Estimation Question

8/11/2019 Estimation Question

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Estimation Question

by Victor Cheng

by Victor Cheng

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I was asked the following management consultingestimation questionby a McKinsey interviewer manyyears ago:

Estimate how long it would take to move or relocate an average size mountain 10 miles using anaverage 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!

To post your reply Click Here.(feel free to use your first name only or your initials)

** Dont cheat by looking at everyone elses answer first!

If you want to know what acceptable answers look like, afteryou post your answer look for theresponses from:

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

{ 290 comments read them below or add one}

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dNovember 11, 2010 at 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,
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time to take them, move them to the other location (depends on the height, take average)

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MedicNovember 11, 2010 at 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 planrstructures 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 complicationsWe have 100 men working

The core peak has a square base of 30 * 30m area and a height of 1000 m. Hence the volumewould 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 mountainpicking up the soilTransportingoffloading at new destination

1 man can move 3.6 metric tons in 100 hoursHence 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.

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MedicNovember 11, 2010 at 9:39 am

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

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BinNovember 11, 2010 at 9:59 am

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

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BobbyNovember 11, 2010 at 10:08 am
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moves 10 miles in 36 seconds relative to the Earths rotations. One could also use the Earthsrotation round the Sun but im not the astrophysicist so ive stuck to what I roughly know. There areINFINITE answers to this question, therefore.

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ahmetNovember 11, 2010 at 10:14 am

My first assumption would be that we are thinking about mountains on the earths 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 ahill. 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 becomesespecially important to calculate an average. Another question is whether the distinction betweenmountains are colloquial or geological. There are mountains which are geologically separate butare 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, onehas to take into account mountains which are laid out on borders and are only partly within aregion.

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, othersare teeming with life. Further aspects in calculating the mass are water (rivers, lakes, undergroundreservoirs) and cavities (caverns, mines etc).

Taking all these factors into account, one can calculate a well defined average. The act ofcalculating 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 themountain 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, onecould also simply move the points of reference, thus changing the coordination system. This would

be moving the map instead of moving the mountain.

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ahmetNovember 11, 2010 at 10:20 am

what I left out, though it is implicated: once we know how much mass we have and how muchforce we need to move it, its simple to map that to number of workers / machines and thus timeneeded. 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)

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ReemNovember 11, 2010 at 10:31 am

I dont know if am right, but i will say it will take minutes, i will draw the mountain and i will rightdown its volume then i will just move it using my mouse on a scal drawing 1 mile or two or

whatever!!! if its not right i still got a point

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DennisNovember 11, 2010 at 10:38 am

Ive 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 trekbeamer 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 youll find a mountain volume of 33km^3Per session 100 trucks transport 23*10-9km^3 each = 23*10-7km^3Therefore 33/23*10-7 ~ 1.5*10^7 = 15 M sessions are requiredIn total 15M sessions*2hr = 30 M hrs are requiredLeading to 30 M hrs/ (300*24 hrs/year) = 30M/7200 ~4000 years

What does that mean: Id build a tunnel rather than moving the mountain

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JakobicekNovember 11, 2010 at 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 usingtrucks. The rest of the work is negligible. So we need to see how much material there is in anaverage 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 vaguedescription, but let us start with modeling an average mountain as follows 500m high, squarewith a side of 10km
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dimension 5m*5m*30m=600 m^3. Assume 100 cartrain, and there are two operating continuously. Realisticallyit might take 1 hour to transport between the sites, soassume a total of 10 loads per day.

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

Solution was typed out within a 5 minutes of seeing thequestion.

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all39November 11, 2010 at 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 quicklyenough such that trucks dont have to queue(4) sufficient earth moving equipment exists at new mountain site to assemble new mountain asquickly as the dirt comes in

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MedicNovember 11, 2010 at 12:15 pm

revising my solution earlier, I should have split the mountain into three upper third would take30o hrs, mid 200 hrs, lower 100 hrs for 3.6m3 per man.

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JavierNovember 11, 2010 at 12:37 pm

From the realistic point of view, the time that will take to relocate a mountain 10 miles, would thetime 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 wholedistance of eacutor(perimeter)/24h..from this calculation we have miles/h..so then we will knowhow many min takes to move itFrom 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..

