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Time and Motion studies are well established.

Here is an easy question:

It takes 8 men 6 months to build 2 houses at a cost of $50,000.

It takes 10 men 5 months to build 3 houses at a cost of $100,000.

It takes 12 men 6 months to build 4 houses at a cost of $80,000.

How much does it cost for 15 men to build 5 houses in 8 months?

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    $\begingroup$ Where can I contract these men? $\endgroup$ – Ian MacDonald Nov 5 '18 at 16:06
  • $\begingroup$ @IanMacDonald; perhaps that was a bit ambitious, even for SE $\endgroup$ – JonMark Perry Nov 5 '18 at 16:09
  • $\begingroup$ I followed the link but didn't find much explanation about how such a puzzle work. Do we assume that the pace is proportionnal to the men * times, i.e. 1 man will do exactly as well in 2 months than 2 men in one month ? Is the cost an helping factor (you give me more money for better materials and I will work faster), or does it includes the men's salary ? $\endgroup$ – Evargalo Nov 5 '18 at 16:37
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    $\begingroup$ Can you confirm whether the cost is increased or decreased by each of the options. Obviously more houses is going to cost more, but does it cost more to have it done quicker because they work harder for more pay, or does it cost less because you're paying people's wages for less time? Also does it mean that more people costs more because you're paying more wages or does it cost less because they do it quicker? $\endgroup$ – AHKieran Nov 5 '18 at 17:22
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    $\begingroup$ @deepthought; 42 by any chance? $\endgroup$ – JonMark Perry Nov 6 '18 at 22:55
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Answer: $82,500 ?

I computed a system of linear equations. Would love to see somebody's logic without using linear algebra.

This probably makes no sense as per Evargalo's comment.. but here it goes:

data <- read.table(text="men months houses cost
8 6 2 50000
10 5 3 100000
12 6 4 80000", header=T)

lm(cost ~ men + months + houses - 1, data = data)

lm(formula = cost ~ men + months + houses - 1, data = data)

Coefficients:
    men   months   houses  
  57500   -35000  -100000 

(15 * 57500) + (-100000 * 5) + (-35000* 8) = $82500
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    $\begingroup$ I suppose the problem is not linear. Otherwise we could deduce than 4 men would build 2 houses for $30.000 instantly... $\endgroup$ – Evargalo Nov 5 '18 at 16:34
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    $\begingroup$ Also, 10 men would make 5 houses in 8 months for -$205,000 $\endgroup$ – Acccumulation Nov 5 '18 at 18:42
  • $\begingroup$ @Acccumulation, seems like an absolute steal to me! $\endgroup$ – Parseltongue Nov 5 '18 at 18:43
  • $\begingroup$ I ended with this result as well. It's the only way that makes sense to me. $\endgroup$ – akozi Nov 6 '18 at 15:42
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My Answer:

$96k

Reasoning:

Now that I'm typing this out I'm finding flaws in my logic so may redo it when I get home.
I decided that an increase in the number of houses would be a linear increase in cost (e.g. double the houses, double the cost).
With this I was able to do simultaneous equations to determine that a 50% increase in workforce means a reduction of 20% of the cost. (And thus, a 25% increase is a 10% reduction)
Using that, I made new simultaneous equations and calculated that a 1/6 reduction in time cost ~1.5 times more. Combining all these, I then adjusted the given values by the calculated amounts to create the situation in the question, and the resulting cost was ~$96k

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I'll go with

$125k

As @justforplaylists suggest,

You can assume they have a certain base rate and a certain overtime rate, and that's why it costs more to do it in 5 months instead of 6.

But:

double the first example and compare to the third

So it looks like

there is no overtime to do it in 6 months: rather, in the first example, they are slacking off and charging you. Bastards.

Therefore

scaling up the third example, 15 men could build 5 houses in 6 months for \$100k, but, if you give them 8 months, they'll overcharge by \$25k.


In more detail:

According to OP, for \$50k, you get 2 houses in 6 months using 8 men. So you'd think that, if nothing else, you hire two entirely separate groups of 8 men, to act completely independently, and pay each group \$50k, and each group makes 2 houses in 6 months. Right? For \$100k, you get 4 houses in 6 months using 16 men.
But according to OP, for \$80k, you get 4 houses in 6 months using 12 men. So it seems that dding extra men is increasing the cost by \$5k for each man, for nothing.
Therefore they charge \$5k for 6 man months (no overtime), at it takes at most 18 man-months to build a house. The base cost of a house is therefore \$5k, plus \$15k labour.
You can work out how much overtime costs to get it done in 5 months (looks like the rush rate is really steep compared to the base rate - like twelve times), but it doesn't matter, because the question is for 8 months, there will be no overtime.
So, for 15 men to build 5 houses in 6 months would cost \$100k.

So it follows that for 15 men to build 5 houses in 8 months would cost

\$100k

But we've already established that

if you give them more man-months than necessary, these guys will happily just charge you more for nothing. So in this case they're going to overcharge you 30 man months. 6 man months costs \$5k, so 30 costs \$25k

For a total of

\$125k.

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Answer: $78,125

Explanation:

I assume that workers are paid at a certain normal rate, and at a certain overtime rate. I assume there is no cost associated with the house itself because that would give 4 variables and 3 equations. We can convert the problem into man-months and thousands of dollars:

men months houses cost cost/mm h/mm
8 6 2 50 25/24 1/24
10 5 3 100 2 3/50
12 6 4 80 10/9 1/18
Then noting that the cost per man-month is different for the three provided examples, at most one can be below the overtime cutoff (in houses/man-month).
So the second two examples have to include overtime.
We can write the equation c/mm = a + b * (h/mm - c)
Subbing in the second two example and solving, we get:
b = 200
c = (a + 10)/200
Subbing into the first example gives a contradiction, so the first example must not include overtime. Finally, we notice that the problem also has houses/man-month = 5 / (15*8) = 1 / 24, so the problem doesn't use overtime and has a cost/man-month of 25/24. Multiplying by 75 man-months and the factor of 1000 we removed earlier gives 5^7 = $78,125.

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47.5k??
8x+6y+2z=50000
10x+ 5y +3z = 100000
12x +6y +4z =80000
deducing the value of x,y, z and putting it we get 47500.

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