Financial Analysis puzzles, solved step by step

No: 20% then 10% off is 28% off, not 30%; buy two get one free is a 33.3% discount per unit. Successive discounts multiply, because the second applies to the already reduced price: 0.80 x 0.90 = 0.72, so you pay 72% of the original. Buy two get one free means paying for two units and taking three, so each unit costs two thirds of its price, a third off, but only if you wanted three.

Why do discounts not simply add?

A Rs 1,000 kurta at 20% off costs Rs 800. The cashier then takes 10% off what you are about to pay, which is Rs 800, so the second discount is worth Rs 80, not Rs 100. You pay Rs 720. Each successive discount is taken off a price that the earlier discounts have already reduced, so discounts combine by multiplying what is left: 0.80 x 0.90 = 0.72, or 28% off. The order does not matter, 10% then 20% also gives 0.72, but the base does.

The shortfall from simple addition is the product of the two discounts: 20% x 10% = 2 points, the discount on the discount. It is small here and grows with the size of the cuts: 50% then 50% off is 75% off, not 100%.

What is buy two, get one free really worth?

You pay for two units and walk out with three, so each unit costs two thirds of the list price. Buy two get one free is a 33.3% discount per unit, deeper than the flat 30%, but only on the condition that you buy three. Buy one, get the second at half price is milder: one and a half prices for two units, 25% off each. If you needed only one, the effective discount of either offer is zero.

Where does this show up at work?

Everywhere rates are applied one after another. A distributor margin taken on a price that already carries a retailer margin, a trade discount followed by a cash discount, a fee charged after another fee: all compound. An FP&A analyst who adds layered discounts to estimate net price will overstate the discount and understate revenue per unit. The same arithmetic runs in reverse for mark-ups: a 20% price rise followed by a 20% cut leaves you 4% below where you started, because 1.2 x 0.8 = 0.96.

The zero falls more: about 8.8% against about 6.4% for the coupon bond. Price sensitivity follows duration, the present-value-weighted average time to each cash flow. The zero's only cash flow is at year 10, so its duration is 10 years. The coupon bond returns part of its value early, so its duration is 7.25 years. Modified duration predicts falls of 9.26% and 6.71%; convexity makes the actual falls slightly smaller.

Why does the timing of cash flows decide the sensitivity?

Two friends each owe you Rs 1,000 in total. One will repay it all in ten years; the other pays Rs 80 a year and the rest at the end. If prices start rising faster and money loses value faster, which IOU loses more value? The one where every rupee is ten years away. The second friend's early payments are already in your hands and can be reinvested at the new higher rates. A bond's price sensitivity to yield depends on when its value arrives on average, not on its final maturity date. That average, weighted by present value, is the Macaulay duration.

How big is each fall?

Use modified duration, Macaulay duration divided by one plus the yield, as the percentage price change per point of yield. For the zero, 10 / 1.08 = 9.26; for the coupon bond, 7.25 / 1.08 = 6.71. So a one point rise should cut prices by about 9.3% and 6.7%. Repricing exactly: the zero goes from 46.32 to 42.24, down 8.80%; the coupon bond goes from par, 100.00, to 93.58, down 6.42%.

Why are the actual falls smaller than duration predicts?

Duration is the slope of the price curve at today's yield, a straight-line estimate. The real curve bows outward, so as yields rise each further step hurts a little less, and as yields fall each step helps a little more. That curvature, convexity, makes a plain bond fall less than duration predicts when yields rise and gain more than predicted when they fall. For a one point move the gap is small, under half a point here; for larger moves, add the convexity term or simply reprice the bond.

Two practical notes. Most Indian government bonds pay coupons twice a year, which shortens the duration slightly but leaves the answer unchanged. And the gap between the bonds widens with maturity and narrows as coupons fall: a coupon bond's duration always sits below its maturity, while a zero's always equals it.

