On a demandé à un Junior Trader...1 mars 2012

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The answer is actually 741. I probably wouldn't have thought of this super quickly during an interview, but just think about it this way: in 10 moves, the most the knight can move to the right (or to the left) is 20 spaces. So you are confined to a 41-by-41 box of spaces. Now just label each column (or row if you'd prefer, as this is obviously symmetric) 0-40 (so the knight begins on the center square of column 20). Then just ask yourself....how many spaces in column 0 can the knight land on in 10 moves? How about column 1? And so on. If you think of it this way the answer will become clear. To "candidate" I suspect you are solving the wrong problem. If you're looking at possible paths that a knight might take then maybe I'd believe your answer. But for spaces to land on this is definitely wrong. By the way, since this is only a confidence interval, my friend had a very good way of estimating this. Each knight moves along the diagonal of a triangle with legs 1 and 2; in other words, the knight moves sqrt{5} away from the center with each move. So the spaces that the knight could land on would be roughly inside a circle of radius 10sqrt{5}, which has area about 1500. But the knight can only land on half of those squares because in an even number of moves he must land on a square that's the same color as the square he started on. So your estimate would be 750, so take some reasonable interval around 750. Moins

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no my bad, the can be reached 741 is correct, it's just in the case 2 that you have to substract these diagonals... Moins

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751

On a demandé à un Junior Trader...23 avril 2015

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I don't believe any of the above have the exact correct answer, which I believe is 192. Summer is closest followed by Janed. Their errors, respectively, seem to be including 1 (1 is a factor for all the numbers!) and including 7 (which is a common factor of 28, also missed out 25, even though it was included in their sum to reach 199). WR describes some of the details I used, but I diverged mostly when using the "remaining" primes, 2, 7, 11 and 13: Lets say we start by looking at 2. Ask, "is 2 more efficiently used in 2^4=16, 2*13=26, 2*11=22 or 2^2*7=28". This is answered by comparing each of these to the individual uses of the primes: (Sum of Individual uses) gt/eq/lt (Combined use) 2^4 +13 = 29 > 2*13 = 26, so 2^4 and 13 is currently the best use of 2 and 13 2^4 + 11= 27 > 2*11 = 22, so 2^4 and 11 is currently the best use of 2 and 11 2^4 + 7 = 23 < 2^2*7 = 28, so 2^2*7 is the best use of 2 and 7. Therefore we keep 28, 13, and 11. The final list is thus 11, 13, 17, 19, 23, 25, 27, 28, 29 giving a sum of 192. As an additional note, above I've found the best use of, say X, by comparing "the sum of individual uses" it should technically be "the sum of the individual use of X and the current best use of the other factors", but this isn't a problem for the example above (and from what I've tried) the numbers up to 30. Moins

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The optimal sequence seems to be 1,11,13,17,19,23,25,27,28,29 which sums to 193. The idea is to list out all odd numbers and temporarily ignore even numbers(since they are all multiples of 2). In the list of odd numbers, we remove the smaller ones since we want to maximize the sum. As such we remove 3,5,7 and 9. We also remove 15(so that we can keep 25) as well as 21(so that we can keep 27). Now we are left with a list of coprime numbers. Next we seek the largest even number that has no common factors with any of the listed odd numbers. The largest such possible number is 28. Moins

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Both of the above answers are incorrect. This might not be optimal, but 7,11,13,16,17,19,23,25,27,29 is better than Nov. 17. Here's how I arrived at my answer. First, any prime > 15 must be included as there is only 1 number with a multiple of that prime between 1 and 30. So 17,19,23,29 are included. Next, we have to use the primes 2,3,5,7,11,13. For 3 and 5, we could "waste" another prime to produce a product of 3 and/or 5 to produce a number under 30, but 3 and 5 have powers (27 and 25) which are very close to 30 that don't "use up" other primes. It seems reasonable that instead of picking say, 30, we could use 25 and 27 and use another number for 2. So let's add 25 and 27. Now the remaining primes are 2, 7, 11 and 13. Multiplying 2 and 13 gives a higher product than multiplying 2 and 7 or 11, but using 16 and 13 is larger than 2*13. So using these four, the highest we can get is 7, 11, 13, 16. All primes have been used, and our sequence is therefore 7, 11, 13, 16, 17, 19, 23, 25, 27, 29 with a total of 187. There might be a better one but I doubt it. Moins

On a demandé à un Junior Trader...11 septembre 2014

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^agree on 1 and 2. For 3) Volume = 4/3pi * r^3 so dVolume/dt (in minutes) = 4/3 pi * 3 r^2 *dr/dt (in minutes) = 4 * pi *r^2 *4 cm/min. Since r = 1/4 cm, we have volume decreasing at a rate of pi cm^3/minute. 4) 2 points on a line can be connected by a triangle with two sides of length r going through the center, and a 'base' that is a chord. We want this chord to be >1. We know the triangle is isosceles since the other sides that connect the center of the circle to the points are length r = 1. Let the angle between the two equal sides be theta. Then splitting our isosceles triangle in half we have a right triangle, and theta is related to the cord and r by: sin(theta/2) = (chord/2) / r from the relation sin = opposite/hypotenuse. r = 1, so we have sin(theta/2) = chord/2 or chord = 2*sin(theta/2). We want chord > 1 so sin(theta/2) must be >1/2. We know that theta/2 has to be >30 degrees (for theta between 0 and 180), and so theta has to be greater than 60. This is easier to visualize with a picture, but basically there is a 60 degree zone to the left and right of our first point where, if our second point is drawn, our chord won't be >1. So, we see that we don't have 120 degrees of our circle to pick a point from if we want a chord of greater than 1 => picking a theta at random gives us a 2/3 chance (360 degrees without 120 available) to get our chord. This question is easy to draw, harder to explain in words. 2nd round: 1) Don't spin. 4 spots to put the pair of bullets in the remaining 5 chambers, but only 1 spot kills you (75% survival). If you spin its just a simple 66% chance since there are 6 spots for the pair to be loaded and 2 result in killing you. 2) Any at the money call/put has a delta of 0.5. Rough price is really subjective, standard pricing models require a volatility as input to determine the premium over parity of an option, not sure what they want. A few dollars (<5) is probably fine. Moins

