Assistant interview questions shared by candidates
You are trying to get to Orlando, which is 800 miles away. You have 2500 apples, and you drive a truck which can hold a thousand at a time. You have unlimited gas, can take as many trips as you'd like, and if you want, you can store apples anywhere on the side of the road and pick them up later. The kicker is- one apple falls out and disappears forever for every mile you drive on the truck How many apples can you transport? Do it in your head- no paper or pencil6 Answers
My roommate just asked me this one earlier today and then I got excited and started to look around for more questions, but since it dosen´t seem to be any answer here and I do belive I did get the right one I´ll post that. There is ofcourse the option to just go 0,9 miles with 1000 apples and then go the last 0,1 mile empty and then go back and get the apples and repeat this all the way to Orlando. But sadly I do believe that would be ruled out. But the alternative solution would be. 866 apples will make it all the way. First transporting with 3 trips 167 miles starting with 750 apples everytime. that would make a loss of 501 apples. That means we now have 1 999 apples left. Meaning we will make 2 trips of 500 miles. Leaving us with 133 miles to go and 999 apples to take that way. After going 133 miles we have 866 apples left.
I think the first 3 trips need to total 2500. Otherwise, looks right.
where do you get numbers like 3 trips 167 miles 750 apples?? is there any mathemaical way of solving this question?
Does anyone have a logical method for this madness?
So i figured out the method. you start with 2500 which needs 3 pick ups. You drive the 3 first pick ups to a distance so that afterwards the new amount will only require 2 pick ups. Because you cannot lose 500 apples exactly with 3 pick ups at the same distance (3 doesn't divide 500) you need to lose 501 apples, so you drive 167 miles for each pick up initially. Now you have 1999 apples which requires 2 pickups, again you want to drive it such a distance so that you lose enough apples to make it the rest of the way in one go. Because u can't lose 999 apples in 2 drives of equal distance, you will need to lose 1000 apples, which brings us to driving 500 miles. Now you have 999 apples a distance of 133 miles from the end. You put it all in your truck and finish the trip and end up with 866 apples in Orlando. Definitely one of the hardest questions I have done.
Since the problem says that you drop an apple every mile and that you can store the apples along the road, I propose the following answer: - load fully the truck with 1000 apples, drive 0.5 miles, drop apples on side of road ( since drove less than 1 mile, no apples are lost). - go back (truck empty) load another 1000 apples, drives 0.5 miles and drop them off-together with the 1000 apples dropped off earlier (again no apples lost since drove 0.5 miles). - go back load 500 apples remaining, drive 0.75 miles this time and drop apples on side of the road. - go back to where have the 2000 apples on side of road, load 1000 apples, drive 0.5 miles and unload on side of road. - repeat sequence until reach destination - Will be able to transport all apples to destination
You roll a die until the sum of the integers rolled is greater than 13. What number are you most likely to stop on?7 Answers
the probability is monotonically decreasing after the first stopping sum, i.e., Smith is correct. 14
Can someone elaborate on this?
Each roll has expected value of 3.5. So you're expected to get 14 after 4 rolls.
We can express the probability of ending up on 14 recursively, in terms of the probabilities of ending up on the numbers from 8 to 13. If we end up on 14, the last roll we made must, of course, have been some number from 1 to 6, each with probability 1/6. We can then express p(14), the probability of hitting a 14, as p(8)*(1/6) + p(9)*(1/6) + ... + p(13)*(1/6). Similarly, we can express p(15) = p(9)*(1/6) + ... + p(13)*(1/6). Note that there is no p(14)*(1/6) term in p(15), because we cannot get to a dice value of 15 through a value of 14, since the game would be over in this case. Thus, p(15) < p(14), p(16) < p(15), ..., and we choose to stay with 14
the question is actually much more tricky than it seems and the last post is very wrong. P(14)>P(8)*(1/6)+...+P(13)*1/6. however they got the answer right the key is to understand that each roll is independent and of course each outcome is equally likely. hence if I am at 13 I'm equally likely to end up at 14-19. if I'm at 12 I'm equally likely to end up at 14-18 since i can either get directly to these points or pass through 13 and get there. in similar fashion at 11 I'm equally likely to end up at 14-17. each each having probability 1/6+1/6*1/6+1/6(1/6+1/6*1/6). the same argument goes down to if I'm at 8 by recursiveness. this means that before the last role I have to be at 8-13 and at none of those positions am i less likely to end at 14 than anything else. in fact I'm mostly likely to end at 14 if I am at 8. since being at 8 has positive probability I'm mostly likely to end at 14.