About 61% from the first 10 years and 94% from the first 30. The perpetuity is worth Rs 100 / 0.10 = Rs 1,000. Payments beyond year N are themselves a perpetuity, starting later, so their share is 1/1.1^N: 38.6% beyond year 10 and 5.7% beyond year 30. Add growth and the far years matter much more: at 5% growth only 37% of the value arrives in the first decade.

Is there a quick way to split a perpetuity by year?

Suppose you are promised a pension of Rs 100 a year forever. Starting ten years from now, what you are owed is again a Rs 100 perpetuity, worth Rs 1,000 at that date; you just have to wait ten years for it. The value of the payments after year N is the full perpetuity value, discounted back N years, so its share of the total is 1/(1 + r)^N and the first N years take the rest. At 10% that is 1/1.1^10 = 0.386 beyond year 10 and 1/1.1^30 = 0.057 beyond year 30.

In rupees, the first ten years are worth Rs 614 of the Rs 1,000. Half the value arrives in the first 7.3 years, because 1.1 to the power 7.3 is 2. A useful mental marker: at 10%, value halves every seven years or so.

Why does this matter for a DCF?

A DCF forecasts a company for perhaps ten years and puts everything after into a terminal value. With no growth, the terminal value would already be 39% of the total. Companies are usually assumed to grow, and growth pushes value further out: at 5% growth and a 10% discount rate, the gap between them shrinks to about 4.8% a year, half the value arrives only after 14.9 years, and the terminal value is 63% of the total. Most of a DCF's value sits beyond the explicit forecast, so the terminal growth rate and discount rate deserve more scrutiny than the detail of year three.

The limitation runs the other way too. The terminal value depends on the gap between the discount rate and growth, so a small change in either swings it sharply. Present a DCF with the terminal value's share stated and a sensitivity table beside it, rather than a single number.

One coffee buyer in three, 33.3%, must also buy a food item. Each cup loses Rs 20: Rs 140 to serve against Rs 120 of price. Each food item adds Rs 60 of contribution. So one food sale covers the loss on three coffees, and the break-even attach rate is 20 / 60. Two checks matter: only food bought because of the coffee counts, and Rs 140 must be the true extra cost of a cup, not a share of rent.

How do you price a product that loses money on purpose?

A shopkeeper who sells milk below cost does it because people who come in for milk also pick up bread and biscuits. The milk is a door, and its price is justified by what walks through it. A loss leader pays its way when the profit on what customers also buy covers the loss on the leader, so the number to manage is the attach rate. Here, each coffee loses Rs 20 and each food item earns Rs 60.

What makes the one-in-three figure misleading?

Two things. First, only incremental food counts. If half the people buying food would have come in for lunch anyway, their food was not brought in by the cheap coffee, and the coffee cannot claim it. The attach rate that matters is the share of coffee buyers who buy food because they came for coffee, which is lower than the attach rate on the till receipts. Measuring it needs a comparison: a price test in a few outlets, or the behaviour of customers before and after the coffee price changed.

Second, ask what is inside the Rs 140. If it includes a share of rent, staff and the coffee machine, those costs are there whether or not one more cup is sold. Suppose the extra cost of a cup, beans, milk and the cup itself, is Rs 70. Then each coffee contributes Rs 50 at the margin and needs no food to pay its way; it only looks like a loss after fixed costs are spread across it. Decisions about one more sale use contribution; fully loaded cost answers a different question, whether the whole cafe covers its overheads.

What would you track after launch?

Attach rate by time of day, average food contribution per coffee buyer, and the number of new customers the coffee brings in. A cafe that hits 40% attach in the morning and 15% in the afternoon may want the cheap coffee only before noon. The limitation is that habit takes time: a promotion judged on its first two weeks can look worse than it will in month three.

Roughly Rs 4.4 crore a year, on the assumptions below. Five shows a day, 200 seats and 30% average occupancy give about 1.1 lakh admissions a year. At an average ticket of Rs 250 that is Rs 2.74 crore; food and beverage at Rs 130 a head adds Rs 1.42 crore; on-screen advertising adds perhaps Rs 0.2 crore. Every input is an assumption to state and then test.

