Estimation Question

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.

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501 comments… add one
  • Jinjin Nov 9, 2017, 5:57 pm

    Time = time/unit x unit
    = [distance x 2/speed] x [mountain’s weight/truck’s tolerance]
    = [10m x 2 / 30m/h] x [10,000t / 5t ]
    = 4000/3 h ≈ 1334 h ≈ 56 days

  • Yang Nov 20, 2017, 6:29 am

    “Estimate how long it would take to move or relocate an average size mountain 10 miles using an average size truck.”

    Here’s my answer.
    First of all I’d like to estimate the volume of the mountain. Talking about an average size mountain, I think there’s one very ordinary mountain in my hometown. Normally it’ll take me 2hrs to walk a full circle around it. I walk fairly fast, say with a speed of 6km/hr. To make it simple, I’ll assume the mountain is cone shaped. The circumference of it is 2*6=12km=2πr ≈ 2*3*r, so r=2km. Since volume = πr^2*h/3 ≈r^2*h (assume h = 0.5km), so total volume would be 2 cubic kilometers = 2 billion cubic meters.

    Then let’s go tackle efficiency of the truck. From my observation, full capacity of most trucks range from 5 cubic meters to 15 cubic meters, so I’ll take an average of 10 cubic meters as its capacity. Since it’s fully loaded, I’ll assume the speed is not that fast, about 60 mph, which makes the come and go of the 10 miles distance 20 mins. I’ll assume that to dig, load and unload, it takes another 40 mins per time. In other words, it takes an hour to completely relocate 10 cubic meters of mountain to a place 10 miles away. Also we will assume that any miscellaneous acts including adding oil, repairing the truck are evenly distributed into the 1-hour pattern of relocation. Plus, we will assume that people take shifts to finish this humongous task that makes the truck operate 24 hours a day, 365 days a year.

    Finally, we can begin our calculation of the time needed. All we need to do now is to divide the volume by the efficiency of the truck: 2 billion cubic meters/10 cubic meters per hour = 200 million hours. To be more client friendly, let’s make it in measurement of years: 200,000,000/(24*365) ≈ 200,000,000/(25*350) ≈ 8,000,000/350 ≈ 22,857 years. (I’m thinking maybe we can round 350 to 400 to further simplify the calculation, because 22,857 years is so close to 20,000 years..)

    So..yes, this is a pretty messed up task if you want my opinion. No matter how many trucks or manpower you’d like to invest, it just seems impossible, and maybe also meaningless. Although it’s not a business case, I just want to say, please don’t do that…

  • hector Dec 2, 2017, 8:46 am

    An average truck speed is 50-60 miles per hour and the operative speed in mountainous terrain is below 55mph, so I will take 50mph as a reference.

    time = distance/speed
    time = 10miles/50mph
    time = 0.2h

    1 hour = 60 min.

    so 0 hours: 12 minutes: 0 seconds.

    what really matters here is how long the truck can take to move 10 miles, after those 10miles the mountain will be relocated cos the driver will see it from another location… my point of view

  • X. Dec 27, 2017, 6:24 pm

    suppose:
    1. mountain size: (1) height:100 meter~300 ft (2) basal area: 3000 ft *3000 ft=9*10^6 sqft; (3) mountain size=27*10^8 cubic ft;
    2. travel and load: (1) assume avg speed of the truct is 40 miles/hr; (2) loading: it takes 15 mins to load/empty the truck; (3) the truck works 24*7; (4) Each load takes in total: 20 miles (round trip)/40 miles/hr+15 mins*2=1 hr; 24 trips could be made in 1 day.
    3. truck size: assume avg. truck is 10 ft* 10ft * 20 ft=2000 cubic ft
    4. time (1) total number of trips: 27*10^8/2000=13.5*10^5 trips; (2) total days: 13.5*10^5/24=56,250 days; which is around 154 years.

