Case Interview Example – Estimation Question and Answer
I was asked the following management consulting estimation question by a McKinsey interviewer many years ago:
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
Below you will see my answer to this estimation question and the process and rational I use to answer this specific question can be used as a template to practice answering other estimation questions as you prepare for case interviews.
The first thing to realize in an estimation question is that an acceptable answer MUST mention a specific number.
This question was how much time it takes to move an average mountain 1 mile (or something along those lines).
If the answer does not include a specific unit of time like X hours, Y days, Z years, then the answer is not acceptable.
By the way, I use the word “acceptable answer” instead of “correct answer” very deliberately. The interviewer’s evaluation in this type of question is in assessing the approach you took, not necessarily the specific answer you gave.
The next thing to the answer must include is that explicit assumptions must be made.
It is not possible to answer this question without making some assumptions. They key is to EXPLAIN to the interview that you are going to make some assumptions. Once you do and once you make a specific assumption, explain your rationale behind that assumption.
For example, when I was given this question. I knew that I needed to estimate the cubic volume of the mountain. And since the mountain loosely resembles a cone, I knew there was a geometric formula to calculate the volume of a cone–except I did not recall the specific formula off the top off my head.
So my interviewer suggested that I estimate the formula of a cone, which in turn I would use to estimate the volume of an average size mountain, which would then be part of a calculation to estimate the average time it would take to re-locate it.
Notice the estimate that is nested within the estimate here. This is very common. Most important thing is to not get mixed up and confused by your own work.
I find it is useful to just write out the formula that will produce the estimate FIRST, THEN go about making reasonable assumptions.
For the move the mountain case, the formula I wrote up on the white board during my interview was:
volume of mountain / volume of a truck * time per truck trip = total time to move a mountain
I would literally write that on the board. That is the amount of time it would take 1 truck to move an average size mountain 10 miles (the 1 truck is an assumption as well)
Then I went about estimating each of those 3 factors.
Assume the average size mountain is 1 mile tall, 1 mile wide, and the shape of a cone. That’s approximately 5,000 ft in height and base.
I forge the formula to calculate the volume of a cone, but if I eye ball it, it is probably a little more volume than half of a cube of similar size height and base.
The volume of a cube that’s 5,000 ft tall, 5,000 ft wide, and 5,000 ft deep is 125,000,000,000 cubic ft.
Since I’m trying to estimate a CONE, and not a CUBE, I’d then take 125,000,000,000 x 50% (my approximate guess as to how much smaller a cone is vs a cube of approximately the same height, and width and length at the base.
With some slight rounding, that gets us 60,000,000,000.
Then underneath my original formula, I would write the following:
60,000,000,000 cubic ft / volume of a truck * time per truck trip = total time to move mountain
Next, I would move on to estimate the volume of a truck.
The carrying capacity of a cargo truck is the width x length x heightof the cargo container.
I said, well I know those big trucks are a little wider than my car, but not by much since they still must be able to fit into a lane on the freeway. My car sits 3 people across, assuming 2 ft in shoulder width per person, that’s 6 ft of interior space. Let’s add on a little more and assume those big trucks are around 8 ft in width.
I know they are about double the length of most passenger sedans. And lets see if I were to lie down in the driver’s seat to take a nap, I cover most of the interior cabin space. And the hood and trunk of the car combined are about the same length as the interior cabin. I’m a little under 6ft tall, so that makes my car around 12 ft long. If I double that, I get the length of one of those trucks to be 24 ft long. I subtract out say 4 ft for the driver compartment, and that leaves me about 20 ft in length for the cargo area.
Last time I looked, I saw a worker standing in the back of one of the cargo areas, and the cargo area was taller than the person. I figure the cargo container is about 8 ft tall. And since most freeway bridges have signs that say “height 13 ft” and I know those trucks can go under those bridges, assuming an 8ft cargo section and a 4ft for the tires and chassis under the cargo area, that gives me 12 ft…which does seem to triangulate with the height of those underpasses. So I’ll say the cargo section is approximately 8 ft tall.
