Investment Banking puzzles, solved step by step

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Showing 1–3 of 3 · filtered from 100Clear filters
  1. 036You must staff a three-person deal team from eight analysts, but two of them refuse to work together. How many different teams are possible?Mental maths and countingCoreBulge bracket IBMiddle market IB

    Try it first

    Quick instinct: how many teams?

    Show the worked solution

    50 teams. Choosing any 3 of 8 analysts gives 8 x 7 x 6 / 6 = 56 teams. The forbidden ones contain both analysts who refuse to work together, and the third seat can go to any of the other six, so there are 6 forbidden teams. 56 minus 6 leaves 50. Count everything, then subtract the cases the rule forbids.

    Why count the teams you do not want?

    Think of seating guests at a dinner when two of them have fallen out. Listing every acceptable table is slow; counting all possible tables and removing those that seat the two together is quick. When a rule forbids a small set of cases, count everything and subtract the forbidden ones, because the forbidden set is usually the easy one to count. Here a forbidden team is the two feuding analysts plus a third person, and there are only 6 choices for the third.

    All 56 teams of three from A to H; A and B will not work togetherA B CA B DA B EA B FA B GA B HA C DA C EA C FA C GA C HA D EA D FA D GA D HA E FA E GA E HA F GA F HA G HB C DB C EB C FB C GB C HB D EB D FB D GB D HB E FB E GB E HB F GB F HB G HC D EC D FC D GC D HC E FC E GC E HC F GC F HC G HD E FD E GD E HD F GD F HD G HE F GE F HE G HF G HBoth A and B: 6, struck outExactly one of A, B: 30Neither: 20Count all: 8 x 7 x 6 / 6 = 56. Remove A + B + any of the other 6: 6.50 teams
    Of the 56 possible three-person teams from eight analysts, only the 6 that contain both A and B break the rule, so 50 remain: 20 with neither of the pair and 30 with exactly one.

    How do you get 56 without a formula sheet?

    Fill the team one seat at a time: 8 choices, then 7, then 6, which is 336 ordered picks. Each team of three turns up in 3 x 2 x 1 = 6 different orders, so divide by 6 to get 56 distinct teams. That is the combinationA selection where order does not matter. The number of ways to choose k from n is written n choose k. count, 8 choose 3. Subtract the 6 forbidden teams and 50 remain.

    The relationship
    (83)−(61)=56−6=50\binom{8}{3} - \binom{6}{1} = 56 - 6 = 50
    8 choose 3every team of three from eight analysts
    6 choose 1teams holding both feuding analysts: one choice of third member from the other six
    What it says in wordsAll teams, less the teams that contain the forbidden pair.

    How do you check 50 a second way?

    Count the allowed teams directly, split by how many of the feuding pair they include. Teams with neither of the two are 6 choose 3, which is 20; teams with exactly one are 2 times 6 choose 2, which is 30; together 50. Two methods agreeing is the answer the interviewer remembers, and the second method is the one you will need when a follow-up adds a second rule.

    Where candidates lose it

    The common wrong answer is 20: candidates take both feuding analysts out of the pool and count teams of three from the remaining six. That throws away the 30 perfectly good teams containing one of them.

    The other slip is subtracting the feuding pair from 56 as if it were one team. The pair plus a third person can be formed 6 ways, so 6 teams go, not 1.

    What the interviewer asks next

    • One senior analyst must be on every team, and the feud still applies. How many teams now?
    • What if the team is four people instead of three?
    • Two separate pairs refuse to work together. How many three-person teams are possible?
  2. 068How many zeros are at the end of 100 factorial, written out in full?Mental maths and countingCoreBulge bracket IBMiddle market IB

    Try it first

    Before you count: how many trailing zeros?

    Show the worked solution

    24 trailing zeros. Each zero at the end is a factor of 10, which is a 2 paired with a 5. Among the numbers 1 to 100 there are 97 factors of 2 but far fewer 5s, so the 5s decide it. The 20 multiples of 5 contribute one 5 each, and the four multiples of 25 contribute a second, giving 20 + 4 = 24. There is no multiple of 125 below 100, so there is no third layer.

    Why is the answer about fives and not about tens?

    Think of making pairs of shoes from a pile of lefts and rights. If you have 97 left shoes and 24 right shoes, you can make 24 pairs; the extra lefts do not help. A trailing zero is a pair of one 2 and one 5 multiplied together, and in 1 to 100 the 2s vastly outnumber the 5s, so the number of 5s sets the number of zeros. Counting multiples of 10 misses the zeros built from a 5 in one number and a 2 in another, such as 5 x 2 = 10 or 15 x 4 = 60.

    Count the fives: every multiple of 5 gives one, every multiple of 25 gives a second510152025525 x3035404550525 x5560657075525 x80859095100525 xone factor of 5 from each multiple of 5: 20a second factor of 5 from each multiple of 25: 4 more125 is above 100, so no third layerFives in 100! = 20 + 4= 24 trailing zerosTwos in 100! = 97: plenty to pairEach trailing zero is one factor of 10,which is one 2 paired with one 5.Twos are everywhere; fives are the scarce part,so count the fives.the four multiples of 25 are the step people miss
    The 20 multiples of 5 from 5 to 100 each contribute one factor of 5, and the four multiples of 25 contribute a second one, so 100 factorial contains 24 fives, each paired with one of its 97 twos to make 24 trailing zeros.

    Where do the extra four come from?

    A number like 50 is 2 x 5 x 5: it carries two fives, not one. Every multiple of 25 hides a second five, and 25, 50, 75 and 100 are the four of them below 100, so the count is 20 from the multiples of 5 plus 4 from the multiples of 25. The next layer would be multiples of 125, which carry a third five, but 125 is beyond 100, so the count stops at 24. The pattern continues for larger factorials: divide by 5, then 25, then 125, and add the whole-number parts.