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SeharNovember 11, 2010 at 12:38 pm

First some clarifying questions where is the mountain, in the middle of a flat plain, in a mountain


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range, get a sense of the geography. Are we taking into account the time for equipment to helpwith the move to arrive?Figure the mountain is attached at its base and must be removed from the base. We could eitherchop away at the bottom while having to make sure the mountain doesnt topple over, perhapshaving 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 takesto drill a mine, which from my work with offshore drilling can be a few weeks. Or use explosivedevices 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 wasstable and then assuming we could get enough vehicle power while taking into account safety, wecould move the mountain at a speed of 5 mph, or about in 20 hours. Or we could airlift themountain by army grade helicopter devices, which I would assume would be able to move themountain, taking into account safety, within a few hours. So the shortest amount of time to movethe entire mountain including preparation for moving, could be a few days plus time to getequipment to the mountain to a few weeks. Physical movement from the mountains origin point tothe point 1o miles away would range from a few hours to about 20 hours.

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GPNovember 11, 2010 at 12:38 pm

(i am using SI units) lets assume that an average mountian is 5000m high and that a average sizemountain is a semisphere with a radius of 5000m. then the volume of it would beV=4/3xpixR^30.5. assuming that the mountainous material has density of 20, its mass is m=dxVsince i am going to push it i assume i will exercise a constant force forever, resulting in a constantacceleration alpha = force/mass. i assume i can exercise force equal to my weight of 100kg, thus aforce of 1000N, thus i know the acceleration alpha=1000/mass.from physics we know thatdistance=0.5xalphaxtime^2, distance in our case is 10 miles, or about 15km (which givest=squareroot(2*distance/alpha)) which equals to roughly 25 years

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GrigoryNovember 11, 2010 at 12:50 pm

The answer depends who (number of people & level of skill) is doing the mountain-moving andwith what quarrying technology. If thats the kind of technology they had in the stone ages, whenthey built the Stonehenge, then very long indeed :). I will assume that were 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


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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 metrictons in the said mountain.

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ArthurNovember 11, 2010 at 1:22 pm

I would take a pyramid 4km long, 3km wide, 2 km high to model my mountain. Thats 12km^3.Lets assume we have 100 trucks to move rocks from the mountain. They can each move 3m^3 atonce, and take one hour to go 10 miles away, dump the rock and come back. So thats 300m^3 anhour. Assuming they work 8hrs/day, 300days/yr, thats 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

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ArthurNovember 11, 2010 at 1:25 pm

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

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AaronNovember 11, 2010 at 1:33 pm

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

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 miles4. 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 fillwithout delay the trucks (say 100 excavator for 400 trucks). The load process would take about 10minutes/truck.

3. A truck can carry about 25 m^3 and we have enough trucks in order to have always a emptytruck 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 (Im assuming that the time will not increase as the mountain heightincreases).

Actually we have to count only the load process and the average time to move the ground of 10miles 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
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m3 in 10^6 hours or about 100 years.

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SMNovember 11, 2010 at 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 roughlycone shaped mountain is somewhere near 3m m3.As I am playing God lets 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 theway back (unloaded), each journey would take one truck 30mins, plus loading time of 30 mins andunloading time of 30 mins (to allow for volume of trucks onsite etc). Also assuming excavators areworking and do not create any delay to loading, and time to load etc does not change with size ofmountain increasing or decreasing.Total journey time of 1.5 hours per truck, total of 300,000 journeys is a total of 450,000 hours.

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Xiaofeng DaiNovember 11, 2010 at 3:11 pm

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

I will segment the problem into the following parts. First is the about the mountain itself, i.e., howbig is the average size mountain, what does the mountain is made of (stone, sand or soil). Second isthe 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 theway during movement (such as weather, houses and land). The third is about the goal, i.e., what isthe precise number of miles, whether the goal is to recover the original look of the mountain aftermovement or just to load the materials to the destination.

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Anna G.November 11, 2010 at 4:31 pm

To estimate the time we have to estimate possible speed we can move the mountain by differentways. (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 examplehelicopter ; or we can pull it with trains; or push it by energy of people; etc depending on whatresourses 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 weneed to pick up the mountain

what speed it can fly when its fully loadedSo 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 helicopters loading (weight it has to take)and its speed we can count what speed we can have if use more and more helicopters.


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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.

