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”
10,000 days or almost 30 years
To solve this questions, we must identify 3 key drivers:
1. size (volume of material) of an average mountain
2. size (volume of material) of an average truck
3. speed of an average truck
Time taken would be = (1)/(2) * 10km/(3) *2 because the truck will have to move back and forth to transport the material.
However, we must add consider loading and unloading time to get a precise estimate. I have assumed that it is negligible in this case.
A mountain can be compared to a cone of height 3 kilometres and base radius of 1 kilometre.
Then the volume of material is 1/3* Pi* 1*1*3 = 3.14 cubic km
A truck can be compared to a cuboid of base 10 m, height 10m, and length 10 m. Then capacity is L*B*H= (1/100) ^3 cubic km.
Assuming the speed is 20 kilomteres per hour,
we have Time taken = (3.14/ (1/100)^3 )* (10/20 )*2
= 3.14*1000000= 31,40,000 hours
The average height of the mountain would be 3 Km (from the sea level) and its radius 0.5 Km.
Its volume would be 2500000 meter cube approx.
The volume of one average sized truck would be=2*1*5=10 m^3
Therefore the no. of times the truck needs to be filled is 250000.
10 minutes will be required to load this truck.
10 minutes to travel (assuming 60 miles an hour)
and 5 minutes to dump
and another say 8 minutes on its return journey (since now the truck is not loaded)
The total time for each cycle would be = 32 minutes
Therefore the time required to shift the entire mountain would be = 250000*32= 8000000 minutes = 15.2 years
Mnt 2000 ft high and 4 sq miles (20,000 ft around) * .7 b/c cone shaped.=30 mil sq ft.
Truck 4/12.5=50 sq ft.-> 6 million trips
Since moving a mountain, drive 20 miles per hr, round trip =1 hr.
=250,000 days
“Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck”
As the problem is to relocate a mountain from point A to point B (i.e. 10 miles apart), my first assumption is that mountain doesn’t have any rock greater than the size of Truck’s container.
Now, estimating the volume of an average mountain :(As the interviewer as said ‘mountain’ which are relatively larger than ‘hill’)
Assume it to be a 7 Km radius and 2 Km high, regular cone, which amount to ~ 25.67 KM3 of debris.
Assuming we are only transferring the visible portion of mountain and not the underground one.
Now, estimating the volume of a truck’s container:
It’s roughly a 3mt wide, 10 mt long & 3 mts high ~ 90M3 space.
Another assumption I would like to make is to estimate the loading & unloading time of truck’s container. A rational assumption to make would be that mountain would be cut away from top and relocated with the earlier summit as base of mountain. As for filling the truck, gravity will help oneself but for creating a new mountain, one has to work against gravity. Over the course of time I would assume an estimated time of ~(10 mins to load the truck and 12 min to unload it).
The journey itself at 60 mile/hr would cost ~ 20 min (to & fro)
Thus, from point A with an empty truck, it would take ( 10 20 12=) 42 mins to be back at work station for next round.
Total rounds needed = 25.67 KM3 / 90M3 ~ 285,222,223
Total time needed in mins = 11,979,333,333 – 10 ( as you don’t have to come back to spot A in end)
=11,979,333,323 mins
450,000,000 min
24 minutes
Lets assume that the size of a mountain is classified as “big” if it is 100 kgs and classified as “small” if it is 40 kgs. So an “average” size mountain will be of 70 kgs.
Now, lets assume an average size truck can run at 60 miles/hour when it is not pulling any additional weight. For every additional 10 kgs of weight, the speed of the truck reduces by 5 miles/hour. So for a mountain of 70 kgs, the speed of the truck will reduce by 35 miles/hour.
So, the effective speed of the truck here is 25 miles/hour.
So the time taken by the truck to move the mountain is (60/25*10) minutes i.e., 24 minutes.
Let us assume an average sized mountain to be a cone…with dia of 1km…and height of 1 km.
So, volume = 1/3*pi*r*r*h= 0.3 cu-km (approx)
Now lets assume dimensions of an average sized truck to be 15x10x7 fts. Excluding ground clearance of 4 ft, cabin dimesions to be 5x6x7 ft, we have the dimesions of container to be 10x6x7 fts….so volume of container = 420 cu-ft. = 730000 cu-inch(approx)= 0.00005 cu-km (approx)
Now lets assume the average speed of the truck to be 60 miles/hr, so it’ll cover a distance of 10 miles in 10 mins.
Now, let’s assume time of loading and unloading to be 10 mins.
No. of times the truck will be loaded = 0.3/0.00005 = 6000 times.
Time taken to load = 6000×10=60000 mins.=1000 hrs.
Time taken to unload = 6000×10=60000 mins.=1000 hrs.
No. of times the truck will move to and fro = 12000 times (assuming truck is initially present at the unloading site i.e. 10 miles from the mountain)
So, time taken = 12000×10=120000 mins = 2000 hrs.
Total time = 4000 hrs= 166 days, 15 hrs, 50 mins and 24 seconds.
Truck=
6 meters x 3 meters x 3 meters
= 54 m3
10 miles
1/3 Area of base x height = 1/3 x pi x2km x 2km x 1km
=4/3 pi km3 = 88/21 km3 = 4.000.000.000
4 B /54 = 75 M times
average speed= 50 miles/hour
2 x 75 Mil x 10 miles x hour/50 miles
= 30 Mil. hours (assume no transit time)
1. height 3000 1 km less than half of Mt Everest (frankly no idea)
2. radius 1000
3. Volume 3141592654 =(volume of cone)
4. volume of Truck 12 ( l, b, h 4 meters, 2 mtr, 1.5 m , guesses based on size of trucks in India)
5. number of trips 261799387.8 (3/4)
6. loading time 0.5
7. Unloading Time 0.083333333
8. Transit Time 0.666666667 (2*10 miles/ 30 miles/hour)
9. Total Time one Round Trip 1.25
Total Time Needed 327249234.7 hours
13635384.78 days
1947912.112 weeks
454512.826 months
Good method.
Is the answer reasonable? Is months a good unit? Convert to years. You get ~40,000 years? Will the truck or the driver work for so many years? What is the systhesis?