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  1. 003A car leaves town A for town B, 100 miles away, at 50 miles an hour. At the same moment a bird leaves B at 100 miles an hour, flies to meet the car, turns back to B, then turns again toward the car, and keeps shuttling until the car reaches B. How far does the bird fly?Logic brainteasersWarm upBlackRockNew York · 2025

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    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.

    Plot position against time: the bird simply flies for as long as the car drivesAB5000.511.52 hoursTimefirst meeting: 40 min, 33.3 miles from Acar, 50 mphbird, 100 mphAsk how long, not how farCar: 100 miles / 50 mph = 2 hBird: 2 h x 100 mph= 200 milesCheck: sum the legs66.7 + 66.7 = 133.3then 22.2 + 22.2, 7.4 + 7.4 ...each pair a third of the last: 200
    The car takes two hours to cover 100 miles at 50 miles an hour, and the bird zig-zags between B and the car for exactly those two hours, so at 100 miles an hour it flies 200 miles; the shrinking legs of 66.7, 66.7, 22.2, 22.2 and so on add to the same total.

    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.

    The relationship
    d=vbird×Dvcar=100×10050=200check: 133.3×11−13=200d=v_{bird}\times\frac{D}{v_{car}}=100\times\frac{100}{50}=200 \qquad \text{check: } 133.3\times\frac{1}{1-\tfrac{1}{3}}=200
    Dthe distance from A to B, 100 miles
    v_{car}, v_{bird}the speeds, 50 and 100 miles an hour
    What it says in wordsThe bird's distance is its speed times the car's travel time; the geometric sum of its legs confirms it.

    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.

    Where candidates lose it

    Candidates start computing the first meeting point, then the second, and lose the room in arithmetic. The interviewer is watching for the moment you ask how long the bird flies; some interviewers stop you once you begin summing legs.

    The second trap is saying infinite because the bird turns infinitely often. A sum of infinitely many shrinking terms can be finite, and here it is.

    What the interviewer asks next

    • What if the bird flew at 150 miles an hour?
    • If both towns sent a car toward each other at 50 miles an hour, how far does the bird fly?
    • How many times does the bird touch the car?

    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 going towards point A at 100 miles per hour