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Sachin SharmaNovember 11, 2010 at 5:05 pm

Break the problem into the following components:

1. Time to break the mountain into moveable pieces2. Time to collect the pieces3. Time to transport4. Time to re-assemble

Assumptions:1. Flexibility of resources available to break, collect, move, re-assemble2. There may be loss of matter during breakage and collection3. How each step is performed will affect the total time (e.g. smaller pieces, easier to collect andmove, longer to re-assemble) estimate may not be optimal4. 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 time3. 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 cubeNo. 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

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Aleksey LeshchankinNovember 11, 2010 at 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.


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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 togetherWe need:

Rock-cutting machine Trucks Workers.Lets consider that the place under new mountain is already cleared away, the mountain flora and

fauna are transferred in parallel

Lets go on this way.

Lets 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 (includingmaintenance, wetting by water, etc.). Simultaneously 1000 machines a used for 24 hours/daywithout 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 truckdrives 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. Lets assume that there are 3000 brigades works for 8hours. 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 (Shipmentof a stone to the first day). Its about 68 years.

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Chandra AryaNovember 11, 2010 at 6:52 pm

Moving a mountain is a monumental task. So lets break it into manageable pieces that is cut it into

small standardized chunks. Once the pieces are made, they can be transported to the destinationand 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 pieces3. Transportation is available to move the pieces4. Optimal labor and equipment available to reassemble the pieces.5. Only road transportation is considered
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High level Activities:1. Cut mountain into manageable piecesa. 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 pyramid3. Mark the necessary pieces with some identifier4. Load into truck5. Transfer the pieces to destination6. 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 metersLet 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 metersTotal pieces = 1000,000/1000 = 1000 pieces

Time Estimations:

Cutting: assuming 5 hours per piece => 5000 hours for 1000 piecesMarking pieces: assuming 12 mins per piece => 200 hours for 1000 piecesLoading: assuming 3 hours per piece => 3000 hours for 1000 piecesTransfer: 10 hrs per truck and two pieces per truck => 500 truck loads, which give us 5000 hoursto 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 additionalhour to assemble the cut piece of mountain in its place. So if it takes 1 hour to place a piece inlayer one it would take 2 hours to place a piece in layer two.

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

Total hours: 21700

If we add inefficiencies of: 10% due to lost productivity etc, we get21700 * 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:


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

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victorNovember 12, 2010 at 4:30 am

Hi Everyone here are a few more tips on this case, 1) assume you move the mountain byTRUCK, 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 willask you to make an assumption or estimate the answer to your question.

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FANovember 12, 2010 at 3:27 pm

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

When I first read the question I had a strategy for finding an estimate and I began working it outwhen 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 orcartoon mountain? Is it a one similar to those we see in real life as part of nature (the ones wetarget 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 whichactually dictates a certain nature? If we can break it, is there a limit on the number of these piecesor 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 traileror 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 theroads 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 tocover this distance.4. What is the required speed that the truck needs to abide by? Does it vary or is it the same allalong 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 about16 Km.


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1. Estimate the total volume of the mountain. Suppose its radius is 3 km and its height is 1km. Thetotal volume will be 1/3*pai3^2*1,which approximately equals 20 km^32. Estimate the total amount of substance that an average truck can carry for one trip. suppose thesize of truck is 3*2*3 (m), so the total amount that the truck can carry for one trip approximatelyequals 20 m^3.3. Estimate how many trips the truck will have to travel in order to dig out the entire mountain. 20km^3/20m^3 equals 1000,000,0004. Estimate how long does it take for a round trip. Suppose it takes the truck half hour to load and i

takes the truck 10 minutes to drive 10 miles and it takes another half hour for the truck to unloadand then 10 minutes to drive back. So it takes approximated 1 hr 20 minutes for the truck to do around trip5. Calculate the time. 1,000,000,000 times (1+1/3) approximately equals 1,300,000,000, whichapproximately equals 50,000,000 days, which approximately equals 150,000 year.

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Hassan MuradNovember 14, 2010 at 6:28 pm

We can divide this task into two parts.

a) number of roundsb) 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 carryingcapacity of an average size truck. This would help us know how many rounds of a truck arerequired. (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 averagespeed of a truck which with distance can give us average time required in one trip.

