Estimation Question

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"

You may NOT ask any clarifying questions. Any answer that does not include a specific amount of time (days, hours, minutes, etc..) is automatically incorrect.

Good luck!

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(feel free to use your first name only or your initials)

** Don't cheat by looking at everyone else's answer first!

If you want to know what acceptable answers look like, after you post your answer look for the responses from:

Bobby, Dennis, Jakobicek, Surya, Aaron, Sachin... there are others like their answers. To see my commentary on why these answers were acceptable when others were not, click here.

 

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466 comments… add one
  • d Nov 11, 2010, 9:27 am

    what ressources do we have?
    What is an average mountain?
    What obstacles are there within the ten mile distance?
    Machine that can pick up certain voulume of mountain, number of those volumes in the mountain, time to take them, move them to the other location (depends on the height, take average)

  • Medic Nov 11, 2010, 9:37 am

    OMG … This is tough! Long winded and probabley incorrect

    I would make a few assumptions to make my life easy:

    The average mountain is barren and consists of only rocks and soil and perhaps minor planr structures … grass and no trees. There will be no animals or ice.
    It is single peaked with peripheries. So it is cone shaped with extra peaks not as high as the core peak. There are two extra peaks, each comprising 10% of the volume of the core peak.
    We can restructure the mountain using the base as the top without any complications
    We have 100 men working

    The core peak has a square base of 30 * 30m area and a height of 1000 m. Hence the volume would be 1/3 *900*1000m = 300,000m3
    other peaks will account 60,000 hence a total volume of 360,000m3

    three tasks involved in moving the mountain
    picking up the soil
    Transporting
    offloading at new destination

    1 man can move 3.6 metric tons in 100 hours
    Hence one man can move 360000 in 36000 hours

    Assuming we have a work force of 100 – it will take 360 hours to move that mountain!

    Assume a working day is 6 hours, it will take 100 men working 60 days!

    And then I get fired.

  • Medic Nov 11, 2010, 9:39 am

    * typed in a hurry
    One man can complete all three tasks for 3.6 m3 in 100 hours (Not metric tons)

  • Bin Nov 11, 2010, 9:59 am

    How can we define move? Can we consider the sun as the original point? In case so, no man is needed, and we could actually find somewhere as the original point, so that the mountain can be moving 1 mile in a second…

  • Bobby Nov 11, 2010, 10:08 am

    I will assume that a mountain is really just a very large pyramid. This means that it has a square base.

    I will assume that an average sized mountain is about 10,000 feet tall.

    I will assume that the slope/incline of the mountain is easily climbable and is about 30 degrees. Since Sin(30) = 1/2 and cos (30) = (sq rt 3)/2, then the length of one of the sides of the base of the mountain/pyramid is 20,000 x (sq rt 3) feet.

    The volume of a pyramid is the (area of the base x height) / 3. Thus the volume of this mountain is 20,000 x 20,000 x 10,000 x 3/ 3 = 4,000,000,000,000 cubic feet.

    Now suppose that the mountain/pyramid is in sections of 1 cubic foot.

    If we think of a single worker going to the top of the mountain, removing one of the cubic sections and putting it in his wheelbarrow and then transporting it a mile to the site of the new mountain and putting it there, we quickly realize that the man will travel approximately the same distance every trip. He would have to go to the top of the mountain to take the first stone, and he would then be able to place that stone at the bottom of the other mountain.

    It is 20,000 up the side of the mountain, 20,000 feet to the bottom and then about 5000 feet (just under a mile) to the new site. This is a total of 45,000 feet. Using 5000 feet as our approximation, this is about 9 miles. We will then assume that the average worker can walk about 1.5 miles an hour while pushing this wheelbarrow (normal walking speed is about 3) which means that it will take him 12 hours to take one cubic foot from one mountain to the other. The time that it takes him to hoist and then dumb his stone we will assume is negligible.

    Therefore, putting this all together, it will take 4,000,000,000,000 x 12 man hours or 2,000,000,000,000 man days to move this mountain.

  • M Nov 11, 2010, 10:09 am

    Seeing as we are not given any information to relate to, this makes it very easy to be as abstract as possible in my opinion, so i’d start relative to Earth’s rotation. We are all moving, but in relation to what (deeeeep I know!)? I think that’s for the person who is answering the question to decide! Anyway, this is just how i’d think about doing it!

    In this case, an ‘average mountain’ is therefore irrelevant. It could be a flea, human or Everest. It doesn’t matter. All objects on earth move with the earth and move at the same speed relative to the earth’s rotational spin. The fact a mountain is stationary makes it that bit easier. If it were a dynamic object like a car then we would have to assume all sorts of things to

    Let’s assume the mountain is at the equator. I’d assume the distance across the earth’s equator is 22,000 miles. It takes 24 hours to complete a full cycle. For the purposes of ease in this question (im simply demonstrating my method not arithmetic ability here!), lets say 24,000 miles:

    This means every 60 mins the earth and any object on it’s equator would travel 1000 miles. So, every 10 miles, it would take 0.6 mins or 36 seconds.

    This method cuts A LOT of corners but the object (“average sized mountain” in this case) still moves 10 miles in 36 seconds relative to the Earth’s rotations. One could also use the Earth’s rotation round the Sun but im not the astrophysicist so i’ve stuck to what I roughly know. There are INFINITE answers to this question, therefore.

  • ahmet Nov 11, 2010, 10:14 am

    My first assumption would be that we are thinking about mountains on the earth’s surface, excluding mountains on the ocean floor and mountains on the moon or other planets.

    Next, we need to think about what qualifies as a mountain and at which point an elevation is just a hill. In addition, one has to define how a mountain is differentiated from another mountain, that is, at which point is it a mountain range, where are the limits / borders of a mountain? This becomes especially important to calculate an average. Another question is whether the distinction between mountains are colloquial or geological. There are mountains which are geologically separate but are regarded as “one” mountain.

    Also, one has to regard what “average” means. Whether the assumption is a global, national or local average defines the mass we will be dealing with. In defining a national or local average, one has to take into account mountains which are laid out on borders and are only partly within a region.

    In regarding the average mountain, it is important to take the surface vegetation (trees, bushes, etc) and wildlife (rabbits etc) into account as well. While some mountains are not very habitable, others are teeming with life. Further aspects in calculating the mass are water (rivers, lakes, underground reservoirs) and cavities (caverns, mines etc).

    Taking all these factors into account, one can calculate a well defined average. The act of calculating the movement by 10 miles is then a simple formula of distance times force necessary to mobilize the mass.

    However, there could be asked further questions complicating the issue, such as: why does the mountain have to be moved and does it have to be moved intact? If we are talking about the parts, then detonation and drilling would be viable means (in which case, we would be thinking of practical methods instead of abstract calculation of force). If we are talking about geolocation, one could also simply move the points of reference, thus changing the coordination system. This would be “moving the map” instead of moving the mountain.

  • ahmet Nov 11, 2010, 10:20 am

    what I left out, though it is implicated: once we know how much mass we have and how much force we need to move it, its simple to map that to number of workers / machines and thus time needed. but again, i think the important points are:
    – what does “average” mean?
    – what does movement / relocation mean?
    the simplest relocation is making a change on the map (there are precedents for this in history, ie. border negotiation in cyprus as a recent example)

  • Reem Nov 11, 2010, 10:31 am

    I don’t know if am right, but i will say it will take minutes, i will draw the mountain and i will right down it’s volume then i will just move it using my mouse on a scal drawing 1 mile or two or whatever!!! if it’s not right i still got a point 🙂

    • Frank Apr 14, 2017, 12:22 am

      Is this the quiz answer page to the mountain question?
      My answer to it is—–9 minutes and 1 second.
      Am I correct?

  • Dennis Nov 11, 2010, 10:38 am

    I’ve used the following assumptions:

    -mountain volume:
    ->Assume cone shape: 1/3*heigth*base area = 1/3*h*pi*r^2
    ->heigth: 2km
    ->width: 4km

    -transport means: I assumed transport by trucks (vs. a big lifting device, cartwheels or a ‘star trek beamer’ thingy… etc)
    ->truckload volume: lxbxh = 5x3x1.5 = 23m^3 = 23*10-9km^3
    ->truck speed: 15km/hr (unpaved road)
    -># trucks: 100
    ->trucks operate at max efficiency (without timeloss due to having to wait in line)
    ->operating hours: 24/7 for 300 days/year
    ->time/truckload: pick-up: 30 min; transport: 60min (10miles/(15km/hr)); delivery: 30min; Total: 120min = 2hr
    ->refueling: happens within 30mins pick-up and delivery

    Which gave me the following results:

    So, based on these assumptions you’ll find a mountain volume of 33km^3
    Per session 100 trucks transport 23*10-9km^3 each = 23*10-7km^3
    Therefore 33/23*10-7 ~ 1.5*10^7 = 15 M sessions are required
    In total 15M sessions*2hr = 30 M hrs are required
    Leading to 30 M hrs/ (300*24 hrs/year) = 30M/7200 ~4000 years

    What does that mean: I’d build a tunnel rather than moving the mountain

  • Jakobicek Nov 11, 2010, 11:17 am

    Oki that is a good one… lets see…

    Lets try to approach it as follows we are going to move the mountain by moving material using trucks. The rest of the work is negligible. So we need to see how much material there is in an average mountain and how much can one truck pick up per load.

    First, we need to somehow get in grasp with what is an average mountain… this is quite a vague description, but let us start with modeling an average mountain as follows – 500m high, square with a side of 10km

    Now lets count how much material that is…

    1/3*500*10000*10000 m^3

    Fine next step is determining what an average size truck looks like…
    yeah a truck is nothing more than a box that is 10m long, 4m wide and 2m high.

    so now we have:
    500*10000*10000/(10*3*2*4) loads –

    Now lets determine how many trips per hours you can make. 10 miles is about 16km so a truck that goes 80km/h should be able to get about 2 trips in per hour.( do not forget to count for return trip) with a work day of 10 hours we get to 20 trips per day.

    500*10000*10000/(10*3*2*4*20)days

    Oki now let us assume we have 100 trucks for the job

    500*10000*10000/(100*10*3*2*4*20)days

    Now lets get this mess sorted
    500*10000*10000/(100*10*3*2*4*20)days = 500*100*100/(4*2*3*2)days ~ 500*2*100 days = 100000 days for 100 trucks ~ 300 years

    • Himanshu Ahuja Mar 27, 2017, 11:45 am

      Why have you taken *3 while calculating the load a truck can carry because the formula should be L*B*H i.e. 10*4*2 instead of 10*4*2*3

  • Surya S Nov 11, 2010, 11:41 am

    I would start with a smile (!) and some clarifying questions:

    1. Where is the mountain? On land or sea or some place else? (Say answer is land)
    2. Does the mountain after move have to be in same shape and size? In other words, can it be broken and moved and the pieces just dumped 10 miles away? (Say answer is it can be broken)
    3. Is there a direction preference and are there any obstacles to consider in that direction? (Say answer is no preference and no obstacles)
    4. Is there any indication on what an ‘average size mountain is? (Say answer is 4000 meters in height, 2000 meters wide and 2000 meters long and assume it’s a pyramid)

    Ok. With these inputs, the total time would be the sum of time of:

    1. Breaking
    2. Loading
    3. Moving
    4. Unloading

    Assumptions:
    1. All tools and eqiupment needed are readily available and there is no lead time involved in procuring or setting up any of these.
    2. We use dynamite to break the mountain and trucks to move the mountain.
    3. One dynamite explosion breaks one truckload worth of mountain.
    4. Truck carrying capacity is 20mx10mx10m = 2000m3.
    5. To optimize time and labor, loading starts every 4 explosions and it takes 1 hour to load trucks (8 of them) and it takes 1 hour to setup the next set of 4 explosions.
    6. Unloading is just dumping the truckload 10 miles away and on mountain regions truck move slowly so assuming speed to be 30 mph
    1. Breaking Time
    One explosion breaks 2000m3 of mountain. To break complete mountain, it takes approx 700K sets of 4 explosions. With each set taking 1 hour, it will take 700K hours to break the mountain.

    2. Loading Time
    Each loading starts after 4 explosions and takes 1 hour each, so to load 700K sets will take 700K hours but as loading happens in parrallel, only last load will take additional time of 1 hour.

    Additional Loading Time = 1 hr

    Moving Time
    3. As trucks will move in a mountain region slowly, let’s assume they take 20 mins one way to cover 10 miles (30 miles per hour speed) so will make a round trip in 40 mins and let’s assume it takes only 20 mins to unload the trucks as unloading is simple dumping of the mountain. So by the time one load is ready for pickup, the trucks are available and so no additional time involved in moving as the move also happens in parallel.

    Additional Moving Time = 20 mins (one way for last load)

    4. Unloading time
    Assumed that unloading is just dumping the load and it takes 20 mins and this happens in parralel to breaking.

    Additional Unloading Time = 20 mins (for last load)

    Total Time = 700K + 1 + 20/60 + 20/60 = 702K hours = 80 years (approx)

  • Erik G Nov 11, 2010, 11:53 am

    So this is a question about a large scale industial project and
    as such we will assume (cost) realistic amounts of available
    labor and equtment and so forth.

    Likely, the bottle neck in this process will be moving the large
    amount of material 10 miles, and not the extracting of rock,
    or unloading etc. Therefore we will estimate this by computing the
    amount of material needing transport, and the rate at which we
    could hope to move it.

    A typical mountain (obviously ambiguous) we will assume to
    be 1km in height, typically sloping at 45 degrees for simplicty,
    and thus a radius 2km at the base. Since I am a physicist I
    actually remember the formula for volume of a cone which
    is (1/3)pi r^2 h, thus about 5 km^3= 5 * 10^9 m^3.

    So I will assume a railway links the two sites, as this would
    be a good investment if not already existing. A typical rail car has
    dimension 5m*5m*30m=600 m^3. Assume 100 car
    train, and there are two operating continuously. Realistically
    it might take 1 hour to transport between the sites, so
    assume a total of 10 loads per day.

    Therefore 10x100x600=6*10^5 m^3/day. Dividing the
    above figure we see that we need about 10^4 days to
    transport the load which translates to 30 years.

    Solution was typed out within a 5 minutes of seeing the
    question.

  • all39 Nov 11, 2010, 12:07 pm

    270 years / number of trucks

    assumptions:

    (1) earth movement by massive mining trucks
    (2) decent road between the mountains
    (3) sufficient earth moving equipment exists at the original mountain to dismantle it quickly enough such that trucks don’t have to queue
    (4) sufficient earth moving equipment exists at new mountain site to assemble new mountain as quickly as the dirt comes in

  • Medic Nov 11, 2010, 12:15 pm

    revising my solution earlier, I should have split the mountain into three … upper third would take 30o hrs, mid 200 hrs, lower 100 hrs for 3.6m3 per man.

