An interesting Maths problem to solve if you want to work at Microsoft

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    • An interesting Maths problem to solve if you want to work at Microsoft

      Good day to you all. First of all, I don't know if this has been posted here, so don't resent me if it did :P .

      Anyway, I saw this rather interesting problem on Quora some time ago. Apparently, one Indian (dots not feathers) guy who went to the job interview at Microsoft had to solve the following problem:

      "Hypotenuse is 10 cm long, and the perpendicular height from the opposing angle to the hypotenuse is 6 cm long. Calculate the surface of the triangle."

      This is the picture:


      Now, I wouldn't be sharing it here, but after I showed it to my friends, family and colleagues, I was surprised that most of them couldn't solve it. So let's see if anyone from here can.

      P.S. He didn't get the job.
      Startdate: October 2016
      Previous xp: OGame.ba UNI1 2006-2009, OGame.org: 2013, summer 2014

      Deut sold so far: 443.000.000
      Metal sold so far: 55.600.000
      as of 31st December 2017
    • It's not that it's not solvable, it's that such a triangle can't exist, since the height can't exceed half of the length of the hypotenuse. In other words, the height can't exceed 5 cm. I should have checked if that was the case, before writing the obvious answer of 30. :D
    • ^this. I've rechecked my math, the only way a triangle with the height of 6 and the given side proportions could exist is if the both smaller triangles are exactly the same. And in that case the total area of the big triangle would be 60 cm^2.


      EDIT: Which makes perfect sense, because Microsoft doesn't test basic geometry, they test your ability to deal with abstracts. The given drawing is distorted, so the majority spend a good hour or more crunching the numbers only to find them not adding up,whereas the minority simply state that the logic does not compute. It's what Bill Gates said: I'd rather hire a lazy person,because they find the easiest solutions.

      The post was edited 1 time, last by LegacyOfDawn ().

    • Blackmass wrote:

      not bad here is another for you how many threes co-exist in the squared quadrant of 7 ?? LOL
      Not being native english speaker nor writer, you're gonna have to provide me some links to that term Squared quadrants to see in the first place what is it X/ .

      Ohh, and that's right, Legacy, seems Microsoft wants us to think outside the box. A simple use of Thales' theorem puts you on the right track.

      Strider, if you happen to play in uni1, I can send you some ress :D :D .

      You can close the thread now, if something interesting comes out, I'll make sure to share it with you again :))
      Startdate: October 2016
      Previous xp: OGame.ba UNI1 2006-2009, OGame.org: 2013, summer 2014

      Deut sold so far: 443.000.000
      Metal sold so far: 55.600.000
      as of 31st December 2017
    • Warrior wrote:

      Blackmass wrote:

      not bad here is another for you how many threes co-exist in the squared quadrant of 7 ?? LOL
      Not being native english speaker nor writer, you're gonna have to provide me some links to that term Squared quadrants to see in the first place what is it X/ .
      Ohh, and that's right, Legacy, seems Microsoft wants us to think outside the box. A simple use of Thales' theorem puts you on the right track.

      Strider, if you happen to play in uni1, I can send you some ress :D :D .

      You can close the thread now, if something interesting comes out, I'll make sure to share it with you again :))
      A squared quadrant is a unit of measure that is depends on the unit of measure for the second number in a problem that revolves around basic space and volume.

      I'm not pretty I'm not graceful I am the inevitable fact of truth.there is no unwinnable situation.
      ogame.support.gameforge.com/en
    • From my knowledge i would add these:

      first of all there can be a perpendicular height from Hypotenuse even bigger then 6 cm, and the resulted triangles do not need to be similar

      second to find the surface=area you need to find the BA and BC(cathetes)

      the formula for the surface(also caled area) is Aof triange ABC= CATHETUS * CATHETUS/2

      I hope my memory is right.

      Also to find out the length of those cathetes BA and BC you need to apply other theorems. but is doable, i think, didnt solve it, maybe i will try to do it but not today

      Great King Pat wrote:

      For the record, no such scrap push hits have ever happened in ASSASSIN NOR would they be tolerated as long as i am leader. Everyone over here cares enough about the game to know right from wrong on this issue.