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BPNovember 16, 2010 at 9:29 am

it depends on how long it would take to dismantle the mountain, piece by piece. you would need ateam of workers with various tools (picks, axes, etc) to essentially break up the mountain intosmaller rocks. then, the rocks can be placed into the truck and the truck will transport them the 10miles down the road. so, the rate limiting step isnt the truck transport, its getting the mountain

broken down into smaller, manageable pieces. this is actually a good allegory for knowing how totackle seemingly insurmountable problems: just break the problem/task down into bite-size pieces,

and soon, it will be accomplished/overcome.

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Rukshan MeegahageNovember 21, 2010 at 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 about5000ft.

What is the weight of an average mountain (This is where Im utterly confused but Ill give it a


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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, youll need 5 of him (I sound ridiculous!) i.e middle maintaining the height, and 4projecting from the central midpoint. This could equate to a BMI (remember the height ismaintained) 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 isnegligible.

Average speed of a truck: On a straight stretch of road with no traffic apprx. 50 miles/hr. And thespeed 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 perhour.

To travel 10 miles = 10 x 625,000 = 6,250,000 hours (Note I havent taken into account the time it

will take to start the mountain moving, to overcome inertia).

Therefore:

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BobbberNovember 22, 2010 at 2:44 am

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

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CalNovember 22, 2010 at 11:07 am

Based on the wording of the question Ive assumed that theres 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 metricas Im from Australia) and an average height of 2000m.This gives a volume of (1/3)x3.142000^3, which is approximately 8 billion cubic metres.

*I made assumptions about the average truck size and decided to go with 4m long, 2m high and1.5m 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.


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*I assumed that there would be a crew at each site to unload and pack the matter, and that they hadthe appropriate machinery.Therefore I assumed roughly 20 mins delay at each site. I approximated this to 0.66hrs, makingtotal 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 canonly afford one truck, they likely wont be staffing crew and lots of equipment at each site for80,000 years)

*hours of operation

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

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MaryamNovember 23, 2010 at 10:42 pm

Why would you want to do that?

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CalNovember 27, 2010 at 12:47 am

How did we do? :p

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MaxNovember 27, 2010 at 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.eassemble and disassemble times are negligible.

You can either make the calculations based on volume or on weight being the limiting factor forthe 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


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-Determine What is an average truck: Assume Volume/weight capacityo 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 dont know much about dump trucks or trucks in general, but Ive seen big truckstransporting 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/dayo 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 assumptionso speed=distance/time -> t=d/so 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 = coneEQ cone: (1/3)*pi*h*r^2 Define mountain

Define mountaino 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. Bymy definition it needs to be smaller than Mt. Everest (h = 9km) and bigger than a hill (highesmountain in Denmark is called a hill; h=0,3km).

Assume h = 1 km Assume r = 1 kmo 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 reallife, we can see that (Mountain weight) >> (other factors), leading to them being of very littleimportance. So the assumptions regarding the volume of the mountain and the density of the

product moved are the central ones.

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MaxNovember 27, 2010 at 5:28 pm


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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

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JessNovember 29, 2010 at 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 thenumber 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 formen to dig and move the earth: as the question seems to focus on the truck I would hope this is anok 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 Id go for anaverage 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 Ill 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, Id guess the depth of the loadingarea to be around 2m, the width around 3m and the length about 5m, giving a total volume of 30mcubed.

Dividing 200,000 by 30 gives 6666.66(recurring), which I will round up to 7000: this is thenumber 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 truckwill 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 isloaded, 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 themountain.

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


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

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@Jess:

Ah! I realise why this is so very small: mistake 1km cubed is not 1000m cubed, but 1000 x1000 x 1000 = 1 bn m cubed! Would re-calculate to fit this

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Jerry December 5, 2010 at 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 ofwork to replicate the original condition in the new site as moving the soil. Now soil part, like somementioned:

1)break2)load3)carry4)unload

I assumed an average size mountain to be 1km high, 2.5 km in base and cone shaped which gives1.3^2*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 100trucks 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 itshould 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

Ive seen some construction site and it took them approximately 1 hr to load the truck full of soilsand 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 replicatethe mountain from one location to another.


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M.Fisher December 6, 2010 at 2:51 pm

If all that I am given to move said average size mountain is an average sized truck the problem isnot 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 dumptruck, heavy construction truck)5. What is the scope of this mountain and its directive to move it? Are we taking it down to sealevel from its current delineation of elevation or simply until we might classify it as part ofanother set, such as a foothill?