The volume of the cargo area of an earth moving truck is:
8 ft wide x 20 ft long x 8 ft tall = 1,280 cubic feet
For sake of simplicity, I’m going to round that down to 1,250 cubic feet and plug this number back into my original formula which now reads as follows:
60,000,000,000 cubic foot mountain / 1,250 cubic foot truck capacity * time for truck trip = total time to move a mountain
The only factor missing in our estimate is figuring out the round-trip time for a trip to move 10 miles, drop its load, and return the 10 miles. Let’s figure out the travel time first. Assume the truck travels on the freeway at 60 miles per hour.
For it to travel 10 miles, it does so in 1/6 and hour or 10 minutes. The drive time is 10 minutes to the new location, and 10 minutes returning to the old mountain for a total of 20 minutes. Assume that the off-loading process has been designed to be pretty quick. The load is just “dropped” and then repositioned while the truck is on its return trip (as opposed to being scooped out of the truck, one scoop at time which seems more time consuming).
That means each round trip takes 30 minutes or 0.5 hours.
Let’s go back to our formula again and update it.
60,000,000,000 cubic ft mountain / 1,250 cubic foot track capacity * 0.5 hours per truck trip = total time to move a mountain
Let me do the math now. For the first 2 components of the formula, that works out to about 50,000,000 (50 million truck loads).
50 million truck loads x 0.5 hours, thats 25 million hours to move a mountain.
If we assume a typical day has 25 hours (to make our math a little simpler), that’s 1 million days to move the mountain using only 1 truck. That works out to a bit under 3,000 years
That is the logic I just presented is a pretty good one that would most likely pass most estimation question interviews.
You will notice that for every little component I explain WHY I felt that was a reasonable assumption.
There is a big difference between making a wild assumption vs. a reasonable one. Your goal is to make as reasonable assumption as you can come up with. When you make such an assumption, it is very important you explain WHY you made the assumption you did.
The math is not that complicated (it’s math we all learned before high school) BUT communicating what you are doing is just as important.
It is also important that you do not make a math mistake. I wrote out this example quickly and hopefully I did not make a math mistake.
If I did make a math mistake, I would full expect to get rejected even if I got the logic and assumptions largely right.
That’s just the way it works. Practice your mental math. You DO use it a lot not just in interviews but with clients as well.
504 thoughts on “Estimation Question”
Assume what is an average mountain look like (in weight or in volume), e.g. 10,000 tonnes
Assume capacity of an average truck, e.g. 10 tonnes
Assume average speed of the truck, full load, e.g. 30 mph
Assume speed of the truck, no load, e.g. 50 mph
Calculate, number of trips required: 10,000 / 10 = 1,000 trips
Calculate, total distance traveled = 10 miles * 1,000 trips = 10,000 miles
Calculate, average truck speed: (30 50 mph) / 2 leg per trip = 40 mph
Calculate, required time in hours = 10,000 miles / 40 mph = 250 hours
Answer: 250 hours
Made a mistake there… the total distance traveled should be double that, thus 20,000 miles (2 legs per trip). Thus, total time will be double that as well, thus 500 hours.
Let’s assume that an average size mountain is 1000meters high and 200 meters wide and the shape of the base is a circle.
An average size truck is 3meters x 5meters x 2 meters.
The average speed of the truck is 50miles per hour.
We will have only one truck
We will have a big shovel to fill the truck. The shovel can take 1m3.
we will work 12 hours non stop per day.
It takes xx minutes to fill the truck.
You have xx to move.
So it will take days.
Time taken = no. of loads*time for each on-load round trip (20 miles) time * no. of round trips time for offloading*no. of loads
no. of loads= size of the mountain/capacity of truck = 200/10 = 20 loads
time for each on-load = 2 hours
[20 loads * 2 hours]
[40 mins(onward heavy load) 30 mins(return)]*20 loads
[20 mins]*20 loads
40 hours 1400 mins 7 hours
40 hours 14 hours 7 hours
=60 hours
200,000,000 minutes or approx 17,000 hours
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
Step 1. Assume that an average size mountain weighs 10,000 tons
Step 2. Assume that an average size truck carries 2 tons per time.