What is the structure before any number?

Estimating what a hall earns is like estimating a restaurant: count the covers, then what each cover spends, then any income that does not depend on covers. Estimate each revenue stream separately: tickets and food both hang off admissions, while advertising depends on the screen, not the audience on the day. So the first job is admissions: shows a day times seats times occupancy times days.

Assume five shows a day on a 200-seat screen, open every day. Occupancy is the soft input: a weekend evening may be nearly full while a Tuesday matinee is mostly empty. Build it rather than guess it: if weekdays run at 20% and weekends at 55%, the weighted average is (5 x 20% + 2 x 55%) / 7, about 30%.

How do you know the number is sensible?

Check it three ways. Per day: Rs 4.4 crore over 365 days is about Rs 1.19 lakh, which is 300 admissions spending about Rs 400 each in total, plausible for a screen with five shows. Per seat: about Rs 2.2 lakh a year. And the mix: food at about a third of revenue is the kind of share multiplex operators discuss, and it is high-margin money. The softest input is occupancy: every 10 points on it moves revenue by about Rs 1.4 crore, because it drives tickets and food together.

Say what revenue is not. Gross ticket revenue includes entertainment taxes in some form, and a large share of net ticket income goes to the film's distributor. Confirm current tax rates and typical revenue-sharing terms before using the figure for anything beyond a size estimate. Food, by contrast, the operator keeps in full, which is why its share matters to profit far more than its share of revenue suggests.

A's overall rate is 9.2%, B's is 3.9%, and A is still the better underwriter in each segment. A: 2 defaults among 100 salaried plus 90 among 900 self-employed, 92 in 1,000. B: 27 plus 12, 39 in 1,000. An overall rate is a weighted average of segment rates, and the weights are the mix of the book. A's book is 90% self-employed, where everyone's rate is high, so its average is pulled up by whom it lends to, not by how well it lends.

How can a branch be better at everything and worse in total?

Two surgeons. One takes the hardest cases in the hospital and loses fewer of them than anyone else would; the other takes mostly routine cases. Count deaths per patient and the first surgeon looks worse, because her patients were sicker to begin with. An overall rate is a weighted average of segment rates, and the weights are the mix of the book, so a branch that lends mostly to the riskier segment carries a higher average even when it is better in every segment. A's rate is 0.1 x 2% + 0.9 x 10% = 9.2%; B's is 0.9 x 3% + 0.1 x 12% = 3.9%.

How do you compare the two branches fairly?

Give them the same customers. Apply each branch's segment rates to one common mix and the mix stops voting. At B's mix, 90% salaried, A's rates give 0.9 x 2% + 0.1 x 10% = 2.8% against B's actual 3.9%. At A's mix, B's rates give 0.1 x 3% + 0.9 x 12% = 11.1% against A's actual 9.2%. At a 50:50 mix, A is 6.0% and B is 7.5%. On any common mix A is lower, so A underwrites better; its higher total is the price of the book it chose to build. This reversal, where a comparison flips when groups are combined, is known as Simpson's paradox.

What should the analyst say beyond the arithmetic?

Three things. First, both statements are true at once: A underwrites better, and A's book loses more money, because capital is charged on actual losses and A chose a riskier book. Whether that was a good choice depends on the pricing: if A earns enough extra yield on self-employed loans to cover the extra 8 points of defaults, its strategy can be sound. Second, ask why the mixes differ, since location, product and targets all shape who walks in. Before judging two rates, ask whether the two populations are the same; if they are not, standardise to one mix or compare segment by segment.

Third, say the limitation. Each segment is itself a mix: of loan sizes, vintages and cities. A's salaried borrowers might be larger-ticket government employees and B's might be gig workers on salary slips, and a finer split could flip the comparison back. Standardising to the segments you can see removes the mix you know about, never the mix you do not. The honest close is that A looks the better underwriter on this cut, and you would want the segment rates by vintage before signing that view.