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

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Min December 9, 2010 at 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 rocks2. assume rocks are desired and only one truck; assume ave. mtn. weigh x tons and measure ycubic 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 rocks3. time to blow up the mtn, assume having super-dynamite, D minutes;4. time to load&unloadA minutes per load, B mimutes per unload

N number of rounds = max(x/m, y/n)total= (A+B)*N5. travel timeassume there is a road, but narrow&winding, C minutes one waytotal =2*C*N6. assume 12-hr per day operation(not safe to drive in the winding road at night), total days neededto move the mtn= (D+(A+B)*N+2*C*N)/60/12.

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SA December 11, 2010 at 9:44 pm

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

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BH January 1, 2011 at 11:01 pm


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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 backand 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 timetaken 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 byexplosives 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.

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Mar January 8, 2011 at 9:52 am

Considering an average size truck needs 2 hours to drive 100 mile, in a flat road with very fewtrafic (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 hurtanybody, or ruin anything. To accomplish this successfully the mountain would also have to have 4wheals to make the transporation easier.

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ariel January 22, 2011 at 1:48 am

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

Volume of mountainAssume cone shapeLength 6kmWidth 6kmHeight 5km higher than surrounding area (bit tall for average, but its a mountain, not a hill, and skimped on the length and width)


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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 volumeLength 3mwidth 2mheight 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)

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ariel January 22, 2011 at 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: 15millionTime required: 15million hoursor 7500 years (regular work week: 40hrs/wk)or 3750 years (double shift)

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Sunanda January 26, 2011 at 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 mountainneeds 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 dirtthat 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 trucko When I moved last, we took an average truck and it could take 2 6 ft sofas lengthwise with room


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to spare so I am going to guess its 20 foot long. I also noticed that 6 ft man could stand straight init 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 followingo 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 fillthe 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 mountainousarea!) I am going to assume a speed of 20 mphso 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 thetrips

o Any rest stops that the driver will take and the length of a work day. I am going to only considerwork 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 workstirelessly

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

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Sunanda January 26, 2011 at 9:22 pm

Sorry: 143,883.18 yrs

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vishal February 6, 2011 at 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 feasibleaccording 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.

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Daryl February 15, 2011 at 9:47 am


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Working in square feet, lets imagine an average sized mountain is 2000 ft tall, with an equalradius (thus, over a mile around). If its a cone, the area will be 1/3 x base x height = 1/3 * [ ( 3.14x 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 thearea would be about 100 square feet. Therefore, well need 80m trips to move the mountain. If weload the truck in an hour, unload it faster in just 30 minutes, and average around 45 miles/hr thereand back (15 min there, 15 min back), the whole trip will be 2 hours. In total, well need 160m

hours, or 18,000 years to complete the task.

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marie April 18, 2011 at 3:18 am

Assume average size mountain would be 2000 mtrs high, with a base of 1000 m. So the totalvolume would be 2*10^3*pi*500^2=500*10^6*pi=1.5*10^9 m3Assume 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 thecomplete mountain.

Traveling 10*2=20 miles each time the truck drives makes a total of 30*10^6*20=600*10^6 milesAssume an average truck drives 60 miles an hour > 600*10^6/60=10*10^6 hrs is the time it willtake 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 millionhours for the truck to transport the mountain

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Can April 21, 2011 at 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 whenloaded and at 80km/h when empty, so one trip of load takes appr. 20min. therefore80bil roundtripsare needed which makes appr. 14,000 years! : )

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BBM May 9, 2011 at 2:58 pm

Assuming only moving time is concerned and the average speed of the truck is about 40miles perhour, 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.e9Time ~ 30min*1.e8 ~ 5000years

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Rahul Priyadarshi May 10, 2011 at 4:06 am

height of mountain 1000mradius 20m


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Volume pi*radius2*height/3 = 1 million mtere3density 1 ton/metre3weight 1 million tonnescapacity of truck- 10tonnesspeed of truck-40miles an hourtotal time taken-50000hrs

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BKD May 11, 2011 at 4:34 am

Assumptions about mountain:-Height = 2000mArea = 500 km^2Shape = conicalRock 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 tonnesAv. speed = 40 mi/hrOperation = 16 hr/day (2 8-hour shifts)Breaks/driver change = 1 hr/daySo, effective operation = 15 hrs/day

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

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

So, it would take about 60 000 years!