    • X. Dec 27, 2017, 6:27 pm

      For the mountain which I assumed to be a cuboid for easier calculations

  • VJ Jan 14, 2018, 2:57 pm

    Mountain is a cone:
    Radius = 100 feet
    Height = 70 feet
    Volume = (22/7*100*100*70)=X
    Truck volume = (22*10*10)=Y
    Number of trips= X/Y=1000
    Trip distance = 20 miles
    Speed = 40 miles per hour
    Time for trip = 1/2 hours
    Time to relocate = 1000*1/2 hours=500 hours

  • Kristoffer Jan 18, 2018, 4:28 am

    I started with the average size truck. I estimate the size of the truck to be 4m x 3m x 5m = 60m3. A truck of that size would probably be loaded in 10 minutes using an excavator. At a speed of 50 km/h., it would take the truck 18 minutes to cover 15 km (I will add 2 minutes for traffic). Now my math is:
    20min+20min+10min = 50-60 min per trip from and back to the mountain, moving 60m2 per trip.
    In Denmark, I estimate the average mountain to have an height of 90 metres (yes, we’re a very flat country). As mountains are cone-shaped, I will use the formula V = 1/3 * H * G.
    Estimating the ground area to be 1,000m2, it translates into:
    V= 1/3 * 90 * 1000 = 30,000m3

    30,000m3 / 60m3 (amount of mountain we can move per hour) comes out at 500 hours to drive 500 trips back and forth – assuming no breaks needed between trips.

  • SL Jan 21, 2018, 1:12 am

    Assume an avg truck can take 100m3 and an average mountain contains 100K m3. To move the mountain, the truck has to first pick up the rocks, carry the rocks 10 miles away, load these rocks and come back. Total Hours = # rounds (each round would finish the above process) x # hours per round.
    # rounds = 100k (mountain) / 100 (m3 per car) = 1K.
    # hours per round = hour to pick up to rock (10 min) + hours to travel with rocks (10 miles/ 20miles per hour = 0.5 hour or 30 min) + hour to move the rock off the truck (10min) and the truck coming back (10 miles/60 miles per hour = 1/6 hour or 10 min) =60 min or 1 hour
    Therefore, to move an avg mountain with avg truck with the assumption above, it takes 1K hours.

  • Vítor Feb 7, 2018, 4:29 pm

    We will use the following formula to compute the estimated time would take to finish the job.
    t = nt * [2d/Vt + tl + tu]
    where:
    t = total time to transfer the mountain of place;
    nt = number of travels;
    d = distance between mountain and place of unload (10 miles);
    Vt = average speed of the truck;
    tl = time to load the truck;
    tu = time to unload the truck;

    We will guesstimate load time to 1/3 hour and unload to 1/6 hour.
    Now to calculate the number of travels, we have:
    nt = (Volume of Mountain) /( Volume of truck load capacity per travel) = Vm/Vt
    If we consider we have a mountain 5,000 meters high x 10,000 meters (base of the mountain) and approximate it to a cone, we will have a Vm = 1.2 x 10^12 m3.
    We will also guesstimate a truck load compartment with dimensions 3mx7mx3m, resulting in a volume of 21 m3.
    So nt = 1.2 x 10^12/21 = approx. 6 x 10^10 travels
    The truck avg speed will be guesstimated using the speed when transporting the load (50mph) and the speed when coming back to the mountain (100mph, which should be faster, as we have no load). So we have:
    Vt = (Vtransp + Vback)/2 = (50 + 100)/2 = 75 mph.
    Finally, we can calculate t as follows:
    t = 6×10^10 x [ 2×10/75 + 1/3 + 1/6 ] = 5×10^10 hours = 6×10^6 years.
    Therefore, we estimate approximately 6 million years to relocate an average mountain 10 miles using an average size truck! This without considering a 9/5 usual week job routine. If we consider it, we would estimate one would need at least 3 times more time to accomplish our objective or 18 million years.

  • Maher Elzein Feb 15, 2018, 1:49 pm

    Is it worth and profitable moving the mountain (by answering why, what, where, when, and how):

    once we know the following information by asking the interviewer the following questions we just need to negotiate, pay and calculate:
    – Is there any residents and facilities on the mountain (houses, humans, animals, etc..)
    – Location, weather, and seasons (winter (as rain and snow)), summer (as temperature and humidity) , etc..)
    – Nature (rivers, trees, etc..)
    – Mountain diameter and height
    – Truck power horse and speed, and its load ability
    – Quantity of trucks and manpower
    – Road obstacles by practically experimenting full load truck and driving to destination (require site survey and measuring)

    1. We must negotiate to the residents and pay for them or relocating them (if all didn’t agreed to relocate, the entire project will stop (even one resident refused will jeopardize the project))
    2. We must catch and relocating the animals .
    3.1 We must move or selling the nature as water and close the water sources (gradually till reach the surface once moving the mountain).
    3.2 We must remove the trees and moving them
    4. We must consider the weather conditions in our calculations and considerations
    5. The remaining can be purely calculated and experimenting
    6. Once moved the mountain we should maintaining

    Result: At end we should calculate the cost, revenue, and profit from moving the mountain.
    This is my humble opinion.

