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”
average size mountain : 1/3*pi*2*(2*sqrt(3))^2, rounding to 25 M3*10^9
density of rock : 10 tons/m3
average size truck : 5 tons
time per trip : 1 hour (load/unload transport time)
total time : 2.1 billion days.
My result is 420 million days of non-stop work. But one of my assumptions was that average truck only loads 5.25 ton of material which I now know was underestimated. Thanks for the case!
volume of mountain = 1/3*pi*rsq*h (avg r, avg h values)
volume of truck= l*b*h (avg l, b, h values)
no of truckloads of mountain material = volume of mountain/ volume of truck
time to move 1 truckload=time to dig time to load truck load time to travel to new site time to unload time to return. we can ignore time to dig as this can be done while truck is in transit.
hence total time to move mountain = (time to load truck load time to travel to new site time to unload time to return) * no of truckloads
-Average size truck: L:0.01km, W:0.002km, H:0.001km–> V:0.00000002km3
-Av Size Mountain: H:0.5km. Radius:10km–> V=75km3
-How many times does the truck need to make the journey? 75/0.00000002= 75*50000000=3,750,000,000 times
-How long would each journey take? 30mins to fill the truck. 1hr to go and come back (assuming 20km/h average speed)
-Total time in Hours=1.5*3.75m=5.6m hours (approx)
-Hrs in a year: 24*350 (approx)=8500 (approx)
-Hrs in a century=850,000
-Total time needed=5.6m hours/0.85m hours in a century=700year approx
answer depends on three factors:
A) carrying capacity of the truck
B) time it takes for the truck to load, drive the dirt 10 miles and drive back
C) weight of the mountain
A) 5 people can lay flat on the bed of the truck, depth of truck bed = 5 ft, 25 bodies can fit in the truck bed (man that just sounds weird) each body weighs 150 lbs, 2 bodies = 300lbs, 300×12=3,600 and dirt can be packed tighter and more compact so lets say each loading can fit 5000lbs
B) avg truck speed = 50 mph, to drive 10m that’s 1/5 of an hour = 12.5 min 5 min for the crane to load/unload the dirt 12.5 = 30 min (there and back)
C) average mountain consists of around 100 football fields
1 sq yard contains 10lbs of dirt, football field around 30yrds across so 300lbs of dirt per yard length, x 100 yards = 30,000lbs/field x 100 fields = 3mn lbs on the base.
avg mt height = 10k feet, basically has a triangle profile so lets practically use 5ft since triangle would be half the rectangle area
5,000ft x 3mn fields (stacked on top of eachother) = 15,000,000,000 lbs of dirt
C (15bn lbs of dirt) / A (5000 lbs of dirt/truck load)
groups of 5k
200 per mn x 1000 = 200,000 per bn x 15 = 3mn truck trips
3mn x 30 min (2trips per hour) = 1.5mn hours
does this make sense to anyone? first attempt at one of these problems so any feedback would be tremendously appreciated, thanks
What do u mean by average size mountain?
What do mean by average size truck?
how many workers do u have?
what is your main intent of moving that mountain?
what is mountain’s rock type? is it rock mountain or sandy.
how do u want ti do with the new mountain?
r u in process of extracting something?
what are the tools used by workers?
which are is the mountain located.
Time taken = (time to make 1 scoop time to deliver soil to new location time to climb up new mountain time to pour soil down time to climb down mountain time return to old location) x #return trips
# Return Trips
= Volume of Mountain / Volume of Truck Scoop
= 1/3π(25km)2km / 5mx4mx2m
= (appox)50e9/40
=1.25e9
Time taken for truck to make one attempt to scoop up the soil off the mountain
= time to lower the scoop time to dig into the soil time to lift the scoop up
= 30s 1min 30s
= 2mins
Time taken for truck to drive soil to destination
= distance/speed
=10miles/25mph
=24min
Time to climb up to peak of new mountain
= distance/speed
= 2.9km/4.8kmph
= 30minutes
i.e. ave time to climb the new mountain = 15mins
Time to pour soil down on new mountain
= Time to lower scoop
=30s
Time to climb down from peak of new mountain
= distance/speed
= 2.9km/9.6kmph
= 15mins
i.e. ave time to climb the new mountain = 7.5mins
Time taken for truck to drive back
= distance/speed
= 10miles/50mph
= 12mins
Time taken to move mountain = (time to make 1 scoop time to deliver soil to new location time to climb up new mountain time to pour soil down time to climb down mountain time return to old location) x #return trips
= (2 24 15 0.5 7.5 12)minutes x 1.25e9
= 61minutes (approx 1 hour) x 1.25e9
=1.25e9 hours
The volume of the mountain can be estimated by assuming that the mountain as a cone. Assuming the height of a mid-sized mountain is 500m and has a radius of 1km.
–> Volume = (r^2 . h . pi)/3 ~ = 280,000,000 m3
Assume the truck is 6 m long, 3m wide and 2m tall.
So the volume of the truck = 6 . 3 . 2 = 36m3 at JUST full capacity (not 80%/120%)
Time required to fill the truck and unload ~= 10 min
Time required to travel forth and back ~ = 20 min
=> each cycle takes the truck 30 min to travel
The number of cycles needed by the truck = 280 e6 / 36 ~= 8e6
=> Total time needed = 8e6 . 30min = 240e6 min
=4e6 hr = 1100 days
(assuming the truck operates 24/7)
For solving this problem we need a series of estimates related to 1. Size of an average mountain 2. Size of an average truck 3. The speed of a process of loading, driving, unloading the truck and driving back.
1. Size of an average mountain. Let’s assume that an average mountain is appx 3000 m tall. (Everest is 8850, so why not). If we imagine that a mountain has a shape of a cone with the height of 3 km and a diameter roughly twice the height – that gives us a volume of the mountain V equals (3.14*3*3*3)/3 = appx. 30 000 cubic meter
2. Let’s say an average truck has dimensions of 2*1.5*2 = 6 cubic meters. Than it is very easy to divide.
3. 30000/6 = 5000, meaning that the truck would need to be loaded and unloaded 5000 times in order to transport this volume of the mountain. It is important here to assume, that we do not take into consideration the weight restrictions for the truck. 7 cubic meters of a mountain can get too heavy to fit into one cycle of transportation.
4. the cycle would be loading – driving – unloading – driving back. Let’s say we have a hulk and a superman loading our truck, so loading 6 cubic meters of massively heavy mountain is a normal exercise for them and they don’t have to take a break, stop, smoke a cig, eat etc and can do that for days, let’s say that loading and unloading takes 15 minutes per operation.
Let’s say an average truck is doing 60 km per hr (sorry, I think in km, not in miles), and 10 miles is roughly 15 km, than to drive this distance the truck needs appx 15 minutes, same for driving back, thus the entire cycle takes exactly 1 hr.
Therefore 5000 cycles will take us 5000 hr, or roughly 209 days.
Average mountain height: 10,000 feet
Assume mountain is a cone with diameter = height which means radius = h/2 – 5,000 feet
Volume: 25 * 10^10 (A)
Average truck size: 15 * 8 * 5 = 600 (B)
Total number of trips required = 4 * 10^8
If we assume speed = 40 m/hr, 10 mile round trip will require 30 minutes each, hence hours required to make this transfer = 2 * 10^8 = 20,000 years