Investment Banking puzzles, solved step by step

About Rs 50. Prices rising 6% a year compound to 1.06 to the power of 12, which is 2.01, so a basket costing Rs 100 today costs about Rs 201 in year 12. Rs 100 then buys what Rs 49.70 buys now. The rule of 72 gets there in one line: 72 divided by 6 is 12 years for prices to double.

Why does the answer land so close to half?

Think of a plate of biryani that costs Rs 100 today. If its price rises 6% every year, each rise is taken on the previous year's higher price, so the rises themselves grow: Rs 6 in the first year, close to Rs 12 by the twelfth. Inflation compounds like interest, so over 12 years at 6% prices roughly double, and the same Rs 100 note buys roughly half as much. To state a future rupee in today's money, you divide by the growth in prices, 1.06 to the 12th.

How do you get 1.06 to the 12th without a calculator?

Square your way up. 1.06 squared is 1.1236; squared again gives 1.2625 for four years; squared again gives 1.594 for eight years. Twelve years is eight plus four, so multiply: 1.594 x 1.262 is 2.012. Repeated squaring turns eleven multiplications into four, and every step can be said out loud. Then Rs 100 divided by 2.012 is Rs 49.70, and the rule of 72 has already told you to expect about 2.

What are the two wrong answers, and why are they tempting?

The straight-line answer, Rs 28, takes Rs 6 off every year as if prices rose by the same rupee amount each time. The subtler one, Rs 48, shrinks the money by 6% a year, 0.94 to the 12th. Inflation raises prices by 6%, which cuts buying power by 6 divided by 106, about 5.7% a year, not by a full 6%. The gap between Rs 48 and Rs 50 is small here and widens with the rate and the years.

Why a deal team cares: a return in rupees is not a return in buying power. Money growing at an assumed 12% a year while prices rise 6% grows in real terms by 1.12 divided by 1.06, minus one, which is 5.66% a year, a little under the 6% you get by simple subtraction. Over a long holding period that gap compounds too.

Six flips on average. Track where you stand: at the start, or one head up. From the start, one flip takes you to one head or leaves you at the start; from one head, a head finishes the game and a tail sends you back. Writing the expected flips from each state as an equation and solving gives 6 from the start and 4 from one head.

Why is the obvious answer of 4 wrong?

The guess of 4 comes from the one in four chance that two given flips are both heads. Think instead of climbing a slippery two-step ladder where any slip drops you to the ground, not one rung down. A tail after a head costs you the head you already had, so progress is lost and the wait is longer than the simple odds suggest. The clean way to handle lost progress is to give each position its own equation.

How do the equations work?

Call E the expected flips from the start and E(H) the expected flips once you hold one head. Every flip costs one. From the start, half the time you move to one head and half the time you are back where you began. From one head, half the time you finish and half the time a tail sends you to the start. Each equation reads: one flip, plus the average of the waits from wherever that flip leaves you. Substitute the second into the first and E = 1.5 + 0.75E, so E = 6 and E(H) = 4. This way of setting up the problem is called a Markov chainA process where what happens next depends only on the current state, not on how you got there..

How do you check 6 a second way?

Wait for the first head, which takes 2 flips on average. Flip once more: half the time it is a head and you are done; half the time it is a tail and you start from scratch. So E = 3 + E/2, which again gives 6. Two routes to the same number is the strongest answer you can give in the room. Simulated 200,000 times, the average comes out at 6.01. Then add the twist interviewers like: waiting for heads then tails takes only 4, because a failed attempt at it, a second head, still leaves you one head up.

Enterprise value is unchanged, equity value rises by Rs 100 crore and EV/EBITDA stays the same. Equity value rises because Rs 100 crore of new shares now exist, and debt falls by the same Rs 100 crore. Enterprise value is equity plus net debt, so the two moves cancel. EBITDA sits above the interest line and is untouched, so the multiple cannot change either.

Why does enterprise value not move?