    The relationship
    Z=⌊1005⌋+⌊10025⌋+⌊100125⌋=20+4+0=24Z = \left\lfloor \tfrac{100}{5} \right\rfloor + \left\lfloor \tfrac{100}{25} \right\rfloor + \left\lfloor \tfrac{100}{125} \right\rfloor = 20 + 4 + 0 = 24
    Zthe number of trailing zeros
    floorthe whole-number part of the division
    5, 25, 125each power of 5 adds a layer of fives
    What it says in wordsAdd up the whole-number parts of 100 divided by 5, by 25 and by 125, because each layer counts one more five per number.

    How do you check it without writing out 100 factorial?

    Test the method on a factorial you can write. 10 factorial is 3,628,800, with two trailing zeros, and the rule gives 10 over 5 = 2, plus 10 over 25 = 0, which matches. Showing the rule on a small case you can verify is the fastest way to earn trust for the big case you cannot. Then say the general version: the number of trailing zeros of n factorial is the whole-number sum of n over 5, n over 25, n over 125 and so on, and for 1,000 factorial that is 200 + 40 + 8 + 1 = 249.

    Where candidates lose it

    The fast wrong answers are 10, from counting multiples of 10, and 20, from counting multiples of 5 once. The second is close enough that candidates say it with confidence and lose the point on the four multiples of 25.

    The other loss is explaining it as a memorised formula. Say why fives are the scarce factor first; the formula then sounds like reasoning rather than recall.

    What the interviewer asks next

    • How many trailing zeros does 1,000 factorial have?
    • How many factors of 2 are in 100 factorial, and why does that number not matter here?
    • What is the smallest n for which n factorial ends in exactly 30 zeros, and is every count reachable?
  3. 088Aptitude test style: you convert Rs 5 lakh to dollars at Rs 83.50 per dollar, and the bank first deducts a 1.5% fee on the rupee amount. How many dollars do you receive, to the nearest dollar?Mental maths and countingCoreBulge bracket IBMiddle market IB

    Try it first

    Pick the answer the test is looking for.

    Show the worked solution

    5,898 dollars. The fee is 1.5% of Rs 5,00,000, which is Rs 7,500, leaving Rs 4,92,500 to convert. Divided by 83.50 that is about 5,898.2. A quick route: 83.5 x 6,000 is 5,01,000, so Rs 5 lakh buys about 5,988 dollars before the fee, and taking 1.5% off 5,988 gives the same 5,898.

    What is the question really testing?

    These questions sit in the timed numerical section of an online assessment, and the arithmetic is easy. What they test is whether you apply each adjustment to the right base under time pressure. Think of a restaurant bill with a 10% service charge: it is 10% of the food, not of the food plus tax, and getting the base wrong gives a number that looks almost right. The test designers put the almost-right numbers in the options, so the base you apply the fee to decides whether you pick the correct one.

    Take the fee off the rupee amount, then convertYou hand overRs 5,00,000less 1.5% fee- Rs 7,500left to convertRs 4,92,500divide by 83.50$5,898Mental shortcut: 83.5 x 6,000 = 5,01,000, so Rs 5,00,000 buys 6,000 - 1,000/83.5 = 5,988 dollars before the fee.Where answers go wrongCorrect: deduct 1.5% of Rs 5,00,000, then convert$5,898Fee forgotten: 5,00,000 / 83.50$5,988Wrong base: 5,00,000 / 1.015, then convert$5,900Order swapped: 5,988 x 0.985, a % fee commutes$5,898
    Deducting the Rs 7,500 fee from Rs 5,00,000 and converting the remaining Rs 4,92,500 at 83.50 gives 5,898 dollars, while forgetting the fee gives 5,988 and dividing by 1.015 gives 5,900, both of which a test will list as options.

    Does it matter whether you take the fee before or after converting?

    Not for a percentage fee. Taking 1.5% off the rupees and then converting, or converting and then taking 1.5% off the dollars, multiplies by the same two numbers in a different order: 5,988 x 0.985 is also 5,898. Order matters only for a flat fee or a fee charged on a different base; a percentage fee on the same amount commutes with the conversion. Knowing this lets you pick whichever order makes the mental maths easier.

    The relationship
    $=5,00,000×(1−0.015)83.50=4,92,50083.50≈5,898\$ = \frac{5{,}00{,}000 \times (1 - 0.015)}{83.50} = \frac{4{,}92{,}500}{83.50} \approx 5{,}898
    5,00,000the rupee amount handed over
    0.015the fee, 1.5% of the rupee amount
    83.50rupees per dollar
    What it says in wordsDollars received equal the rupees left after the fee divided by the rupee price of one dollar.

    Where does 5,900 come from? Dividing by 1.015 answers a different question: how much could you convert if the 1.5% were charged on top of the converted amount. That base is smaller than Rs 5 lakh, so the fee comes out as about Rs 7,389 instead of Rs 7,500. On a test with options two dollars apart, that is a wrong answer.

    Where candidates lose it

    The trap is speed. Candidates divide 5,00,000 by 83.5, see 5,988 in the options and move on, or remember the fee and divide by 1.015 because it feels like the same thing. Both are listed because both are common.

    Read what the fee is charged on, write the base down, and check the order of magnitude: about six thousand dollars, a little under.

    What the interviewer asks next

    • What if the bank also charged a flat Rs 500 per transaction?
    • What exchange rate are you effectively getting after the fee?
    • How would you convert the dollars back, and what do the two fees cost you in total?