  • S Feb 17, 2018, 11:05 am

    Various approaches that can be followed to move a Mountain :
    1. Move mountain as a whole
    2. Move mountain in parts
    Before deciding on the approach let us analyze the mountain & the truck:
    1. Avg size truck : Average size mountain = 1:1000
    Therefore a truck can carry 1/1000 of the load of the mountain in one trip.

    Total Time = Time taken to break the mountain into 1000 parts + Time to load + Time to transport (to and back) + Time to unload
    Time to break the mountain:
    The transport can start when the first load is available. Let us assume that at anytime 2 people are working on breaking the mountain. 2 people work for 10 hours a day on the mountain.
    Time taken to break one part = 5 hours so in one day , 2 parts are broken.
    Loading time = 2 hours
    Transport time = Avg. of 40 mph – it travels 10 miles in 15 mins.
    2 people work on transporting & unloading.

    Unloading = 2 hours
    Loading + Unloading + Transport (to & back) = 4.5 hours

    The time taken to transport 2 loads in one day : 5 hrs ( breaking the first part) + 5 hrs (breaking the second part, by the time this is done the truck is back for second load)+ 4.5 hours (for the second load) = 14.5 hours
    For 1000 loads= 500 days or 14.5 *1000 hours = 14500 hours

  • Steve Feb 25, 2018, 12:13 pm

    – First of all, I would like estimate the size of an average mountain.
    A mountain typically has the shape of a pyramid, with almost square-shape base, and converge to a point on the top. Let’s say the base of a mountain is of a size of 5 km x 10 km, so the base has a size of 50 kmˆ2. About the height, we have an average mountain around 3000-5000 m, with the upper limit about 8000 m. So let’s assume average mountain height is 4500 m, thus the volume will be 1/3 x 50 x 4.5 = 75 kmˆ3 = 75 x 10ˆ9 mˆ3.

    – Estimating size of container.
    Let’s now assume average truck size has one container, with, length between 5-10 m, so I’ll pick 7.5 m. Then the wide and height, say 2 m each, so the size of a container is 7.5 x 2 x 2 = 30 mˆ3.

    – # of delivery needed.
    divide the volume of mountain by the container, as we have one average-size truck, and assuming only one container for the truck.
    = (75 x 10ˆ9) / 30 = 25 x 10ˆ8 times.

    – time needed per delivery.
    I’m familiar using km, so let’s transpose 1 mile to 1.6 km so 10 miles is equivalent to 16 km for one delivery. Assuming the truck will travel at 50km/h, then it takes about 20 minutes per delivery, 40 minutes including the return travel. As we have no limitation on number of employees we can hire to move the mountain, I would assume reasonably the employees are able to load the truck in 10 minutes, another 10 to unload. So, 1 hour for 1 delivery.

    -total time required.
    So total time required is 1 hour x 2,500,000,000 delivery = 2,500,000,000 hours

    ~obviously much faster if we get more trucks.

  • Gus Mar 27, 2018, 9:07 am

    Main proxies to drive us to the results here:
    – size of an ‘average’ mountain
    – size of an ‘average’ truck
    – truck’s trip

    Assuming, first of all, that we’re gonna break down the mountain, digging it’s dirt and moving it to the truck, and then
    moving it 10 miles away to pile it up again. Mountain first:

    We could assume that an average size mountain is 1.000m tall. Most mountains, for example,in sky/summer resorts are around 1.500m tall, and less than 1.000m is a small mountain, or even a hill. Also, we must consider that mount Everest is ~8.500m tall, so 1.000m seems a good guess.
    Usually a mountain is flatter near its base and very steep at its summit, so we could assume that its average steepness is 45 degrees. And why do we use it? To calculate the horizontal distance between the base and its peak. Since its a triangle with 45 degrees, we know that in a right triangle with its degrees being 45, its height is equal to its base. So the base would measure 1.000m as well. To calculate the mountain’s volume, we could consider that it has a cone-like shape. The volume of a cone is 1/3*pi*r²*h. Since r = h = 1.000m and considering pi =~3, the volume would be 1.000³ = 10^9 m³.

