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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
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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.
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
$D = v_{bird} \times \frac{d}{v_{car}} = 100 \times \frac{100}{50} = 200, \qquad \frac{400/3}{1 - 1/3} = 200$
v_{bird} the bird's speed, 100 mph
v_{car} the car's speed, 50 mph
d the distance from A to B, 100 miles
400/3 the 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

v_{bird}	the bird's speed, 100 mph
v_{car}	the car's speed, 50 mph
d	the distance from A to B, 100 miles
400/3	the first round trip, 133.3 miles

Try it first

What is the question really asking?

Can you prove it the long way, too?

Where candidates lose it

What the interviewer asks next