I read this odd spot* some time ago: If the population of China was to walk by you in single file, the line would never end due to the rate of population growth.
I thought, “No. Really? Really?” I’ve looked up China’s birth rate in Wikipedia and with a population of 1,298,847,624 people and a growth rate of 0.57%, and assuming that growth is mainly due to births, then there are 14 new Chinese babies every minute, or one birth every 4.3 seconds.
Now, I reckon that in 4.3 seconds, about three people could walk pass me in single file. That means that if I wait long enough, then I will eventually get to the end of the line.
It would take a while and I suppose I can work out how exactly how long. Bother. It’s late but now I have to do it. Hey, this would make a good short answer problem in Unit 3 of Maths Methods.
Okay, it’ll take 58 years based on the assumptions above (and 365.24 days a year). If someone wants to check this, I’d appreciate it.
Of course, the growth rate 0.57% won’t be constant. China’s population profile is onion-shaped. I’m not sure what that means in terms of rate of population growth. The profile is probably dependent on the Government’s population policy too. Perhaps the current state of affairs isn’t all that significant in the grand scheme of the 58.4 years that I’m sitting there watching them all walk by.
*Girls will probably know this trivia already and also the ones that go ‘Until 1990, sausages were still legal tender in East Germany’ and ‘No piece of paper can be folded more than 7 times’.
China’s population onion (Source: NationMaster.com)
I totally overanalysed the problem and got a different answer. I did it the way I learnt in maths modelling last semester. Working is presented here in all its glory.
Let P(t) = population which has not walked past you at time t. t is in years.
Now, the derivative of P with respect to t, denoted as P’, is:
P’ = -{ amount of people walking past you in one year } + { population growth among people who have not walked past you }
If we assume that exactly 3 people can walk past you in 4.3 seconds, then that means one person walks past you every ( 4.3 / 3 = 1.4333 ) seconds. Per year, then, the amount of people who walk past you is:
1.4333 * ( seconds in a minute ) * ( minutes in an hour ) * ( hours in a day ) * ( days in a year )
= 1.4333 * 60 * 60 * 24 * 365.24
= 4.5230e+07 (scientific notation; e+07 means times 10 to the power of 7)
Therefore:
P’ = -4.5230e+07 + .0057P
Rearrange to give:
P’ – .0057P = -4.5230e+07
Write in D operator form (you can also solve in different ways, but this is the one I’m most comfortable with):
(D – .0057)[P] = -4.5230e+07
Multiply each side by e to the power of -.0057t, which is written as exp(-.0057t) to avoid confusion:
exp(-.0057t)(D – .0057)[P] = -4.5230e+07 exp(-.0057t)
Using the D operator theorem D[exp(at)f(t)] = exp(at)(D+a)[f(t)], we can write the equation as:
D[exp(-.0057t) P] = -4.5230e+07 exp(-.0057t)
Integrate both sides:
exp(-.0057t) P = 4.5230e+07 / .0057 exp(-.0057t) + C1
Where C1 is an arbitrary constant. Simplify:
exp(-.0057t) P = 7.9351e+09 exp(-.0057t) + C1
Now multiply each side by exp(.0057t).
P = 7.9351e+09 + C1 exp(.0057t)
We can remove the constant by using our knowledge that P(0) = 1298847624 . Note that exp(0) becomes 1.
1298847624 = 7.9351e+09 + C1
1298847624 – 7.9351e+09 = C1
-6.6363e+09 = C1
Therefore:
P = 7.9351e+09 + -6.6363e+09 exp(.0057t)
We want to know at what time t is P equal to zero; this will mean that everyone has walked past us. So, set P = 0 and solve for t.
-7.9351e+09 = -6.6363e+09 exp(.0057t)
7.9351e+09 / 6.6363e+09 = exp(.0057t)
.0057t = ln( 7.9351e+09 / 6.6363e+09 )
.0057t = 0.17875
t = 0.17875 / .0057
t = 31.359
My answer: it’d take 31.359 years for them all to walk past you. Phew!