    Now the truck:
    It’s loading part has a rectangular shape. It’s height can be considered 1.5m. Its width is practicaly the width of a street lane, say 3m. To estimate its lenght, we could imagine a real situation where we are driving with our car and see right next to us an average size truck. Its lenght would be around 2 times the lenght of a car. Considering that a car has ~4m of lenght, the lenght of the loading part of the truck would be 8m. So its volume would be 1.5 (height) x 3 (width) x 8 (lenght) = 36 m³. Since its height could be a little higher than that, we could round this number to 40 m³.

    Cool. So we have 1 billion cubic meters to be moved by a 40 cumbic meters ‘container’. So, the total number of trips that we’d need to make is 1.000.000.000/40 = 25.000.000 trips.
    Alrighty. So now let’s see how much time we would need to go there, take 40 m³ from the mountain and move it 10 miles away. Since it’s a mountain, we can consider that the road from and to the mountain is a nice, traffic-free rural road. So while it’s rolling, we can consider that the truck travels at 50 mph. To move 10 miles, it would need 0,2 hours = 12 min. To load the truck, let’s say we could use a bulldozer to do that (without a bulldozer, we’d have to do it by hand. Ugh!). So let’s say it would take somewhere close to 10 min to do it. Let’s say 9 minutes. Moving 10 miles away from the mountain would take the same 12 minutes, and unloading the weight is faster, the truck would just need to lean its loading part. So say 3 minutes to do that.
    So the total amount of time to go there, load the truck, move it and unload it would be 12+9+12+3 = 36 minutes.
    Since we’d need to make 25.000.000 trips, the total time would be 25.000.000*36, in minutes. Instead of multiplying it already, since we know it’s a huge amount of minutes, we know that it would take months to do it.. maybe years! So let’s divide this number by 60(minutes to hours)*24(hours to days)*30 (days to months). So that would be 25.000.000*36/(60*24*30). 25mi/24 = ~1mi. So 36.000.000/60*30 = 360.000/18 = 20.000 months. So, that’s a lot of years. More precisely, that’s 20.000/12 = 5.000/3 = ~1670 years.

    Answer: 1.670 years.

  • RUOYU Mar 27, 2018, 7:52 pm

    First of all, how long to move a mountain includes the time to fill the trunk, move to somewhere 10 miles away, and unload it. The first part to be considered is the volume of the average size mountain, how long it takes to fill the trunk. The average size mountain is cone shaped with 100 meters high, and 200 meters radius. Therefore, nearly 4,000,000 m3. The excavator fills 1 m3 every time using 3 mins. It takes about 1,300,000 mins to excavate the mountain. In addition, the average size of trunk is 4m*10m*2m, 80 m3. The trunk will carry 50,000 times in total. To move somewhere 10 miles away, it takes an average size trunk 10 mins on one way, totally 20 mins every time. Also, it takes 2 mins each time to unload the soil each time, 100,000 mins in total. In essence, it will take 1,000,000 mins to move the mountain 1,300,000 mins to fill the trunk, 100,000 mins to unload. The answer is 2,400,000 mins, equal to 40,000 hours, 1500 days, 50 months.

  • Philip Apr 10, 2018, 11:51 am

    1. define average size of mountain: 1km x 1 km square base and 100 m high.
    so total volume is 100 M cubic m
    2. an average truck can take 2m x 5m x 2m = 20 cubic m
    3. to relocate entire mountain, 100m / 20 = 5m times
    4. each time, loading + transportation + unloading + movement = 1 hours
    5. total time needed = 1 * 5m = 5 m hours = ~ 600 years