  • Javier Nov 11, 2010, 12:37 pm

    From the realistic point of view, the time that will take to relocate a mountain 10 miles, would the time the earth rotates from the first point where you measure the place till it reaches 10 miles. Saying that it will be different where the mountain is located..say that it is in the ecuator, so like that takes shorter time to do 10 miles..So, saying that, to rotate a mile the earth takes the whole distance of eacutor(perimeter)/24h..from this calculation we have miles/h..so then we will know how many min takes to move it
    From the urealistic way of being able to move it with resorces, it will depend on how big it is, the path we have to cross to move it, etc..

  • Sehar Nov 11, 2010, 12:38 pm

    First some clarifying questions – where is the mountain, in the middle of a flat plain, in a mountain range, get a sense of the geography. Are we taking into account the time for equipment to help with the move to arrive?
    Figure the mountain is attached at its base and must be removed from the base. We could either chop away at the bottom while having to make sure the mountain doesn’t topple over, perhaps having some sort of removable support devise put underneath, noting some of the mountain might be lost in the chopping, as in it crumbles away. This would take as much time as I assume it takes to drill a mine, which from my work with offshore drilling can be a few weeks. Or use explosive devices to try to break the mountain from its base which is a much faster technique that could be done in a matter of days, but we risk a possibly higher percentage of losing some of the mountain. Then we could either have the mountain rest on a plank and move it over land by pulling it with powerful automotive vehicles. It would probably take a few days to make sure the mountain was stable and then assuming we could get enough vehicle power while taking into account safety, we could move the mountain at a speed of 5 mph, or about in 20 hours. Or we could airlift the mountain by army grade helicopter devices, which I would assume would be able to move the mountain, taking into account safety, within a few hours. So the shortest amount of time to move the entire mountain including preparation for moving, could be a few days plus time to get equipment to the mountain to a few weeks. Physical movement from the mountain’s origin point to the point 1o miles away would range from a few hours to about 20 hours.

  • GP Nov 11, 2010, 12:38 pm

    (i am using SI units) let’s assume that an average mountian is 5000m high and that a average size mountain is a semisphere with a radius of 5000m. then the volume of it would be V=4/3xpixR^3×0.5. assuming that the mountainous material has density of 20, it’s mass is m=dxV.
    since i am going to push it i assume i will exercise a constant force forever, resulting in a constant acceleration alpha = force/mass. i assume i can exercise force equal to my weight of 100kg, thus a force of 1000N, thus i know the acceleration alpha=1000/mass.
    from physics we know that
    distance=0.5xalphaxtime^2, distance in our case is 10 miles, or about 15km (which gives t=squareroot(2*distance/alpha)) which equals to roughly 25 years

  • Grigory Nov 11, 2010, 12:50 pm

    The answer depends who (number of people & level of skill) is doing the mountain-moving and with what quarrying technology. If that’s the kind of technology they had in the stone ages, when they built the Stonehenge, – then very long indeed :). I will assume that we’re talking top-of-the-range modern excavators, drilling machines and trucks.

    Independent variables needed to make a time estimation:
    – Number and type of quarrying machines and their capacity

    Dependent variables:
    – The average time it takes to quarry 1 metric ton (drill, excavate, load on truck)
    – The average time it takes to deliver 1 metric ton of mountain to 10 miles away

    Constant:
    – Volume of an average-sized mountain. (can be estimated by finding the volume of a cone with a radius of 1000 metres and a height of 1000 metres).

    The formula:
    Time it takes to quarry and transport 1 metric ton of mountain multiplied by the number of metric tons in the said mountain.

  • Arthur Nov 11, 2010, 1:22 pm

    I would take a pyramid 4km long, 3km wide, 2 km high to model my mountain. That’s 12km^3. Let’s assume we have 100 trucks to move rocks from the mountain. They can each move 3m^3 at once, and take one hour to go 10 miles away, dump the rock and come back. So that’s 300m^3 an hour. Assuming they work 8hrs/day, 300days/yr, that’s 720000m^3/yr, or 0.72 x 10^(-3) km^3.
    12 / 0.72 x 10^(-3) = 1/6 x 10^6, so about 150 000 years. Better get started right now…

  • Arthur Nov 11, 2010, 1:25 pm

    oops volume of a pyramid is a third of that, so divide this by 3: 50 000 years. Still quite long…

  • Aaron Nov 11, 2010, 1:33 pm

    I would subdivide the problem into 4 parts and make some assumption on the number of trucks available:

    1. Calculate the size of the moutain (m^3)
    2. Calculate the time needed to demolish it.
    3. Calculate the average time needed by a truck to cover 10 miles
    4. Calculate the time needed to dump it.

    1. Assuming the mountain as a cone 1500 m high with a base radius of about 4 km, volume is: PI*R^2*h/3 = 3.1415*4000^2*1500/3= about 25*10^9 m3

    2. Assume that a excavator can demolish continually and that we have enough excavators to fill without delay the trucks (say 100 excavator for 400 trucks). The load process would take about 10 minutes/truck.

    3. A truck can carry about 25 m^3 and we have enough trucks in order to have always a empty truck in the queue. So we need 1 billion trip.
    A truck moving at 40 mph would take 15 minutes in order to cover 10 miles.

    4. Another 10 minutes (I’m assuming that the time will not increase as the mountain height increases).

    Actually we have to count only the load process and the average time to move the ground of 10 miles because the dump process is negligible as we have enough trucks.

    Every excavator can load 6 camion/hour or 250 m^3 hour. 100 excavator would cover the 25*10^9 m3 in 10^6 hours or about 100 years.

  • SM Nov 11, 2010, 2:21 pm

    Average mountain….
    when does a hill become a mountain? say 1,000m. Tallest mountain 10,000m. Average mountain guessing 4,000m. Say with that height, base is 8,000m diameter which means volume of roughly cone shaped mountain is somewhere near 3m m3.
    As I am playing God let’s assume we have 100 trucks capable of carrying 10m3 each in one journey, and that we want to move the mountain 10km.
    Travelling at 30km per hour on the way from the mountain (loaded), and 60km per hour on the way back (unloaded), each journey would take one truck 30mins, plus loading time of 30 mins and unloading time of 30 mins (to allow for volume of trucks onsite etc). Also assuming excavators are working and do not create any delay to loading, and time to load etc does not change with size of mountain increasing or decreasing.
    Total journey time of 1.5 hours per truck, total of 300,000 journeys is a total of 450,000 hours.

  • Xiaofeng Dai Nov 11, 2010, 3:11 pm

    I would like to first segment this question into the several parts, then give an assumption for each segmented problem and give the corresponding estimation, and finally summarize and give a conclusion.

    I will segment the problem into the following parts. First is the about the mountain itself, i.e., how big is the average size mountain, what does the mountain is made of (stone, sand or soil). Second is the transportation, i.e., what transportation method is used and its speed, how much resources is allocated on this project (including people and money), what kinds of obstacles are there on the way during movement (such as weather, houses and land). The third is about the goal, i.e., what is the precise number of miles, whether the goal is to recover the original look of the mountain after movement or just to load the materials to the destination.

  • Anna G. Nov 11, 2010, 4:31 pm

    To estimate the time we have to estimate possible speed we can move the mountain by different ways. (as we know t = s/v and s = 10 miles)
    We can move it using different sourses of energy and by different ways: by air using for example helicopter ; or we can pull it with trains; or push it by energy of people; etc depending on what resourses do we have.
    Lets estimate time we need to relocate it by air. We need to know:
    – avarage weight of this avarage mountain
    – maximum weight that 1 helicopter can pick up – >so we can count how many helicopters do we need to pick up the mountain
    – what speed it can fly when it’s fully loaded
    So we can estimate time to move the mountain for 10 miles with minimum speed of helicopters. But if we can get the data shows the depending between helicopter’s loading (weight it has to take) and it’s speed we can count what speed we can have if use more and more helicopters.
    The same way we can estimate speed for people power, or trains, or horses etc.
    All we need is specified data for each way I indicated.

  • Sachin Sharma Nov 11, 2010, 5:05 pm

    Break the problem into the following components:
    1. Time to break the mountain into moveable pieces
    2. Time to collect the pieces
    3. Time to transport
    4. Time to re-assemble

    Assumptions:
    1. Flexibility of resources available to break, collect, move, re-assemble
    2. There may be loss of matter during breakage and collection
    3. How each step is performed will affect the total time (e.g. smaller pieces, easier to collect and move, longer to re-assemble) – estimate may not be optimal
    4. The relocated mountain will differ from original state

    Factors affecting the estimate:
    1. Terrain (additional effort required to clear the area)
    2. Climate – harsher conditions will increase time
    3. Road conditions – bad conditions will affect the transport time

    Calculation:
    Size of mountain: A cone (D=50m, Height = 200m), volume = 1/3*Pi*25*25*200=125k m cube

    Max size possible to move = 2*2*2 = 8 m cube
    No. of pieces = 15k apprx.

    1. Time to break = avg time to break 1 piece (using explosives)* no. of pieces
    = 10 s*15K = 40 hrs

    2. Time to collect = avg time to collect one piece * no. of pieces* 5%loss
    = 20 s * 15K*95% = 75 hrs

    3. Time to transport = avg speed * distance
    = 10 miles/ hr * 10 miles = 1 hr

    4. Time to re-assemble = avg time to re-assemble * no. of pieces
    = 30s*14K = 110 hrs

    TOTAL = 40+75+1+110 = 226 hrs

  • Aleksey Leshchankin Nov 11, 2010, 5:08 pm

    For estimations we need to think about: motivation, method, means.
    Motivation – very strong : )
    Method: 1) to drag entirely; 2) to drag in parts.
    Means: depends on method

    1) To drug entirely.
    We need:
    – Special prime movers
    – Special towlines
    – Workers.
    And…. First and second options Are not invented yet : ). We draw X on this method.

    2) To drag in parts.
    We should
     To saw on blocks;
     To move;
     To stick together
    We need:
    – Rock-cutting machine
    – Trucks
    – Workers.
    Let’s consider that the place under new mountain is already cleared away, the mountain flora and fauna are transferred in parallel

    Let’s go on this way.

    Let’s assume that the volume of mountain is 600 million square meters (6000-high, 300square km – the basis, pyramid form). Rock-cutting machine makes block of 1 cubic meter/hour (including maintenance, wetting by water, etc.). Simultaneously 1000 machines a used for 24 hours/day without interruption. So we have 1x100x24 = 2400 cubic meters of volume per day.

    To transport this volume we use 100 trucks, each can transport 2400 blocks a day. And every truck drives with the speed of 20 miles per hour. Then to assemble 1 cubic meter of new mountain we need a 1 hour and 1 brigade of workers. Let’s assume that there are 3000 brigades works for 8 hours. So we have 24000 cubic meters in one working day.

    Thereby, we need 600mln cubic meters / 24000 cubic meters per day = 25 000 days + 1 (Shipment of a stone to the first day). It’s about 68 years.

  • Chandra Arya Nov 11, 2010, 6:52 pm

    Moving a mountain is a monumental task. So let’s break it into manageable pieces that is cut it into small standardized chunks. Once the pieces are made, they can be transported to the destination and reassembled to complete the mountain again.

    It takes 1000 days or close to 2 years and 9 months with the following assumptions.

    Assumptions:
    1. Mountain is pyramid in shape with 200 m (length), 150 m (width) and 100 m (height)
    2. Modern tools are available to cut it into pieces
    3. Transportation is available to move the pieces
    4. Optimal labor and equipment available to reassemble the pieces.
    5. Only road transportation is considered

    High level Activities:
    1. Cut mountain into manageable pieces
    a. Optimize the size of each piece by the size (width and length) of truck.
    2. Cut into standard pieces based on the volume of the pyramid
    3. Mark the necessary pieces with some identifier
    4. Load into truck
    5. Transfer the pieces to destination
    6. Unload the individual pieces
    7. Reassemble the pieces

    Math Calculations:

    Pieces:
    Volume of pyramid = 1/3 * area of base * height
    = 1/3 * 30,000 * 100 = 1000,000 cubic meters
    Let us say we cut the pieces into size of rectangular solids of dimensions 10 x 10 x 10
    = volume of each piece is 1000 cubic meters
    Total pieces = 1000,000/1000 = 1000 pieces

    Time Estimations:

    Cutting: assuming 5 hours per piece => 5000 hours for 1000 pieces
    Marking pieces: assuming 12 mins per piece => 200 hours for 1000 pieces
    Loading: assuming 3 hours per piece => 3000 hours for 1000 pieces
    Transfer: 10 hrs per truck and two pieces per truck => 500 truck loads, which give us 5000 hours to transfer all the pieces (Assuming single truck is used)
    Unload: assuming 3 hours per piece => 3000 hours for 1000 pieces

    Assemble the pieces:

    We take a layered approach and divide the pyramid into 10 layers. Each layer takes 1 additional hour to assemble the cut piece of mountain in its place. So if it takes 1 hour to place a piece in layer one it would take 2 hours to place a piece in layer two.

    Each layer has 100 pieces
    Total time is: 100 (1+2+3+….+10) = 5500 hours
    We have all the pieces now. Let us sum up the hours we spent on all the activities above.
    Cutting: 5000
    Marking pieces: 200
    Loading: 3000
    Transfer: 5000
    Unload: 3000
    Reassemble: 5500
    —————————
    Total hours: 21700

    If we add inefficiencies of: 10% due to lost productivity etc, we get
    21700 * 1.10 = 23800 or 24000 hours approximately.

    Dividing 24000 hours into days we get 1000 days or close to 2 years and 9 months.

    Recommendation:

    It takes 2 years and 9 months to move a mountain between point A to point B that are a mile apart.

  • victor Nov 12, 2010, 4:30 am

    Hi Everyone… here are a few more tips on this case, 1) assume you move the mountain by TRUCK, 2) the correct answer should be a specific number (as in hours, or days)… as opposed to an idea.

    You are not permitted to ask questions. If you do have a legitimate question, the interviewer will ask you to make an assumption or estimate the answer to your question.

  • FA Nov 12, 2010, 3:27 pm

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

    When I first read the question I had a strategy for finding an estimate and I began working it out when I found out that this would not be the case. I realized that are a lot of questions to answer before trying to solve the case and these are:

    Regarding the “mountain”:

    1. What does “an average size mountain” mean?
    2. What is the nature of the mountain? Is it soil or rocks, or just a plastic mountain, a paper or cartoon mountain? Is it a one similar to those we see in real life as part of nature (the ones we target when planning on a mountain-climbing trip) or are they just “hand-made” demonstrations?
    3. Can this mountain be broken into smaller parts or should it be held as one whole part which actually dictates a certain nature? If we can break it, is there a limit on the number of these pieces or not?

    Regarding the “truck”:

    1. What does “an average size truck” mean?
    2. What are the characteristics of that truck? Is it a six wheel or a road train (does it have on trailer or more than one)? Could it be a fork for example?
    3. Where are we moving the mountain to? How is the relief of that place and to that place? Are the roads easy and straight or are they bumpy roads with a lot of ups and downs? Are all the toads asphalt roads (all the way along the 10 miles) or are there both?
    This is important because geography and characteristics of that place affect the time it takes us to cover this distance.
    4. What is the required speed that the truck needs to abide by? Does it vary or is it the same all along the way to the end of the 10 miles?

