Mutual Fund Mastery puzzles, solved step by step

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All topicsCompounding and time value9Statistics, correlation and diversification8Bond maths and duration10Performance measurement and returns8Costs and fee drag8Valuation riddles11Logic and numeracy brainteasers6Estimation and market sizing7Probability and expected value8NAV, units and fund mechanics7Risk, volatility and drawdown8Behavioural traps6Withdrawals and after-tax arithmetic4
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  1. 023A car leaves A for B, 100 miles away, at 50 miles an hour. At the same moment a bird leaves B toward the car at 100 miles an hour. Each time it meets the car it turns, flies back to B, then turns again toward the car, until the car reaches B. How far does the bird fly in total?Logic and numeracy brainteasersCoreBlackRockNew York · 2025

    Try it first

    How far does the bird fly?

    Show the worked solution

    200 miles. The car covers 100 miles at 50 miles an hour, so the journey lasts two hours. The bird flies the whole time at 100 miles an hour, so it covers 200 miles however many times it turns. The long way agrees: the bird's round trips are 133.3, then 44.4, then 14.8 miles, each a third of the last, summing to 133.3 times 1.5, 200.

    What is the question really asking?

    Think of a dog running back and forth between you and your front door while you walk home. You could trace every turn, or you could notice the dog runs the whole time you are walking. When something moves at a constant speed for a known length of time, distance is speed times time, and the path it takes in between does not matter. The bird's speed is given; the only thing you need is how long it flies, and that is the car's travel time.

    Ask how long the bird flies, not where it turnsAB25 mi50 mi75 mi0.5 h1.0 h1.5 h2.0 hfirst meeting, 33.3 mibirdcar, 50 mphcar arrives at B, 2 hThe shortcutCar: 100 mi / 50 mph= 2 hours of flyingBird: 2 h x 100 mph200 milesSeries check: 133.3 + 44.4+ 14.8 + ... = 133.3 x 1.5= 200
    The bird zigzags between the car and B in trips that shrink by two thirds each time, but it flies for exactly the two hours the car takes to cover 100 miles, so it covers 200 miles at 100 miles an hour.

    Can you prove it the long way, too?

    Yes, and doing so briefly earns credit. The car and bird close a 100 mile gap at 150 miles an hour, meeting after two thirds of an hour, 66.7 miles from B. The bird flies 66.7 miles back to B, arriving at 1 hour 20 minutes, when the car is 66.7 miles from A. Each new chase starts with a gap a third of the last, so each round trip is a third as long: 133.3, 44.4, 14.8 and so on. A geometric series with ratio one third sums to the first term times 1.5, which is 200.

    The relationship
    D=vbird×dvcar=100×10050=200,400/31−1/3=200D = v_{bird} \times \frac{d}{v_{car}} = 100 \times \frac{100}{50} = 200, \qquad \frac{400/3}{1 - 1/3} = 200
    v_{bird}the bird's speed, 100 mph
    v_{car}the car's speed, 50 mph
    dthe distance from A to B, 100 miles
    400/3the first round trip, 133.3 miles
    What it says in wordsThe bird's distance is its speed times the car's travel time, and the infinite series of shrinking trips sums to the same number.

    Why would a fund interviewer ask this? It tests whether you look for the quantity that is fixed before diving into detail, the same habit that makes you ask what an investor actually earned before reconciling every trade. The limitation is that it is a pure puzzle; turning instantly at each meeting is an idealisation that makes the infinite number of turns harmless.

    Where candidates lose it

    The common failure is starting to sum the legs without noticing the shortcut, then getting lost in the second or third term under time pressure. The interviewer is watching whether you step back and ask what is constant.

    The opposite failure is saying infinitely far because the bird turns infinitely many times. Infinitely many terms can have a finite sum when they shrink fast enough; here each is a third of the last.

    What the interviewer asks next

    • If the bird flew at 150 miles an hour instead, how far would it fly?
    • How far from A is the car when the bird reaches B for the second time?
    • How many round trips does the bird complete before the car is within one mile of B?