  • Mukesh Apr 21, 2018, 3:45 am

    I have assumed that the average mountain circumference would be 1000 m and its height would be 200 m and from here we can calculate the volume of the mountain i.e. 1/3 * pi* r2*h = ~ 5 million cubic m.
    Now let’s calculate the volume carrying capacity of an average size truck ( why I am not bothered about the weight is because the trucks are meant to carry objects such as metals which are more dense than soil and trees):
    With a length of 10 m and a width of 2.5 m and a Height of 1 m, the carrier of an average size truck can carry 25 cubic meter of load.
    Total No. of Trips required = Volume of the Mountain / Truck carrying capacity, i.e. 5 *1000*1000 / 25 = 200,000 Trips.
    Time required for the entire process (T) = Time required for one trip (t) * Total number of trips (N)
    t = time required for loading (a) + journey period to and fro (b) + time required for unloading (c)
    a = 15 minutes
    b = at a speed of 30 miles/hour (keeping in view all the ups and downs of the road), it would take 40 minutes.
    c = 5 minutes
    So, t = 15+40+5 = 60 minutes or 1 hour
    T = 1 * 200,000 = 200,000 hours or ~8000 days

  • AA Apr 27, 2018, 10:21 am

    The volume of the mountain can be thought of in terms of the height and width/diameter of base. Thinking of a very typical mountain, let’s assume that it is canonical or pyramid-like in shame (i.e. it has a circular or square base and the shape tapers as you reach the top).

    I climbed a mountain in China that was 5000 ft — I think this was neither remarkably high nor tiny so let’s use this as a proxy for an average mountain. I might assume that the base of a mountain is 1 km in diameter. As I am no geometry expert, I’m going to think of the mountain encased in a cube with the same dimensions. That cube would be 1000m x 1000m x 1500m (1ft being 30cm, which gives 150,000cm, which gives 15,000m). The volume of the cube would therefore be 1,500,000,000 cubic metres. Thinking of how much space a pyramid would take up of that cube, I think it would be around a third. So that gives me a mountain of 1,500,000,000 / 3 = 500,000,000 cubic metres.

    The next considerations are the logistics of moving 500k cubic metres 10 miles down the road in an average sized truck and the labour involved.

    An average truck (I’m thinking of a lorry that might pass you on the motorway which has a square end and is long) might hold 3 x 3 x 10 = 90 cubic metres.

    I’m going to assume that the truck might drive along a motorway, given we’ll be in a rural area. So let’s assume tame speed of 60 miles per hour. This means the truck could do the journey in 10 mins. It would have to do this 2-way, which totals 20 minutes.

    Then we need to think about the loading and de-loading of the truck. I’m going to assume that we have a decent supply of workers, say 20. Each might be able to collect, load and unload around 2 cubic metres per hour, so 90 cubic metres in 2.5 hours. Taking into account travel, let’s assume that an entire round trip/process with 90 cubic metres takes 3 hours.

    Let’s assume there’s a push to get this project completed and the workers are employed Monday to Friday, 8 to 6 including an hour’s lunch break. So each day 3 trips could be made which is equivalent to 210 cubic metres per day. Let’s round this to 200 cubic metres for ease of calculation. 1000 cubic metres could be moved per week. So moving 500,000,000 cubic metres would take 500,000,000/1000 = 500,000 weeks. Or 10,000 years. Seems like an impossible task….

  • Denzil May 22, 2018, 10:37 am

    Assumptions:
    1 – Full capacity of dump truck is 10 tons of material
    2 – Dump truck speed when empty 30 MPH
    3 – Dump truck speed when full 10 MPH
    4 – From visual memory, 10 loads would probably represent a decent sized mound, so a mountain would be at least 1,000 times that.
    5 – So moving the mountain would be 10,000 loads
    6 – 30 minutes to load the truck
    7 – 10 minutes to unload the truck

    Workout:
    Total roundtrip for one load would be = loading time + transport time + unloading time + return time
    This would be = 30 Min + transport time + 10 Min + return time
    1 – 10 miles at 10 MPH = 1 hour trip forward = 60 Min
    2 – 10 miles at 30 MPH = 1/3 hour return trip = 20 Min
    Total roundtrip = 30 + 60 + 10 + 20 = 120 Min
    Total mountain move = 10,000 * Total rountrip
    Therefore 10,000 * 120 – 20 = 1,200,000 Min; 20,000 Hours

  • Parth Jul 21, 2018, 1:43 am

    How big is the average mountain?
    Let’s assume that the average mountain is shaped as a cone with height 10,000 feet and radius 20,000. The total volume of this mountain is then 3.14×20,000×20,000×10,000/3 =
    = 20,000 x 20,000 x 10,000 = 4 x (10,000)^3

    How much can the average truck carry?
    If I went wrong, this is likely where for two reasons.
    1. I am assuming pick up truck.
    2. My only constraint is volume and not load.
    Compact trucks are probably 2x4x5 = 40 ft^3 whereas bigger trucks are probably 2x5x6 = 60 ft^3, so let’s go with 50 ft^3.
    The amount of trips the truck would have to take if filled to the brim is 4 x (10,000)^3 / 50 = 4x2x1000x(10,000)^2 = 8000 x (10,000)^2.

