Portfolio Management puzzles, solved step by step

200 miles. The car needs 100 miles at 50 miles an hour, which is two hours. The bird flies without stopping for those two hours at 100 miles an hour, so it covers 200 miles, however many times it turns. Summing the legs gives the same answer: 66.7 there and back, then 22.2 there and back, each pair a third of the one before, which adds to 200.

Why is summing the legs the slow way?

Picture a dog on a walk that runs ahead to the gate and back to you, over and over, until you reach the gate. Nobody counts the dog's sprints; you just ask how long the walk took and how fast the dog runs. When something moves at a constant speed for a known time, distance is speed times time, whatever path it traces. The bird's zig-zag looks like the hard part of the question. It is a distraction. The only thing that matters is when the flying stops, and that is when the car arrives.

How do you check 200 the long way, in case the interviewer asks?

The bird and the car close at 150 miles an hour, so they first meet after 100 / 150 of an hour, 40 minutes, when the car is 33.3 miles from A and the bird has flown 66.7 miles. The bird flies 66.7 miles back to B. By then the car is 66.7 miles along, 33.3 miles from B, and the same geometry repeats on a gap a third as large. Each round trip is a third of the one before, so the legs form a geometric series: 133.3 times one over one minus a third, which is 200.

Why ask this on a quantitative research desk? Because the same move, stepping back from the path to the total, is how you price anything path-dependent in your head: ask what is conserved or fixed before tracing every step.

5 with one trade and 7 with any number. With one trade, buy at the lowest price that comes before a higher one: buy at 1 on day 2 and sell at 6 on day 5, for 5. With unlimited trades, collect every day-to-day rise and sit out every fall: 1 to 5 earns 4, 3 to 6 earns 3, a total of 7. The general answer is the sum of the positive daily changes.

How do you find the best single trade without checking every pair?

Walk through the prices once, like a shopper who notes the cheapest price seen so far and asks each day how much they would make selling today. The best single trade is the largest gap between today's price and the lowest price seen before today, found in one pass. At day 2 the low is 1. Day 3 offers 5 minus 1, which is 4; day 5 offers 6 minus 1, which is 5; no later day beats it. The trap is subtracting the lowest price from the highest overall: the high of 7 comes before the low of 1, so it cannot be sold after buying.

Why is the unlimited answer just the sum of the rises?

Any profitable trade from a low to a later high can be split into daily steps, and it gains only on the up days inside it while paying for every down day it sits through. With no limit on trades and no shorting, the best strategy holds the stock on every day it rises and nothing on every day it falls, so profit equals the sum of positive daily changes. Here that is 4 plus 3, which is 7. One pass through the prices gives the answer, which is the point of asking a coding-flavoured candidate.

Say the limitation as a portfolio manager would: this is perfect hindsight with no costs. Add a transaction cost per trade and the two answers move toward each other, because catching a small rise is no longer worth paying for.

Light rope 1 at both ends and rope 2 at one end, together. When rope 1 burns out, light the other end of rope 2; when rope 2 burns out, 45 minutes have passed. Rope 1 lasts 30 minutes because two flames share its 60 minutes of burning. At that moment rope 2 has 30 minutes left, and lighting its other end halves that to 15. 30 plus 15 is 45.

Why does uneven burning not spoil the halving?

Two people eating one plate of food from opposite sides finish it in half the time one person would take, whether the food is piled high on one side or spread evenly. They simply meet off-centre. A rope holds 60 minutes of burning in total, and two flames consume it twice as fast, so it is gone in 30 minutes wherever the flames happen to meet. The unevenness decides the place, never the time. That is the only fact the puzzle needs.

How do you know rope 2 has exactly 30 minutes left at the half-hour?

Rope 2 has been burning from one end for 30 minutes, so it has used 30 of its 60 minutes of material, whatever length that turned out to be. You never measure length; you only ever track time used and time left. Lighting the far end of what remains halves the 30 minutes left, and the rope finishes 15 minutes later. Say the timeline in that order and the answer is audible.

Interviewers often extend it: with the same two ropes you can also time 15, 30, 60 and 90 minutes, each by deciding which ends are burning at which moment. Answering one extension shows you own the principle rather than a memorised trick.

Why would a real asset desk ask this? It is a clean test of whether you separate the thing you can measure from the thing that is noisy. A rent roll is lumpy month to month; the annual total is what the valuation rests on. Saying that link in one sentence costs nothing.

1,000 x (1 + 1/2 + 1/3 + ... + 1/n) miles, which grows without limit but only like the logarithm of n. Drive all n cars 1,000/n miles; together they have burned one full tank, so one car can refill the rest and be abandoned. Repeat with n minus 1 cars for 1,000/(n minus 1) miles, and so on. Four cars reach 2,083 miles, 100 cars about 5,187, and the distance tracks 1,000 x (ln n + 0.577).

When should a car drop out of the convoy?

Think of friends sharing water on a long walk, where every bottle is full at the start and nobody can carry more than one. The moment the group has drunk exactly one bottle's worth, one friend can pour the rest of theirs into everyone else's bottles and head home. A car should drop out the instant the convoy has burned exactly one tank in total, because that is the first moment its remaining fuel exactly fills the others. With n cars that happens after 1,000/n miles, since n cars burn fuel n times as fast as one.

