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.

531,441. Rewrite 9 to the 6th as the square of 9 cubed. 9 cubed is 729. Then square 729 by splitting it into 700 plus 29: 700 squared is 490,000, twice 700 times 29 is 40,600, and 29 squared is 841. They add to 531,441. As a check, it should be a little over half a million, because 0.9 to the 6th is about 0.53.

How do you keep every step in your head?

Think of carrying a heavy box up six floors: you take it in two flights of three with a rest between, not six lifts in a row. Rewriting a power as the square of a smaller power cuts the work to one easy cube and one square you can split. 9 squared is 81, and 81 times 9 is 729, so 9 cubed is 729. Then 9 to the 6th is 729 squared, and squaring a number near a round one is the most practised move in mental maths.

How do you check it before saying it?

Three quick checks. An even power of 9 ends in 1, because 81 does, and 531,441 ends in 1. Any power of 9 is divisible by 9, so its digits must sum to a multiple of 9: 5 + 3 + 1 + 4 + 4 + 1 is 18. And the size is right: 9 is 10% below 10, and six 10% trims leave about 53% of a million. A digit check and a size check together catch nearly every slip in the cross term, which is where most mental errors happen.

A second route, if you prefer it: 9 to the 6th is also 81 cubed. 81 times 81 is 6,561, and 6,561 times 81 is 6,561 times 80 plus 6,561, which is 524,880 plus 6,561, giving 531,441 again. Saying one route and confirming with the other is worth more than speed.

About Rs 450 crore. At full run-rate the synergies add Rs 75 crore of after-tax earnings, which the market capitalises at 8x: Rs 600 crore. Two things the headline ignores come off. Year 1 delivers only half, a shortfall of Rs 37.5 crore after tax, and the one-off cost of Rs 150 crore is Rs 112.5 crore after its tax saving.

Why can you not just multiply Rs 100 crore by 8?

A shop owner expects a new supplier to cut her costs by Rs 1 lakh a year. Before celebrating, she has to pay tax on the extra profit, wait for the new contract to start, and pay the one-off cost of switching. A synergy is worth the after-tax earnings it adds, capitalised, less everything it costs to achieve and every rupee that arrives late. The 8x multiple applies to after-tax earnings, so the pre-tax Rs 100 crore first becomes Rs 75 crore at 25% tax, and 75 x 8 is Rs 600 crore.

What comes off the Rs 600 crore?

Two things. Phasing: the market value assumes the full run-rate, but year 1 delivers only half, so Rs 50 crore of pre-tax savings, Rs 37.5 crore after tax, never arrive. Cost to achieve: Rs 150 crore of severance, systems work and site closures, assumed tax deductible, so it costs Rs 112.5 crore after tax. Net of phasing and the cost to achieve, the synergies are worth about Rs 450 crore, a quarter less than the after-tax figure and 44% less than the pre-tax headline.

What does this rough answer leave out?

Three refinements, each worth naming. The year-1 items fall a year from now, so discounting them would trim the deductions slightly. The 8x assumes the market gives synergies the same multiple as the core earnings, while in practice cost synergiesSavings from combining two businesses, such as closing duplicate offices or merging purchasing. earn more credit than revenue synergies and both are discounted for execution risk. The deductibility of integration costs is an assumption to confirm for the jurisdiction. The answer to give is the method and Rs 450 crore, then one line on which assumption you would test first.

The number matters because it caps what a buyer can sensibly pay away. A premium above roughly Rs 450 crore hands the target's shareholders more than the whole value the deal creates, before any risk that the savings slip.

It adds about Rs 18.5 crore, or 4.9%, taking the value from Rs 379.1 crore to Rs 397.6 crore. The mid-year convention assumes each year's cash arrives evenly through the year, so on average halfway through rather than at the end. Every cash flow is then discounted for half a year less, which multiplies each present value by 1.10 to the power of one half, 1.0488.

Why would cash arrive in the middle of the year?

A shop does not collect its year's takings in one lump on the last day of March; money comes in every day. A DCF that discounts each year's cash flow as if it all arrived on the final day is too harsh. The mid-year convention treats each year's cash flow as arriving, on average, halfway through the year, so it is discounted for t minus one half years instead of t. For a business with steady receipts that is the more accurate assumption, and it is the default in many banks' models.

Why is the uplift exactly the square root of 1.10 minus one?

Because the shift is the same for every cash flow. Moving a payment six months earlier multiplies its present value by 1.10 to the power of one half, whatever year it sits in. Since every cash flow gets the same uplift, the whole value rises by the same 4.88%, and neither the size nor the pattern of the cash flows changes that. Half the discount rate, 5%, is a fair first approximation to say out loud; at a 20% rate the exact uplift is 9.5%, against the 10% the approximation would give, so the shortcut weakens as rates rise.