Think of a house worth Rs 1 crore with a Rs 40 lakh mortgage. If you pay down Rs 10 lakh of the loan with your savings, your stake rises to Rs 70 lakh and the loan falls to Rs 30 lakh, but the house is worth exactly what it was. Enterprise value measures the business, and swapping one kind of claim for another changes how that value is split, not how big it is. At first order, issuing shares to repay debt is exactly that swap.

Then what does change?

Everything below the interest line. Take EBITDA of Rs 100 crore, D&A of Rs 20 crore, Rs 400 crore of debt at 10% and a 25% tax rate, with 60 crore shares at Rs 10. Interest falls from Rs 40 crore to Rs 30 crore, so net income rises from Rs 30 crore to Rs 37.5 crore, and with equity now worth Rs 700 crore, P/E falls from 20.0x to 18.7x. Equity multiples such as P/E move with capital structure; enterprise multiples such as EV/EBITDA do not, which is why bankers compare companies on enterprise multiples.

EPS also rises, from Rs 0.50 to Rs 0.54. The reason is a comparison of two costs: the new shares were sold at an earnings yieldNet income divided by equity value, the inverse of the P/E. It is what each rupee of equity earns. of 5%, and the debt they retired cost 7.5% after tax. Replacing a 7.5% cost with a 5% cost lifts earnings per share.

What could make enterprise value move after all?

Three second-order effects, worth one sentence each. Fees on the share issue leak value out to advisers. Less debt means a smaller interest tax shield, which in Modigliani and Miller's analysis with taxes lowers value slightly. Against that, lower leverage reduces the risk and cost of financial distress. In an interview, give the first-order answer, unchanged, and then name the effects that could nudge it.

7 races. Race five heats of five, then race the five heat winners: the winner of race 6 is the fastest overall. Now strike out every horse that has three horses known to be faster. Only five can still be second or third: the second and third from the winner's heat, the first two from the runner-up's heat and the winner of the third-placed heat. Race them; the top two are second and third overall.

What does each race actually tell you?

Without a stopwatch, a race tells you the order only among the horses in it. Think of five classrooms each running their own sprint: you know the fastest in each room, but nothing about how one room's second best compares with another room's winner. Every horse must race at least once, so five heats are unavoidable, and they give five separate rankings that a sixth race between the heat winners stitches together. Name the heats A to E in the order their winners finished race 6, so A1 is the fastest horse of all.

Which horses can you strike out after race 6?

Any horse with three horses known to be faster cannot finish in the top three. Heats D and E go entirely, because D1 and E1 already finished behind A1, B1 and C1, and everything in those heats is slower still. In heat C only C1 survives: C2 trails C1, B1 and A1. In heat B, B1 and B2 survive, but B3 trails B1, B2 and A1. In heat A, A2 and A3 survive and A4 trails A1, A2 and A3. That leaves exactly five: A2, A3, B1, B2 and C1.

Race 7 puts those five on the track, and its first two finishers are second and third overall. A1 sits out, because it is already known to be the fastest. Could six races ever be enough? No: the five heats alone cannot name the fastest horse, and the race that does name it leaves five horses still unranked for second and third. That is the reasoning to say out loud, because it shows the 7 is a minimum rather than just a method that works.

Two thirds. You are not choosing among cards; you are looking at a face. There are three red faces you could be seeing, all equally likely: two belong to the red-red card and one to the red-white card. In two of the three cases the hidden side is red. The tempting answer of one half treats the two red-capable cards as equally likely and ignores that the red-red card shows red twice as often.

Why is one half so tempting, and where does it go wrong?

Imagine two families at a party: one with two daughters, one with a daughter and a son. If you meet one of the girls at random, she is twice as likely to come from the two-daughter family, because that family brought two girls. The red-red card is twice as likely to be the one showing red, because it has two red faces to show. Treating the two remaining cards as equally likely throws that information away, which is exactly what the question is built to catch.

How do you prove it with Bayes' rule?

Write it as a conditional probability using Bayes' ruleA formula for updating a probability after seeing evidence: the chance of the evidence given the cause, times the prior chance of the cause, divided by the overall chance of the evidence.. The chance of drawing the red-red card is one third, and if you did, you are certain to see red. The chance of seeing red at all is the share of red faces, three of six, one half. So the chance the card is red-red, given a red face, is one third times one, divided by one half: two thirds.