4 times, since expectation number = 3.5/each time
you have five coins. One is double-headed. Pick one coin at random without looking and throw it 5 times. Suppose the outcome are five heads, what is the probability that the coin picked is the double-headed one?6 Answers
use Bayesian's formula (1/5*1) / (1/5*1 + 4/5 * 1/32) = 8/9
This is basic Bayes theorem. Let X be the event that the double-headed coin is picked. Pr(X | Outcome) = Pr(Outcome | X) * Pr(X) / Pr(Outcome) Pr(Outcome | X) = 1 Pr(X) = 1/5 Pr(Outcome) = 1/5 + 4/5*1/32 = 36 / (5*32) Plug everything in, you've got: Pr(X | Outcome) = 32/36 = 8/9.
sorry... its 8/9 (cant del my prev post)
use Bayesian's formula (1/5*1) / (1/5*1 + 4/5 * 1/16) = 4/5
Toss 4 coins. For each head shown, the play gets 1000$. Given a 2nd chance to play the game again. what's the strategy of the game and what's the expectation of the game.5 Answers
1. the expectation for the number of head is 2 which means the expected return from the game is 2000$. 2. continue the game unless there are 3 or 4 heads in the first game. 3. chance of continue the game is 11/16 4. the expectation of the 2nd game is 2000$ 5. chance of stop at the first game is 5/16 6. the expectation of the 1st game is 1/5*4000+4/5*3000=16000/5 7. the expectation of the game is 2000*11/16+16000/16 =38000/16 = 2375
I will answer the all the questions posed: 1) 15% of 175 (100+50+ 25)(.15)=15+7.5+3.75=22.50+3.75=26.25 2) Degree of angle between hour and minute hand at 7:30 The hour hand progresses at a rate of (360/12) degrees/hour=30 degrees/hour. So, at 7 o'clock, the hour hand is at 7 hours * 30 degrees/hour=210 degrees. The minute hand progresses at a rate of (360/60) degrees/minute=6 degrees/minute. So, at 7:30, the minute hand is at 30 minutes *6 degrees/minute=180 degrees. Therefore, the angle is 30 degrees. 3) Let N be the random variable representing the total number of heads when flipping a coin four times. That is, N~Bin(4,1/2). The problem asks to calculate P(N=3| N>1). First, P(N>1)=1-P(N=0)=1-(1/16)=15/16. P(N=3)= (4 choose 3)* 1/16=4/16. Therefore, P(N=3| N>1)=4/15. Note that this problem appears again in this forum, but with different wording. 4) Bob Barker says he will flip four coins and give you $1000 dollars for each head. How much on average will Bob pay you? Drew Carey now comes along and says he will flip four coins and will give you the choice to obtain $1000 dollars for each head or flip the four coins again and give you $1000 dollars for each head on the second flipping. What should your strategy be for Carey's game? Given that strategy, how much on average will Drew pay you? Let N be the random variable given in Problem 3) above. The expected value of N is E(N)=(1/16)(0* (4 choose 0) + 1 ( 4 choose 1) + 2 (4 choose 2) + 3 (4 choose 3) + 4 (4 choose 4))= (1/16)(0*1+ 1*4 + 2*6 + 3* 4 + 4* 1)=(1/16)* 32=2. This was obvious, but it can be calculated if necessary. So, you can expect Bob to pay you $2000 dollars. In Drew's game, the winning strategy is to make him flip the coins again if you get either 0,1, or 2 heads in the first flip. If you adopt this strategy, then we can divide the expected payoff into that which comes from the first flipping and that which comes from the second flipping. On the first flipping, the expected payoff given the strategy is: 3000* (4 choose 3)/16 + 4000 * (4 choose 4)/16= 3000* 4/16 + 4000 * 1/16= 16000/16. Note that there is a 5/16 chance of stopping the game at the first flipping, and hence an 11/16 chance of continuing onto the second flipping. On the second flipping, the expected payoff given the strategy is: 2000* 11/16=22000/16. This is obvious because it is just the expected pay out of Bob's game times 11/16. Think about that for a second and make sure you understand. So, you can expect Drew to pay you (16000+22000)/16=38000/16=2000+ (6/16)*1000=2000+(3/8)1000=2000+(3* .125)1000=2000+375=$2375. if you adopt the above strategy. Toss 4 coins. For each head shown, you receive $1000. Then Given a 2nd chance to play the game again. what's the strategy of the game and what's the expectation of the game.
(3) is wrong. P(N>1)=1-P(N=0) -P(N=1)=1-(1/16)-(1/4)=11/16. Hence the final answer is 4/11.
2) is wrong... hour hand will be half way between 6 and 7 so answer is 15 degrees...
Thanks for fixing (3) MMXMW. I agree. In regards to (2), the correct answer is 45 degrees. Indeed, at 7:00 the hour hand is at 210 degrees, but after 30 minutes, the hour hand traverses 30/2=15 degrees. So, the correct angle is 45 degrees.
How many digits are there in 2^50?5 Answers
log2(2^50) / log2(10) = 50 / 3.3 ~= 15
2^50 / 10 = 16 * 2^46 *1.6
16 digits: 2^50 = (2^10)^5 = 1024^5
every 3 powers of 2 the number of digits increases by 1. 50/3 = 16.67, so there are 16 digits.
