Brainteaser Interview Questions

To estimate the number of petrol stations in the UK, I will use a top-down approach based on consumer demand. I will start with the UK population, which is approximately 67 million people. I will assume an average household size of roughly two to three people, giving about 25 to 30 million households. Since many households own multiple vehicles and corporations run fleets, let us estimate there are roughly 33 million active fuel-burning cars on UK roads. Next, I will estimate consumer behaviour. Assume the average car fills up its tank once every week. This creates a total national demand of 33 million petrol station visits per week. Now, let us look at the supply side to find out how many visits a single average petrol station can handle in a week. Suppose a standard station has 8 fueling pumps. Assume it operates for 16 hours a day, 7 days a week. That is roughly 110 hours of operation per week. If it takes an average of 5 minutes for a driver to pull in, fill their tank, pay, and leave, each pump can handle 12 cars per hour. Multiplying 12 cars by 8 pumps gives 96 cars per hour for the entire station. To simplify the mental maths, I will round this to 100 cars per hour during active operational times. Multiplying 100 cars per hour by 110 operating hours gives a weekly capacity of 11,000 car visits per petrol station. Finally, to find the number of stations, I divide the total weekly demand by the capacity of a single station: 33,000,000 total weekly visits divided by 11,000 visits per station yields exactly 3,000 stations. Since petrol stations often experience peak-hour congestion and down-time where pumps sit empty, the true operational throughput is likely lower, meaning we need more stations to meet demand. If we assume a 40 per cent utilization rate, we adjust our baseline up to roughly 7,500 stations. This matches the low thousands order of magnitude, which aligns with the actual UK market reality of around 8,000 stations.

To solve this problem, I need to calculate the interior volume of a standard school bus, calculate the volume of a single golf ball, divide the two figures, and then apply a packing factor to account for empty space and internal fixtures. First, let us estimate the dimensions of a standard school bus. I will assume it is roughly 10 metres long, 2.5 metres wide, and 2 metres high. Multiplying these dimensions together (10 x 2.5 x 2) gives a total external volume of 50 cubic metres. However, the bus is not an empty box; it contains rows of seats, a driver console, and internal structural framing. Let us assume these fixtures take up roughly 20 per cent of the space, leaving 80 per cent of the volume open. This gives us a net usable volume of 40 cubic metres. Second, let us estimate the volume of a golf ball. A golf ball has a diameter of roughly 4 centimetres, which means its radius is 2 centimetres. The volume of a sphere is calculated using the formula V = (4/3) pi r cubed. Substituting 2 centimetres for the radius, r cubed is 8 cubic centimetres. Multiplying 8 by (4/3) pi (approximating pi as 3) gives roughly 4 x 8 = 32 cubic centimetres per golf ball. Third, I must convert the usable bus volume into cubic centimetres so the units match. Since 1 metre equals 100 centimetres, 1 cubic metre equals 100 x 100 x 100, which is 1,000,000 cubic centimetres. Therefore, 40 cubic metres is equal to 40,000,000 cubic centimetres. Now, I divide the usable bus volume by the golf ball volume: 40,000,000 divided by 32. To do this mentally, 40,000,000 divided by 8 is 5,000,000, and 5,000,000 divided by 4 is 1,250,000 balls. Finally, I must apply a packing factor for solid spheres. Spheres cannot be packed perfectly without air gaps; even random close packing leaves about 36 per cent empty space. If we multiply 1,250,000 by a packing efficiency of roughly 64 per cent, we get 800,000 balls. Factoring in the awkward shapes around the seats and windows, a conservative estimate would place the final total between 250,000 and 500,000 golf balls.

When I am presented with an entirely unfamiliar estimation question, my priority is to establish a structured process rather than rush to a final number. I follow a consistent four-step framework to ensure accuracy and clarity. The first step is framework selection. I decide whether to approach the problem top-down or bottom-up. A top-down approach is ideal for large national figures, where I start with a macro metric like the UK population of 67 million or the US population of 330 million and narrow it down via demographic filters. A bottom-up approach works best for specific businesses or localized assets, where I calculate the revenue or output of a single unit over an hour or day and then scale it up across time or geography. The second step is stating assumptions clearly. Before I perform any mathematical operations, I lay out my assumptions out loud for the interviewer. For instance, if I am estimating the size of the UK coffee market, I will state that I assume there are roughly 50 million adults, that 60 per cent of them drink coffee, and that an average takeaway coffee costs GBP 3 (roughly USD 4). This allows the interviewer to see my operational logic. The third step is simplified computation. I perform the actual calculations using rounded, clean numbers to minimise mental fatigue and avoid mechanical errors. I keep my units uniform throughout the process, converting between volumes, currencies, or timeframes explicitly as I go. The final step is the sanity check. Once I arrive at the final figure, I step back and evaluate whether the order of magnitude makes sense in the real world. If my calculated market size implies that every citizen buys ten coffees a day, I will openly point out the flaw in my initial assumptions and adjust my calculations on the spot.

