I have absolutely no idea

So, calculators at the ready, and please round to the nearest whole number. “A sphere has a diameter of 2,160 meters. How many meters long is unit X if the surface of the sphere, measured in square units X, is equal to the volume of the sphere measured in cube units X?”


21 thoughts on “I have absolutely no idea”

  1. Google says…

    The volume of a sphere is defined as 4/3πR³

    The area is defined as 4πR²

    Equating both, we have: 4/3πR³ = 4πR²

    Dividing both sides by 4R² : (1/3)R meters = 1 Unit x “R”, in this case, is 1080 meters (2160/2)

    So, 1 Unit X is 1 third of the radius, or 360 meters.

  2. 360, innit?

    It’s what you get when cancelling the formula for the volume (four thirds pi r cubed) by that of the surface area (four pi r squared). The four, the pi, and an r-squared evaporate from each side and you get one third r left.

    D is 2160, so r is 1080, so one third r is 360.
    No calculator needed, and the “nearest whole number” bit is misdirection

  3. Yikes. This is what happens when you crowdsource math and then publish it on Forbes without verifying it.

    For the surface area to be set equal to the volume, you end up with (1/3)r = 1, or r = 3.

    r = 3 the only situation in which that statement is true. So when r is 1080, it simply isn’t true that the volume of that sphere is equal to its surface area. This is a trick question.

  4. But Nina Curley, r is equal to 3, because your unit is not the metre but X (=360m). It’s not a trick question and it’s not impossible to answer.

  5. Nina, it’s asking for what X- basically, a scaling factor- will produce the correct result, which is 360. Not knowing the general case, I actually did all the algebra with pencil and paper in two different ways and got the correct result.

  6. Why should I care? To anyone whoasks “have you tried the calculation?”, my answer is the the same. WTF should I care? Seriously, I’m interested. How can answering this question make me richer?

  7. Witchsmeller Pursuivant

    Wow. Perhaps educational standards haven’t fallen so far after all. This is a GCSE (Higher) level question. And I thought it was just the Guardianista arts graduates who had trouble with basic maths. Shame on you Tim.

  8. “The volume of a sphere is defined as 4/3πR³” – the volume of a sphere is not DEFINED as, the volume of a sphere IS , there is no defined about it.

    For instance 3 cats in a box is 3 cats in a box, definitions don’t come into it.

    I could define the above box as containing four cats, that would be a bad or preverse definition.

    I could even make it into law that there are four cats in the box, and prosecute anyone who behaves as thought there are 3 cats in the box, but is still remains that there are 3 cats in the box, regardless of definitions or laws.

  9. From Tim at Forbes: “They only want to invest with people who do have that sort of math knowledge. Although why they do I’m not sure.” Why would they want to hire only people with the maths knowledge of a fourteen year-old? You got me there.

  10. Witchsmeller Pursuivant

    Johnny Bonk – ” “The volume of a sphere is defined as 4/3πR³” – the volume of a sphere is not DEFINED as, the volume of a sphere IS , there is no defined about it.”

    Sorry to be a pendant about this, but “defined” means to state or describe exactly the nature, scope, or meaning of. This is a general definition, as in “The volume of any sphere is defined as 4/3πR³” and is describing the relationship between the volume and the radius.

    Perhaps you’d like to state the volume of this particular sphere, since you’re being so definite about it?

  11. I suppose Thiel is interested in investing in enterprises he can understand, which would be a geeky ones. And he’s using this question to eliminate non-geeks.

    The formulas are GCSE level, but you don’t actually need to know them if you understand the concepts.

    1) The volume of a sphere is proportional to the cube of its radius. (More generally, the volume of a solid of any defined shape is proportional to the cube of its linear dimension. To see this, imagine it as made up of a billion billion tiny cubes, each of side x, volume x^3.) So call the volume k.r^3 for some constant k.

    2) If you increase the radius by a small amount dr, the volume will increase by dr times the differential of k.r^3 with respect to the radius, which is 3.k.r^2 . But the increase in volume is the volume of the thin shell you’ve added to the sphere, which is dr times its surface area. So the surface area is 3.k.r^2

    3) So if the surface area and the volume are numerically the same, r=3 in the units being used.

  12. V = 4 π r³/3, A = 4 π r², plug in r = d/2 = 1080. Then solve V/x³ = A/x², x = A/V = r/3.

    It’s basically a restatement of the fact that the surface area (in square units) of the unit sphere is three times greater than the volume (in cubed units). It’s generally true for a unit n-sphere that V_n =S_(n-1)/n, so V_3 = 4 π/3. One interesting result is that the volume and area of unit hyperspheres doesn’t increase indefinitely as the dimension increases.

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