    Returning back to the solution, I made a lot of assumptions and what I found out is the following:

    • First, I prefer using Km; so, since 1 mile is about 1.61 Km, 10 miles are equal to 16.1 Km about 16 Km.

    • Second, what is the size of an “average size mountain”?

    Let us assume that a mountain has a pyramid shape and it is made up of soil (one similar to that found in nature)
    The volume of a pyramid is therefore {area of the base*height *(1/3)}.
    Let us also assume that a mountain has a rectangular base. Hence, Vp= L*W*H *(1/3).
    To solve for that let us assume L= 3Km, W=1Km, H=5Km
    Vp= 3*1*5*(1/3) = 5 km3

    • Third, we need to know what an “average size truck” means.
    If assuming that truck is a six wheel truck, then the back holding of the truck will have a rectangular shape (more of a rectangular prism with 5 instead of 6 rectangles i.e. it is open from the top). So to find its area let us assume again that on average L= 6m, W=3m, H=2m so its volume = 36m3

    • So, 5 km3 / 36 m3 = 5,000,000,000 m3 / 36 m3 is about 138,888,889 trucks needed to completely move the mountain.
    Assuming that each truck is required to move at 60km/hr then we will need:

    V=d/t
    t= d/V
    = 16 km / 60km/hr
    = 0.3 hr per truck

    Thus for 138,888,889 trucks we will need 37,037,037 hrs which is about 1,543,210 days which is about 4,228 years to move the mountain 10 miles.

  • KSEguy Nov 12, 2010, 5:55 pm

    Well, this is tricky one, main hit is that question was given by guy who hold PhD in Astrophysics. Thus, if we put the origin in to the Sun. As far as I remember the orbital speed of the Earth averages about 30 km/s. thus to cover 10 miles (which is 16 km) we need about 0.5 second

  • ankur kumar Nov 14, 2010, 6:18 am

    The time depends on many factors as follows:

    1. The average speed of truck. Higher the speed, lesser would be the time required to cover 10 miles.
    2. The number of rounds required. Its very unlikely than an average size mountain can be shifted in one attempt. It may be required to break it into pieces and reassemble after relocation. Larger the volume ratio of mountain to truck, more trips will be required and hence more time.
    3. The time required to dismantle and reassemble the mountain also contributes to the total time in question.

  • Cynthia Nov 14, 2010, 1:08 pm

    1. Estimate the total volume of the mountain. Suppose its radius is 3 km and its height is 1km. The total volume will be 1/3*pai3^2*1,which approximately equals 20 km^3
    2. Estimate the total amount of substance that an average truck can carry for one trip. suppose the size of truck is 3*2*3 (m), so the total amount that the truck can carry for one trip approximately equals 20 m^3.
    3. Estimate how many trips the truck will have to travel in order to dig out the entire mountain. 20 km^3/20m^3 equals 1000,000,000
    4. Estimate how long does it take for a round trip. Suppose it takes the truck half hour to load and it takes the truck 10 minutes to drive 10 miles and it takes another half hour for the truck to unload and then 10 minutes to drive back. So it takes approximated 1 hr 20 minutes for the truck to do a round trip
    5. Calculate the time. 1,000,000,000 times (1+1/3) approximately equals 1,300,000,000, which approximately equals 50,000,000 days, which approximately equals 150,000 year.

  • Hassan Murad Nov 14, 2010, 6:28 pm

    We can divide this task into two parts.
    a) number of rounds
    b) time required in each round

    Total time required = a x b

    To answer this I need to know what is the average size of mountain and what is the carrying capacity of an average size truck. This would help us know how many rounds of a truck are required. (We will divide total size by capacity of the truck)

    Then we want to know how long each round takes and to calculate that we have to know average speed of a truck which with distance can give us average time required in one trip.

  • BP Nov 16, 2010, 9:29 am

    it depends on how long it would take to dismantle the mountain, piece by piece. you would need a team of workers with various tools (picks, axes, etc) to essentially “break up” the mountain into smaller rocks. then, the rocks can be placed into the truck and the truck will transport them the 10 miles down the road. so, the rate limiting step isn’t the truck transport, it’s getting the mountain broken down into smaller, manageable pieces. this is actually a good allegory for knowing how to tackle seemingly insurmountable problems: just break the problem/task down into bite-size pieces, and soon, it will be accomplished/overcome.

  • Rukshan Meegahage Nov 21, 2010, 7:13 am

    Firstly, how big is the average size mountain:
    The tallest being apprx.9000 ft and the shortest being apprx. 1000ft. Thus, average size is about 5000ft.

    What is the weight of an average mountain (This is where I’m utterly confused but I’ll give it a go):
    I will take a common measurement; namely the BMI for humans, is weight (Kg)/height (m) squared. The BMI for a very an average person is about 25, if you are to make a pyramid from the person, you’ll need 5 of him (I sound ridiculous!) i.e middle maintaining the height, and 4 projecting from the central midpoint. This could equate to a BMI (remember the height is maintained) of about 125.

    Assuming that the [weight:height squared] of a mountain is about 125, the weight of a 5000 foot mountain would be 3125million Kg which is about 3,125,000 tonnes.

    Weight of truck: A medium sized truck would weigh apprx. 5 tons and the weight of the driver is negligible.

    Average speed of a truck: On a straight stretch of road with no traffic apprx. 50 miles/hr. And the speed decreases as the weight increases by a constant (5 tons – 50miles/hr and 10tons 25miles/hr).
    Therefore to carry 3,130,000 tons the average speed would be 5/3,125,000 = 1/625,000 miles per hour.

    To travel 10 miles = 10 x 625,000 = 6,250,000 hours (Note I haven’t taken into account the time it will take to start the mountain moving, to overcome inertia).

    Therefore:

  • Bobbber Nov 22, 2010, 2:44 am

    If it is an average mountain of shit, it would take about 10 minutes of driving time.

  • Cal Nov 22, 2010, 11:07 am

    Based on the wording of the question I’ve assumed that there’s one truck.

    First I split the problem into # loads and time per load.

    # trips:
    *I assumed that the average mountain can be approximated as a cone of radius 2000m (went metric as I’m from Australia) and an average height of 2000m.
    This gives a volume of (1/3)x3.14×2000^3, which is approximately 8 billion cubic metres.

    *I made assumptions about the average truck size and decided to go with 4m long, 2m high and 1.5m wide, giving a volume of 12 cubic metres (marked this assumptions to test later).

    Dividing the volume of the mountain by the truck size gives the required number of trips. 8billion/12 is approximately 700 million trips.

    Time per load:
    *I assumed that the average speed limit in the area would be 60m/h (rural)
    Therefore travel time would be 20miles (return)/60 = 0.33hrs of driving.

    *I assumed that there would be a crew at each site to unload and pack the matter, and that they had the appropriate machinery.
    Therefore I assumed roughly 20 mins delay at each site. I approximated this to 0.66hrs, making total time per load 1 hr.

    This gives a total time of 700million hours.
    *I assumed 24/7 operating hours.

    This figure should be divided by # hours in a day and # days in a week –
    I divided by 7, then by 25 (up from 24 for simplicity) to give 4 million weeks.

    Dividing by 50 (down from52 for simplicity) yields 80,000 years.

    Sense check – I would check the following assumptions:
    *average mountain size
    *average truck capacity
    *equipment and crew available at the sites (would affect answer dramatically – and if they can only afford one truck, they likely won’t be staffing crew and lots of equipment at each site for 80,000 years)
    *hours of operation

    Overall, the number is too high for any feasible plan; but then again, nobody would do this with one truck.

  • Maryam Nov 23, 2010, 10:42 pm

    Why would you want to do that?

  • Cal Nov 27, 2010, 12:47 am

    How did we do? :p

  • Max Nov 27, 2010, 5:18 pm

    Note: being Swedish I did my math in metric while keeping the distance in miles as stated.

    I assume the mountain is to be moved piece by piece.

    Assume data give represents bottlenecks – everything else runs as smoothly as possible i.e assemble and disassemble times are negligible.
    —————————————————————————————–
    You can either make the calculations based on volume or on weight being the limiting factor for the truck.
    • {kg/(kg/day)}
    • {m^3/(m^3/day)}
    Density of rock is assumed to about 2,5 grams/cc = 2,5kg/l = 2,5 tonne/m^3
    —————————————————————————————-
    Determine What is an average truck:
    • Assume Volume/weight capacity
    o Volume = 20m*2,5m*2m=100m^3 -> 250 tons (seems too high in my opinion i.e weight is the bottle neck chosen here)
    o Assumed Weight capacity = 20 tonnes (corrected to 25 to make the math easier)
    o Motivation: I don’t know much about dump trucks or trucks in general, but I’ve seen big trucks transporting 10 or more cars and a car weighs ~1 ton. Assuming a dump truck is more specialized for its purpose I assume 20 tonnes
    • Weight moved/day = (weight capacity) X (runs/day)
    —————————————————————————————-
    Runs/day
    o Runs/day = (hours/day) / ({[(distance) X 2] / (speed)} + [(loading t) X 2])
    Runs/day = 10/({[(10) X 2] / (40)} + [1 X 2] = 10/({20 / 40} + 2) = 10/2,5 = 4
    – Runs/day = 4
    • Motivation and assumptions
    o speed=distance/time -> t=d/s
    o Runs= hours/(miles/[miles/hour])
    o Assume average Speed = 40 mph
    o Assume: Loading time = unloading time = 1 hour
    o Assume work day = 10 h (we want to move it fast 🙂
    —————————————————————————————-
    Determine What is an average mountain:
    • Assume Size (Volume)
    o Shape = cone
    EQ cone: (1/3)*pi*h*r^2 Define mountain

    • Define mountain
    o For me it is something you can climb or ski on, something bigger than a hill, in terms of abstract size it is something that takes you some effort to cross or climb – say more than an hour or so. By my definition it needs to be smaller than Mt. Everest (h = 9km) and bigger than a hill (highes mountain in Denmark is called a hill; h=0,3km).
    – Assume h = 1 km
    – Assume r = 1 km
    o Volume = (1/3)*pi*1*1^2 = pi/3 km^3 ~ 1 km^3
    – 1km^3 = 10^9 m^3
    – Weight = 2,5 ton * 10^9 m^3 = 2.500.000.000 tonnes
    —————————————————————————————-
    Estimated time to relocate:
    • (Mountain weight) / [(runs/day) X (weight moved/day)] {kg/(kg/day)}
    • 2.500.000.000/(4*25) = 25.000.000 days -> 250000/365 = ~700 years

    Conclusion: Except for taking a very long time and not being a remotely plausible project in real life, we can see that (Mountain weight) >> (other factors), leading to them being of very little importance. So the assumptions regarding the volume of the mountain and the density of the product moved are the central ones.

  • Max Nov 27, 2010, 5:28 pm

    Edit:
    Last EQ is missing /100:
    • 2.500.000.000/(4*25) = 25.000.000 days -> 250000/365 = ~700 years

    should be:
    • 2.500.000.000/(4*25) = 25.000.000/100 days -> 250000/365 = ~700 years

  • Jess Nov 29, 2010, 3:22 pm

    The following bits of information are the key ones I need to work out this answer:

    – An estimation of the volume of the mountain; how many metres cubed of earth there is to shift
    – An estimation of the volume of the ‘average truck’
    – Use these to estimate how many times the truck must be filled to move all the earth
    – Using this info about how many trips, work out how long it would take the truck to do the number of 20 mile return trips needed to shift the mountain.

    This misses out any other time that would be needed for actually filling the truck, time needed for men to dig and move the earth: as the question seems to focus on the truck I would hope this is an ok assumption and check with the interviewer.

    So, to start with the mountain volume.
    I lived in Geneva last year, I was told 3000m is considered a high mountain, so I’d go for an ‘average’ height of 2000m.
    For the base, I think 10km seems like a reasonable diameter.
    As this is an estimation, I want to avoid pi, so instead of calculating a cone, I will calculate a pyramid. For the square base I’ll go for 10km x 10km, which gives me a base of 100km squared (hopefully not too far off what a diameter of 10km would give me). Times this by the height, and I get 200,000m cubed.

    Now for the truck: estimating based on open trucks I have seen, I’d guess the depth of the loading area to be around 2m, the width around 3m and the length about 5m, giving a total volume of 30m cubed.

    Dividing 200,000 by 30 gives 6666.66(recurring), which I will round up to 7000: this is the number of times the truck would need to be filled to transport all the earth.

    So there need to be 7000 return trips. Each trip is 20 miles (10 there and back). So in total the truck will need to travel 7000 x 20 miles, 14000 miles.

    A truck probably travels at about 50mph: they are usually slower than cars max speed, plus it is loaded, and may be travelling on rural roads.

    A truck travelling 50mph will take 14000/50 = 280 hours.

    So, in terms of truck travelling time, it would take 280 hours or about 12 days to shift the mountain.

    Sanity check: this seems way too quick, and does not take into account the actual time of digging the earth, and loading the truck, nor reconstructing the mountain. I have oversimplified…

  • Jess Nov 29, 2010, 3:27 pm

    Oops, forgot to turn the cuboid into a pyramid by /3…

    • Jess Nov 29, 2010, 7:47 pm

      @Jess:

      Ah! I realise why this is so very small: mistake 1km cubed is not 1000m cubed, but 1000 x 1000 x 1000 = 1 bn m cubed! Would re-calculate to fit this

  • Jerry Dec 5, 2010, 6:02 pm

    I pretty much got a process similar to what you guys had. But even before tackling the soil part, I took trees and cottages into consideration and assumed that it would take equal amount of hours of work to replicate the original condition in the new site as moving the soil. Now soil part, like some mentioned:

    1)break
    2)load
    3)carry
    4)unload

    I assumed an average size mountain to be 1km high, 2.5 km in base and cone shaped which gives 1.3^2*pi (substitute with 3)*1(height)/3 = 1.7km^3 or 1.7billion m^3

    an average truck may hold 4m(length)*1m(depth)*1(width) = 4m^3 and assume there are 100 trucks available for this project

    4million travels needed altogether

    now assume no traffic jam, no accidents, and trucks travel at 50km/hour, a mile is about 2km so it should take 0.3 appx one way and 1hr for both way.

    and because 4million travels are needed, it takes them 4 million hours just by transportation

    I’ve seen some construction site and it took them approximately 1 hr to load the truck full of soils and again assume that it takes equal amount of time to unload = 2hrs. additional labour such as breaking and rebuilding would take another 2hrs.

    4*400million truck loading and unloading = 1600 million hours

    added altogether for soil = 4million + 1600 million = 1604 million appx 1600 million (gosh, transportation hours are non-material after all)

    another 1600 million for trees and all other misc activities = 3.2 billion hours to move and replicate the mountain from one location to another.

