Financial Analysis puzzles, solved step by step

17 minutes. 1 and 2 cross (2 minutes) and 1 brings the torch back (1). Then 5 and 10 cross together (10), and 2 brings the torch back (2). Finally 1 and 2 cross again (2). The obvious plan, where the 1-minute person escorts everyone, takes 19 minutes because the two slowest walkers cost 15 minutes on separate trips but only 10 together.

Why does the obvious plan lose two minutes?

The escort plan sounds efficient: the fastest person walks each one across and runs back with the torch. It costs 2, 1, 5, 1 and 10, which is 19 minutes. The waste is that the 5 and the 10 cross on separate trips, so you pay both their times, 15 minutes, when one shared trip would cost only 10. Think of two slow friends who will hold up any group they walk with; the cheapest thing is to let them hold up the same group, once.

How do you make sure the torch is on the right side?

If 5 and 10 cross first, one of them has to bring the torch back, which costs at least 5 more minutes. So before the slow pair goes, a fast walker must already be waiting on the far side to carry the torch back. That is why 1 and 2 go first and 1 returns: it parks the 2 on the far side. After the slow pair crosses, the 2 brings the torch back, and the fast pair finishes together.

How do you show 17 cannot be beaten?

There must be three forward trips and two returns. The 10 must cross on some forward trip, costing 10. If the 5 crossed on a different forward trip, the forward trips alone would cost at least 10 + 5 + 2, which is 17, and the two returns add at least 2 more, so 5 and 10 must share a trip. The slow pair cannot go first, or one of them walks the torch back. So 1 and 2 go first, costing 2, and one of them returns. After the slow pair crosses, the only fast walker on the far side is whichever of 1 and 2 stayed, so the two returns cost 1 + 2 between them, and the last trip, 1 and 2, costs 2. That gives at least 10 + 2 + 2 + 1 + 2 = 17. A brute-force search over every schedule confirms 17.

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. The 4 already holds 3 litres, so it takes only 1 more. That is four moves. A longer route fills the 4 first and ends with 2 litres in the 4-litre bottle after six moves.

How do you avoid getting lost halfway through?

Think of following a recipe that says pour half the milk into the other pan: if you do not write down how much is in each pan, two steps later you have no idea. Track the pair of levels after every single move, written as (3-litre, 4-litre), and the puzzle becomes a short walk through a few states instead of a juggling act. There are only three kinds of move: fill a bottle from the tap, empty a bottle, or pour one into the other until the source is empty or the target is full.

Say the states out loud as you go: (0, 0), (3, 0), (0, 3), (3, 3), (2, 4). The interviewer can check each one in a second, and if you slip, you can see where.

Which amounts can these two bottles measure at all?

Every pour moves water in steps of 3 and 4, so the amounts you can isolate are the combinations of 3 and 4 with whole numbers in front: 4 minus 3 is 1, 3 times 2 minus 4 is 2, and so on. Because 3 and 4 share no common factor bigger than 1, these bottles can measure every whole number of litres from 1 to 4, and up to 7 if you count both bottles together. With a 4-litre and a 6-litre bottle you could never get an odd number, since every amount would be a multiple of 2. Saying that rule turns a party trick into reasoning.

200 km. The car takes 100 / 50 = 2 hours to reach B, and the bird flies without stopping for exactly those 2 hours at 100 km/h. Distance is speed times time, so the bird covers 200 km. How many trips it makes and where each turn happens do not matter; only the time in the air does.

Why is the zigzag the wrong thing to count?

Picture a dog running back and forth between you and your front gate while you walk home. You could chart every lap, or you could notice that the dog runs for exactly as long as you walk. When something moves at a constant speed for a known time, its distance is speed times time, and the route it takes is irrelevant. The puzzle is built to pull you into the laps.

Here the clock belongs to the car. It has 100 km to cover at 50 km/h, so the whole episode lasts 2 hours. The bird flies at 100 km/h throughout, so it flies 200 km. That is the whole answer, and it takes one sentence.

How do you check it the long way if the interviewer asks?

Sum the laps. On the first leg the bird and the car close at 150 km/h across 100 km, so they meet after 40 minutes, 33.3 km from A. The bird has flown 66.7 km and flies the same back to B, a round trip of 133.3 km. By then the car is 33.3 km from B, so the next round trip is 44.4 km. Each round trip is one third of the one before, so the laps form a geometric series that sums to 133.3 / (1 - 1/3) = 200 km.

Why would a finance interviewer ask this?

It tests whether you look for the quantity that controls a problem before you start calculating. Most analysis questions hide one controlling number under a lot of detail, and the skill is finding it first. In a model review that might be the single assumption the valuation hangs on; in a variance analysis it might be one volume line. Give the two hour answer first, then offer the series as a check. Speed plus a second method is exactly what the interviewer is listening for.

36 ways for 10, 45 for 11, and (n minus 1)(n minus 2) / 2 in general. Write 10 as a row of ten ones. Three positive parts means two dividers in two different gaps between the ones, and there are nine gaps, so the count is C(9, 2) = 36. For 11 there are ten gaps: C(10, 2) = 45. For any n it is C(n minus 1, 2). If order did not matter, 10 has only 8 splits, and if zero were allowed it would have 66, so say which version you are answering.

How do you turn 'three numbers add to 10' into something you can count?

Ten sweets in a row, three children, each must get at least one. You hand them out by cutting the row in two places, and every pair of cuts is one way to share. Each ordered triple is one choice of two distinct gaps out of the nine between ten objects, so counting triples is the same as counting pairs of gaps: 9 choose 2. That is 9 x 8 / 2 = 36. The same picture gives 11 at once: eleven sweets have ten gaps, and 10 x 9 / 2 = 45.

Check the small cases before trusting the formula. A total of 3 can only be 1 + 1 + 1, one way, and C(2, 2) = 1. A total of 4 gives 1 + 1 + 2 in three orders, and C(3, 2) = 3. A total of 5 gives six, C(4, 2). The counts 1, 3, 6, 10, 15 are the triangular numbers, which is what (n minus 1)(n minus 2) / 2 produces.

What changes if the interviewer meant something else?

Three variants are common, and the right first move is to ask or to state your reading. If zero is allowed, the dividers may share a gap or sit at the ends, and the count becomes C(n + 2, 2), which is 66 for 10. If order does not matter, you are counting partitions of 10 into three parts: 1+1+8, 1+2+7, 1+3+6, 1+4+5, 2+2+6, 2+3+5, 2+4+4, 3+3+4, 8 in all, and there is no neat formula, so you list them from the smallest part upwards. Say which version you are answering, then answer it; the count is easy once the question is fixed, and most lost marks here come from solving a different question than the one asked. If negative numbers were allowed the count would be infinite, which is worth one sentence to show you noticed.

Why would a trading desk ask this?

Counting outcomes is the floor under every probability question, and the follow-up about 11 tests whether you hold a method or a memorised answer. The natural next step is three dice: how many of the 216 rolls add to 10? Start from 36 ordered triples and remove those with a part above 6: an 8 with two 1s gives 3 orders, a 7 with 1 and 2 gives 6, so 36 minus 9 is 27, and the chance is 27/216, 12.5%. The limitation of the gap method is exactly that it has no upper bound on any part, so a capped problem needs the subtraction.