How do you check it without any formula?

Imagine drawing 600 times, each card about 200 times. The red-red card shows red all 200 times, the red-white card shows red about 100 times, and the white card never does. Of the 300 red sightings, 200 have red on the back. Counting outcomes in a large imagined sample is a reliable check whenever a conditional probability feels slippery. The puzzle is a version of Bertrand's box paradox, and the same reasoning sits under the Monty Hall problem.

Net income falls Rs 7.5, cash rises Rs 2.5, inventory falls Rs 10 and retained earnings fall Rs 7.5. The write-down is an expense, so pre-tax income falls 10 and, after the Rs 2.5 tax saving, net income falls 7.5. No cash left the business, so the cash flow statement adds the 10 back and cash ends Rs 2.5 higher. Assets fall 7.5 and equity falls 7.5, so it balances.

Why does a loss leave the company with more cash?

A shopkeeper finds a carton of biscuits past their date. They cost Rs 10 and are now worth nothing. No money changes hands today; the cash went out when the biscuits were bought. What changes is that the loss lowers this year's taxable profit. A write-down is a non-cash loss, so its only cash effect is the tax it saves, and cash rises by 25% of Rs 10. Inventory is carried at the lower of cost and net realisable valueWhat the stock can be sold for, less the costs of selling it., which is why the carrying amount is cut when goods lose value.

How does each statement move?

Income statement: the write-down usually sits inside cost of goods sold, so pre-tax income falls 10, tax falls 2.5 at 25%, and net income falls 7.5. Cash flow statement: start from net income of minus 7.5, add back the 10 because no cash left, and operating cash flow is plus 2.5. Balance sheet: inventory is down 10 and cash is up 2.5, so total assets are down 7.5, and retained earnings are down 7.5. The check is minus 10 plus 2.5 on the asset side equalling minus 7.5 in equity.

What assumption should you say out loud?

That the write-down is deductible for tax now, as the question states. Whether a tax system allows the deduction when the stock is written down or only when it is sold depends on its rules, which you would confirm. If the deduction comes later, cash does not move this year and the company records a deferred tax asset of Rs 2.5 instead, with the balance sheet still balancing. Offering that variant in one sentence shows you understand why the cash moved in the first place.

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.

About 8 years: 72 divided by 9 is 8.0, and the exact answer is 8.04 years. The exact doubling time is the log of 2 divided by the log of 1.09. The rule is a shortcut built for moderate rates: it is almost exact around 8%, slightly long at low rates and increasingly short at high ones. At 40% it says 1.8 years against an exact 2.06.

Why does 72 work at all?

Think of a sapling that grows 9% taller each year; the question is how many of those steps multiply up to 2. The exact answer uses logarithms: years equal ln 2 divided by ln(1 + r). For small r, ln(1 + r) is close to r, and ln 2 is 0.693, so the exact rule is close to 69.3 divided by the rate in per cent. 72 replaces 69.3 because it divides cleanly by 2, 3, 4, 6, 8, 9 and 12, and because ln(1 + r) sits below r at the rates people actually meet, which pushes the true constant up. At 9%, ln 1.09 is 0.0862, and 0.693 / 0.0862 is 8.04.

Where does the rule break down?

At high rates. ln(1 + r) falls further below r as r grows, so the true doubling time is longer than 72 divided by r: at 20% the rule says 3.6 years against 3.80, and at 40% it says 1.8 against 2.06, an error of 12.6%. At very low rates it errs the other way: 36 years against 35.0 at 2%. The rule is within about 1% of the exact answer only between roughly 6% and 10%, and should be adjusted outside that band.

A common adjustment for high rates adds one to the 72 for every three points of rate above 8%. At 20% that gives 76 divided by 20, 3.80 years, and at 40% about 82.7 divided by 40, 2.07 years, both within a hundredth or two of the exact figures. For deal work, where target returns of 20% to 30% are common, that adjustment is worth knowing.