      Uni 1 will be like a prison & Urza will be the big black guy that wants to bum u in da ass. Avoid at all costs. ;)
    • i found the easy-est solution, having all the theorems on right triangle with an 90 degree angle

      first you need to note the intersection point of the height and hypotenuse with letter D

      then the area in a right triangle is = AC*BD/2, and in this case its 6*10/2=30

      Great King Pat wrote:

      For the record, no such scrap push hits have ever happened in ASSASSIN NOR would they be tolerated as long as i am leader. Everyone over here cares enough about the game to know right from wrong on this issue.

      Uni 1 will be like a prison & Urza will be the big black guy that wants to bum u in da ass. Avoid at all costs. ;)
    • Bibas wrote:




      first of all there can be a perpendicular height from Hypotenuse even bigger then 6 cm, and the resulted triangles do not need to be similar
      I need to scratch this almost non sense :D

      Great King Pat wrote:

      For the record, no such scrap push hits have ever happened in ASSASSIN NOR would they be tolerated as long as i am leader. Everyone over here cares enough about the game to know right from wrong on this issue.

      Uni 1 will be like a prison & Urza will be the big black guy that wants to bum u in da ass. Avoid at all costs. ;)
    • Bibas wrote:

      Bibas wrote:

      first of all there can be a perpendicular height from Hypotenuse even bigger then 6 cm, and the resulted triangles do not need to be similar
      I need to scratch this almost non sense :D
      Well, if it was 6 cm long, then hypotenuse must be at least 12 cm long :D . In this particular case, such a triangle is impossible to exist. I admit, this problem does force you to think outside the box.


      I gave this to a maths student to solve it, he came up with a quadratic equation which yielded negative roots, lol! :headbanging: :rofl: .
      Blackmass, heh, just tell the solution, please, I should look deeper into this :(
      Startdate: October 2016
      Previous xp: OGame.ba UNI1 2006-2009, OGame.org: 2013, summer 2014

      Deut sold so far: 443.000.000
      Metal sold so far: 55.600.000
      as of 31st December 2017
    • i was thinking at something else when i wrote that, only after i re-read my post i realized what i wrote

      i think that the outside the box thinking is about the missing connection point and the know how on the right triangle(with 1 angle of 90 degrees)

      i dont think they give you a very heavy problem at a interview, and ask for a super ridiculous long solution that ends in minus infinite limits :D

      but also who knows what they wanted to see on the person who solves this

      Great King Pat wrote:

      For the record, no such scrap push hits have ever happened in ASSASSIN NOR would they be tolerated as long as i am leader. Everyone over here cares enough about the game to know right from wrong on this issue.

      Uni 1 will be like a prison & Urza will be the big black guy that wants to bum u in da ass. Avoid at all costs. ;)
    • Bibas, you only need inner angles to get to the same conclusion as I have.
      The trick lies in the right angle at point B, lets call it Beta angle. Lets call the right angle on the hypotenuse the X angle.
      The sum of inner angles of a triangle is 180 degrees. If its a right triangle, it means the other two angles are 45 degrees each.
      Since the Beta and the X angle are both 90 degrees, it means the Alpha and the Gamma angles are also 45 degrees. For this to be true, both smaller triangles need to be exactly the same. That's due to the fact that if you change the sides as it is shown here, then the angles would change aswell, so the Beta angle wouldnt be 90 degrees anymore. That's the source of the distortion - the drawing is possible if the 90 degrees Beta is omitted, so the ensuing triangle would have all 3 different angles.
    • It stands to reason that Microsoft would utilize a logical problem via geometry versus a raw mathematical problem. Coding as a general concept doesn't use the rigid concepts of mathematics, and is a higher application of logical statements. Of course, the actual solution is an imaginary concept that only has applications in calculus and beyond.

      Trouble wrote:

      Sounds like an expert clicker to me :D

      Doc Brown wrote:

      I have read several of your posts elsewhere over a number of years. I find your mental state to be disturbing and you probably need professional help.
      What you write in the spam section doesn't have much impact on the game as a whole ... But I don't like to see you attempting to influence normal players in universe 1.
    • This isn't 1,2,3 simple it's 4,5,6 simple!


      There was a good one on the news from US SAT's


      If classroom with P amount of students has an average of 70
      & classroom with N amount of students has an average of 92
      How many people are in these classrooms if the overall average is 86


      What is the values of P & N


      Simple really math made complicated!
      Obscurium per Obscurious
      Uni 1 retired, Uni 35 retired(DMA)