About 10.4% a year, forever, with the company paying out 48% of its earnings. A forward P/E equals the payout ratio divided by the cost of equity less growth, and with a fixed ROE the payout is set by growth: payout = 1 minus g/ROE. So 30 = (1 minus g/0.20) / (0.12 minus g). Multiply out: 3.6 minus 30g = 1 minus 5g, so 25g = 2.6 and g = 10.4%. Check: a dividend yield of 48% / 30 = 1.6% plus 10.4% growth is the 12% cost of equity.

Where does a multiple hide a growth forecast?

If someone offers to sell you a shop for thirty years of its current profit, they are not quoting a price; they are telling you how fast they expect the profit to grow. The multiple is the forecast. Under the Gordon growth model a share is worth next year's dividend over (r minus g), so dividing by next year's earnings gives P/E = payout / (r minus g), and a given P/E can be solved for g. The one thing people forget is that the payout is not free to choose: growth has to be paid for with retained profit.

With a fixed return on reinvested profit, growth equals ROE times the retention rate, so retention is g/ROE and payout is 1 minus g/ROE. At a 20% ROE, growing at 10% means keeping half the earnings. Put that into the multiple: 30 = (1 minus 5g) / (0.12 minus g). Cross-multiply: 3.6 minus 30g = 1 minus 5g, so 2.6 = 25g and g = 10.4%. The implied payout is 1 minus 0.104/0.20 = 48%, and the implied dividend yield is 48% of a 3.33% earnings yield, 1.6%.

Why does the ROE matter as much as the growth rate?

Because the ROE sets how much growth costs. At a 15% ROE the same 30x needs 11.1% growth, since a bigger share of earnings has to be retained to fund each point of it. At an ROE equal to the cost of equity, 12%, the multiple is 1/r = 8.3x whatever the growth rate, because every retained rupee earns exactly what shareholders could earn elsewhere. Growth adds value only when the reinvested profit earns more than the cost of equity; the multiple prices the spread between ROE and r as much as it prices g. The curve is also steep near the answer: at 10% the multiple is 25x, at 11% it is 45x, so a 30x stock is one point of growth away from either 25x or 45x.

Now say the limitation. 10.4% growth forever, only 1.6 points below the discount rate, is not a forecast anyone would defend; no company outgrows the economy indefinitely. The honest reading is that 30x prices a long period of fast growth that will fade, and a two-stage model, say 10.4% for a decade and a lower rate after, is the next thing to build. The one-line solve is still worth doing, because it converts a multiple into a sentence you can argue with: the market expects this company to compound earnings at about 10% for a very long time while earning 20% on what it retains.

134,547. Split 301 into 300 and 1. Three hundred times 447 is 447 x 3 with two zeros: 1,341, then 134,100. Add one more 447: 134,547. Check before you say it: 300 x 450 is 135,000, so the answer should sit a little below that; the last digit must be 1 x 7, a 7; and the digit sums agree, since 301 reduces to 4, 447 to 6, 4 x 6 = 24 reduces to 6, and 1+3+4+5+4+7 = 24 also reduces to 6.

Why split 301 rather than attack both numbers?

If a shop sells 301 items at Rs 447 each, nobody in the shop multiplies 301 by 447 digit by digit. They price 300 items and add one more. Break the awkward factor into a round number and a small remainder, multiply the round part first, and the remainder is one easy addition. 447 x 3 is 1,200 plus 141, which is 1,341; two zeros make it 134,100; plus 447 is 134,547. Say each intermediate number aloud as you reach it, so the interviewer can follow and a slip is caught before it compounds.

The split can go the other way too. 447 is 450 less 3, so 301 x 447 = 301 x 450 minus 301 x 3. The first is 300 x 450 plus 450, 135,450; the second is 903; the difference is 134,547. Having two routes is what turns a hopeful answer into a checked one.

