i.e. 90 - 60 = 30 degree APPROX. Hope this helps! Good luck in your career search fellow aspiring actuary! Menos

↳

you can calculate it easily, no need exact answer, just tell them your step

### We need you to answer by tomorrow morning as our classes are filling up and then they moved the class date!

2 respuestas↳

It's within their control to change their dates and you must conform.

↳

This firm has been doing this since 2003. It is run by a certain Hal Butler of Mount Joliet, Tennessee. He sells the exam prep course and claims to be able to place candidates who have passed the first exam. He operates under numerous names as he is quickly identified to be a fraud. I have had his red listed at all local universities where he often attempts to lure unsuspecting students through the on campus career center. I have been in this industry for 13 years now and not once have I heard of a candidate who has been referred by his firm. He has no industry contacts, and is for all practices just selling an exam prep course. Better options for exam prep existed elsewhere. Menos

### You Have a 5 gallon bucket and a 3 gallon bucket. How do you measure out 4 gallons

3 respuestas↳

1. Dump 3 gal into 5 gal 2. Fill 3 ga and empty into 5 gal again 3. This will leave 1 gal in the 3 gal 4. Dump the 5 gal 5. Add the 1 gal from the 3 gal to the 5 gal 6. Fill the 3 gal again 7. Dump the 3 gal into the 5 gal (where there is all ready 1 gal) 8. Now there are 4 gallons of water in the 5 gallon bucket. Menos

↳

get 2 gal first by 5-3 dump 2 gal into 3 gal bucket fill 5 gal, dump 1 gal in the 3 gal bucket youhave 4 gal left in the 5 gal bucket , solved Menos

↳

1. Dump 3 gal into 5 gal 2. Fill 3 ga and empty into 5 gal again 3. This will leave 1 gal in the 3 gal 4. Dump the 5 gal 5. Add the 1 gal from the 3 gal to the 5 gal 6. Fill the 3 gal again 7. Dump the 3 gal into the 5 gal (where there is all ready 1 gal) 8. Now there are 4 gallons of water in the 5 gallon bucket. Menos

### Suppose you have four people that need to cross a bridge in the night. The bridge can hold the weight of two people at the same time and each person takes 1,2,5,10 minutes. What's the minimum time that they need to cross the bridge?

3 respuestas↳

10 mins

↳

1omin. Let the person that will take the longest time go first, the others go sequentially and so will take 1+2+5 so 8 minutes which is smaller than 10 min Menos

↳

Is the answer 12?

### You have seven balls and a two sided scale. all weigh the same but one. In two measurements find th odd bal

3 respuestas↳

It took me like 2 minutes to figure this out. I'm not fond of word problems either lol. I'm assuming this is correct. Okay put 3 balls on each side, hold onto 1. If one side is already leaning down with more weight, I know the heavier ball is that group of 3. If both sides are equal on the scale, then I’m already holding onto the heaviest ball and their is no need to continue. Okay, assuming one set of 3 balls is heavier, clear off the other lighter side and repeat the process with the heavier set of 3 balls. Again hold onto 1 ball, put 1 ball on the left and 1 ball on the right. If one ball is leaning down, that’s the heaviest ball. If they are both equal on the scale, then I’m holding onto the heaviest ball. Menos

↳

Just wanted to state that you can do the same thing / process for finding the lighter ball, you would just keep the lighter balls on the scale and remove the heavier ones or if there equal you know you're holding onto the lighter ball. Menos

↳

I'm pretty sure that you have to know if the odd one is heavier or lighter than the other ones to do this in 2 measures Menos