Well, if T is total time to build, D is the time that can be distributed equally among any number of workers, and C is constant, indivisible time extra time that goes along with construction, and X is the number of workers, then:

T - C = D / X

so, since T is 12 and 6 is half of 12, then:

T/2 - C = D/X * 1/2

T/2 - C = D/2X where X > 0, C = 0, T=12, and D = (T - C) / X

which is both the answer it's looking for (twice as many workers) and the correct answer (it depends on at least two things we don't know), while assuming what they're assuming, which is C = 0

(Stupid ass junior high math problems piss me off, junior high is a traumatic experience)

[–] jj4211 2 points 3 weeks ago (1 children)

Well, arguably still "incorrect" in real world terms since it fails to have an adjustment for divisibility of D as a function of how many people. If theoretically a task is "perfectly divisible" at two people and halves the time, it will not be the case that a million people will cause it to happen in one millionth of the time. Improvement by expressly pointing out "C" and declaring your assumption of zero for math to work. Also assumption than for any increment of X, the time impact is equal.

In math this is pedantic, but it sure impact project planning in very disastrous ways, and business people love to assume C is zero, any change to X is linear and with linear impact, and make embarrassingly bad calls as a result.

[–] [email protected] 1 points 3 weeks ago* (last edited 3 weeks ago)

yup, but that answer was based entirely on the assumptions present in the question. D is all divisible work, and C is everything else, because that's literally all you can assume to make the math work. D has to therefore be 12 months worth of divisible work minus C. C could very well be 12 months of work, meaning D is zero and adding more workers won't matter.