  • M.Fisher Dec 6, 2010, 2:51 pm

    If all that I am given to move said average size mountain is an average sized truck… the problem is not solvable. The unknown or unallocated variables are significant:
    1. What equipment do I have to load this truck? (shovel, backhoe, excavator, etc…)
    2. What type of workforce to I have? (just myself, a team, an army)
    3. What distance am I to move said mountain? (10 feet, 10 miles, cross country)
    4. What class of truck are we talking about? (dual axle pick up truck, 2 ton or greater sized dump truck, heavy construction truck)
    5. What is the scope of this mountain and its directive to move it? Are we taking it down to sea level from its current delineation of elevation… or simply until we might classify it as part of another ‘set’, such as a foothill?

    Without additional information, any answer is arbitrary in nature and lacking necessary premises… therefore I submit my answer in similar fashion: 4

  • Min Dec 9, 2010, 11:26 pm

    1.clarify what is the end-product the interviewer wants at the new location – still a mtn (hope not, may take very long) or pile of rocks
    2. assume rocks are desired and only one truck; assume ave. mtn. weigh x tons and measure y cubic feet; assume an ave. truck capacity is m tons and n cubic feet; assume no need to build a road between the two sites; assume no need to dig a hole to dump the rocks
    3. time to blow up the mtn, assume having super-dynamite, D minutes;
    4. time to load&unload
    A minutes per load, B mimutes per unload
    N number of rounds = max(x/m, y/n)
    total= (A+B)*N
    5. travel time
    assume there is a road, but narrow&winding, C minutes one way
    total =2*C*N
    6. assume 12-hr per day operation(not safe to drive in the winding road at night), total days needed to move the mtn= (D+(A+B)*N+2*C*N)/60/12.

  • SA Dec 11, 2010, 9:44 pm

    My main issue here is the definition of average size..if the largest truck can take the largest mountian in one trip and the smalles one can take the smallest mountain in one trip then the average size truck will take one trip to locate the average sized mountain. So if the spead of the truck is 60 m/h then it will take about 10 minutes to relocate the mountian plus the load and unload time.

  • BH Jan 1, 2011, 11:01 pm

    Average Size mountain:
    1km diameter, round base, height 0.5km.
    So ballpark estimation of Volume is 1/3*pai*0.5*0.5*0.5=0.1km3=1E8m3.

    Once truck is loaded (5*2*2=20m3 much of material), it will take 1 hr to deliver and come back and refill (assuming 20 miles/hr).

    So it will minimally consume 1E8/20=5E6 hrs=500 yrs

    The rest will be compounding factors (in comparison to the 1 hr per round) that will make the time taken longer in a linear fashion for simplicity.

    1. loading and unloading. 2X

    2. truck driver shifts. Assuming 3 shifts of 8 hours and unlimited supply of workers. 1X

    3. gas refill, mechanical breakdown, tire change (only 1 truck available), 2X.

    4. Increment weather. 2X.

    5. for the hundreds of years merely to transport, the time taken to break down the mountain (say by explosives in rounds of 5-10 seconds) can be neglected.

    Considering the above, the estimated time taken will be :
    500*2*2*1*2=4000 yrs.

  • Mar Jan 8, 2011, 9:52 am

    Considering an average size truck needs 2 hours to drive 100 mile, in a flat road with very few trafic (no mountains in between, or curves). To transport the mountain, I would add an extra hour (i.e. it would take the truck 3 hours) as the truck driver also needs to pay attenion not to hurt anybody, or ruin anything. To accomplish this successfully the mountain would also have to have 4 wheals to make the transporation easier.

  • ariel Jan 22, 2011, 1:48 am

    Transit time for one load:
    10 mi @ 40 mi/hr = 15 min one way, 30 min round trip
    + 20min loading time (they’re professionals with good equipment)
    = 50 min, round to 60 min = 1 hr per trip

    Volume of mountain
    Assume cone shape
    Length – 6km
    Width – 6km
    Height – 5km higher than surrounding area (bit tall for average, but it’s a mountain, not a hill, and i skimped on the length and width)

    Volume of a cone: 1/2 pixr^2 x h (off the top of my head)
    = 1/2 * pi*3km^2 * 5km (round 3^2 to 10)
    = 1/2 * 3* 10km * 5km = 75000 cubic meters

    Pickup truck load volume
    Length – 3m
    width – 2m
    height – 1m

    volume of load: 6 cubic meters – round to 5 cubic meters

    Number of truck loads to move the mountain:
    75000 / 5 = about 15000 loads

    15000 loads @ 1 hr/ trip = 15000 hrs

    2000 hrs in regular work year = 7.5 years (3.75 yrs if they work a double shift)

  • ariel Jan 22, 2011, 2:02 am

    Oops – lost track of my zeroes. Amend the above to be:

    Volume of the mountains:
    1/2 pi * 10,000m * 5000m = 75,000,000 cubic meters

    Volume of truck load: 5 cubic meters

    Number of truck loads: 15million
    Time required: 15million hours
    or 7500 years (regular work week: 40hrs/wk)
    or 3750 years (double shift)

  • Sunanda Jan 26, 2011, 9:12 pm

    1. Well first I would first like to explore what the volume of an average mountain is:
    o Mt Everest is the tallest mountain and is approx 30,000 ft and I read somewhere that a mountain needs to be roughly over 10,000 ft to qualify as a mountain, I am going to assume that the average mountain is roughly 20,000 ft high and has a slope of 45 degrees. So the vol= 1/3 * radius^2*ht= 1/3*(15000)^2*30000= 2,250,000,000,000 cubic feet (because ht= 2*radius at 45 degrees)

    2. And then understand the level of effort required to move the mountain:
    o For example are we digging the mountain and then moving it using the truck or is it loose dirt that can be easily divided into equal loads?
    o Since I heard the word “move”, I am going to assume that the mountain has already been prepared for relocation, i.e its dug out and pulverized into equal loads

    3. Thereafter, I would calculate the volume of an average truck
    o When I moved last, we took an average truck and it could take 2 6 ft sofas lengthwise with room to spare so I am going to guess its 20 foot long. I also noticed that 6 ft man could stand straight in it and that it could fit a sofa and half in width. So the volume of the truck = length*breadth*height =1800 cubic feet

    So an average truck will take 2,250,000,000,000/1800 = 1,250,000,000 trips

    4. When we consider the time, we should think abt the following
    o Time to load the mountain into the truck part by part

    Assuming we use a crane with dimensions 5 ft* 3ft* 4ft = 60 cubic ft. It will take 30 moves to fill the truck per trip. Assuming that it takes 10 mins to per move, it will take 300 mins to load a truck per trip and 300* 1,250,000,000 = 37,500,000,000 mins for all the trips

    o Time to drive 10 mi. Since its going to be rugged train (after all we are moving thru mountainous area!) I am going to assume a speed of 20 mph
    so 10 mi/20 mph = .5 hrs = 30 mins per trip= 625,000,000 mins

    o Time to unload the mountain should be the same as loading = 37,500,000,000 mins for all the trips

    o Any rest stops that the driver will take and the length of a work day. I am going to only consider work hours. But if assume manhours we need to think abt the no of drivers, length of shift and break between shifts. Keeping it simple here so assuming a super worker with no break who works tirelessly

    5. So total time = 37,500,000,000 + 625,000,000 + 37,500,000,000 = 75,625,000,000 mins = 143,883.18 days

    • Sunanda Jan 26, 2011, 9:22 pm

      Sorry: 143,883.18 yrs

  • vishal Feb 6, 2011, 12:53 pm

    Before answering this question, I would ask following questions to interviewer:
    1. What according to them is an average size truck and mountain?
    2. Is the mountain uprooted from ground and already loaded on that truck?
    Now, answer to above questions would prove that this phenomenon is “actually” feasible according to interviewers. So, I will do the calculations after I get answer to further 3 questions:
    1. What is the maximum or minimum speed of the truck?
    2. How is the infrastructure quality all throughout the way to destination?
    3. How is the weather and traffic in the way?
    I assume that I will do simple math after I get the above questions answered.

  • Daryl Feb 15, 2011, 9:47 am

    Working in square feet, let’s imagine an “average” sized mountain is 2000 ft tall, with an equal radius (thus, over a mile around). If it’s a cone, the area will be 1/3 x base x height = 1/3 * [ ( 3.14 x 2000^2 ) x 2000 ] , which is about 8b square feet.

    A pickup may be 6 feet long, about 3 feet wide, and a mound of rock could get 4 feet tall, so the area would be about 100 square feet. Therefore, we’ll need 80m trips to move the mountain. If we load the truck in an hour, unload it faster in just 30 minutes, and average around 45 miles/hr there and back (15 min there, 15 min back), the whole trip will be 2 hours. In total, we’ll need 160m hours, or 18,000 years to complete the task.

  • marie Apr 18, 2011, 3:18 am

    Assume average size mountain would be 2000 mtrs high, with a base of 1000 m. So the total volume would be 2*10^3*pi*500^2=500*10^6*pi=1.5*10^9 m3
    Assume average trucksize to be 8*2*3=48m3; make this 50m3.
    so the truck will need to drive 1.5*10^9/50= 30*10^6 times back and forth to transport the complete mountain.
    Traveling 10*2=20 miles each time the truck drives makes a total of 30*10^6*20=600*10^6 miles. Assume an average truck drives 60 miles an hour –> 600*10^6/60=10*10^6 hrs is the time it will take the truck to transport the complete mountain.
    Take an additional 20 minutes per transport to load and unload the truck and you get to 20 million hours for the truck to transport the mountain

  • Can Apr 21, 2011, 1:51 am

    Avg height 5km, base 5 km. a mountain is acone. with density twice of water. so apprx. it weights 40 bil. tones. average truck carries 10tonnes, travels the 10 mile at avg speed of 50km/h when loaded and at 80km/h when empty, so one trip of load takes appr. 20min. therefore80bil roundtrips are needed which makes appr. 14,000 years! : )

  • BBM May 9, 2011, 2:58 pm

    Assuming only moving time is concerned and the average speed of the truck is about 40miles per hour, then it takes about 30 minutes for one trip.
    Average size of a mountain divided by average size of a truck gives the total number of trips the truck has to make.
    Average size of a truck ~ one-bdr moving truck by 3x2x2.
    Average size of a mountain ~ 2.e9
    Time ~ 30min*1.e8 ~ 5000years

  • Rahul Priyadarshi May 10, 2011, 4:06 am

    height of mountain – 1000m
    radius – 20m
    Volume – pi*radius2*height/3 = 1 million mtere3
    density – 1 ton/metre3
    weight – 1 million tonnes
    capacity of truck- 10tonnes
    speed of truck-40miles an hour
    total time taken-50000hrs

  • BKD May 11, 2011, 4:34 am

    Assumptions about mountain:
    ——————————————-
    Height = 2000m
    Area = 500 km^2
    Shape = conical
    Rock density = 2.5 g/cm^3 = 2.5 * 10^9 tonne/km^3

    Therefore, we can now calculate the total mass of the mountain:
    V = 2*500/3 = 3 km^3 (approx.)
    M = V*density = 3*2.5*10^9 = 7.5*10^9 tonnes

    Assumptions about truck:
    ——————————————-
    Capacity = 25 tonnes
    Av. speed = 40 mi/hr
    Operation = 16 hr/day (2 8-hour shifts)
    Breaks/driver change = 1 hr/day
    So, effective operation = 15 hrs/day

    Now, # return journeys = 7.5*10^9/25 = 3*10^8
    Return time = 20 mi/40 mi/hr = 0.5 hrs
    Loading/unloading time = 0.25 hrs (assumption)

    So, total time/journey = 1 hr
    Total time = 3*10^8 hrs
    Total days = 3*10^8 hrs/15 hrs/day = 2*10^7 days
    Assuming 350 working days/year,
    Total years = 2*10^7/350 = 6*10^4 years

    So, it would take about 60 000 years!

  • Achlesh Sood May 12, 2011, 2:42 am

    Size of mountain: Assume It is in shape of pyramid. with radius of 300 metres. Height 2000 metres, Hence volume = 1/3*3.14*300*300*2000= 180,000,000 m3 (approx). Assume Density = 1m3 = 1 kg.
    So, in nutshell mountain is piece of rock 180,000T.
    We have to break the mountain in small parts to transport it to 10 mile.
    Requires 1h to break it to 1T of rock.
    180,000h to break the mountain in 1T of rock(1)

    Capacity of tuck: 49 Tonne truck (1 of the biggest in the industry) carry 50 T
    Loading 50 T material takes 1 h
    Travelling 10 miles, deloading and coming back also takes 1 h.

    So loading 180,000T takes = 180,000/50 = 3600 h(2)
    Similarly travelling 10 miles, deloading and coming back takes =3600 h(3)

    Adding (1) +(2) + (3) = 187,200 h

    P.S. My first post at this form. I hope i didnt made a mess 😛

  • Nicholas Luby May 16, 2011, 9:04 am

    The tallest mountain on earth is about 29000 feet tall. Small mountains have incline requirements to be considered mountains, taller mountains have a height minimum (I think 9000 feet?). Otherwise its just a big hill. Also, I’m guessing the distribution of mountains is skewed right, with a big fat right tail bringing up the average, so I’ll go against my gut instinct to use a very small height of a couple thousand feet. So I’m going to say height is 10000 feet, the mountain is shaped like a cone and the base diameter is 20000 feet, so radius is 10000 feet.

    1/3*pi*r^2*h = aprrox. 1/3*3*(100 million)*10,000=1 trillion cubic feet.

    The average truck bed is 20 feet long * 10 feet wide * 5 feet deep = 1000 cubic feet

    I assume we have some other large bobcat device to load the truck as soon as it gets back, so loading time is neglible.

    The truck goes 60 mph for 10 miles which takes 10 minutes. Then it has to drive back another 10 minutes, for a 20 minute total round trip transit time.

    It has to make 1 trillion / 1000 = 1 billion trips. 1 billion * 20 = 20 billion minutes.

    20 billion / 60 = 333.333 million hours / 24 = about 13.5 million hours (its 13,888,889 hours) but i’m trying to do this in my head.

    So a long time. Let’s teleport that sucker.

  • Nikhil M Jun 15, 2011, 5:28 am

    Assumptions
    1) Avg size mountain has a height of 100 mts
    2) shape of the mountain is that of a Pyramid
    3) Mountain is made up of soil and rocks of uniform density – say 3 (for simplistic calculations)
    4) Base of the pyramid is a square – 100X100mts^2
    Volume of Pyramid = 1/3* 100* 100*100
    Total mass of pyramid =( 100*100*100/3)*3=100*100*100
    5) Avg truck carrying capacity is say 100
    so number of trips the truck would have to make = 100* 100
    now say in each trip truck would take 4 hrs ( 2 for loading, 1 for unloading, 1 for traveling 10 miles)
    so total time required = 100*100* 4=40000 hrs

  • TA Jun 28, 2011, 8:26 pm

    16-17 d

    Avg Size Mountain x 10 miles/ Avg Size Truck
    Avg Size Truck-Can move 2 couches, 100 lbs each—200 lbs
    Avg Speed of Truck-40 mph
    Pounds/Hr-Truck= 200 lbs*40 mph/10 miles= 800 lbs per hr
    Kg/Hr-Truck= 800/2.2 = 800*5/6 = 133*5= 665 kg / hr ~ 6.5 * 10^2 kg/hr
    Avg Size Mountain-ht 500 m, r 500 m
    Volume = 1/3*r^2*h*pi~r^2*h~125*10^6 m^3
    Assume: Density 1 g/cm^3 10^6 g/m^3 = 1000 kg /m^3
    Stone 2x more dense than water2000 kg/m^3
    Weight Avg Mountain = 125*10^6 m^3 * 2 * 10^3 kg/m^3
    =250*10^9 kg
    Time to Move Mountain with Truck = (2.5 * 10^11)/(6.5*10^2)= 25*10^8/6.5 ~ 4*10^8 hrs = 400 mill hrs
    400*10^6 / 24 = 16.67 d

  • TA Jun 28, 2011, 8:31 pm

    Correction-not 16-7 d. 27000 yrs approximately. Copied incorrect version of my calculation.