    Now with all this load, we can assume that the truck will go at an easy 20 miles per hour to the relocation, making each of those trips 30 minutes. On the way back, the truck will go double the speed at 40 miles per hour, returning in 15 minutes. Note, this does not account for time in loading and unloading. So with 3/4 hour to move each load of mountain, it will take 8000 x (10,000)^2 x 3/4 hours = 6000 x (10,000)^2 hours.

    In days this looks like 6000 x (10,000)^2 / 24 =
    = 250 x (10,000)^2 = 250 x 100,000,000 = 25,000,000,000 days

  • MA Jul 22, 2018, 6:42 am

    In hours, the total amount of time it would take is = (transport timex2 +loading time) x volume of mountain in cubic feet/carrying capacity of truck in cubic feet

    Assume that the average truck that can carry a load has dimensions 10 feet by 5 feet by 10 feet. That means it has a capacity of 500 feet.

    I know that mountains range anywhere from a couple hundred metres high to thousands of metres high, so for simplicity I will assume that the average mountain is 1000 metres high or 3000 feet high. The height of the mountain is probably less than its length and width, so let’s say that each is twice the height. That yields cubic volume of 108,000,000,000 cubic feet.

    Divided by 500, that is equal to 216,000,000 truck loads.

    If the average truck travels at 60miles per hour, then a trip both ways is 20 minutes. If the average loading time is 40 minutes, then that’s a total time of 60mins per truck load or 1 hour per truck load. The total time is then 216,000,000 hours to move the mountain 10 miles.

  • Bogumil Jul 24, 2018, 2:13 am

    9 years

  • Abhishek Singh Sep 10, 2018, 2:42 pm

    Size of the mountain – 1,00,000 tons
    The capacity of an average truck – 30 tons
    Trips required – 3,333
    Time to drive to & fro per trip – 50 mins ( 30 mins loaded & 20 mins unloaded)
    Time to load and unload – 3 hours
    Total time – 10,000 + 17000 = 27000 hours

  • Cope Sep 29, 2018, 1:45 am

    assuming by “truck” we mean pickup

    pick up bed aprox 8ft by 5ft and assume we can pile dirt 4ft high
    aprox truck volume= 1,500 ft^3

    mountain “average” well say 2000ft high cone with a 3000ft diameter giving us an aprox volume = 9,000,000,000 ft^3

    assuming truck travels at 60mph to go to the new sight and then back to get more dirt would take 12 min.

    truck volume x mountain volume x round trip time
    aprox = 12,000,000 minutes
    or
    aprox 24 years

  • AC Oct 11, 2018, 10:25 pm

    15 day 8 hours 30 minutes

  • Cameron Hector Oct 15, 2018, 11:10 am

    So, conceptually, I believe there’s three things we need to determine at a big picture level. Size of an average mountain, how much dirt we can fit in one load for a truck, and the length of time it takes for a truck to move a load 10 miles away and return.

    The ultimate answer should be (Size of Mountain/Size of Load/Hours it takes to move the load). Below are my justifications and answers:

    Size of Mountain:
    Really no reference point here to be honest. I think back to when I traveled Yosemite and thought about how large Half Dome was. I remember seeing elevation levels reach 10,000+ feet. I think? An average mountain is obviously much smaller, so my assumption is that an average mountain is 5,000 feet tall.

    The size of the mountain I consider a circular based with a triangular elevation to the tip. So, I determined the diameter of the mountain being roughly a mile, as it seemed appropriate to walk a mike to get from one side of the base to the other. Again, based on personal experience, so this may be totally off in reality. There’s 5,280 feet in a mountain, so I’m estimating a diameter of 5,000 (round to 5400 since I rounded down the height of mountain for simplicity). Therefore, I’ll use the volume equation for a cone, which is V=(pie, roughly 3)r^2 x H/3.

    Equation: 3 x 2500^2 x 5000/3 = 3 x 6,250,000 x 18/3 = 18,750,000 x 6 (round up 19 million) = 19,000,000 x 6 = 94,000,000 sqft of dirt.