115.6. Notice that 34% of 170 is the same as 17% of 340: halving one number and doubling the other leaves a product unchanged. So the sum is 17% of 340 twice, which is 34% of 340. That is 30% of 340, which is 102, plus 4% of 340, which is 13.6, giving 115.6.

What is the trick hiding in the numbers?

Two plots of land, one 34 metres by 17 and one 17 metres by 34, have the same area; turning a rectangle on its side does not change it. Percentages behave the same way, because x% of y is x times y divided by 100. The percentage and the base can trade places, so 34% of 170 equals 17% of 340: each is 17 x 340 / 100, which is 57.8. Once you see that, the question is one product, not two.

How do you do 34% of 340 in your head?

Split it into easy pieces: 30% of 340 is 102 and 4% of 340 is 13.6, so the total is 115.6. Or notice that 34% of 340 is 34 x 3.4, and 34 x 34 is 1,156, so the answer is 115.6. Spotting a structure first and calculating second is the habit the interviewer is checking, and it usually turns two awkward products into one easy one.

How do you check it before you say it?

Estimate first: 17% is a little more than a sixth, and a sixth of 340 is about 57, so each term is near 57 and the sum near 115. Then check each term directly: 10% of 340 is 34 and 7% is 23.8, so 17% is 57.8; 30% of 170 is 51 and 4% is 6.8, so 34% of 170 is 57.8. The two terms match, which confirms the swap.

Expected wealth is Rs 9.31 lakh, but the most likely outcome is Rs 1 lakh, exactly where you started. Each round multiplies the stake by 1.25 on average, and 1.25 to the tenth is 9.31. Yet the wealth you actually hold is 2 to the power of heads minus tails, and the likeliest split is five and five, which multiplies to one. The median player goes nowhere; the mean is carried by a handful of lucky runs.

Why is the average so far from the typical result?

Ten friends each put Rs 1 lakh into this game. Most finish near where they began, a few lose most of their money, and one, with a long run of heads, finishes with hundreds of lakh. Add it all up and divide by ten and the average looks wonderful; ask the friend in the middle how it went and the answer is that nothing happened. Gains and losses compound, and a doubling followed by a halving lands exactly back where you started, so the mean grows 25% a round while the median stays flat. The arithmetic average of 2 and 0.5 is 1.25; their geometric average, the square root of 2 times 0.5, is exactly 1, and compounding follows the geometric one.

How do you get both numbers in two lines?

For the mean, use the fact that the expectation of a product of independent rounds is the product of the expectations: 1.25 to the tenth, 9.31. For the typical outcome, count heads. After ten flips the wealth is Rs 1 lakh times 2 to the power of heads minus tails, which is 2 to the power of 2H minus 10. Five heads is the single likeliest count, 24.6% of games, and it gives a multiplier of exactly one; it is also the median, because the outcomes sit symmetrically on either side of it. 37.7% of players finish below Rs 1 lakh and the same share above, and you can say all of that without touching the mean.

Where does the Rs 9.31 lakh come from, then?

From the tail. A run of ten heads happens once in 1,024 games and turns Rs 1 lakh into Rs 1,024 lakh, which on its own adds a full Rs 1 lakh to the mean. The three best outcomes, eight or more heads, occur in 5.5% of games and contribute Rs 6.31 lakh of the Rs 9.31 lakh average, about 68% of it. Nothing is wrong with the expected value; it is the wrong statistic for a question about what will probably happen to you. The right one is the geometric meanThe growth rate that compounding actually delivers: the nth root of the product of n multipliers. It is never above the arithmetic mean. return, which here is zero.

The lesson a desk draws is about sizing. Bet only half your stake each round and the multipliers become 1.5 and 0.75, whose geometric mean is 1.0607: the typical player now grows about 6.1% a round and finishes ten rounds near Rs 1.80 lakh, even though the expected value per round has fallen from 1.25 to 1.125. Half is the fraction that maximises the typical growth rate in this game. The same arithmetic sits under volatility drag in fund returns: a strategy that gains 50% and then loses a third has a flattering average and nothing to show for it.