How do you check an answer you cannot write down?

Three checks, each a few seconds. The estimate: 300 x 450 is 135,000, and you used one factor slightly above and one slightly below, so the answer should be close to it, which 134,547 is. The last digit: 1 x 7 ends in 7, and so does the answer. The digit sums, an old bookkeeper's test: reduce each number to a single digit by adding its digits, 301 gives 4 and 447 gives 15 then 6; the product 24 reduces to 6, and the answer's digits 1+3+4+5+4+7 = 24 also reduce to 6. A wrong answer rarely survives all three checks, and running them aloud shows the interviewer a habit, not a lucky number. The digit-sum test has a blind spot: it misses errors that are exact multiples of 9, including two digits swapped, which is why the estimate sits beside it.

What is the interviewer marking?

Pace, an audible route and a check, in that order. A right answer after a long silence scores below a right answer reached in three spoken steps, because the desk wants to hear how you handle a number, not whether you can. The limitation is that this trick suits factors near a round number. For something like 347 x 683, say that you would estimate, 350 x 700 less a bit, give about 237,000, and label it an estimate; pretending to exact arithmetic you cannot check is the real failure.

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.

Six tosses for HH on average, four for HT. For HH, a tail after a head sends you back to the start with nothing. For HT, once you hold a head, every further head keeps you one toss from finishing, so nothing is lost. Solve HH with two unknowns: from the start E0 = 1 + E0/2 + E1/2, and after a head E1 = 1 + E0/2, which gives E1 = 4 and E0 = 6. For HT, E1 = 1 + E1/2 = 2 and E0 = 4.

Why is waiting for two heads different from waiting for a head then a tail?

Suppose you will go trekking after two dry days in a row. One wet day after a dry one puts you back to zero: you need two fresh dry days. Now suppose instead you are waiting for a dry day followed by a wet one. Extra dry days cost nothing; you stay ready, and the first wet day finishes the job. A failed attempt at HH throws away the progress you had, while a failed attempt at HT keeps it, so HH pays for its failures twice: the wasted toss and the lost head. That asymmetry is the whole answer; the algebra just prices it.

Write the states. E0 is the expected tosses still needed from nothing, E1 from one head in hand. From nothing, one toss is spent and you land back at E0 with a tail or at E1 with a head: E0 = 1 + E0/2 + E1/2, so E0 = 2 + E1. For HH, from one head a head finishes and a tail drops you to E0: E1 = 1 + E0/2. Substitute: E1 = 1 + (2 + E1)/2, so E1/2 = 2, E1 = 4 and E0 = 6. For HT, from one head a tail finishes and a head keeps you at E1: E1 = 1 + E1/2, so E1 = 2 and E0 = 4.

Is there a rule for longer patterns?

Yes, and it is quick enough for the room. For each length k at which the pattern's first k symbols equal its last k symbols, add 2 to the power k. HH matches itself at k = 1 (H and H) and at k = 2 (the whole thing): 2 + 4 = 6. HT matches itself only at k = 2: 4. HHH gives 2 + 4 + 8 = 14; HTH gives 2 + 8 = 10; HTT gives just 8. Patterns that overlap with themselves take longer to appear from a fresh start, because their occurrences come in clusters, and the same long-run frequency spread into clusters means longer gaps between them. Every pattern of length n shows up once in 2 to the n positions on average; what differs is how bunched those appearances are.

The average hides a wide spread. The chance that HH has not appeared after ten tosses is 14.1%, against 1.1% for HT, so one game in seven is still running at toss ten when you wait for HH. And the figures assume a fair, independent coin: with heads at 60%, the HH wait falls to 1/p + 1/p squared, about 4.4 tosses. The finance link is any rule that needs consecutive results, such as a bonus or a covenant test that requires two good quarters in a row: one bad quarter resets the count, so the rule is harder to satisfy than its two-in-four odds suggest.