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Achlesh Sood May 12, 2011 at 2:42 am

Size of mountain: Assume It is in shape of pyramid. with radius of 300 metres. Height 2000metres, 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.


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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 TLoading 50 T material takes 1 hTravelling 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

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Nicholas Luby May 16, 2011 at 9:04 am

The tallest mountain on earth is about 29000 feet tall. Small mountains have incline requirementsto be considered mountains, taller mountains have a height minimum (I think 9000 feet?).

Otherwise its just a big hill. Also, Im guessing the distribution of mountains is skewed right, witha big fat right tail bringing up the average, so Ill go against my gut instinct to use a very smallheight of a couple thousand feet. So Im going to say height is 10000 feet, the mountain is shapedlike 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, soloading time is neglible.

The truck goes 60 mph for 10 miles which takes 10 minutes. Then it has to drive back another 10minutes, 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) butim trying to do this in my head.

So a long time. Lets teleport that sucker.

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Nikhil M June 15, 2011 at 5:28 am

Assumptions1) Avg size mountain has a height of 100 mts2) shape of the mountain is that of a Pyramid3) 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^2Volume of Pyramid = 1/3* 100* 100*100Total mass of pyramid =( 100*100*100/3)*3=100*100*100


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5) Avg truck carrying capacity is say 100so number of trips the truck would have to make = 100* 100now say in each trip truck would take 4 hrs ( 2 for loading, 1 for unloading, 1 for traveling 10miles)so total time required = 100*100* 4=40000 hrs

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TA June 28, 2011 at 8:26 pm

16-17 d

Avg Size Mountain x 10 miles/ Avg Size TruckAvg Size Truck-Can move 2 couches, 100 lbs each200 lbsAvg Speed of Truck-40 mphPounds/Hr-Truck= 200 lbs*40 mph/10 miles= 800 lbs per hrKg/Hr-Truck= 800/2.2 = 800*5/6 = 133*5= 665 kg / hr ~ 6.5 * 10^2 kg/hrAvg Size Mountain-ht 500 m, r 500 mVolume = 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^3Stone 2x more dense than water2000 kg/m^3Weight Avg Mountain = 125*10^6 m^3 * 2 * 10^3 kg/m^3=250*10^9 kgTime to Move Mountain with Truck = (2.5 * 10^11)/(6.5*10^2)= 25*10^8/6.5 ~ 4*10^8 hrs = 400mill hrs400*10^6 / 24 = 16.67 d

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TA June 28, 2011 at 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 = 400mill hrs400*10^6 / 24 = 16.67 *10^6 d = 1667/365 * 10^4 yrs ~ 2.7 * 10^4 yrs = 27000 yrs

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RS June 29, 2011 at 10:58 am

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

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

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1. Estimate the volume of average mountain:Mountains are usually kinda cone-shape, so lets 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 loadAssume the average speed is 50 miles/hourThe loading time is 20 minutesThe unloading time is 10 minutesTotal time for each load is 60X10/50+20+10=42 minutes

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

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AOA August 8, 2011 at 1:43 pm

Estimating:1) a mountain have 10,000,000 TN2) a Truck load can move 10 TN3) the time to load, move and unload the truck is about 2 hr

10,000,000tn / 10tn, truck load = 100,000 truck loads100,000 truck loads/ 4 truck loads, day = 250,000 days

Moving a mountain to an nearby location will take approximately250,000 days

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Pietro August 23, 2011 at 7:33 am

If we assume the mountain to be a cone, high 2000 metres, and large 14000, its volume is28000*pi = about 90000 m^3. If we assume that an average truck can load 3*2*1.5 = 9 m^3, itwould 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 newmountain, and half an hour to unload it, each travel is 2 hours long. Therefore the total amount oftime is 20000 hours = 3 years, 238 days, 8 hours. That is, almost 4 years.

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Pietro August 23, 2011 at 7:55 am


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

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ChrisF September 5, 2011 at 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).

