- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
I would have thought that the round trip would be theoretically the same. The speed lost flying into the headwind would be compensated by the speed gained on the reverse journey.
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
I agree that the ratio will remain approximately constant throughout the generations. A further downside to the edict is that it caps the female population which would likely cause a slow decline in the total population as every so often a couple will fail to produce a girl.
As to Q2. I do not know an optimal solution but believe that if each couple were to reproduce until they have one less boy than girls, then the proportion of women in the populace would gradually increase.
- Joined
- 18 September 2008
- Posts
- 4,041
- Reactions
- 1,185
Nope. It will always take longer if there is a constant wind speed in one direction over the complete trip than if there is no wind speed. You could take some theoretical air speeds and wind speeds and work out, but the reason I said that no math is needed is to use values that require no calculation (or minimal) and see where that leads.
If the Air Speed is A and the Wind Speed is W and the distance between Perth and Sydney is D, then in the case where there is no wind, the time to complete the return journey is 2D/A. In the second example, where there is a constant wind speed between Perth and Sydney, the total time is D/(A+W) + D/(A-W). Note that the divisor of all three elements (A, A+W, A-W) is the ground speed of the plane.
As I said you can work out for various combinations of A and W, but it should be obvious if you take some limiting examples logic will override the need for complex computation. Just imagine if the Wind speed were the same as the Air speed of the plane. Obviously the outbound trip to Sydney would only take half the time as previously as your ground speed is double what it was when there was no wind. But for the return journey, your ground speed is effectively zero (A-W) where A=W. So technically, once airborne, the plane stays still relative to the ground and it would take an infinite time to get back from Sydney to Perth.
Getting a bit more mathematical, it also should be obvious that the two elements to compute the time in the second return trip change at different rates, so that their sum is not constant.
When the Wind speed is zero, the out bound and return times are the same: D/(A+W) = D/(A-W) = D/A (as W=0)
But as the Wind speed increases from zero to the Air speed, the outbound time DECREASES by a limited amount, from D/A to D/(A+W)= D/2A. But the return journey time INCREASES at a vastly higher rate, from D/A to D/(A-W) which is infinity when A=W. So if it took 5 hours when no wind, then as the wind speed increases from 0 to the Air speed, the outbound time decreases from 5 to 2.5 hours and the return increases from 5 to infinity.
If you want practical proof, the next time you are at an airport with a long travelator, you can time yourself walking beside the travelator but not on it and then walking on it in the same direction and on it in the opposite direction. If you can maintain a constant walking pace and are not arrested, the sum of the latter two will be more than twice the former. In this case the travelator represents the wind speed and your walking pace is the air speed. It is also obvious if your pace is the same as that of the travelator (A=W), when walking against it, you get nowhere.
- Joined
- 18 September 2008
- Posts
- 4,041
- Reactions
- 1,185
I find these sentences contradictory. If the probability of a boy (or girl) is exactly 50%, then there will always be the same amount of boys and girls as keithj correctly showed.
Again not possible as keithj's example demonstrates. The answer to Q2 is that there is NO edict of the type given in Q1 that can change the ratio. If we assume for simplicity that each couple can produce one child per year, then if we begin with an equal number of males and females and the probability of a boy (or girl) at birth is exactly 50%, then after the first year 50% of new babies will be boys and 50% girls. This means the overall ratio in the population is still 50%, so at the beginning of year 2 we are exactly where we were at the beginning of year 1. So there is no edict you can give at the start of year 2 (which is what your Q2 solution implied) that could not have been given at the start of year 1.
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
I'm okay with your windspeed explanation, but I think I'm viewing things from a slightly different angle on the other problem.
If a couple produces an odd number of children there will always be an imbalance in the numbers of boys and girls. All the edict has to do is demand that each couple reproduce until that imbalance favours girls in their particular family and then stop. Or perhaps I'm misunderstanding something!?
