Financial Analysis puzzles, solved step by step

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100
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All topicsAccounting flow riddles10Valuation and multiples riddles10Ratio and margin riddles8Cost of capital, leverage and rates8Compounding and time value8Mental maths8Probability and expected value9Working capital and cash riddles6Percentages and averages7Estimation and market sizing7Logic and counting brainteasers7Pricing, costing and unit economics6Data and statistics intuition6
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  1. 011You have twelve coins that look identical. One is counterfeit and is either heavier or lighter than the rest; you do not know which. Using a balance scale only three times, find the counterfeit and say whether it is heavy or light.Logic and counting brainteasersWarm upConsulting-style caseBank credit

    Try it first

    How many coins go on each pan in the first weighing?

    Show the worked solution

    Weigh four against four with four left off, then use known good coins to mix the suspects. If the first weighing balances, the fake is among the four left off, and two more weighings find it. If it tips, the fake is one of four possibly heavy or four possibly light coins; weighing 1, 2 and 5 against 3, 6 and a good coin splits those eight into groups of three or fewer, and the third weighing names the coin and its direction.

    How do you know three weighings can be enough?

    Count the possible answers before touching a coin. Any of 12 coins could be the fake, and it could be heavy or light, so there are 24 answers. A balance has three outcomes, left down, right down or level, so three weighings give 3 x 3 x 3 = 27 outcomes. Twenty four answers fit inside twenty seven outcomes, but only just, so every weighing must split the remaining possibilities into three nearly equal groups. Think of the game of twenty questions: a question with three possible answers is only worth asking if all three answers are about equally likely.

    Why four against four, and what happens next?

    Six against six can never balance, which throws away one of the three outcomes. Four against four with four left off splits the 24 answers into 8, 8 and 8. If it balances, the eight coins on the scale are proven good, and you can use them as reference weights in the next weighing. Weigh three of the suspects, 9, 10 and 11, against three good coins. Level means coin 12 is fake and one last weighing against a good coin says heavy or light. A tip tells you both that the fake is among 9 to 11 and which way it is off, so weighing 9 against 10 finishes it.

    Three weighings, 27 outcomes, 24 possible answersWeigh 1 2 3 4 v 5 6 7 8Balances: fake is in 9 to 12Next: 9 10 11 v 1 2 3 (good)Left heavy: 1-4 heavy or 5-8 lightNext: 1 2 5 v 3 6 9Left lightMirror imagebalancesonly 12 leftweigh 12 v 112H or 12Lleft heavy9-11 heavyweigh 9 v 109H 10H 11Hleft light9-11 lightweigh 9 v 109L 10L 11Lbalances4H, 7L or 8Lweigh 7 v 87L 8L 4Hleft heavy1H, 2H or 6Lweigh 1 v 21H 2H 6Lleft light3H or 5Lweigh 3 v 93H 5LSame tree,heavy and lightswapped8 more answers12 coins x heavy or light = 24 possible answers. 3 x 3 x 3 = 27 outcomes from three weighings.So the first weighing must split the 24 into 8 / 8 / 8: four coins a side, four left off.
    Weighing four coins against four splits the 24 possible answers into three groups of 8. Each branch then uses a second weighing that mixes suspects with known good coins, so every one of the 24 answers ends at a single coin and a heavy or light verdict after three weighings.

    If the first weighing tips with the left side heavy, the fake is one of 1 to 4 and heavy, or one of 5 to 8 and light. Now move coins between pans. Weigh 1, 2 and 5 against 3, 6 and good coin 9: each outcome leaves at most three suspects, each with a known direction. Level leaves 4 heavy, 7 light or 8 light, and weighing 7 against 8 settles it. Left heavy leaves 1 heavy, 2 heavy or 6 light; weigh 1 against 2. Left light leaves 3 heavy or 5 light; weigh 3 against a good coin.

    The relationship
    2N≤3n−3⇒N≤33−32=122N \le 3^{n} - 3 \quad\Rightarrow\quad N \le \frac{3^{3}-3}{2} = 12
    Nnumber of coins
    nnumber of weighings, here 3
    2Npossible answers: each coin, heavy or light
    What it says in wordsThree weighings can handle at most twelve coins when you must also say heavy or light, so twelve is the hardest version that still works.