    Time to Move Mountain with Truck = (2.5 * 10^11)/(6.5*10^2)= 25*10^8/6.5 ~ 4*10^8 hrs = 400 mill hrs
    400*10^6 / 24 = 16.67 *10^6 d = 1667/365 * 10^4 yrs ~ 2.7 * 10^4 yrs = 27000 yrs

  • RS Jun 29, 2011, 10:58 am

    Assume an average sized truck can max load 1 ton at a time. Average sized mountain has 50 tons of rocks. So that it takes 50 loads for one truck to move the mountain.

    Now let’s assume truck runs 20 miles/hour on average when fully loaded (include acceleration and deceleration time on a short distance trip of 10 miles), and 40 miles/hour when empty. So 10 miles each way, average driving time per load is 45 (30mins+15mins) minutes. Also assume loading and unloading the 1 tons of rocks take half an hour each time, so that’s an hour. In total, one load round-trip takes 1 hour and 45 mins, or 1.75 hours. 50 loads then take 50*1.75=87.5 hours

  • tomhouston Jul 4, 2011, 12:39 pm

    For the purposes of the discussion, I am going to assume the following..
    – Volume (or capacity) of the truck is the limiting factor rather than weight (no super heavy rocks).
    – Loading / unloading is the limiting factor rather than breaking the mountain apart or clearing the mountain of trees, access issues et al
    – truck when loaded or unloaded will travel at 60mph or travel 10 miles in 10 mts
    – Typical mountain is a pyramid… at 500 m radius and 1000 m height. so volume is 1/3 * 3.14 * 500*500* 1000
    – Truck is a 10 m * 10m * 15 m
    – So # of truck loads in the mtn is 1.6 * 10^5
    – Total time to move the mountain = (loading time in mts + 10 mts transportation + unloading time) * # of truck loads
    – assuming 10 mts for loading and unloading.. this turns out to be 20 mts * 1.6 * 10^5 = 3,200,000 mts… or 53,000 hrs or 2,200 days

  • Raj Jul 11, 2011, 7:01 am

    Average mountain is cone shaped.
    Truck (dumpster) has a rectangular box-shape of holding area, that will be used to move the rocks/rubble from the mountain.
    Assumptions:
    Mountain size: height = 150m, radius of base = 100 m. Therefore volume of rubble – 1/3*pi*r^2*h
    Volume = 1554,300 cu metres.

    Truck size = 10 x 6 x 3 = 180 cu metres. In one trip, truck can move 180 cu metres of mountain

    Ignoring the skillset/time required to re-build the mountain, and assuming that for one trip, truck takes 30 mins (area would be hilly) to cover the 10 miles and that 1.5 hours are taken to load and unload the truck, the truck can make 6 trips in a 12 hour day.
    Moving 180 cu metres per trip, truck needs 8635 trips in total, to move all 1,554,300 cu metres of rubble.
    At the rate of 6 trips per day, truck needs 1439 days to make 8635 trips.

  • Sunando Jul 31, 2011, 1:06 am

    Average Weight of the Mountain = 10000 tons
    An average truck can move at one time = 1 ton
    Time taken to load truck = 30 min
    Time taken to unload truck = 30 min
    Time taken to travel a dist of 10 miles = 30 min
    Time taken to move 1 ton of the mountain = 30*2 (travel time) + 30 (load) + 30 (unload) = 2hrs.
    Time to move mountain ~ 20000 hrs

  • wz Aug 2, 2011, 11:34 am

    1. Estimate the volume of average mountain:
    Mountains are usually kinda cone-shape, so let’s assume that we can caculate its volume as a cone. For average mountain, I would assume the height is 400 meters and the diameter is 1000 meters. Then the volume would be 3.14X (1000/2)2(square) X 400 X1/3=~100,000,000 cube meters

    2. Estimate the loading capacity of an average truck:
    The truck trunk is usually a rectangle with a length of ~10 meters and a width of ~6 meters. Assume that the truck can load a height of ~3 meters. Then for 1 load, the volume is 10X6X3=180 cube meters

    3. Calculate the number of loads the truck need to move the mountain:
    100,000,000/180=555556 loads

    4. Estimate the average time per load
    Assume the average speed is 50 miles/hour
    The loading time is 20 minutes
    The unloading time is 10 minutes
    Total time for each load is 60X10/50+20+10=42 minutes

    5. Total time needed:
    42 minutes X 555556=233352 minutes = ~4.43 years

  • AOA Aug 8, 2011, 1:43 pm

    Estimating:
    1) a mountain have 10,000,000 TN
    2) a Truck load can move 10 TN
    3) the time to load, move and unload the truck is about 2 hr

    10,000,000tn / 10tn, truck load = 100,000 truck loads
    100,000 truck loads/ 4 truck loads, day = 250,000 days

    Moving a mountain to an nearby location will take approximately
    250,000 days

  • Pietro Aug 23, 2011, 7:33 am

    If we assume the mountain to be a cone, high 2000 metres, and large 14000, its volume is 28000*pi = about 90000 m^3. If we assume that an average truck can load 3*2*1.5 = 9 m^3, it would need 10000 travels to move the mountain.
    Assuming half an hour to load the car, an hour to go from the top of the old to the top of the new mountain, and half an hour to unload it, each travel is 2 hours long. Therefore the total amount of time is 20000 hours = 3 years, 238 days, 8 hours. That is, almost 4 years.

  • Pietro Aug 23, 2011, 7:55 am

    lol. I would like to point out my errors:
    1) in the initial multiplication I’ve forgot 3 zeros! Then the final result is going to be something close to 3700 years! This type of error is the safest way to fail an interview.
    2) I’ve forgot the way back for the truck, another hour. Then I should multiply 3700 * 1.5 = 5550 years!

  • ChrisF Sep 5, 2011, 5:20 pm

    Assuming:
    – avg mountain height: 4 km
    – avg slope: 45 deg
    – mountain shape can be approximated as a cone
    – body of the truck: 2m (w) x 8m (l) x 1.5m (h)
    – avg speed of the truck when full: 30 mph
    – avg speed of the truck when empty: 50 mph

    Since the avg slope is 45 deg, radius and height are the same.
    The volume of the mountain is radius*height*pi ~= 50 km^3.
    The volume of the body of the truck is ~25m^3.
    Total round-trips: 50km^3/25m^3 = 2000

    If we assume to have machinery so that loading/unloading times are negligible, we end up having: 2000*(10mi/30mph + 10/50mph) ~= 10^3 hrs ~= 40 days (or 125 days working 8 hours per day).

  • Kakarrot Nov 1, 2011, 1:33 pm

    First of all we need to Know what is the mass of an average mountain.

    The average density of a rock is Higher than water’s which is d=1
    So we can suppose it is roughly d=2tn/m^3

    We can suppose that a mountain’s figure is a cone. So V= π* r^2 * h

    An average mountain’s height is about 1000m, and radius r= 5000m .

    So the volume of our average mountain is V= 78500000000m^3

    m=d*V=157000000000tn

    the average truck carries 10tn.

    Lets suppose that the average time to load, transfer and unload is 5hours

    the truck has to repeat the action 15700000000 times
    So it needs 31400000000 hours= 3584474 years

  • Artem Nov 27, 2011, 11:09 am

    Lets assume, that we can explode the mountain, thus time for breaking the mountain into carryable pieces is zero. Then lets estimate:

    volume of average mountain= 2.5km*1/3*((1.7*2.5*km)^2)*Pi=
    =50 cubic km. (approximately)=50 000 000 000 cubic m

    volume of averagge truck is 2m*2m*6m=24 cubic m
    time for the 10miles trip = half hour (assume there are no good roads near the mountain)

    time for filling the truck (by another machine) = (24/1/7)*15seconds= 45 minuts

    + overhead near 10 minuts every trip

    TIME= 1.4 hours * (50 000 000 000 /24)= (200 000 000 000 ) /96 *1.4=
    = 280 000 000 0 hours approximatley = more then 100 million days

  • v Dec 14, 2011, 3:43 pm

    Volume of Mountain: 1,000,000 cubic meters
    Volume of truck: 20 cubic meters
    # of one way trips: 50,000
    Trip time:10mp for 10 km 1 hour
    Total one way trip time: 50,000 hours (A)
    Total Return trip time: @20mph: 25,000 hrs (B)
    Total Trip Time:75000 Hours
    Dwell time per trip on each end: 5 minutes (Load Unload etc)
    Total Dwell: 5*2*50000= 500,000 minutes= 8500 hours say 9000 hours
    Mountain Breaking and assembling time: 0 minutes assume that work is done while truck is in transit
    total time: 75000+9000 hours=84000 hours= 3500 days

  • Viddu Dec 31, 2011, 1:54 am

    300 days

  • Gaurav Jan 7, 2012, 1:11 pm

    I assume that an average mountain is a 1 km long, 200 m and 50 m wide. It tapers uniformly as we move up its height. So that gives me a volume = .5*50*200*1000 cubic meters = 5000000 cubic meters.
    Now I would estimate the volume that an average truck could carry. I assume that loading portion of an average truck is 5m long, 3 m wide and 2.5 m high. That gives me 37.5 cubic meters.
    Now I would assume that, keeping in mind the obstacles and jams, the average speed of the truck is 20 miles/hr. So for an up and down trip of 20 miles (10 miles up and 10 miles down) it will take an hour. It will takes about 30 min to load and 30 min to unload. I will assume that mountain is cut and the truck does not have to wait just because the mountain is uncut at any moment. So to carry 37.5 cubic meters of the mountain it takes 2 hours.
    Assuming the work continues 24*7 in three shifts. So daily 12*37.5 cubic m of the mountain could be moved. That gives me 450 cubic meters of mountain per day. So number of days it will take to transfer the whole mountain would be 5000000/450, i.e. 11111 days.

  • Elise Jan 7, 2012, 8:12 pm

    Suppose the mountain consists of solely stones and soil, moreover, it has a cone shape. Assume the radius of bottom is r=25m, and the height is h=300m. Using the formula V= (Pi)x(r^2)xh , we have the volume of the mountain being 3.14 x 625 x 300= 588,750m^3. Let’s approximate again the capacity of a truck is V=length x width x height = 10 x 5 x 10 = 500m^3. Suppose we have 100 labor, it will take them 30 minutes to fill up the truck at one time, now 10miles=16.09km. Using common sense, it takes me 1.5h to get from my apartment to CDG airport in Paris by bus, the length is 29km. The time is roughly 16.09×1.5/29=50 minutes for the truck to reach destination. Total time for transportation in one go is 50+30+50=130minutes

    There are 1,178 repeated times to clear the mountain. (588,750/500=1178) This means it’d take 1178×130/60= 2553 hours. The average working hour per day is 8, so 2553/8 = 320 days

  • Anne Jan 8, 2012, 12:23 pm

    First we need to determine how much material there is in the mountain and how much the truck can hold. Let’s assume that an average mountain is approximately cone-shaped and is 1,000 feet high with a radius of 500. If the cone formula is 1/3 base * height, this gives us an area of approximately 500,000 feet squared (1/3 * 1,000*500). An 18-wheeler truck could hold perhaps 500 square feet of material. [Note: I chose these numbers because they are relatively easy to work with but they could be way off — I would verify that my assumptions are reasonable with the interviewer.] This means that it would take 1,000 trips for the truck to move the mountain.

    Each trip can be broken into 4 parts: loading the truck, traveling the 10 miles, unloading the truck, and traveling back. We don’t know how many people we have doing the loading — let’s assume that we have a big crew and they can load up the truck in 1 hour. Since the truck is driving to a mountain, it’s probably not on a highway, so I’ll assume it’s going 40 miles an hour and can make the 10-mile trip in 15 minutes. Unloading the truck would also take 1 hour, and traveling back would be another 15 minutes. This gives us a total of 2.5 hours per trip. Multiplied by 1,000 trips = 2500 hours, or approximately 104 days. (2400 hours = 100 days, 100 leftover hours = ~4 days.)

  • Jap Jan 9, 2012, 3:38 pm

    To determine the duration of the move, we need to know the duration of each trip and the number of trips needed:

    Total duration = (# trips) * (duration of a trip)

    The first element; the # of trips can be calculated by dividing the size of the mountain by the size of the truck:

    # trips = (size of mountain) / (size of truck)

    The second element; the duration of a trip, is the sum of several components:
    – Duration of filling the truck
    – Duration of the trip to the new location
    – Duration of unloading the truck
    – Duration of the trip back
    – Overhead (truck breaks down, truck has to get gas, getting the materials ready in the morning, preparing for the night after the work day etc.)

    So let’s look at each of the two main components individually to get an estimate.

    1. # of trips necessary to move the mountain
    a. Size of truck
    Let’s say the bed of the truck has a volume of 5m*3m*2m = 30m^3. For efficiency, we will pile up high and get about 50m^3 of earth into the truck.
    b. Size of the mountain.
    Let’s assume the mountain is a cone shape. The height is about 1400m and the radius is therefore 1000m. The volume of a cone is 1/3*pi*r^2*h, which is 1/3 * 3.14 * 1E6 m^2 * 2E3 m. This is about 2E9 m^2.

    ===> Total number of trips necessary is 2E9 / 50 = 40 million

    2. Duration of a trip
    Here we have to make a couple of strong assumptions. First of all, I’m going to assume that the new location of the mountain is a little under an hour’s drive away from the old one. I am going to assume a 8-hour work day. I will assume the following numbers for the different parts of the trip:
    – Duration of filling the truck: 0.1 workday
    – Duration of the trip to the new location: 0.1 workday
    – Duration of unloading the truck: 0.1 workday
    – Duration of the trip back: 0.1 workday
    – Overhead: 0.1 workday
    The total of a trip then comes to 0.5 workday, which means we can make 2 trips per workday.

