Derivatives Foundation interview preparation

Questions

100

Traced to a firm

29

Firms

19

Updated

September 2026

Asked at

All firmsMSMorgan Stanley4Nomura4Akuna Capital2Amundi2HSBC2PIMCO2Bank of America1Barclays1Citadel1DRW1Goldman Sachs1Jane Street1Millennium Management1Mizuho1Old Mission Capital1RCRBC Capital Markets1Scotiabank1UBS1Wells Fargo Securities1

Topic

All topicsForwards and futures10Options basics8Option pricing7The Greeks10Volatility7Option strategies9Swaps and rates7Credit derivatives4Market structure and clearing6Indian derivatives8Trading and markets9Brainteasers6Fit9

Level

AnyCoreIntermediateHard

Type

AnyTechnicalCaseMarket viewBrainteaserFit

1. 089There are n cars on a circular track and between them just enough petrol for one car to complete a lap. Show that there is a car that can complete the lap by collecting petrol from the others as it goes.BrainteasersHardsuperdayMillennium ManagementInvestments · London · 2024

   Say this

   Yes, such a car always exists. The cleanest proof: imagine a phantom car with enough fuel to complete the lap anyway, start it anywhere, and track its fuel level as it picks up each deposit. The point at which its fuel is at its minimum is a valid starting car — from there the cumulative balance never goes negative.

   Then walk it

   1. Set it up as a sequence of partial sums. Going around the circle, each car contributes a gain of its petrol and each gap costs fuel. Total gains minus total costs is exactly zero, because there is precisely one lap's worth.
   2. The argument: define the running balance starting from an arbitrary car. It ends at zero. Take the position where the running balance is at its global minimum, and start there instead. Relative to that point, every partial sum is non-negative, because you subtracted the most negative value from all of them.
   3. So the starting car is the one immediately after the minimum of the cumulative balance. That is a constructive answer, not just an existence proof, which is what makes it satisfying.
   4. There is an induction proof too: with n cars, there must exist some car that has enough petrol to reach the next one — otherwise the total would be insufficient. Merge those two into a single car and you have the same problem with n minus 1. Induct down to one car, which trivially works.
   5. One line for n equals 2 to show the mechanism: if car A has 0.7 laps of fuel and B has 0.3, and the gap from A to B is 0.4, then A cannot reach B directly if it only had 0.3 — the partial-sum argument tells you which one to pick without checking cases.
   6. Why this gets asked at a fund rather than in a maths class: it is the same structure as a cash flow or margin problem. You know the total is sufficient and you need to know whether the path ever goes negative. That is exactly a funding liquidity question, and the answer is always about the minimum of the cumulative balance, not the total.

   Where candidates lose it

   Trying small cases and asserting a pattern. The interviewer wants the partial-sum or induction argument. And the move that impresses is connecting it to cash flow timing — total sufficiency does not imply path feasibility, which is the whole of liquidity risk.

   Expect next

   • Give me the induction version of the proof.
   • Is the starting car unique?
   • What financial problem has exactly this structure?

   Reported by candidates at Millennium Management (Investments, London, 2024). Source: Wall Street Oasis.

Firm tags come from public, anonymous candidate reports on Wall Street Oasis: strong signal, not sworn testimony. Firms are named as the places a question was reported, not as partners of Fin Maverick. Answers are written for this page to show how to think out loud; they are not scripts to recite.