I’d imagine the discrepancy between our answers comes from one of:
– you didn’t take into account that, as the amount of people who have not yet walked past you decreases, so does the amount of children they have; i.e., it’s not as simple as just adding one new person to the end of the line every 4.3 seconds.
– I made a mistake in my maths. Very possible… I’m pretty rusty. Blame the holidays.
– we’re both totally wrong.
Of course, this answer still ignores the changing growth rate over time. In any case, this problem was a good opportunity for me to brush up on my maths for the upcoming semester. Thanks, Joan 🙂
Oh, by the way, that was me. I accidentally chose ‘anonymous’ 🙂
Hahaha, wonderful. You have pleased me.
Admittedly, I have not checked through your maths. However, are you assuming that once someone has walked past me, they won’t have any more kids? I don’t believe this would be the case. I reckon they’d walk past me then go back to their lives and produce more kiddies.
The way I’ve done it, I haven’t cared whether people have more kids once they walk past you. I’ve just assumed that, regardless of whether or not they have kids, the kids don’t go and join the end of the line (it seemed most logical to do it this way – but, of course, it’s silly to try to apply logic to such a contrived thought problem 🙂
I suspect (meaning I can’t be bothered to do the maths this time) that, if you assume the children of the people who have walked past you will indeed join the end of the line, then you would get your original answer of about 58 years.
By the way, I know what you’re referring to when you say “Girls will probably know this trivia already”. I find that quite amusing.
Gasp! Who initiated you into the secret realm of Odd Spots?!
Yes, I did assume that the kids joined the end of the line, which seemed to me the intention of the bit about ‘due to the rate of population growth’.
Okay, so here’s a problem that my dad asked the family. Mum and Jason revealed how similar they are: they thought about it for two minutes, gave up then demanded an answer.
I revealed my similarity with dad. Like him, I plodded quietly along and got a workable solution after not long.
I know this blog has clever readers. I’m sorry, I don’t have any brain teasers more difficult than this not difficult one (except maybe the one about the sheep crossing the ferry…James?)
THE PROBLEM
There are ten bags of money. Each bag contains ten coins. Each coin weighs ten grams. However, in one of the bags, there are ten fake coins. We know they’re fake because they weigh 11 grams.
You have a weighing scale. You get one weighing (one reading) to identify which bag has the fake coins. How do you do it?
A guess: take a different amount of coins from each bag, then put them all on the scale at the same time. e.g., 1 coin from bag 1, 2 coins from bag 2, … all 10 coins from bag 10. Then the last digit of the weight tells you which bag contains the fake coins (if the last digit is 0, then it was bag 10). Correct?
Correct! Rohan, you are a thinking machine, brrrmm brrrmmm!
🙂
An extension to THE PROBLEM, for the thinking machines:
There are still ten bags of money, but now any number of them could have fake coins. You still get only one weighing. What do you do?
Note: For this version, I believe some bags needs to have more than 10 coins in them (at least, the solution I have in mind requires this). Let’s assume there are a large number of coins in each bag.
Is the weight still 10 grams real coins, 11 grams fake? If so, you can’t fool me! An answer sprung to mind straight away, because I’m so used to using bitmasks while programming. However, it only works if there are indeed lots and lots of coins.
Just double the amount of coins from each bag, i.e. 1 coin from bag 1, 2 coins from bag 2, 4 coins from bag 3, 8 coins from bag 4, etc. Then take the full weight. Then subtract the weight if all the coins were real (i.e., subtract 10 + 20 + 40 + 80 + … ). You should then be left with a number which is the weight of the fake coins. Write it in binary. Now the digits counting from right to left tell you which bags have fake coins, i.e., if the n’th digit from the right is 1, then bag n contains fake coins; if it’s 0, bag n contains real coins.
Have I overlooked anything?
Oops… I made a mistake. “You should then be left with a number which is the weight of the fake coins” – that is not correct; you should be left with 1 gram per fake coin. But that doesn’t affect the final answer 🙂
Yes, that’s the solution I had in mind. Imagine how many coins you’d have in that single weighing:
2^0 + 2^1 + 2^2 + … + 2^9
= 2^10 – 1
= 1023
Wouldn’t it be easier if we just weighed ten coins, one from each pile? 🙂