    Asked at BlackRock, Quantitative Research, New York, 2025 (Wall Street Oasis): A car starts at point A going 50 miles an hour towards point B, and a bird starts at point B

  2. 074A gold fund's vault holds nine coins that look identical, but one is slightly lighter than the others. Using a balance scale only twice, how do you find the light coin?Logic and numeracy brainteasersCoreIndian AMCsGlobal asset managers

    Try it first

    What should the first weighing be?

    Show the worked solution

    Weigh three coins against three. The pan that rises holds the light coin; if they balance, it is among the three left aside. Then weigh two of those three suspects against each other: the rising pan holds it, and a balance means it is the third. Each weighing has three possible outcomes, so two weighings can separate 3 x 3 = 9 coins.

    Why split into three groups rather than two?

    A shopkeeper asking a customer "yes, no, or not sure?" learns more from one answer than one asking only "yes or no?". A balance scale gives three answers, left pan rises, right pan rises, or the pans balance, so the best weighing splits the suspects into three equal groups and lets the scale tell you which group holds the light coin. Put coins 1 to 3 on the left and 4 to 6 on the right. If the left pan rises, the light coin is 1, 2 or 3. If the right pan rises, it is 4, 5 or 6. If they balance, it is 7, 8 or 9.

    Each weighing has three outcomes, so it splits the suspects into thirdsWeighing 1coins 1-3 vs coins 4-6Left pan risesLight coin is 1, 2 or 3Weigh 1 vs 2132Pans balanceLight coin is 7, 8 or 9Weigh 7 vs 8798Right pan risesLight coin is 4, 5 or 6Weigh 4 vs 5465Nine end points for nine coins: every coin is found in exactly two weighings.Why not halves? Weighing 4 against 4 wastes the third outcome, the balance.Two weighings tell apart at most 3 x 3 = 9 cases; with halves you need 3 weighings for 9 coins.
    Weighing three coins against three leaves three suspects whatever the scale shows, and weighing two of those three against each other identifies the light coin, so two weighings cover all nine coins.

    How do you know two weighings is the minimum, and how far does this go?

    Count outcomes. One weighing has 3 outcomes, two have 3 x 3 = 9, and there are 9 possible light coins, so two weighings are just enough. The number of weighings needed is the smallest w with 3 to the power w at least the number of coins: 9 coins need 2, 27 coins need 3. One weighing can only handle 3 coins, so you cannot do 9 in one. Splitting into halves wastes the balance outcome and needs three weighings for nine coins.

    The relationship
    3w≥N  ⇒  w=⌈log⁡3N⌉,N=9:  w=23^w \ge N \;\Rightarrow\; w = \lceil \log_3 N \rceil, \qquad N = 9:\; w = 2
    wthe number of weighings
    Nthe number of coins, one of them light
    3the outcomes of one weighing: left rises, right rises, balance
    What it says in wordsEach weighing can at most divide the suspects by three, so the weighings needed grow with the logarithm to base three of the number of coins.

    The desk lesson is about information, not coins: a test with three outcomes is worth more than one with two, if you design it to use all three. The limit of the puzzle is that you are told the odd coin is lighter. If it could be heavier or lighter, each coin has two possible states, there are more cases to separate, and the strategy needs more care, which is the classic twelve-coin follow-up.

    Where candidates lose it

    The trap is splitting in halves: four against four, then two against two, then one against one. It works but takes three weighings, and the interviewer asked for two. Candidates who think in halves have missed that a balance is an answer too.

    The second miss is getting the method but not the reason. Say that a weighing has three outcomes and two weighings give nine, which is why nine coins is the most two weighings can handle.

    What the interviewer asks next

    • How many coins can you handle with three weighings?
    • Twelve coins, one odd, and you do not know whether it is heavier or lighter. Can you find it in three weighings?
    • You have a digital scale that shows exact weights instead. How many weighings do you need for nine coins?