With these brain teasers, is it allowed to use scratch paper? Is there a favored approach between trying to work quickly and mentally as opposed to thoroughly and methodically? I can see the pro's and con's of both, and I'm just curious whether I should practice one way or another.
Calculate 68% of 132 up to two decimal numbers without calculation4 Answers
a = (132 * 70)= 9240 b = (132* 2) = 264 (a-b)/100 = (9240-264)/100 = 89.76
68*132 = 100^2-32^2 = 10000-1024 = 8976
The idea is: 1) (68+32)^2=100^2 2) (68+32)^2= 68^2 + 2*68*32+ 32^2 =68(68+2*32)+32^2=68(100+32)+32^2=68(132)+32^2. Letting 1) =2), yields 100^2-32^2=68(132). This works for multiplications of the form
Even simpler, 132*68=(100+32)(100-32)=100^2-32^2
Expected value of the products of a six-sided die and a eight-sided die.4 Answers
Write down the table with 48 entries (6 rows x 8 columns), sum them up and divide by 48: 756 / 48 = 15.75.
EV(6) = 3.5 EV(8) = 4.5 3.5 * 4.5 = 15.75
Last answer is right. Never ever try to tell them to write the whole table, its better to say u cant solve it.
Have a truck starting that can carry a maximum of 1000 apples, and starts with 2000 apples at a depot. What's the maximum number of apples that can be carried across a 1000 mile desert assuming 1 apple falls off the truck for every mile travelled?4 Answers
Not sure if this is optimal, but: Load up 1000 apples, drive 1/2 of the way there. Repeat. [The number 1/2 comes from the observation that this gives you your maximum carrying capacity at the 1/2 way point. You pay 2 apples per mile when you have to take two separate trips to move a load. So you want to take as few trips as possible.] You now have 1000 apples at the 1/2 way point. Now drive the remainder of the way: You lose 1000/2 apples en route, but keep 500. So you get that many to your destination. What if the numbers were different? Say that you were trying to move 3000 apples instead. Then, following the strategy above, you would load 1000 apples and drive 1/3 of the way there. Take two more such trips. So now you have 2000 apples at the 1/3 point. Now load up 1000 apples and drive 1/2 of the total distance. Repeat. You now have 1000 apples at the 1/2 + 1/3 = 5/6 point. Now load up those 1000 and drive the remaining distance. You keep 5/6 of the 1000 apples, or roughly 833. Again, I'm not sure if that's optimal. It's just one strategy. If someone thinks of a better thing: Please post it!
I don't think above is correct. You drop an apple when you make your way back. I got 1000/3
I haven't gone completely through the math, but in the case where there are a total of two stops along the way.... Let the first stop be A and the second be B (given two points we'll assume the trip goes from left to ri. The picture should then look like this. start-------A---------------B--------------------------finish The number of applies that are left at stop A are 1000-A and 1000-B respectively, assuming that each trip is made separately. Let's now assume that we're going to go to A, pick up all of A's apples, go to B, and finally then go to the rest of the way. Let's pick up 1000-A apples at point A, and then lose B-A of these applies to yield 1000-B apples still left in the truck once we arrive at B. Then pick up the 1000-B apples at B. ....At this point its clear that we have 2000-2B apples in the truck. This value is less than or equal to 1000. The optimal value is to have all of the trucks space filled, or have 2000-2B= 1000. This means that B=500 From here, a total of 500 apples are delivered. Of the 2000 apples, A+B are used. Given that B=500, and you want the minimal value used with the 500 delivered. Since 1000 are wasted total, given the loss of one apple per mile, A=500 as well. Generalizing this strategy. you can make 2 starting trips of 1000 apples and drop them both off at the 500 mile marker. However, as Blah above me concluded, the maximum number of apples you can deliver is 500. This is based on the condition that the truck is loaded at the start point twice, each time up to its maximum capacity of 1000 apples.
There are a couple different shortcuts that work here, all need to be memorized. 1) If you know the square of a number above: (X+1)^2 - (X*2)+1 = 400 - (38+1) = 361 2) If you are multiplying two numbers between 10-19 Add the second digit of the smaller number to the larger and add a 0. 19+9 = 28+0 = 280 Times the 2 second digits and add to original 9*9 = 81 81+280 = 361
I would do; 20 x 19 = 380 380 - 19 = 361 (to get back to 19 x 19)
Use (x+y)^2=x^2+2xy+y^2 to see that (10+9)^2=100+180+81=361
Make me a 2-point wide market on the chance that a randomly selected number between 1 and 100 does not contain a 7.3 Answers
we have 7, x7 and 7x, 77 is overlapped once, so that's 10 + 10 -1 = 19. 81%
Could someone give an explanation of what is meant by a "2-point spread"?
80 & 82?
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