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First Question: The value of the call option (assuming no discount) should be 35/36. Moins

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Agreed with the second post. 2nd question of the 2nd round: I think they were asking about trader's rule of thumb. If time to maturity is short and interest rate is low, ATM option price is close to 0.4* sigma * spot price * sqrt(T) Moins

On a demandé à un Junior Trader...26 juin 2012

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Hi Athos, could u plz elaborate this..can swap be implemented here..

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Go short on futures with expiration at delivery date

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you might also have to hedge the exchange rate risk between the currencies...

On a demandé à un Junior Trader...20 décembre 2012

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If you're referring to a five cent wide butterfly trading at an underlying price of 50 and another trading at an underlying price of 100, then with all things constant, the one trading at an underlying price of 50 would be more expensive. 5 cents is a much wider spread relative to an underlying price of 50, compared to 100. You'd be buying much more vol. (and premium) by buying atms and selling wings. The wings would be cheaper relative to the body, therefore the price of the spread would be more expensive. Now, assuming these are the same products, the max gain on the strategy is the same regardless of where the underlying is trading (5 cents). You'd just be paying more premium if underlying is 50. Moins

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Easiest way to think of it: The option makes money if you stay within the spread. The spread is the same size for both, however it is relatively half as small in the case of the 100 butterfly (in terms of percentage). Therefore, the 50 butterfly is a better deal. Moins

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Depends where the underlying is trading at. This question is unclear. The more expensive butterfly will be the one closer to the money and closer to expiration. Moins

On a demandé à un Junior Equity Trader...7 juillet 2012

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Isn't it 90 + 15 = 105 degrees?

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Yea its 105 degrees

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It's 90 degrees between 9 and 6 o'clock but the hour hand will move half way towards 10. Since there's 3 hours between the 9 and 6 we can say 1 hour = 90/3 = 30. The initial 90 degrees + the additional (1/2)*30 = 105 degrees. Moins

On a demandé à un Junior Trader...25 octobre 2020

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Hi, which books did they give you to read?

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Correction 30 KM apart*

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Answer 1) 20 km (calculate the time it takes to collide and use the fly's speed) Answer 2) 27 days (calculate the number of pads) Answer 3) 6,6, and 1 (make combinations of 36, add all combinations, and out of the two combinations that give the same answer when added select the combination where there is only 1 smallest number) Moins

On a demandé à un Junior Trader...25 juillet 2015

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7 degrees, from 12:00, Minute hand = 204 degrees, Hour hand =197 degrees

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because of hour hand would not stay on 6 at 6:34 ; 360 degree / 12 hour= 30 degree/hour and 34 minutes would have 30 degreeX 0,34 hour = 10,2 degree 24-10,2 = 13,8 degree Moins

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22 degrees

On a demandé à un Junior Trader...19 septembre 2012

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Crazy is craziness

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I found 1 of the questions on their test here: http://www.math.ku.dk/~rolf/teaching/mfe03/mfe03.soln.pdf Moins

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Is this 10? 2^10 > 1000 and 2^9 < 1000?

On a demandé à un Junior Trader...27 mars 2012

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It's (N-1)! There are N! ways of situating N people in a line, but since the ends are now connected we have translational invariance, i.e. person 1 could be in any of N spots. Therefore, we are actually overcounting by N, so there are N!/N=(N-1)! combinations. To Sherry, just because it works for one case, doesn't mean it's the general formula! Moins

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I guess to guarantee that two certain people sit beside each other, we have to bundle them to each other. That means the possible combinations around the table would be (N-2)!. However, for the bundled group of two people, there is a 2! combination. Therefore the final answer would be (N-2)!*2!. Moins

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Seems incorrect. Suppose we have two people and of course they are required to sit together, there would be only 1 way to arrange which is not (2-1)!*2. To complete the process: get one of the "friends" seated on an arbitrary chair, his or her friend can be seated at the next chair either clockwise or counter clockwise if the two chairs are not identical (which is true when N=2). After this, seats are no longer indistinguishable, and the number of ways to arrange the rest of the people would be (N-2)! Hence, result =1, if N=2, and result=2*(N-2)!, if N>2. Simple check, suppose there are three people A, B, C, and A B must be together. Permutation: ABC ACB BAC BCA CAB CBA Because table is round, ABC=BCA=CAB, ACB=CBA=BAC. And in this case, both ABC and ACB allow AB sitting together. Hence, result=(3-2)!*2=2, the answer is correct. Moins

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