If a couple produces an odd number of children there will always be an imbalance in the numbers of boys and girls. All the edict has to do is demand that each couple reproduce until that imbalance favours girls in their particular family and then stop. Or perhaps I'm misunderstanding something!?
- Joined
- 18 September 2008
- Posts
- 4,041
- Reactions
- 1,185
Cynic, if you list possible outcomes of babies born like keithj did, you would see that is is not possible to produce the imbalance you suggest with an exact 50% probability of a girl.
Think of what I said in my last response. At the end of each year we are back to square one, so there is no edict to use in year 2 that couldn't have been used in year one.
Your example. Starting point: #Males = #Females. At end of first year, all couples will have one baby, half of which are boys. They all now have an odd number of children and 50% of those will be girls. But #Males still = #Females. Those with girls stop producing (your solution) and remainder (those who had boys) have a baby the second year. Half of these will again be girls (so #Males still = #Females), but as all these couples have an even number of children now, they reproduce once more. Again you will have 50% girls with no change to the ratio. Each will now have an odd number of children, so those with two girls and one boy are to stop reproducing. But we are still in exactly the same situation as we were at the start of year 1, when those with 1 girls and 0 boys were to stop reproducing.
It's a bit like flipping a coin. The outcome is always 50% heads (or tails) and is never dependent on what came before. So if you have a thousand people flipping continuously, you can never change the probably outcome overall by telling some flippers that they have to stop flipping because of a particular sequence that they have experienced. At the end of the day, 50% of the flips, no matter who has done them should be heads and 50% tails.
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
19 minutes.
First A crosses with D taking 10 minutes.
A returns with flashlight in 1 minute then accompanies C across taking 5 minutes.
A again returns with flashlight in 1 minute then accompanies B across taking 2 minutes .
They are all now safely across taking a total of 10 + 1 +5 + 1 + 2 = 19 minutes.
First A crosses with D taking 10 minutes.
A returns with flashlight in 1 minute then accompanies C across taking 5 minutes.
A again returns with flashlight in 1 minute then accompanies B across taking 2 minutes .
They are all now safely across taking a total of 10 + 1 +5 + 1 + 2 = 19 minutes.
- Joined
- 17 August 2006
- Posts
- 7,811
- Reactions
- 8,429
That's what I thought unless there is some sneaky formula
My first thought however was .... A) is obviously fit if he can do it in 1 minute ..... He could just piggy back the other three guys across = 5 minutes + a bit of extra time to accommodate for exhaustion
A & B cross taking 2 minsThat's what I thought unless there is some sneaky formula
My first thought however was .... A) is obviously fit if he can do it in 1 minute ..... He could just piggy back the other three guys across = 5 minutes + a bit of extra time to accommodate for exhaustion
A returns in 1 min
C & D go across in 10 mins
B returns in 2 mins
A & B go across in 2 mins
Total of 17 mins.
This minimises the return times, and also minimises the time taken for the slowest 2.
- Joined
- 18 September 2008
- Posts
- 4,041
- Reactions
- 1,185
Correct keithj.
Alternatively you could swap when A and B returns and also get the same answer of 17 minutes.
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
What's a couple of minutes between friends?!
Congratulations again Keith.
Congratulations again Keith.
- Joined
- 17 August 2006
- Posts
- 7,811
- Reactions
- 8,429
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
3 links of one of the 3 link lengths need to be opened. Each link can then be used to join the remaining 3 chain lengths into a circle.
- Joined
- 18 September 2008
- Posts
- 4,041
- Reactions
- 1,185
Correct.
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
Q1. 5 cms
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
Q2. 20 sq cms
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
Q3. 10 cms
pixel
DIY Trader
- Joined
- 3 February 2010
- Posts
- 5,359
- Reactions
- 345
pixel
DIY Trader
- Joined
- 3 February 2010
- Posts
- 5,359
- Reactions
- 345
Q5: 35 cm^2
- Joined
- 25 February 2011
- Posts
- 5,688
- Reactions
- 1,231
+ 1 and congratulations on beating me to the tougher ones.
Similar threads