    ===> T0tal number of working days necessary to move the mountain (with 1 truck): 20 million

  • Djordje Jan 10, 2012, 6:58 pm

    Let’s assume the following:
    average size truck:
    Load 8m3
    speed 30 mph

    average size mountain:
    Piramid shape
    height 1000m
    basis square 1000m x 1000m
    Volume to be transported:
    V = 1/3 (B x H) = 1/3 (1.000.000.000) m3 = 333.333.333 m3

    Time needed:
    Load/Unload = 10 min
    Load + unload + go to + go from = 10 + 10 + 20 + 20 = 60 min = 1h
    Time needed is one hour per one truck load

    Total time needed with one truck = (1h x 333.333.333 / 8) = 41.666.667h

  • Tara Jan 22, 2012, 6:59 am

    I’ve tried to do this in as few moves as possible. As no follow up questions are allowed, and little detail is provided (and also from reading the above responses) it is clear that endless variations in assumptions are available, and is it normal to to sit in front of an interviewer for 15 minutes in silence while you mull over these in your head? Plus I am not a maths or science person!

    Remembering that MC is the best solution available in the time available with the information available, I assumed that anything we haven’t been given details on is outside scope, and all that is at your disposal is one average truck. Therefore all that really matters is the volume of the truck, the volume of the mountain, and the amount of time it takes to move that volume 10 miles.

    Average mountain – 3,500 metres.
    As its triangular, I guessed its volume from halving that of a cube with sides all equalling 3,500m – i.e. 3,500 cubic metres / 2 = 1,750 cubic metres.
    Assumed each cubic metre of dirt/gravel would weigh 1 tonne i.e. 1000 kg.
    Your average moving truck has a load of 3 tonnes.
    1,750 / 3 = approx 580 truckloads.
    10 miles equals approx 20 kilometres.
    Say the truck travels at 60km/h, making a one way trip 20 minutes and return 40 minutes.
    580 x 40 minutes = 23,200 –> i.e. approx 400 hours.
    Taking into account an average 8 hour work day – 400/8 = 50 days.

    I.E. 50 working days of one truck in constant movement, not taking into account loading time, quarrying, labour constraints etc etc etc.

  • Sudhanshu Feb 24, 2012, 2:54 am

    I would do the following:
    1. Estimate the size of an average mountain and divide it into smaller geometrical shapes- rectangles or smaller triangles
    2. Estimate the size of an average truck and estimate the volume it can transport. Volume would be in iterms of rectangles or triangles

    Make assumptions:
    1. Assume an avg speed of the truck
    2. Assume the time of loading of the truck
    3. Assume there are no roadblocks in the path of the truck
    4. Assume there is enough resource availability

    Answer would be: Time for breaking the mountain for the first delivery+ time for loading

  • Sudhanshu Feb 24, 2012, 2:56 am

    Oops- Sent submit too early for the previous post.

    Answer would be: Time for breaking the mountain+ time for loading the truck+time for truck’s transportation and back

  • Jermia Mar 5, 2012, 9:00 pm

    It’s only 10 miles. So you can move the mountain just by moving part of its bottom. Let’s cut the mountain into horizontal layer, and then shift the most bottom layer 10 miles to any direction. The layer shape would be cylinder. All we have to do is to move half of its side to the opposite side. Let’s guess the volume of such part. Let’s say the mountain have r=500m. Half of the cylinder length is (pi)*r = 3.14*500 = 1,570 m. Say the height of the cylinder is 20m, and the depth (towards inside the cylinder) is around 5 m. Thus, 1,570*20*5 = 157,000 m3 to move.
    Now let’s guess the volume of a dump truck, let’s say it’s 5 m * 3 m * 1.5 m = 22 m3.
    It depends on how many trucks do we have. Given 10 trucks, it’ll take 714 times to move the soil volume.
    If one truck can move back and forth 10 times in one day (include load and unload, and smooth road, and rest time – we have relief crew, it’ll take 71 days to move the mountain 10 miles.

  • MA Mar 21, 2012, 10:32 pm

    Average size of the mountain = 1000^3 ft
    Average size car’s carrying load = 5^3 ft
    Distance for one trip = 10 miles * 2 (for return trip) = 20 miles
    Estimated speed = 60mph
    Estimated time to travel for one trip = 20 minutes
    Time to unload = 10 minutes
    Total time for one trip = .5 hours (20 + 10 minutes)

    One trip bring 5^3 feet, so taking 1000^3 feet would take (1000/5 )= 200 trips
    Each trip takes a half hour, so it would take about 100 hours of consistent work. Factoring in 10 hour work days, it would take 10 days to complete

  • Duncan Mar 28, 2012, 9:59 pm

    730 days

  • Paulina Apr 2, 2012, 3:39 pm

    Average mountain has the height of 1000 m and the base in the shape of circle with radius 50 m. Then, the volume of the mountain is equal to 1/3*3,14*50^2*1000. When we round it, it’s about 2500000m 3. he average truck has dimension: 10x3x2=120m3. Then, we know that we need around 21000 trucks. To load and unload the truck we will need 20 minutes and to go back and forth(so it’s 20 miles) we need 40. So it’s one hour for one truck, hence it’s 21000 hours/24=875 days = 2 years and 4 months (rounded)

  • marie-paule Apr 8, 2012, 8:55 am

    Assume a truck can charge a max of 5x3x2= 30 m3
    Assume a mountain ,1000m high, with cone shape, radius 5,000m
    > volume mountain = 1/3 pi r2 h = (5,000)2 x 1000= 25,000,000,000 m3
    Number of truck loads needed, c.a. 1B
    Assume material to be moved is prepared while truck travels back and forth
    Assume truck takes 1 hr to load, travel, unload, travel back
    > 1,000,000,000 hours
    Assume 10 hrs work per day
    > 100,000,000 days
    300,000 years!

  • Leonardo May 1, 2012, 6:31 am

    Each two-way trip: 20miles
    At an average speed of say 40MPH, each trip would take 30min.
    Now, considering an average mountain to have a volume of 1 cubic mile and a truck to have a volume of 1000 cb ft, it would take 1cbmi /1000 cbft *5280^3 cbft/cbmi = 125 *30min= 3750 min.

  • Patrick May 18, 2012, 4:14 pm

    Total time needed in hours = time for one trip * number of trips

    where
    time for one trip = time to load a normal size truck + time to move 10 miles with full load + time to unload a full truck + time to move 10 miles without load

    number of trips = total volume of rocks (and sand, soil, etc) in a typical mountain / truck’s load capacity
    ——————————————-
    1. time for one trip = time to load a normal size truck + time to move 10 miles with full load + time to unload a full truck + time to move 10 miles without load

    1.1 The truck can be either loaded manually or automatically using assisting specialized machines/vehicles. A normal truck (e.g. a pickup truck) has loading capacity of approximately 3m*2m*1m=6m^3. Let’s suppose on average it will take about 5 minutes to load the truck.

    1.2 Fully loaded trucks move slower, let’s suppose 40 mph, to takes about 15 minutes to reach 10 miles.

    1.3 To simplify, we suppose the unloading time is the same as the loading time. So another 5 min.

    1.4 Finally, trucks without load can move faster, so suppose 50 mph, takes about 12 minutes to go back to the original site.

    Therefore the total time for one round trip is about 5min+15min+5min+12min=37 minutes. Let’s say another 3 minutes are wasted on average per trip for gas refill and emergency. So 40 minutes per trip.
    ——————————————-
    2. number of trips = total volume of rocks (and sand, soil, etc) in a typical mountain / truck’s load capacity

    2.1 Suppose the mountain is pyramid-shaped with a square base. The volume of the mountain = the height*the area of the base*(1/3). Let’s suppose the typical mountain to be 200 meters tall, and 200 meters wide on each size at the base. so the volume is 200m*200m*200m/3=8000000/3 m^3=2.7million m^3.

    2.2 the loading capacity of the truck is about 6 m^3 as estimated in section 1.1.

    so the number of trips needed is 2.7million cubic meters/6 cubic meters per trip = 0.45 million times = 450,000 times.

    3. total time = 40 minutes * 450,000 times = 18,000,000 minutes = 300,000 hours

    Suppose a 8-hour long working day, 5 days a week. One year has 51 weeks, suppose workers take one week off per year, so let’s say 50 weeks per year, then that is 300,000/8/5/50 year = 150 years.

  • Dave May 19, 2012, 4:31 am

    Let’s assume a cone shape for the mountain:
    h = 2000 m average
    r = 4000 m average

    Thus the volume we’ve got to relocate is:
    V = 1/3*pi*r^2*h = 32 km^3

    Assuming a density of about ro = 3000 kg/m^3, the mass to be moved is:
    M = 96 000 000 000 tons

    A truck can take along a volume of about:
    Vt= 5*2*2 m^3 = 20 m^3

    which correspond to a mass of:
    Mt = 20*3 tons = 60 tons

    This is quite a heavy load, therefore we cannot fill th truck up to the limit, then the mass is a constraint. Let’s assume the max weigth is about 30 tons, then the number of loads the truck needs to bring the rocks 10 miles away is:
    n = M/Mt,limit = M/30 tons = 3 200 000 000

    The overall distance is then:
    d = n * 2 * 10 miles = 64 000 000 000 miles

    Assuming an average speed of 40 miles/h (the truck will be faster on the way back, let’s say 50 miles/h, than the outbound, let’s say 30 miles/h), the time required is:
    t = d/40 hours = 1 600 000 000 hours

    Assuming we have 1 truck and 1 man working 8 hours/day, 220 days/year (and rounding it down to 200 because we spend time also for loading the truck and unoading it), the time it takse to relocate the mountain is:
    T = t/(200*8) years = 1 000 000 years (!!!)

  • Pramod May 22, 2012, 9:11 am

    wt of small mountain: assuming 10,000 tons

    1 ton – 1000 kg
    proxy – ship 10000 tons – 10,000,000 kg

    avg truck carries: 2000 kg – 2 tons

    1 truck carrying 2 tons per trip will have to make total 5000 trips. 10,000 trips to and fro.
    5000 * 2 = 10,000

    distance = 16 km ( 10 miles)
    v = 60 km/hr

    t = 16/60 = 16 min per trip

    Total time: load time + travel time + unload time

    Assuming one chunk weighing 2 tons at a time. a crane is used to load truck. takes 10 min to load. similar time to unload.

    load time = 5000 * 10 = 50,000

    16 min per trip. 10,000 trips

    travel time = 16 * 10,000 = 16,000 min

    unload time = 5000 * 10 = 50,000

    total: 50,000 + 16,000 + 50,000 = 116000 min

    1933 hrs

    81 days

    Assumption: worked non stop for 81 days

  • kalare May 31, 2012, 5:33 pm

    estimate the size of the mountain, and an average sized truck:

    truck: 2m*2m*1m
    mountain: 1000m^2*2000m

    Suppose the truck moves at a speed of 40miles/hr. It takes 10/40*2=0.5 hr for a round trip

    1000*2000/2/2/1 = 1000*500=500000

  • kalare May 31, 2012, 5:36 pm

    Finally, 0.5hr * 500,000 = 250,000 hrs

    250,000/24/365 is approximated to be 30 years.

  • Nikita Jun 2, 2012, 2:04 pm

    You need to load the truck, move it, and then unload it.

    Let us assume that a truck is able to move 10,000 kg in one take.

    An average mountain is about 1000 by 1000 metres in basis and is about 1000 metres high. Assume that its volume is the volume of a cube with same dimensions divided by 10. So its volume is 100 mil. cubic metres, which should weight rougly 100 mil. kg.

    That means that a truck will have to make its journey 10,000 times to move a mountain.

    One journey take:

    1. Assume 1 hour to load the truck.
    2. 10 minutes to drive, if the truck drives with an average speed of 60 miles per hour. Also 5 minutes to start the journey and finish it.
    3. Assume 15 minutes to unload the truck.

    So in total it is 90 minutes per journey, multiplied by 10,000, which means 900,000 minutes or 15,000 hours, which is roughly 600 days, provided the work is 24/7

  • kar sr Jun 4, 2012, 11:29 am

    It would take 30 hours for moving the mountain using 1 truck.

    Lets consider average size of mountain is 3000 mts.
    Average size of truck can carry 100 kgs on a single visit. And it would take approximately 6 minutes to cover 1 mile with such weight.
    lets assume 1 meter of mountain would weigh 1kg
    So it would take 1 hour for covering 10 miles.

    Time taken to relocate average mountain(3000 mts) = Time it takes to cover 10 miles * number of trips it takes for truck to complete

    time it takes to cover 10 miles = time taken to travel 1 mile * 10
    = 6 mins * 10 = 60 mins

    Number of Trips it takes
    Average Truck can carry 100 kgs in one trip
    1 hour = 100 kgs
    As 1 mt = 1kg
    3000 mts = 3000 kgs
    Hence 30 hours = 3000kgs
    No. of trips = 3000/100 = 30

    To relocate 3000 mts = 60 mins * 30 trips = 1800 mins or 30 hours it would take for relocating average size mountain

  • RG Jun 5, 2012, 3:46 am

    I am assuming the average mountain to be movable and fits in an average sized truck.
    Since the mountain is heavy the truck will only be able to move with an average speed of 10 miles per hour and hence it will take an hour to relocate or move the mountain.

  • Vaibhav Jun 8, 2012, 2:14 am

    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

    • Raj Oct 13, 2013, 11:12 pm

      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?

  • Muhammad Jun 13, 2012, 1:56 am

    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)

  • Vaibhav Nagar Jun 28, 2012, 4:00 am

    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.

  • S.G. Jun 30, 2012, 2:13 pm

    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.

  • haa Jul 2, 2012, 10:50 am

    450,000,000 min

  • Ahmad Jul 3, 2012, 5:11 pm

    “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

  • Bob Jul 19, 2012, 11:18 am

    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

  • VDS Aug 11, 2012, 5:22 am

    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

  • Shambo Aug 13, 2012, 2:33 am

    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

  • Priti Aug 24, 2012, 3:38 am

    10,000 days or almost 30 years

  • SR Aug 30, 2012, 4:14 am

    2400 yrs
    Mountain Height=1000m
    Mountain radius= 1000m
    Volume = 1/3 * pi * 10^9 ~= 10^9 m^3

    Each truck size = 2m (height) *2.5m (width) *10m (length) = 50 m^3
    Number of truck loads= 2*10^7 truck loads
    Speed of truck = 60 miles/hr
    each way= 10 miles ( 10 minute)
    number of loads per hour = 3 ( going back and forth)
    hours of work per day = 8
    number of loads per day = 24 ~= 25

    Number of days requires= 2*10^7/25= 8* 10^5 days
    Number of years= 8*10^5/365 ~= 8* 10^5/ (1000/3) = 2400 years

  • Linnea Sep 16, 2012, 9:18 am

    1. Estimated size of an average size mountain:
    Height: 30m
    Width: 30m
    Depth: 30m

    Size in square meters: (30*30*30)/3=9000 square meters.

    2. Estimated time for breaking the mountain using dynamite in order to enable transporting on the truck:
    Estimation: 100 square meters of mountain exploded/ hour.
    9000/100= 90 hours spent on breaking the mountain.

    3. Estimated time for transport on truck:
    Estimation: each truck load can take 100 square meters of stone. 9000/100= 90, meaning we need to make 90 trips to transport the whole mountain.