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KakarrotNovember 1, 2011 at 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 waters which is d=1So we can suppose it is roughly d=2tn/m^3

We can suppose that a mountains figure is a cone. So V= * r^2 * h

An average mountains 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 timesSo it needs 31400000000 hours= 3584474 years

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ArtemNovember 27, 2011 at 11:09 am

Lets assume, that we can explode the mountain, thus time for breaking the mountain into carryablepieces 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 mtime 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

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v December 14, 2011 at 3:43 pm

Volume of Mountain: 1,000,000 cubic metersVolume of truck: 20 cubic meters# of one way trips: 50,000Trip time:10mp for 10 km 1 hourTotal one way trip time: 50,000 hours (A)Total Return trip time: @20mph: 25,000 hrs (B)Total Trip Time:75000 HoursDwell time per trip on each end: 5 minutes (Load Unload etc)

Total Dwell: 5*2*50000= 500,000 minutes= 8500 hours say 9000 hoursMountain Breaking and assembling time: 0 minutes assume that work is done while truck is intransittotal time: 75000+9000 hours=84000 hours= 3500 days

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Viddu December 31, 2011 at 1:54 am

300 days

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Gaurav January 7, 2012 at 1:11 pm

I assume that an average mountain is a 1 km long, 200 m and 50 m wide. It tapers uniformly as wemove up its height. So that gives me a volume = .5*50*200*1000 cubic meters = 5000000 cubicmeters.

Now I would estimate the volume that an average truck could carry. I assume that loading portionof 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 truckis 20 miles/hr. So for an up and down trip of 20 miles (10 miles up and 10 miles down) it will take


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an hour. It will takes about 30 min to load and 30 min to unload. I will assume that mountain is cutand the truck does not have to wait just because the mountain is uncut at any moment. So to carry37.5 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 totransfer the whole mountain would be 5000000/450, i.e. 11111 days.

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Elise January 7, 2012 at 8:12 pm

Suppose the mountain consists of solely stones and soil, moreover, it has a cone shape. Assume theradius of bottom is r=25m, and the height is h=300m. Using the formula V= (Pi)x(r^2)xh , we havethe volume of the mountain being 3.14 x 625 x 300= 588,750m^3. Lets approximate again thecapacity of a truck is V=length x width x height = 10 x 5 x 10 = 500m^3. Suppose we have 100labor, it will take them 30 minutes to fill up the truck at one time, now 10miles=16.09km. Usingcommon sense, it takes me 1.5h to get from my apartment to CDG airport in Paris by bus, thelength is 29km. The time is roughly 16.091.5/29=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 itd take1178130/60= 2553 hours. The average working hour per day is 8, so 2553/8 = 320 days

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Anne January 8, 2012 at 12:23 pm

First we need to determine how much material there is in the mountain and how much the truckcan hold. Lets assume that an average mountain is approximately cone-shaped and is 1,000 feethigh 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 perhaps500 square feet of material. [Note: I chose these numbers because they are relatively easy to workwith but they could be way off -- I would verify that my assumptions are reasonable with theinterviewer.] 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 dont know how many people we have doing the loading lets assumethat we have a big crew and they can load up the truck in 1 hour. Since the truck is driving to amountain, its probably not on a highway, so Ill assume its going 40 miles an hour and can makethe 10-mile trip in 15 minutes. Unloading the truck would also take 1 hour, and traveling backwould 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.)

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Jap January 9, 2012 at 3:38 pm

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

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


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The first element; the # of trips can be calculated by dividing the size of the mountain by the sizeof 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 lets look at each of the two main components individually to get an estimate.

1. # of trips necessary to move the mountaina. Size of truckLets say the bed of the truck has a volume of 5m*3m*2m = 30m^3. For efficiency, we will pile uphigh and get about 50m^3 of earth into the truck.

b. Size of the mountain.Lets assume the mountain is a cone shape. The height is about 1400m and the radius is therefore1000m. The volume of a cone is 1/3*pi*r^2*h, which is 1/3 * 3.14 * 1E6 m^2 * 2E3 m. This isabout 2E9 m^2.

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

2. Duration of a tripHere we have to make a couple of strong assumptions. First of all, Im going to assume that thenew location of the mountain is a little under an hours drive away from the old one. I am going toassume 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 workdayThe 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

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Djordje January 10, 2012 at 6:58 pm

Lets assume the following:average size truck:Load 8m3speed 30 mph

average size mountain:Piramid shapeheight 1000m

basis square 1000m x 1000m


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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 minLoad + unload + go to + go from = 10 + 10 + 20 + 20 = 60 min = 1hTime needed is one hour per one truck load

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

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Tara January 22, 2012 at 6:59 am

Ive tried to do this in as few moves as possible. As no follow up questions are allowed, and littledetail is provided (and also from reading the above responses) it is clear that endless variations inassumptions are available, and is it normal to to sit in front of an interviewer for 15 minutes insilence 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 informationavailable, I assumed that anything we havent been given details on is outside scope, and all that isat 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.