To calculate the UK coffee market size, I will use a bottom-up model centered on out-of-home consumer spending. Start with a UK population baseline of 67 million, and isolate the adult population at roughly 50 million. Let us segment these 50 million adults into three consumer profiles. First, the daily premium consumers: assume 20 per cent (10 million people) buy one premium coffee every working day (5 times a week). At GBP 3 (USD 4) per cup, they spend GBP 15 per week. Second, the occasional consumers: assume 50 per cent (25 million people) buy an average of 2 coffees a week, spending GBP 6 per week. Third, the remaining 30 per cent (15 million people) consume coffee exclusively at home or do not drink coffee at all, spending GBP 0 in the commercial retail market. Now, calculate the total weekly spend. The daily group spends 10 million multiplied by GBP 15, which equals GBP 150 million. The occasional group spends 25 million multiplied by GBP 6, which equals GBP 150 million. Combined, this results in a total retail spend of GBP 300 million per week. To annualize this market size, multiply GBP 300 million by 52 weeks. To simplify the mental maths, 300 x 50 = 15 billion. Adding the remaining two weeks (300 x 2 = 600 million) brings the total to roughly GBP 15.6 billion. However, this model assumes maximum consumer behaviour every single week without factoring in holidays, home working, or seasonal drops. If we apply a realistic adjustment factor assuming people travel or reduce takeaway purchases for part of the year, a commercial cafe market size of GBP 3 billion to GBP 4.5 billion emerges, which aligns accurately with industry data.

To find the probability that the sum of two fair six-sided dice is 7, I first establish the total sample space. Since each die has 6 independent outcomes, rolling two dice creates a total of 6 x 6 = 36 equally likely possible outcomes. Next, I systematically count the number of combinations that sum to exactly 7. These pairs are: 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, and 6 and 1. There are exactly 6 winning outcomes. The probability is the number of successful outcomes divided by the total outcomes, which is 6/36. Reducing this fraction yields 1/6, or approximately 16.7 per cent.

This is a conditional probability problem. A standard flip of two coins creates an unconstrained sample space of four equally likely outcomes: Heads-Heads (HH), Heads-Tails (HT), Tails-Heads (TH), and Tails-Tails (TT). The prompt introduces a condition: we know that at least one coin is heads. This condition completely eliminates the Tails-Tails (TT) outcome from our calculations. The remaining updated sample space consists of only three valid outcomes: {HH, HT, TH}. Out of these three remaining possibilities, there is only one outcome where both coins are heads: HH. Therefore, the probability is 1 divided by 3, which is 1/3, or 33.3 per cent.

This problem is resolved using the pigeonhole principle. We have two distinct categories of items: red socks and blue socks. To guarantee a pair, we must look at the absolute worst-case scenario for our draws. If we draw the first sock, it will be either red or blue. If we draw a second sock, the worst-case result is that it is the opposite color of the first draw, leaving us with one red sock and one blue sock. At this stage, we have exhausted all unique categories. When we pick a third sock, it must be either red or blue. Whichever color it is, it will match one of the socks we already have, guaranteeing a pair. Therefore, the answer is exactly three socks.

To find the expected value of a single roll of a fair six-sided die, I use the expected value formula, which is the sum of all possible outcomes multiplied by their respective probabilities. For a fair die, each face from 1 to 6 has an equal probability of 1/6. The equation is set up as follows: E[X] = (1 x 1/6) + (2 x 1/6) + (3 x 1/6) + (4 x 1/6) + (5 x 1/6) + (6 x 1/6). This can be simplified by adding the face values together and dividing the total by 6: (1+2+3+4+5+6)/6 = 21/6. Dividing 21 by 6 results in exactly 3.5.

To multiply 23 by 27 mentally, I will use the difference of squares shortcut instead of standard long multiplication. I notice that both numbers are centered exactly around 25; 23 is 25 - 2 and 27 is 25 + 2. I can rewrite the expression as (25 - 2) x (25 + 2). The difference of squares identity states that (a - b)(a + b) = a squared - b squared. Substituting our values gives 25 squared - 2 squared. I know that 25 squared is 625, and 2 squared is 4. Subtracting 4 from 625 results in exactly 621.

To find the square root of 2,000 mentally, I will bracket the number between known perfect squares. I know that 40 squared = 1,600 and 50 squared = 2,500, so the answer lies between 40 and 50. Let us narrow this range down. I know 45 squared = 2,025 (using the trick 4 x 5 = 20 followed by 25). Since 2,025 is just above 2,000, the answer must be slightly less than 45. Next, let us calculate 44 squared, which can be broken down as 44 x 40 + 44 x 4 = 1,760 + 176 = 1,936. The number 2,000 sits between 1,936 (44 squared) and 2,025 (45 squared). It is 64 units away from 1,936 and 25 units away from 2,025, placing it roughly two-thirds of the way up the interval. Therefore, a precise decimal estimate is 44.7.

To measure exactly 45 minutes using two unevenly burning 60-minute ropes, I need to create a 30-minute interval followed by a 15-minute interval. I will start by lighting rope one at both ends and lighting rope two at one end all at the same time. Even though rope one burns unevenly, lighting it from both ends guarantees it will consume itself completely in exactly half its normal time, which is 30 minutes. At the exact moment rope one burns out, 30 minutes have elapsed. Because rope two was lit at one end at the start, it now has exactly 30 minutes of total burn time left within its structure. To measure the final 15 minutes, I immediately light the other end of rope two. Lighting its second end doubles its consumption rate, cutting its remaining 30-minute burn time down to exactly 15 minutes. When rope two burns out completely, the total elapsed time will be 30 minutes plus 15 minutes, which is exactly 45 minutes.