Investment Banking puzzles, solved step by step

Fill the 3-litre bottle, pour it into the 4, fill the 3 again and pour until the 4 is full: 2 litres are left in the 3-litre bottle. That is four moves. Because 3 and 4 share no common factor, fills and pours can build every whole number of litres, so any total from 1 to 7 is possible across the two bottles.

What does each move actually measure?

Think of paying for something with only Rs 3 and Rs 4 coins. You can pay Rs 2 by handing over two Rs 3 coins and taking a Rs 4 coin back as change. The bottles work the same way. Every fill adds a full bottle and every pour into a full bottle sets one bottle's worth aside, so the volumes you can make are sums and differences of 3s and 4s. Two litres is 3 plus 3 minus 4: you draw two small bottles from the tap, the big bottle swallows 4 of those 6 litres, and 2 are left.

Is there a second route, and which volumes are out of reach?

Start from the big bottle and it still works, only slower. Fill the 4, pour into the 3 to leave 1, empty the 3, move the 1 litre across, fill the 4 again and top up the 3, which takes 2 litres and leaves 2 in the big bottle. That is six moves, 4 plus 4 minus 3 minus 3. Saying both routes and choosing the shorter one shows the interviewer you searched the problem rather than remembered it. Write each state as a pair, small bottle first, so neither of you loses track.

For the second half of the question, use BezoutA result in number theory: whole-number combinations of two numbers produce exactly the multiples of their highest common factor, and nothing else.'s rule: two bottle sizes can measure exactly the multiples of their highest common factor. The highest common factor of 3 and 4 is 1, so every whole number of litres can be built, limited only by capacity: up to 4 in one bottle and 7 across both. Contrast a 4 and a 6: their common factor is 2, so 3 litres is impossible however long you pour.

Light rope A at both ends and rope B at one end at the same moment. When A burns out, 30 minutes have passed: light B's other end, and B burns out 15 minutes later, at 45 minutes. Lighting a rope at both ends halves whatever burning time it has left, however unevenly it burns, so B's remaining 30 minutes take 15.

Why can you not just cut a rope into pieces?

Because uneven burning means length tells you nothing about time. Think of a candle with a thick base and a thin top: half its height might burn in ten minutes or in forty. The only thing you know for certain about each rope is its total burning time, 60 minutes, so every step must work with time, never with length. Cutting, folding or marking a rope uses length, which is why every answer built on them fails.

Why does lighting both ends halve the time?

Picture the rope as a row of short segments, each with its own burning time, adding up to 60 minutes. Two flames eat into the row from both ends at the same moment and meet somewhere. When they meet, each has burned for the same length of time, and between them they have consumed all 60 minutes of segments. Two flames sharing 60 minutes of burning finish in 30, wherever along the rope they happen to meet. The same holds for a rope with any amount of burning time left: lit at both ends, it lasts half that.

How does that build 45 minutes?

Light A at both ends and B at one end at time zero. A is gone at 30 minutes, which is your signal. B has burned for 30 minutes from one end, so exactly 30 minutes of burning time remain in it, wherever the flame has reached. Light B's other end at that moment and those 30 minutes burn in 15. B goes out at 30 plus 15, which is 45 minutes, and no step relied on length. Say the answer as three events with clock times, 0, 30 and 45, so the interviewer can follow each one.

22 times. The minute hand goes round 12 times in 12 hours and the hour hand once, so the minute hand gains 11 laps and passes the hour hand 11 times. The overlaps come every 12/11 hours, about 65 minutes 27 seconds apart, and the one you expect between 11 and 12 is the one at 12:00 itself. Two 12-hour cycles make 22, counting midnight once.

Why is it not once an hour?

Think of two runners on a track, one doing 12 laps while the other does 1. The faster runner passes the slower one only 11 times, because the slower runner's single lap is subtracted from the faster one's total. Overlaps count the laps the minute hand gains on the hour hand, and it gains 12 minus 1, which is 11, every 12 hours. The hour hand keeps creeping forward, so each catch-up takes a little longer than an hour.

When exactly do the hands meet?

The minute hand moves 6 degrees a minute and the hour hand half a degree, so the minute hand gains 5.5 degrees a minute. A full lap of 360 degrees takes 360 over 5.5, which is 65 5/11 minutes, so the meetings come at 12:00, about 1:05:27, about 2:10:55 and so on. The eleventh gap lands exactly back on 12:00, which is why there is no meeting between 11 and 12: the hands meet at about 10:54:33 and next at 12:00.

What do you say about the edges?

State how you count midnight. Counting from one midnight up to but not including the next gives 22; counting both midnights gives 23. That one sentence shows you saw the edge case rather than stumbling into it. The same lap method answers the follow-ups in seconds: the hands point in exactly opposite directions 22 times a day as well, and form a right angle 44 times, because a right angle happens twice per lap gained.

Two weighings. Put three bars on each side and leave two off. If the scale balances, the light bar is one of the two left off, and weighing them against each other finds it. If one side is lighter, the light bar is among those three: weigh one against one, and if they balance it is the third. A weighing has three outcomes, so two weighings separate up to 3 x 3 = 9 bars.

Why is halving the wrong instinct?