    Size of Mound per load:
    Okay, so thinking about the average size of a truck. If I were to stand on a truck bed, I feel that a mound could reach up to roughly 4 ft. The length of the trucked, if I were to lay down, seems much longer than me. I would guess a truck bed to be about 10 ft long. Again, using the cone formula of V= (pie) r^2 x (h/3) I’d come up with size.

    Equation: 3 x (5)^2 x 4/3 = 3 x 25 x 4/3 = 100 sq ft.

    Time it takes to load a truck:
    I assume to move a mountain we’re using some heavy equipment and plan to have a number of workers on-site at both the new location and old location. Let’s assume that while the truck is driving, the operating teams are breaking down the mountain and putting the dirt in piles to load the truck as soon as it arrives. I’m using these assumptions basing it on an effective team.

    Time it takes to load pile into car = 10 minutes
    Time it takes to drive to back and forth based on average speed of 60 miles per hour = 1 minute x 20 miles = 20 minutes
    Time it takes to unload pile of car = 10 minutes

    Total time per load = 40 minutes.

    Back to the original equation:

    (Size of Mountain)/(Size of Load)/(Time takes to move a load) = Total time needed to move mountain

    94,000,000 sqft/100 sqft/ (2/3) hours = Roughly 600,000 hours

    Let’s say the crew is working 10 hours per day, six days a week because they take off Sundays.

    600,000 hours/ 10 hours/six days = 10,000 days + 1700(seventh day that is off) = 11700 days in total

  • LL Dec 6, 2018, 12:51 pm

    55,000days

  • Deborah Chima Dec 9, 2018, 4:07 pm

    The proxy for how long it will take our average sized truck to move the average sized mountain is the speed of the truck. An average trucks moves at 50-60 mph, so given the fact that we’re hailing a mountain, our truck’s average speed is 35mph.

    However the speed of the truck is an imperfect proxy as it is also dependent on the road that the truck will travel on. We broke our proxy into parts– uphill travel, flat travel, and downhill travel.

    Given that there are 3 different types of road, we split our 10mile journey into thirds, giving each mode of travel 3.3 miles.

    1. Uphill journey
    * round 3.3 miles down to 3 miles, and adjust the trucks speed down to 30mph (given it will slow down travelling uphill).
    * When travelling at 30mph, it will take a truck 1/10th of an hour to travel 3 miles. This means it will take the truck 6 minutes to complete the uphill journey (60/10)

    2. Flat journey
    * round 3.3 miles up to 4 miles, and note that the truck travel’s at its average speed of 35mph given the flat road.
    *It will take a truck travelling at 35mph roughly 1/9th of an hour to travel 4 miles, roughly equating to 6.2 minutes. The flat journey takes about 6.2 minutes (60/9)

    3. Downhill Journey
    * To offset our rounding up of .7 miles in the flat journey, we will also round the downhill journey down to 3 miles, as we did for the uphill journey. We estimate that the truck moves at 40mph as it will be moving faster going downhill.
    * A truck moving at 40mph will take roughly 1/13th of an hour to travel 3 miles. Estimating this to be 1/12th of an hour, we take 60/12 to get 5 minutes. This means our downhill journey is estimated to take 5 minutes long.

    The total minutes of the whole journey are found by adding the minutes of each “different” journey up.
    Total journey mins = uphill mins. + flat mins. + downhill mins
    + 6 + 6.2 + 5 = 17.2 minutes.

    The total journey will take roughly 17.2 minutes

  • SCA Dec 29, 2018, 9:32 am

    I broke down the problem as into two first: how many truckloads a mountain would be equal to in terms of volume and how long would it take for one truck with 100% utilized capacity to go 10 miles (I used 16 km as I feel more comfortable)

    I further broke down the volume part into two: volume of a mountain and volume of one truck.

    I first tried to find the height. Everest is around 9 km high, the highest, and I assumed a small mountain to be 1 km high. However, most of the mountains are small, so found 1/4 of (9-1) and added this to 1 in order to find height of average size mountain. I assumed a rectangular pyramid volume to be a good method to find volume of a mountain. Because if we look from above, in general we would find it like placed on a square with sides approximately equal to double the height. Giving us a volume of 36 km^3.

    I then assumed that a truckload would roughly equal 90m^3 if it has sides 3*3*10.