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Sudhanshu February 24, 2012 at 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 triangles2. Estimate the size of an average truck and estimate the volume it can transport. Volume would bein iterms of rectangles or triangles

Make assumptions:1. Assume an avg speed of the truck2. Assume the time of loading of the truck3. Assume there are no roadblocks in the path of the truck


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4. Assume there is enough resource availability

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

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Sudhanshu February 24, 2012 at 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 truckstransportation and back

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Jermia March 5, 2012 at 9:00 pm

Its only 10 miles. So you can move the mountain just by moving part of its bottom. Lets cut themountain into horizontal layer, and then shift the most bottom layer 10 miles to any direction. Thelayer shape would be cylinder. All we have to do is to move half of its side to the opposite side.Lets guess the volume of such part. Lets say the mountain have r=500m. Half of the cylinderlength 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 lets guess the volume of a dump truck, lets say its 5 m * 3 m * 1.5 m = 22 m3.It depends on how many trucks do we have. Given 10 trucks, itll take 714 times to move the soilvolume.If one truck can move back and forth 10 times in one day (include load and unload, and smoothroad, and rest time we have relief crew, itll take 71 days to move the mountain 10 miles.

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MA March 21, 2012 at 10:32 pm

Average size of the mountain = 1000^3 ftAverage size cars carrying load = 5^3 ftDistance for one trip = 10 miles * 2 (for return trip) = 20 milesEstimated speed = 60mphEstimated time to travel for one trip = 20 minutesTime to unload = 10 minutesTotal 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 tripsEach trip takes a half hour, so it would take about 100 hours of consistent work. Factoring in 10hour work days, it would take 10 days to complete

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Duncan March 28, 2012 at 9:59 pm

730 days

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Paulina April 2, 2012 at 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, its about2500000m 3. he average truck has dimension: 10x3x2=120m3. Then, we know that we needaround 21000 trucks. To load and unload the truck we will need 20 minutes and to go back andforth(so its 20 miles) we need 40. So its one hour for one truck, hence its 21000 hours/24=875days = 2 years and 4 months (rounded)

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marie-paule April 8, 2012 at 8:55 am

Assume a truck can charge a max of 5x3x2= 30 m3Assume 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. 1BAssume material to be moved is prepared while truck travels back and forthAssume truck takes 1 hr to load, travel, unload, travel back

> 1,000,000,000 hoursAssume 10 hrs work per day> 100,000,000 days300,000 years!

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Leonardo May 1, 2012 at 6:31 am

Each two-way trip: 20milesAt 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 avolume of 1000 cb ft, it would take 1cbmi /1000 cbft *5280^3 cbft/cbmi = 125 *30min= 3750 min

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Patrick May 18, 2012 at 4:14 pm

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

wheretime 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 / trucks loadcapacity

-1. time for one trip = time to load a normal size truck + time to move 10 miles with full load + timeto unload a full truck + time to move 10 miles without load

1.1 The truck can be either loaded manually or automatically using assisting specializedmachines/vehicles. A normal truck (e.g. a pickup truck) has loading capacity of approximately3m*2m*1m=6m^3. Lets suppose on average it will take about 5 minutes to load the truck.


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

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 goback to the original site.

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

2.1 Suppose the mountain is pyramid-shaped with a square base. The volume of the mountain = theheight*the area of the base*(1/3). Lets suppose the typical mountain to be 200 meters tall, and 200meters wide on each size at the base. so the volume is 200m*200m*200m/3=8000000/3m^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 milliontimes = 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 takeone week off per year, so lets say 50 weeks per year, then that is 300,000/8/5/50 year = 150 years.

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Dave May 19, 2012 at 4:31 am

Lets assume a cone shape for the mountain:h = 2000 m averager = 4000 m average

Thus the volume weve 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 aconstraint. Lets assume the max weigth is about 30 tons, then the number of loads the truck needs


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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, lets say 50miles/h, than the outbound, lets 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 downto 200 because we spend time also for loading the truck and unoading it), the time it takse torelocate the mountain is:T = t/(200*8) years = 1 000 000 years (!!!)

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