Where should you not apply it?

Two cautions. A terminal valueThe value of all cash flows beyond the forecast years, usually from a growing perpetuity or an exit multiple. built on a perpetuity of flows is usually discounted on the same mid-year basis, but one built on an exit multiple is not, because a sale happens at a point in time, the end of the final year. Shifting an exit-multiple terminal value half a year earlier overstates the answer, and an interviewer who asks about the convention is often fishing for exactly that. For seasonal businesses that collect most cash late in the year, the convention itself flatters the value.

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.

About Rs 8. With near-zero rates, an at-the-money call is worth roughly 0.4 times the price times the volatility times the square root of time: 0.4 x 100 x 0.20 x 1 = 8.0. The full Black-Scholes formula gives Rs 7.97. The intuition: the option collects the upside of a bell-shaped spread of outcomes and nothing on the downside, and the average of that upside half is about 0.4 of one standard deviation.

What is Black-Scholes actually doing?

Picture a ticket that pays you whatever a stock gains over the next year and nothing if it falls. You would pay something for it, even though an at-the-moneyAn option whose strike price equals the current stock price, so exercising it today would pay nothing. ticket is worth nothing if cashed in today. Black-Scholes prices that ticket as the average payoff across all the ways the stock could move, with the spread of those moves set by volatility and time. Its inputs are the stock price, strike, time, volatility and interest rate, and with zero rates the formula reduces to S times N(d1) minus K times N(d2), where the two N terms come from the bell curve.

Where does the 0.4 come from?

Over a year at 20% volatility, the stock's likely outcomes form a bell curve around Rs 100 with a standard deviation of about Rs 20. The call keeps the right half and treats the left half as zero. The average of the positive half of a bell curve, counting the negative half as zero, is the standard deviation divided by the square root of 2 pi, about 0.4 of it. So the call is worth about 0.4 x Rs 20, which is Rs 8. Halve the time to six months and the spread shrinks by the square root of 2, so the option is worth about Rs 5.6, not Rs 4.

What does the shortcut leave out?

It holds for at-the-money options with modest volatility and low rates. Away from the strike, the payoff is no longer the clean right half of the curve, and with meaningful rates the forward price drifts above the spot, which raises a call's value. The rule is a sanity check on a quoted price, not a replacement for the formula. Black-Scholes itself assumes constant volatility and smooth price moves, which is why traders quote options in implied volatility and adjust for the skew they see in the market. Doubling volatility to 40% roughly doubles this call's value, to about Rs 15.9.

Give B heavy depreciation and more debt, so much less of its EBITDA reaches shareholders. Both earn EBITDA of Rs 100 crore. A has D&A of 2 and interest of 18, keeps net income of 60, and at equity value 720 trades on 12x; its EV of 900 is 9x EBITDA. B has D&A of 25 and interest of 35, keeps only 30, and at equity value 450 trades on 15x, though its EV of 800 is 8x.

How can one company be cheaper on one multiple and dearer on another?

Two shops each take in Rs 1 lakh a month before equipment and loan costs. One works from a simple counter with a small loan; the other leases costly machines it must keep replacing and carries a large loan. Priced on takings before those costs, they look alike; priced on what the owner keeps, they do not. EV/EBITDA values the business before depreciation and before financing, while P/E values what is left for shareholders after both, so a company with heavy D&A or heavy interest can look cheap on one and dear on the other.

Which line explains how much of the gap?

Work down from EBITDA. B carries Rs 23 crore more D&A and Rs 17 crore more interest than A, which after 25% tax is Rs 17.25 crore and Rs 12.75 crore of net income: together exactly the Rs 30 crore gap between A's 60 and B's 30. Here depreciation does more of the work than debt, which you can show with EV/EBIT, a multiple that charges D&A but ignores financing: A is 9.2x and B 10.7x, so A is cheaper there too.

So which company is actually cheaper?

It depends on whether B's D&A is a real cost. If it reflects equipment that must be replaced, which capital expenditureCash spent on buying or replacing long-term assets such as machines and buildings. running close to depreciation would confirm, then EBITDA flatters B and A is better value on every measure that counts the cost of staying in business. If B's D&A is mostly amortisation of acquired intangibles with no cash replacement, EBITDA is the fairer lens and B may be the cheap one. The answer the interviewer wants is the diagnosis: check capex against D&A and check leverage before trusting either multiple.