Think of guessing a number with questions that can be answered higher, lower or spot on. Each answer splits the possibilities three ways, not two. A balance scale gives three outcomes, left lighter, right lighter or level, so each weighing should split the suspects into three groups, not two. Halving, four against four, uses only two of those outcomes and needs 3 weighings for eight bars.

Why can it not be done in one weighing?

One weighing has only three possible results, and any of the eight bars could be the light one. Three outcomes cannot point to eight different answers, so one weighing is never enough, and two weighings, with 3 x 3 = 9 outcome paths, are the fewest that can cover eight bars. That counting argument is the proof the interviewer wants: it shows two is a floor, not just a method that happened to work.

What does the puzzle teach beyond the scale?

The same counting sits behind any search. The fewest questions you need is set by how many answers each question can give, not by how clever the questions are. With nine bars, two weighings still suffice; with ten, you need a third. Saying where the method breaks shows the interviewer you understand why it works, which is the point of asking it.

A always tells the truth and B always lies. Suppose A is a liar. Then A's statement is false, so neither of them is a liar, which means A tells the truth: a contradiction. So A is truthful, the statement is true, at least one of the two is a liar, and since A is not, B is. B never needed to speak.

Why does assuming A is a liar fall apart?

Picture someone who says, with a straight face, "I am lying right now." If that is a lie, they are telling the truth, and if it is the truth, they are lying; the sentence eats itself. A's statement is a milder cousin. A liar can never truthfully announce that there is a liar among us, because the announcement would then be true. So test the liar case first: if A lies, the statement is false, which means neither person is a liar, which means A is truthful. The assumption contradicts itself and is thrown out.

How does B get classified without saying a word?

Once A is truthful, A's sentence becomes a fact about the pair: at least one of them lies. A has been cleared, so the only person left who can be the liar is B, and B's silence changes nothing because the information came from A. The question is built this way to see whether you notice that evidence about a person can come from someone else. Say the two steps in order, A first and then B, and name the contradiction out loud rather than skipping to the answer.

Why do interviewers like this shape of puzzle?

It rewards the method you need for any logic test: assume a case, chase it to a contradiction, and move on. The same discipline is how you check a model: assume the circular reference resolves, trace it, and if it cannot, the assumption was wrong. Candidates who fail it usually try to argue from B's silence, which carries no information, instead of from A's sentence, which carries all of it.

The 10 perfect squares: lockers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Locker n is toggled once by every student whose number divides n, so it ends open if n has an odd number of divisors. Divisors come in pairs, d and n over d, so the count is even unless one divisor pairs with itself, which happens only when n is a perfect square. Those ten lockers are open; the other 90 are shut.

What decides whether a locker ends open or closed?

Think of a light switch flicked by a series of people walking past. Only the number of flicks matters: an odd count leaves the light on, an even count leaves it off. Locker n is flicked by student d exactly when d divides n, so its final state depends only on how many divisors n has. Locker 12 is touched by students 1, 2, 3, 4, 6 and 12, six flicks, and ends closed. Locker 16 is touched by 1, 2, 4, 8 and 16, five flicks, and ends open.

Why do only perfect squares have an odd number of divisors?

Divisors arrive in pairs. If d divides n then so does n over d: for 12 the pairs are 1 and 12, 2 and 6, 3 and 4. Pairs give an even count, and the only way to break a pair is for a divisor to be its own partner, d equal to n over d, which means n is d squared. So 16 has the pairs 1 and 16, 2 and 8, and then 4 standing alone. Every perfect square has exactly one such unpaired divisor, and no other number has any, which is why the open lockers are exactly the squares.

How should you present it in the room?

Do not simulate 100 students. Work one small locker aloud, say the pairing rule, name the exception, and list the ten squares: that is a ninety second answer that shows structure rather than stamina. Then offer the check: the number of open lockers is the whole-number part of the square root of 100, which is 10, and for 1,000 lockers it would be 31. Interviewers use this puzzle to see whether you look for the rule behind a process instead of running the process, which is the same instinct a model builder needs when a spreadsheet has a thousand rows and one formula.

75 km. The trains close the 100 km gap at a combined 100 km/h, so they meet after exactly one hour. The bird flies without stopping for that hour at 75 km/h, so it covers 75 km. You never need to track the individual legs, though they do add up: 60 km, then 12, then 2.4, each a fifth of the one before, summing to 75.

What question should you ask first?

If a friend walks a dog in the park for an hour and the dog runs at 10 km/h the whole time, the dog covers 10 km however it zig-zags between trees. Nobody measures each dash. Distance is speed times time, so find how long the bird flies and the shape of its path stops mattering. The trains settle the time: they approach each other at 50 plus 50, which is 100 km/h, and 100 km apart means one hour.

If the interviewer insists on the legs, how do you sum them?

On the first leg the bird and the oncoming train close at 75 plus 50, which is 125 km/h, so they meet after 100 / 125 = 0.8 hours; the bird has flown 60 km. In that time the trains have closed 80 km, leaving 20 km between them. The next leg is the same problem a fifth the size: 12 km, then 2.4, and so on. Each leg is a fixed fraction of the last, which makes the total a geometric series that sums to 60 / (1 - 1/5) = 75 km.

Why ask this in a banking interview? It is a test of whether you look for the simpler quantity before diving into the mechanics, which is what a good analyst does with a messy model: find the number that settles the question and compute that. Answer the short way, then offer the series as a check if there is time.