    10 miles is approx. equal to 16 kilometers. If the truck has an estimated average speed of 80km/h it will takes us 80/16 = 1/5 of an hours in each direction, which is equal to 12 minutes (60/5= 12). Considering both directions we spend 12*2=24 minutes on the road .

    4. Estimated time for on- and off loading of truck:

    Off-loading: Assuming the truck has a platform that can be lifted allowing for the stones to slide off, off-loading is considered taking minimal time.

    On-loading: 10 square meters of stone is estimated to take approx. 3.6 minutes to load. Therefore it will take 36 minutes to load one truck load of 100 square meters of stone ((100/10)*3.6=36).

    5. Total time for transport:
    On the road =24 minutes
    Loading = 36 minutes
    24+36=60 minutes or 1 hour.

    From point 3: We need to make 90 trips. If each trip takes 1 hour it will take us 90 hours in total time for transport.

    6. Total time for moving the average size mountain with an average size truck:
    90h+90h=180 hours
    or 180/24 = 7.5 days if we work 24 hours a day
    or 180/8 = 22.5 days if we work 8 hours a day.

  • Will Sep 16, 2012, 6:34 pm

    Ok, first off, I’ll start with the size of the mountain and break that down. I’ve been to a few mountain ranges, and I know that typical peak elevation can be about 5,000 or 6,000 feet in altitude. Some can be a lot higher, but we’re saying average mountain, so let’s assume 5,000 feet. Also, the actual ground level in mountain ranges isn’t at sea level by any means – so let’s say the ground level is at 2,000 feet. Which means, the mountain is only 3,000 feet tall.
    (5,000 – 2,000 = 3,000).

    Next, the width of the mountain. I don’t really know how wide mountains are, but I think it’s safe to say they’re at least bigger than a football field. A football field is 100 yards, or 300 feet, so I guess a typical mountain would be about 500 feet.

    Let’s assume the mountain is cone-shaped. Therefore the volume of the mountain is equal to the height of the mountain x half of the width squared x pi, or: 3,000 feet x 3.14 x (250 ft)^2

    3,000 feet x 3.14 x 6,250 sq. feet
    I’ll simply my numbers a bit to make the math easier:
    3,000 ft x 3 x 6,000 sq. ft.

    3×3 is 9, so that’s basically like saying 9×6, which is 54, and the two groups of thousands gives us another million, so
    54 million cubic feet is the volume of our mountain.

    Now, I need to know the volume of the truck. I would say the typical truck bed is about 9 feet long and 6 feet wide. That would give us a surface area of 54 squared feet (9×6), and we can probably load the dirt from the mountain about 2 feet high in the truck bed.
    54 x 2 = 108.
    So, I think we can feet 108 cubic feet of mountain into the truck at any one time.

    with a 54 million cubic foot mountain and a truck bed of 108 cubic feet, we need 500 thousand truckloads in order to move the mountain.

    10 miles isn’t too far a trip, but we also have to load and unload the truck bed each time. I’m going to assume each trip takes a total of one hour to load, move the mountain, unload and come back.

    That means it takes us 500 thousand hours to move the mountain 10 miles.

    If we assume there’s 25 hours in a day, then 500k/25 = 20k.

    So, it takes us 20 thousand days.

  • Raghav Dasgupta Sep 27, 2012, 9:41 am

    It is not specified whether or not the mountain is actually on the ground or not. Assuming it is floating in mid air (as this is an assumption based question), then it could take any amount of time varying from one minute or one hundred thousand days. The idea of a mountain of ‘average’ mass. There is no such thing as an average mountain as we cannot count the mass and size of every mountain in the world. We have no definition of a mountain, so a mountain could be as small as a grain of sand. The mountains we are aware of could be outliers in terms of what we know. An average size truck, by the same token, might not mean what we think it means. There is no time frame for such a question. This question could be as old as trucks are, or it could be a question meant for people in the future. Trucks come in a variety of sizes, and although there is a ‘common’ truck, there are trucks that could be quite small or quite large in comparison. This is all surrounded by the idea of an ‘average’. An average is all things considered and divided by the amount of all things. We cannot know ALL things, and therefore, we cannot judge. There is no wrong answer if the answer is in a time, so the time it would take is five minutes, simply because, until the answer is given, there is no way of getting it wrong because the answer is UNKNOWN knowledge. TOK’d.

    • George Oct 8, 2012, 3:35 pm

      You miss the point completely. If you are honestly trying to get a consulting job watch the videos on the main page to understand the whole point of asking a question like that. The goal is to demonstrate your reasoning and estimating process and skills. The exact numbers don’t matter. Just your process and skills.

    • Victor Cheng Oct 8, 2012, 4:00 pm

      Raghav,

      George is absolutely correct. Although your answer is the most true, if you provide that answer in a case interview it is an automatic rejection.

      Victor

  • GP Sep 27, 2012, 3:44 pm

    I’ll assume the average mountain has the following dimensions:

    Base area (ba): 10,000 m^2
    Height (h): 100 m

    For ease of calculation, I’ll use the formula to find a pyramid’s volume:

    h.ba.1/3 = 100 . 10,000 . 1/3 = 1,000,000/3 ~ 330,000 m^3

    I’ll also assume the truck travels at an average speed of 40 mph when it’s empty and at an average speed of 20 mph when it’s full and that it’s load capacity is of 200 m^3 (20m x 2m x 5m). It also takes 15 minutes to load and unload the truck.

    Therefore, each time the truck travels to the mountain and back, it takes 1 hour and moves 200m^3 of it.

    It’d take a total of 5,000 straight hours to move the mountain to somewhere 10 miles from it, but, considering it has only one driver, working 8 hours shift, it would take a total of 625 days, or 1 year 8 months 2 weeks and 6 days.

  • GP Sep 27, 2012, 3:47 pm

    I should have divided it by 3. Busted

    😀

  • VM Oct 3, 2012, 12:19 pm

    Hypothesis:
    1) The container of an Average size truck is long 7 meter x 3meter which is an Volume of 21 cube-meter. To semplify let’s use 20 cube-meter.

    2)An average size mountain is 1000 meter long x 1000 meter high. His volume is 1,000,000 cube meter.

    3) The average speed of the average truck is 80 km/h. Consequently, the truck will need 1/4 hour (15 minutes) to do one transport of 20 cube meter for 10 miles (or 20 km). Time needed:Distance/Speedrate–> Time needed: 20 km/ 80 km/h–> 1/4 h

    Considering the following hypothesis the number of transportation-trip the truck will do in order to transport the whole mountain is : Volume of the mountain / Volume of the truck. Which is : 1,000,000 / 20. Which is 50,000.

    Consequently, the time needed to transport the mountain for 10 miles (or 20 km) is equal to the time needed to do one transportation-trip times the total number of transportation-trip.

    Total Number of Transportation-trip: 50,000
    Time needed to do one transportation-trip: 1/4 h (15 min)

    Total time needed to transport the mountain for 10 miles ( or 20 km) : 50,000 times 1/4 h : 12,500 h.

    VMMCS

  • George Oct 8, 2012, 3:30 pm

    First we have to figure the approximate square footage of material (mountain) to be moved. What is the height of an average mountain? This varies greatly. I took the tallest mountain in the world, Everest, which I estimated at 24,000 ft. But then, it doesn’t start at sea level. What is the base camp? I think I’ve heard something like 14,000ft. So the largest mountain we could guess has 10,000ft of vertical. From hiking I know that gaining 1000 ft to 2000 ft of vertical is a normal long hike, but I’m not sure I’d call that an average mountain. More a small mountain. From the range of possibilities from the hills in the center of America which they consider mountains, to the cascades or sierra nevadas or Olympics. I would estimate that the average vertical rise would be somewhere near 3,000ft.

    Now that have this estimate, I want to estimate the volume of a mountain this large. If a mountain has proportions approximating an isoscolese triangle, then draw a line down from inside the tip of the triangle to the base labeled 3,000ft. The approximate proportions of long side to hypotenous in a right angle triangle put the hypotenous about 30% longer. this could be the measure of the length down the slope. This equals the other sides of the triangle, so the base is also the same. If our height is 3000ft then, we can guesstimate the base is 4000ft across. If we were to draw a cube around this basis of height, width and length, we would know how much volume was in that cube. 4,000×3,000×4,000= 48billion square feet. If we take that cube and draw our pyramid inside it, we can guestimate from looking at the proportions that our pyramid makes up about 1/3 of that cube. So take 1/3 of 48 billion and we have 16 billion square feet to move.
    Now how long to move it? The question says average truck. It doesn’t specify what kind of truck. Dump truck? Pickup truck? I would say that the most “average” kind of truck is a consumer pickup. Some would argue that one would more likely use a dumptruck to move a mountain, but I would say that the question is already so absurd, perhaps best to go with the exact wording of “average”. So I choose pickup.

    I estimate an average pickup to be 4 feet wide (can just fit a 4×8 piece of plywood, widthwise), and 7 feet long (I just fit with about 1 foot lee-way laying down in the back with the tailgate up.) The side height I’m estimating at 2 feet. So our cubic feet on one pickup load full is 4x7x2=56 square feet. A load can be piled up a bit in the middle so we’ll round up to 60. So how many loads do we need to move 16billion sq feet? 16billion/60= 266,666,666. We’ll round up to 270,000,000 total loads in the pickup truck.

    How long will each load take?

    10 miles each direction. Assume a normal highway and we can go 60mph. 10 minutes each direction. Add loading and unloading time. Let’s assume 5 minutes on each end for that. So one round trip will take about 30 minutes.

    So now we can just calculate down our number of loads and load time. 270,000,000 loads x 30 minutes gives us total minutes, but we are about to divide that back down to get years, so let’s simplify and say 30 minutes is 1/48 of a day and divide 270,000,000 by 48. This equals about 5,400,000 days, divided by 365 gives us about 150,000 years.

    This, of course, doesn’t take into account all kinds of variables like the trucks breaking down and driver issues, etc. etc. etc. (Which I imagine would SIGNIFICANTLY affect the final answer.)

  • Rob Oct 16, 2012, 9:37 am

    2) Let’s assume it has a base area of 10 hectares
    3) 2,000 meters tall*100,000 m2 base = 200,000,000 m3 (if square). Account for the cone shape, say ½, thus 100,000,000 m3 is the average volume of a mountain.
    4) Assume an average truck bed is 2 meters x 3 meters and can be filled vertically to carry 10 m3
    5) Thus, it would require 1,000,000 loads to move the mountain
    6) Assume 30 minutes to cut and load 10m3 in a truck
    7) Assume the truck travels at 60 mph X 10 miles
    8) Round trip would be 20 minutes
    9) Unloading time 10 minutes
    10) Thus, we require 60 minutes to load and deposit 10m3 and return
    11) 8 hour work day
    12) Thus 8 loads / day
    13) 252 work days in a year
    14) Thus 2,016 loads transported / year (round down to 2,000)
    15) 2,000, 10m3 loads per year would require 500 years to move 1,000,000 m3.

    • Rob Oct 16, 2012, 9:37 am

      Step one did not paste in the initial post.

      1) Let’s assume an average mountain is 2,000 meters tall
      2) Let’s assume it has a base area of 10 hectares
      3) 2,000 meters tall*100,000 m2 base = 200,000,000 m3 (if square). Account for the cone shape, say ½, thus 100,000,000 m3 is the average volume of a mountain.
      4) Assume an average truck bed is 2 meters x 3 meters and can be filled vertically to carry 10 m3
      5) Thus, it would require 1,000,000 loads to move the mountain
      6) Assume 30 minutes to cut and load 10m3 in a truck
      7) Assume the truck travels at 60 mph X 10 miles
      8) Round trip would be 20 minutes
      9) Unloading time 10 minutes
      10) Thus, we require 60 minutes to load and deposit 10m3 and return
      11) 8 hour work day
      12) Thus 8 loads / day
      13) 252 work days in a year
      14) Thus 2,016 loads transported / year (round down to 2,000)
      15) 2,000, 10m3 loads per year would require 500 years to move 1,000,000 m3.

  • sri Oct 23, 2012, 5:40 am

    Assume radius of an average mountain – 1 km (which makes the circumference 6.2 km which is too big – but what the heck, how to define an ‘average size’ mountain)

    Assume height of the mountain – 1/3 rd of its radius ~ 330m ~ 1/π km . Therefore volume of the mountain – 1/3 * π * 1*1*1/π = 0.33 km^3 ~ 300 million m^3

    Assume the density of the mountain as a mountain and as disintegrated mud (rocks, shrubs included) in the truck are same

    Assume the size of the truck container be 3m * 1.7 m * 1.7m (an average truck used in quarries are normally deep, but 1.7m may not be accurate). Hence the volume of the truck is approx 9 m3. Considering 1 m3 of sand can be heaped above the truck’s rectangular container , total transportable volume is 10m3

    So in order to transport 300 million m3 of rock n sand with a 10m3 truck, we need to use the truck 30 million times.

    Now we need to estimate how long it takes – Assuming a truck carrying a load of sand and rocks cannot travel more than 30 miles per hour. It takes 20 minutes to travel 10 miles and 20 minutes to travel back to the mountain. Take 20 minutes to load and unload (assuming we have some high performing loader) . So for 1 movement it takes 1 hour.

    Hence for 30 million movements it would take 30 million hours. Assuming we work only 20 hours a day (4 hours for changing shift drivers, filling gas, stuck in traffic etc). we would take (30 million/20) days = 1.5 million days.

    Assuming we work 350 days a year ( 15 days for Christmas holidays, Superbowl, Mitt Romney debate etc). It would take 1.5 million / 350 ~ 4200 years

  • Akshay Oct 31, 2012, 1:20 am

    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

  • Anya Nov 16, 2012, 6:36 am

    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.

  • Ben Lo Nov 25, 2012, 2:56 pm

    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)

  • Krishna Dermawan Nov 25, 2012, 9:26 pm

    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

  • Arash Dec 16, 2012, 2:32 pm

    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.

  • Matthew Dec 24, 2012, 3:24 am

    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

  • Petros Jan 7, 2013, 2:31 pm

    -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

  • Ram Jan 20, 2013, 12:03 am

    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

  • Wojtek Jan 21, 2013, 7:09 pm

    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!

  • jc Jan 22, 2013, 8:20 am

    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.

  • steph Jan 23, 2013, 10:06 am

    assumptions:
    a truck can hold 100 kilos of soil and an average mountain has 10,000 kilos of soil
    a truck will travel the same rate (time and distance) with or without the load from point A to point B. Let us assume that it takes 10hrs for the truck to travel

    we would like to find out how many hrs it would take
    Thus our equation should be
    10,000 / 100 = number of trips
    number of trips x (HRStotravelpoint1-2 (x2))
    i multiplied it with x2 because it goes back and forth thus doubling the time

  • aimee Jan 25, 2013, 4:47 pm

    53 hours total

  • Konami Jan 26, 2013, 8:13 pm

    I know the tallest mountain is 27k feet high. So I would estimate that the “avg” size mountain is about 8k feet high. Assuming a pyramid shape to the mountain, I would put the length and width of the mountain at about 2k feet each. I think we can probably cut the pyramid in the middle and put the two sides together to form a rectangular shape of volume 8k x 2k x 2k = 32b cubic feet.