Step 3. Estimate time spent on loading and unloading. Let’s say that a machine will load the truck full in 2 minutes, and the truck will unload itself in 1 minute. So overall per one time full cycle takes 3 minutes.
Step 4. Estimate time spent on the two 10 miles. Assume that the truck drives at a hourly speed of 40 miles constantly, in all it takes 30 minutes.
Step 5. Determine total time span of a complete cycle: 3 30=33 minutes.
Step 6. The mountain requires the truck to run 10,000/2=5,000 times commute.
Step 7. Total time spent: 33*5,000=165,000 minutes = 2,750 hours = 114.3 days
To travel 16 Kms (10 miles), firstly I have to start the analysis from the very basic level.
Approximately, 1 brick weighs 2 kgs
Assuming, a average mountain will be equal to 5000 bricks, hence 10000 kgs.
weight of the truck is approximately 2000 kg.
It is general physics concept that increasing the weight, will increase the reaction and hence increase friction, which slows up speed.
firstly the weight was 2000 kg. After loading, the weight will become 12000 kgs. by linear interpolation, speed should become 1/6th of the average empty truck speed.
Assuming an empty truck move with 30 Kmph. now the loaded one will move with 5 kmph.
To travel 16 Kms, It will approximately take about 3 hrs in such scenario.
625 millions hours
Assumptions: “average” size mountain is 5km long, circle base, 500m high, cone shaped. That’s 35000000 m3. But what’s an average mountain… I’m assuming here it’s not a hill, it’s a mountain.
Avg truck 75m3 (2,5 by 3 by 10). Time to load it 30min, time to drive 10 miles 10-15min, time to find the spot and unload 10min, drive back 10min. So total of 75min for 75m3. that’s 1500m3 per day. (also assuming working day and night here, with people tacking shifts, and negligeable time for the truck to fill up on petrol)
So… with these guesses and assumptions. I get to around 2500 days, i.e around 6 to 7 years for this decently sized mountain and one truck. haha
actually made a stupid mistake there: 3,5 km3 is 3500 000 000 m3 so it’s not 6 years but 6000 years according to my scenario. Bam, out the door, next candidate!
I’m going to use the metric system since its easier to work with.
10 miles is around 16 km.
To move the mountain you’ll need to see:
– The total weight of the mountain
– The average weight carried by a truck
– Average speed of the truck
The time to move the mountain will be = time of one trip * number of trips of the truck
= 16 km/(70 km/h) * number of trips
number of trips = (volume of mountain/volume of truck carrying rocks)* 2 (because the truck comes back empty)
Assume volume of mountain to be cone shaped = (1/3)*pi*r^2*h
Height = 1 km, radius to be 2 km
Volume of mountain = (pi=3)/3 * 2^2*1
Volume of mountain = 4km^3
Volume of one truck full of rocks is 6m truck length * 2 m width * 3m height =36m^3
Number of trips: 4*10^9 m3 / 36 m3 * 2 = around 1.5*10^9
1.5*10^9 * (16/70) = 1.5 * 15/75 = 1.5* 1/5 =
0.3*10^9 hours
Typical truck size: 20 MT
Average truck speed : 30 miles per hour
so 20 MT can be carried across 10 miles X 2 in 40 mins= 2/3 hr
Typical mountain = 5km radius & 5 km high
Volume = 1/3*22/7 *5*5*5=135 cubic km
= 135*10^9 cubic metres
Density = 5MT/ cubic metre
Mass = Vol * Density
= 135 *5 *10^9
=675*10^9 MT
20 MT = 40 mins
675 *10^9 MT = 40 *675 *10^9 /20
=1350 *10^9 mins
~22.50 *10^9 hrs
~0.95 *10^9 days
~950 million days