Why does the distance grow only like a logarithm?

The total is 1,000 times the harmonic sum 1 + 1/2 + ... + 1/n. Each extra car adds the shortest leg of the journey, 1,000/n miles, so the gains shrink as the convoy grows, and the harmonic sum rises like ln n plus about 0.577. It never stops growing, so any distance is reachable in principle, but slowly: 10 cars reach about 2,929 miles, and getting past 5,000 miles takes 83 cars. Doubling the fleet adds only about 1,000 x ln 2, roughly 693 miles.

Why a hedge fund asks it: the structure is the same as scaling a strategy. Each extra unit of capital or effort buys a smaller increment, and a candidate who sees the diminishing returns and names the rate of decay is showing the instinct the desk wants. Also be ready to argue optimality in one sentence: any plan that drops a car earlier wastes fuel it cannot hand over, and dropping later wastes the fuel spent carrying a car that is no longer needed.

Two weighings. Put three bags on each side and leave three aside. Whichever side rises holds the light bag; if the pans balance, it is among the three on the table. Take that group of three and weigh one bag against another: the one that rises is light, and if they balance, it is the third. A balance gives three outcomes per weighing, and three times three covers nine bags.

Why is halving the pile the wrong instinct?

Halving is what you would do with a question that answers yes or no, like guessing a number between 1 and 100. A balance tells you more than that. A balance scale has three outcomes, left pan rises, right pan rises, or level, so each weighing should split the suspects into three equal groups, not two. The group left on the table is not wasted: a level balance is information too. With four and four plus one aside you would learn little from a level result except that the odd bag is the spare one, and an unlucky run needs three weighings.

How do you prove two is the minimum?

Count outcomes. One weighing produces at most three different results, and three results cannot point to nine different bags. With k weighings you can separate at most 3 to the power k bags, so nine bags need at least two weighings, and the method above shows two is enough. The same count answers the natural follow-up: three weighings handle up to 27 bags, four handle 81.

Why would an asset manager ask this? It is a test of whether you use all the information a measurement gives you. A risk report, a performance attribution or a data screen can be read in more ways than pass and fail, and the candidate who spots the third outcome here is the one who reads the level result as a finding.

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Solve from the end. With two pirates, D keeps all 100 because his own vote is half. With three, C buys E for 1 coin. With four, B buys D for 1. With five, A needs two votes and buys the two pirates who get nothing under B's plan, C and E, for one coin each, keeping 98.

Where do you start?

At the end, where there is no choice left. Think of planning a train journey with connections: you start from the time you must arrive and work back to when you must leave. A sequential game is solved backwards, because each pirate's vote depends only on what he would get if the current proposal failed. With two pirates left, D proposes 100 for himself; his own vote is half, which is enough. So E gets nothing if it ever comes to that, and E knows it.

How does each step follow from the one below?

With three pirates, C needs two votes, his own and one more. E gets 0 if C dies, so one coin buys E: C proposes [99, 0, 1]. With four, B needs two votes; under C's plan D gets 0, so B buys D for one coin: [99, 0, 1, 0]. A vote is worth exactly one coin more than that pirate's fallback, so a proposer always buys the cheapest voters, the ones left with nothing in the next round. With five, A needs three votes. Under B's plan C and E get nothing, so one coin each buys them, and A proposes [98, 0, 1, 0, 1].

State the assumptions, because the answer rests on them. If an indifferent pirate voted yes, A could buy votes for zero coins. If the rule needed a strict majority, the counts change. Interviewers often change one rule as a follow-up to see whether you rebuild the chain or reach for a memorised answer. The buy-side lesson is about incentives: what someone will accept depends on their alternative, not on fairness.

n cubed minus (n minus 2) cubed, which expands to 6n squared minus 12n plus 8. Everything not on the surface forms a hidden cube with one layer peeled from each side, so its side is n minus 2. For n = 3 that is 27 minus 1, or 26; for n = 10 it is 488. If the interviewer means the coloured squares instead, the answer is 6n squared.

Why count what you cannot see?

To count the tiles round the edge of a courtyard, it is easier to measure the whole yard and subtract the inner lawn than to walk the border without counting corners twice. The cubes not on the surface form one clean block of side n minus 2, so subtracting it from n cubed avoids every double-counting trap at the edges and corners.

How do you check the formula a second way?

Count by type. There are always 8 corner cubes, 12 edges each carrying n minus 2 cubes between the corners, and 6 faces each with an (n minus 2) by (n minus 2) centre. That is 8 plus 12(n minus 2) plus 6(n minus 2) squared, which expands to 6n squared minus 12n plus 8. Two routes agreeing is the check the interviewer wants to hear.

Two edge cases show care. The formula needs n of at least 2; a single cube, n = 1, is all surface, and the formula would wrongly give 2. And the question is sometimes asked as surface area: the number of coloured squares is 6n squared, 54 for a standard cube. Ask which is meant before answering.