    If we assume average sand/rock carrying trucks, they seem to have about a one-car length x 5ft high x one-car width sized box in the back. One car length is about 12ft and width is about 6ft so total box volume is 360 cubic ft.

    32/36 = 0.88; 1b/10 = 100m; 0.88 x 100m = 88m round trips. If each truck goes 30mph avg speed start to finish then 10 miles will take 20min to go there and 20min to come back = 40min, and let’s assume another 20min to load & unload = 1hr per round trip. So it will take 88m hours.

  • Fernando Jan 31, 2013, 8:19 pm

    This is my first real estimation problem so I hope at least I get some points in favor. Thank you Victor!

    1-Finding average truck capacity: I remember seeing average cargo trucks lift cars therefore I used the estimated weight my car has.
    Truck capacity at 100%: 6 tons
    2-Mountain area: I live in a big hilltop in Colombia. the height difference btn Bogota to it is from 1.700-2.400 mts (700 mts). The length of the mountain is approx 24 blocks, about 2 km.
    Note: I took halft of this mountain into consideration due that the problem states average mountain.
    (So height 700/2=35o mts and length from 2000 mts to 1000 mts.)
    Mountain area in mts= 1000mts x 350 mts= 350.000 mts2
    2,1-Converting mts2 to kg: I know that I litter is equal to 1 kg, therefore I estimated the area 1 litter milk carton.
    Milk carton: 10 cm length x 20 cm width= 200 cm2 which equals to 2mts2.
    *Mountain Area in tons: 350.000 mts2 / 2 mts2= 125.000 kg
    or 12.500 tons.

    3-Truck travel distance relocation: 10 miles
    avg speed of truck at full capacity: 10 miles /hour (as I´ve recalled truck speed in my farm).

    *Total distance time per load: 2 hours

    4-Total working hours: 8 hours per day (mon-fri)
    4,1-Lunch time per day: 1 hour
    4,2-Hours taken to load and unload dirt per journey: 1 hour

    *Total hours per journey: 3 hours

    *Total available working hours: 7 hours

    *Journeys per day: 7 hours / 3 hours = 2 journeys carrying 12 tons per day.

    *Total number of days needed: 12.500 total tons / 12 tons per day= 1.040 days

    Note: I want to take a wild hypothesis; 1 hour was “given” away per day, the problem could also solve which would be more cost efficient; hier this person for full time or per day.

  • andrushka84 Feb 2, 2013, 7:33 am

    One way would be to assume the average truck has a speed of 20 miles/hour , it can carry 10 tons and the mountain has 1 million rocks, each of them weighs 10 tons.
    time=distance/speed= 2*10/20=1 hour/rock; 2 because the truck goes 10 miles and comes back 10 miles for another rock.
    So it takes 1.000.000 hours.
    Another way would be to assume that the mountain is a 10 tons rock, so the truck can move it in half an hour.

  • Saurabh Arora Feb 9, 2013, 5:26 am

    Time taken to move the mountain 10 miles =

    Assuming that the mountain is already broken down into pieces which are ready to be put on the truck, the time taken to move the mountain can be written in the following form:

    (Number of trips required to move the mountain)*(Time taken to make a round trip of 10 miles + Time taken to load and unload the truck)

    Taking each of the three required numbers one by one,

    Number of Trips required to move the mountain:
    The number of trips required to move the mountain are the trips that the truck needs to make to transfer the entire volume, when it is filled up to its capacity in each trip. Hence, it can be written as:

    (Volume of the mountain/Volumetric Capacity of the truck)

    Volume of the Mountain:
    The mountain can be thought of as a pyramid. Since the volume of the pyramid as 1/3*A*h (A, being the area of the base and h being the height of the pyramid), I will now try to estimate both A and h

    Assuming base is a square of side 100 m, A = 10000 m2
    Also let’s assume the height of pyramid to be 100 m

    Volume of the pyramid is 100*10000 = 1 mn cu.m.

    Volumetric Capacity of Truck:
    Assuming the interior of truck is a cubiod, volume of its interior can be taken as = x*y*z (x and y being the sides of the cross-section and z being the length). Now I will like to estimate the dimensions to calculate the volume.

    An average truck is around 2 times the size of a human being in cross-section sides, hence I would estimate both x and y to be around 3 m (twice the average human size of 1.6 m)

    From my experience, I feel that the length of the truck is slightly more than the twice of its cross-section side; hence I estimate it to be around 7 m.

    Volume of the interior of the truck = 3*3*7 = 63 cu.m. Rounding it off to 60 cu.m for the ease of calculation.

    The number of trips is hence = 1 mn. cu. m/60 cu. m. = 16,667 trips. Rounding it off to around 16,000 (since we reduced the volume of truck in previous rounding off)

    Time taken to make a round trip:

    Time taken to make a round trip = 20 miles/Average Speed

    Assuming the average speed of around 40 miles per hour (average over loaded and unloaded condition) for the round trip.

    Time taken to make a round trip = 30 minutes = 0.5 hours

    Time taken to load and unload the truck

    Though time taken to load could be more than the unloading time, I will take the average time of around 2 minutes to load and unload 1 cu. m. of mountain.

    Time taken to load and unload the truck = 63 cu. m. * 2 minutes = 126 minutes = 120 minutes = 2 hours

    Hence the total time comes out to be =
    (Number of trips required to move the mountain)*(Time taken to make a round trip of 10 miles + Time taken to load and unload the truck)

    16000*(0.5+2) = 40,000 hours = Around 1700 days or around 5 years

  • Rakesh Shivran Feb 9, 2013, 6:39 am

    Time to break the mountain + (Time to load+ time to move 10 miles*2 + Time to unload)*No. of rounds

    No. of rounds = volume of mountain/ volume space of truck

    Volume of mountain= volume of square pyramid = 1/3 volume of cuboid
    Side of base of average mountain = 200 mtrs
    Height of average mountain = 75 mtrs
    Volume of mountain = 200*200*75/3 = 1000000 cube mtrs

    Volume space of truck = length (6 mtrs)* breadth (3 mtrs)* height (3 mtrs)= 54 cube mtrs ~ 50 cube mtrs

    Time to break mountain = only time while breaking mountain to fill first truck is to be considered as other parts can be broken when truck is moving, so it will be negligible.

    time to load (assuming loaded by crane) = (volume of truck/volume of crane loader)*time per loading = {50/(2*2*2)}*1 minute=~ 6 minutes= 0.1 hour

    Time to move 10 miles = (10 miles/average speed i.e. 50 miles/hr.)*2 = 0.4 hour

    Time to unload = 1 minutes (assuming truck has in-built crane unloader)

    Total time = Negligible time to break + {0.1 hr to load + 0.4 hrs to move + 1 minute to unload}*20,000 rounds = 10,000 hrs + 20,000 minutes = 10,333 hrs

  • GVL Kasturi Feb 9, 2013, 6:41 am

    Here I am assuming that we break the mountain into rocks like in a quarry and carry the load.
    To calculate the time, we can say
    Time taken = (No. of loads required to carry the whole mass * no. of trips per load * 10 miles/speed of truck) + (Time to break the mountain mass) + (unloading + loading time)

    So, to calculate the number of loads needed,
    One load is limited by the weight the truck can carry

    Average mountain: say 50m high, 20 m in diameter and conical in shape (assumption)
    volume of the mountain = 1/3 * 50 * pi * 100 = 5000 cu. m
    Density of mountain mass
    A brick which is approximately 15 cm by 5 cm by 5 cm weighs 1 kg. Since the mountain rock would have higher density, assuming twice that of a brick, we can say
    density of mountain mass = 1 kg/ (15*5*5*2) cu.cm =1 kg per 750 cu. cm = 1000 kg/ cu. m (approx.)
    Weight of the mountain =volume * density

  • GVL Kasturi Feb 9, 2013, 7:03 am

    Here I am assuming that we break the mountain into rocks like in a quarry and carry the load.
    To calculate the time, we can say
    Time taken = (No. of loads required to carry the whole mass * no. of trips per load * 10 miles/speed of truck) + (Time to break the mountain mass) + (unloading + loading time ) * no. of loads

    So, to calculate the number of loads needed,
    One load is limited by the weight the truck can carry

    Average mountain: say 200m high, 400 m in diameter and conical in shape (assumption)
    volume of the mountain = 1/3 * 40000 * pi * 200 = 6 million cu. m (approx.)
    Density of mountain mass
    A brick which is approximately 15 cm by 5 cm by 5 cm weighs 1 kg. Since the mountain rock would have higher density, assuming twice that of a brick, we can say
    density of mountain mass = 1 kg/ (15*5*5*2) cu.cm =1 kg per 750 cu. cm = 1000 kg/ cu. m (approx.)
    Weight of the mountain =volume * density = 8 million*1000 =8 billion kg

    Now, coming to the truck,
    to get the amount of weight it could carry: atleast 20 ppl of 50kgs could be carried in the truck making it 1000 kg; taking an upside for concentrated mass, we could say 1200 kg per load per trip.

    no. of loads needed = 8 billion kg/1200 kg
    = 6 million trips (approx)

    Loaded truck would travel at say 40 miles per hour and 60 miles per hr when empty so on an average say 50 miles per hr

    So, our first term in the equation, will lead us to
    = no.of loads *2* 10 miles/ speed of truck =( no. of loads*2*10/50 ) hrs = (0.4 * no. of loads) hrs

    For the breaking of the rock and putting it back using cranes and other equipment, time taken will be negligible as it will be carried out simultaneously as well.

    Now, time taken to load and unload
    say unloading takes 10 mins
    loading however takes, maybe 3 mins per 40 kgs in a crane, that would make it 1.25 hrs

    In total for loading and unloading lets say it takes 1.5 hrs. to round off

    Now calculating the time required,
    = (0.4* no. of loads + 1.5* no. of loads)
    = 2(appx)*6 million = 12 million hours

  • Dion Feb 21, 2013, 5:41 am

    I would estimate 1.2 million days

    Let’s compare the mountain with a cone with base a circle with a diameter of 900 m and a height of 1000 m. The volume of this cone is comparable to the volume of a cylinder with as a base a circle of 300 m. So we would have to move approximately 280000000 cubed metres. I guess you can about 20 cubed metres a time. So that would be 14000000 trips. Every trip will take about 2 hours (including loading and unloading the truck). So 28000000 hours, which about 1.2 million days.

  • Saransh Feb 22, 2013, 8:41 am

    Estimating volume of the average mountain to be a cone of 4km height and 4 km base. V=(1/3)*3.14*64 km3=6*10^10 m3

    A typical truck has a rectangular load of dimensions 5*3*2m=30m3

    No. of turns required by truck=2*10^9

    Time to pick up load at deposit at destination = 10 mins=0.16hours
    Time to make two way journey from source to desitination=20miles/(60miles per hour)=0.33 hours

    Total time for each turn=0.49=0.5hours.
    Assuming a 40 hour working week, in one week the truck would do 40/0.5=80 turns=100turns.

    No. of weeks required to do 2*10^9 turns= 2*10^7 weeks

    Assuming 1 year has 50 weeks, total years=4^*10^5 years=
    400,000 years.

  • Jeanette Mar 3, 2013, 2:46 pm

    60 years

  • alex Mar 11, 2013, 7:18 am

    1 truck :
    2m*2m*2m = 8m3

    average size moutain, in average : 100m * 100m * 50m = 500.000m3

    => in 62500 times with one truck

    for each time : fill the truck, drive 10 miles, but it down, drive back : 1h.
    so we need 62500 hours.

  • AM Mar 17, 2013, 10:27 am

    4 million days

  • miao Apr 2, 2013, 6:11 pm

    The concept equation is
    number of hours = number of hours per trip * number of trips

    assumptions made
    (1) From visual imagination of the mountains I saw before, an average size of mountain is about 1,000 typical trucks. Thus it need a truck to run 1,000 round trips to move a mountain
    (2) Assuming a truck’s speed is 40 miles per hour. For a round trip of 20 miles, it will need 0.5 hour on the road.
    (3) Assuming it take 1.5 hour to load and dump.
    Total number of hours = 1000 * (1.5 + 0.5) = 2000 hours

  • Daphne Apr 12, 2013, 5:56 pm

    I estimate, based on an average speed of 50 mph, average truck volume of 10 cubic meters vs. average mountain volume of 1000 cubic meters, that it would require 100 back and forth drives (100 times 20 miles) to relocate the mountain – which would take, assuming no stops, breaks and sleeping time, roughly 40 hr.

  • vishal Apr 21, 2013, 9:47 pm

    I would like to approach this problem via simple maths. i would calculate area of mountain first. lets assume mountain as an equilateral triangle. with the help of formula 1/2 area of base * height we can calculate area to be relocated and now calculate the area of an average truck with the help of l*b*h. now by dividing both terms we”ll get no of turns truck has to make to relocate it. It comes out to be 7892100. Now calculate the toatal time taken. Lets say time taken for each loading to be 30 minutes and to reach the area where it has to relocated is 20 minutes and for unloading is 10 minutes. Total time for each single ferry is 1 hour. TOTAL time will be 7892100* 1 hours.

  • wen Apr 24, 2013, 5:39 am

    Suppose that an average truck car can hold 5 cubic meters of soil and the height of the mountain is about 1000 meters. The dimension shall be around 800 million cubic meters. This is equivalent to 160 million times of transportation(800 million/5). The driving rate of a truck car is about 70 km per hour. 10 miles equal to 16 km. It will take the truck about 15 minutes to arrive. 2*15 minutes*160 million=4 800 million minutes=9041 years

  • Mike Apr 25, 2013, 8:56 am

    The first step is to determine the area of an “average” size truck bed. I estimate the dimensions of average to be approximately 5ftx6ftx2ft=60ft(square footage). The next step is to determine the area of an average sized mountain. Using the 1/2(pi)(r^2), I estimated the height of the mountain to be 5,000 feet and the radius to be 5,000 feet as well. The area for half of a circle, or a mountain, would be 187,500ft(square footage). The last step is to divide the area of the mountain by the area of the truck bed. My estimated figures allowed to me to deduce that roughly 3,125 trips were needed to move an “average” sized mountain with an “average” sized truck, assuming that you are only using the truck bed to transport the materials and not the inside of the truck as well.

  • Mike Apr 25, 2013, 9:08 am

    Edit to previous comment: I didn’t answer the question with a time frame before.

    Assume that the truck is traveling at 30mph because there is a heavy load and it’s dangerous to drive on normal roads any faster with this load in an average truck. This would take 40 minutes round trip in traveling alone. Also assume that loading the truck (with a machine) would take 10 minutes and unloading the truck (dumping it or using a machine to pull the load off) would take ten minutes, this would be approximately one hour for one full trip. In my previous answer I calculated that there would be 3,125 trips which would mean that the transportation process would take approximately 3,125 hours or just a little over 130 days. This is assuming that the truck consistently makes trips throughout the time frame and is not including stops for gas.

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