    Dividing the mountain with truckload would give nearly 400 million trucks if I did not do any calculation mistakes.

    Assuming a truck with fully utilized capacity would go slow at 60 km/h, 16 km requires 16 mins. With 400 million trucks needed, this would give a result of 6.4 billion minutes. Converting this months would give 1.5 million months.

    Most probably there is a calculation mistake because this is a very very long time 😀

  • IF Dec 29, 2018, 10:46 am

    assuming mountain can be relocated:
    10 miles = 16km
    average truck speed = 60km/h
    time taken per travel (one-way) = 15 min
    no. of trips required = 200
    hence = 200 x 15 min = 3000min = 50 hours

  • Renata Jan 7, 2019, 2:45 pm

    Assuming the truck has a cable to tie around the mountain and to pull it, and the truck is capable of pulling 1 mph in 2.5 hours with a velocity of 15 mph; thinking the mountain weighs about 11000 pounds or 5 tons, the truck will take about 25 houras to move it 10 miles, or at least more than an entire day without any obstacles or stops.

  • Ryan Mar 21, 2019, 1:27 am

    Total time= cutting the mountain for first time + n × (loading into truck by worker+ truck travel + truck unload + truck return)

    Time for further cutting of mountains is assumed to be equal to the time taken in bracket. So only first time is considered.

  • Asd Apr 1, 2019, 4:38 am

    What I did was I made assumptions about the mountain.
    The mountain contains only sand and rocks and its total weight of these elements is 10,000Kg. And our truck has a carrying capacity of 1,000Kg

    My frame work was:
    Loading time x number of rounds
    Unloading time x number of rounds
    Time from point A to point B x number of rounds
    Time from point B to point A x number of rounds

    Number of rounds is: Mountain weight/Carrying capacity=10
    Loading time x 10
    Unloading time x 10
    Time from point A to point B x 10
    Time from point B to point A x 10

    Loading time: I assumed that we have an advanced truck that can load 100 Kg per shovel per minute. 100×10=1000Kg in 10 minutes
    10 Minuts x 10 Rounds
    Unloading time x 10 Rounds
    Time from point A to point B x 10 Rounds
    Time from point B to point A x 10 Rounds

    Unloading time: unloading is much faster than loading so we’ll assume that 2 minutes is the time required.
    10 Minutes x 10 Rounds
    2 Minutes x 10 Rounds
    Time from point A to point B x 10 Rounds
    Time from point B to point A x 10 Rounds

    Time from point A to Point B: let’s assume that the average truck moves at 50 mph but since we are carrying 1000Kg of sand and rocks we’ll take 20% from its speed and say its 40 mph.
    10/40 x 60= 15 minutes
    10 Minutes x 10 Rounds
    2 Minutes x 10 Rounds
    15 Minutes x 10 Rounds
    Time from point B to point A x 10 Rounds

    Time from point B to point A: Here we’re not carrying any loads so we’ll stick with the average truck speed.
    10/50 x 60= 12 Minutes
    10 Minutes x 10 Rounds
    2 Minutes x 10 Rounds
    15 Minutes x 10 Rounds
    12 Minutes x 10 Rounds

    Our formula is:
    (10×10)+(3×10)+(15×10)+(12×10)= 6.5 hours or 6 hours and 30 minutes.

    now I know this is somewhat unrealistic but for simplicity sake I tried to simplify each step.

  • Ivan Apr 1, 2019, 11:03 am

    time = time required to move 1 mile * 10
    = weight/weight per hour
    =(assume the mountain is a triangle)
    =1/2*(100 mile * 100 mile) * density / weight per hour
    =(assume density is 20 kg/mile^3, weight per hour = 20 killograms)
    =2*10^4 hours

  • Est Apr 19, 2019, 5:53 am

    Mountain size : 1000high and 500 radius
    Mountain volume = between 1/3*pi*(r^2)*h (cone) and 1/2*pi*(r^2)*h (1/2 of cylinder)= 100,000,000
    Truck size : 5*5*8
    Truck volume= 200
    Truckloads = 100,000,000/200=50,000
    Time per ride= load (automatically) + drive+ unload (automatically) =20+20+20=60 min= 1hour
    Working hour: 8 hours per day
    Duration = 50,0000 loads x 1 hour/8 hours = 6,250 days=25 years (250 working days per year)

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