    Where candidates lose it

    The usual loss is opening with six against six, or splitting into halves out of habit. The scale tips, you have twelve candidates and two weighings left, and nine outcomes cannot cover them.

    The second loss is keeping the suspects in fixed groups after the first weighing. The solution only works because you move coins between pans and bring in coins already proven good. Say that idea out loud even if the details of a branch take you a moment.

    What the interviewer asks next

    • What if you only had to find the fake, not say whether it is heavy or light?
    • With four weighings, how many coins could you handle?
    • If you are told the fake is heavier, how many coins can three weighings handle?
  2. 081You have two ropes. Each takes exactly 60 minutes to burn from one end to the other, but they burn unevenly: half the rope does not take half the time. With a lighter and nothing else, how do you measure exactly 45 minutes?Logic and counting brainteasersWarm upConsulting-style caseCorporate finance

    Try it first

    If you light one rope at both ends at the same moment, when does it finish?

    Show the worked solution

    Light rope A at both ends and rope B at one end, together. When A burns out, 30 minutes have passed; light B's other end, and B finishes 15 minutes later, at 45 minutes. Lighting both ends halves any rope's remaining burn time whatever its speed profile, because two flames use the time up together. B has 30 minutes of burn left at the half hour, and two flames clear that in 15.

    Why does lighting both ends give exactly half, when the rope burns unevenly?

    Picture two friends eating one plate of biryani from opposite sides. One side is all rice and goes fast, the other is full of bones and goes slowly. They will meet somewhere off centre, but if the plate takes one person 60 minutes, two people eating at the same moment finish it in 30, because between them they are always doing two minutes of eating every minute. The rope's total burn time is fixed at 60 minutes; two flames consume it twice as fast, so it lasts 30 minutes regardless of where the slow sections are.

    That is the whole trick. Measuring by length fails, which is why cutting a rope in half or marking its midpoint gets you nothing: the midpoint of the string is not the midpoint of the time. Measuring by burn time is exact.

    Light A at both ends and B at one; when A dies, light B's other endRope Aboth ends lit: out at minute 30burnt outRope Bone end lit: 30 min of burn usedboth ends: 15 min015304560stop: exactly 45 minutesminute 30: light B's other endminutesRope A as a length: the flames meet off centre, each after 30 minutesleft flame, 30 minright flame, 30 min: this stretch burns fasterflames meet at minute 30physical midpoint, irrelevant
    Rope A, lit at both ends, burns out at minute 30 wherever its flames meet, and that is the signal to light rope B's other end, which turns B's remaining 30 minutes of burn into 15 and stops the clock at exactly 45 minutes.

    Where do the last 15 minutes come from?

    Rope B was lit at one end at the same moment as A. When A dies at minute 30, B has burned for 30 minutes, so it holds exactly 30 minutes of burn time, although nobody knows how long the remaining piece is. Lighting B's unlit end at that moment halves its remaining 30 minutes to 15, so B goes out at 30 plus 15, or 45 minutes. Both ropes are doing the same job: A is a 30 minute timer, and B is a 60 minute timer that you shorten halfway through.

    Say the steps in order and say the start condition: all three ends must be lit at the same instant. An interviewer listening to a phone screen is checking that the sequence is clean and that you explain why it works, not just that you have heard the riddle before.

    What else can two ropes measure?

    With one rope you can time 60 or 30 minutes. With two you can add 15, 45, 75 and 90: 90 by burning one rope at both ends and then the other at one end, 75 by burning A at both ends, then B at one end for 30 minutes and lighting its second end. Each rope gives you only halving moves on whatever time it has left, so the times you can build are sums of 60, 30 and 15. You cannot get 20 minutes with these ropes, and saying why is a strong close.

    Where candidates lose it

    The common wrong answer folds or cuts the rope in half to get 30 minutes. That assumes even burning, which the question rules out in its first sentence. The interviewer is checking whether you noticed that the only reliable quantity is burn time, not length.

    The other way to lose the point is to give the right recipe with no reason. If you cannot explain why two flames take exactly half the time, the answer sounds memorised, and the follow-up about 75 minutes or 20 minutes will expose it.

    What the interviewer asks next

    • How would you measure exactly 15 minutes with the same two ropes?
    • Can you measure 20 minutes? If not, why not?
    • With three such ropes, what is the longest time you can measure, and what new times become possible?