Well, why not. First person to, blah blah yawn.
1) You have a working analogue watch with a second hand. How many times a day do all three hands overlap?
2) You have a 4 minute hourglass and a 7 minute hourglass. How do you measure exactly 9 minutes?
3) In the land of Nevernever, all parents only want sons. Every family keeps having children until they get a boy, then they stop. What's the proportion of girls to boys in Nevernever land?
4) Using standard arithmetic signs, make this equation true:
3 1 3 6 = 8
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1. Don't understand.
2. With the working analogue watch you gave me in 1.
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No takers eh? Perhaps I should have added these were interview questions at Google.
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The answers are freely available if you google. YOu can find most things on the web. Was going to cheat and claim the answers as my own work but thought better of it!
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1/ Answer - all my own work - Honest.
22 times a day if you only count the minute and hour hands overlapping. The approximate times are listed below.
2 times a day if you only count when all three hands overlap. This occurs at midnight and noon.
am
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
pm
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55
A really simple way to see this is to imagine that the two hands are racing each other around a track. Every time the minute hand 'laps' the hour hand, we have the overlaps we want.
So, we can say that the number of laps completed by the minute hand every T hours, Lm = T laps. Since there are 12hours in a full rotation of the hour hand, that hand only rotates Lh = T/12 laps.
In order for the first 'lapping' to occur, the minute hand must do one more lap than the hour hand: Lm = Lh +1, so we get T = T/12 + 1 and that tells us that the first overlap happens after T = (12/11) hours. Similarly, the 2nd lapping will occur when Lm = Lh + 2.
In general, the 'Nth' lapping will occur when Lm = Lh +N, which means every N*(12/11) hours (for N = 0,1,2,3...). In other words, it will happen approximately every 1hr5mins27secs, starting at 00:00. In 24hours, this occurs a total of 24/(12/11) = 22 times.
==============================================
The same thing is explained below in a slightly less intuitive manner, using angular frequency:
==============================================
So we are looking at two rotating hands. Ultimately, its just the angles we care about. Let θH represent the angle of the hours hand and θM represent the angle of the minutes hand. You could also introduce the seconds hand but that makes the problem more complicated. For now, lets assume the question only cares about the minute and hour hands. Initially we might think we are looking for:
θH=θM
But this doesn't take into account that if one hand has "gone around" a few times, its angle will be different from a hand in the same position that hasn't "gone around" the same number of times. So we have to modify our goal. Instead we let the angles differ by an integer multiple of 2π (360°). Let us call this arbitrary integer z. Now our condition is:
θM-θH=2πz
You could subtract the two angles in either order but the reason I chose to subtract hours from minutes is because it will result in positive integers which is just simpler. The minute hand goes around more times, thus its angle is bigger, thus this order of subtraction is positive. Now we have to find out how these angles depend upon the time. Let us call our time t and measure it in hours. I omit units for simplicity. The hour hand goes around a full rotation (2π) once every 12 hours. So:
θH=(2π/12) t
For those more versed in mathematics, 2π/12 is the "angular frequency" for the hour hand (usually denoted by ω).
Similarly the minute hand goes around a full rotation (2π) once every hour. So:
θM=2π t
Plugging back in:
θM-θH=2πz
2π t - (2π/12) t = 2πz
t - t/12 = z
(11/12) t = z
Now we are ready to solve. The two hands overlap at every solution of this equation, so we want to know the number of solutions of this equation. But remember, we want to know how many times this happens in a single day, so t cant be bigger than 24 (remember we are measuring t in hours), and technically no smaller than 0 (assuming we start our clock at 0 hours). Since t and z are proportional, each solution for z corresponds to exactly one solution for t, and accordingly exactly one solution of the equation.
Also, remember than z must be an integer. So if we wanted all the times we would just let z go from 0 (when t=0) up and solve for t and stop as soon as we passed t=24. Then of course we'd have to convert that into hour and minute format. However, we only care about the number of times this happens. So we can notice that as t increases, z is just keeping track of how many times the two hands have overlapped. When z=0 we get the first time, when z=1 we get the second time, and so on. Since t and z are directly proportional, t increases with z, thus z represents the number of times the hands have overlapped up until time t minus 1 (and starting from t=0). Since we don't want t to go past 24, we plug in 24 and solve for z which will tell us how many times this event has occurred from t=0 to t=24 (one day).
(11/12)*24 = z
22 = z
So this happens 22 times in a day. Technically this has 23 solutions (0 through 22) but the last one is for t=24 which has begun the next day. If we don't count that solution we are left with 22.■
If we want the second hand to overlap as well, we have to go a bit further. First we note that the second hand makes a full rotation once every minute, thus 60 times an hour. From this we have:
θS=(2π*60) t
We want the second and hour hands to overlap AND the minute and second hands to overlap. Those conditions can be summarized as follows, where x and y are positive integers:
θS-θM=2πx
θS-θH=2πy
Plugging in our functions of t for the θ's and solving for t we are left with:
t=x/59
t=12y/719
We want our integers x and y to produce the same time (making all hands overlap at that time). So we want to set the two equations equal. Simplifying, we get;
x=708y/719
708 and 719 are coprime (719 is prime and 708 is decomposed into 2^2*3*59). In fact 708y and 719 are coprime except for when y is an integer multiple of 719. Thus 708y/719 can only be reduced when y=719k for some integer k. In this case we have:
x=708k
The first solution is when k=0. Then x=0 and t=0 corresponding to midnight. The next solution is k=1. Then x=708 and t=12 corresponding to noon. The next solution is k=2 but this corresponds to t=24 which is (midnight for) the next day and due to the direct proportionality of t and k, every k from here on up will produce t's higher than 24.
In summary, all three hands only overlap twice a day: at noon and midnight. ■
All of this assumes that the hands sweep continuously. So the math is more(?) complicated for those with fake Rolex's (or any ticking handed clocks).
Last edited by: Zero on Tue 1 May 12 at 10:39
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I wonder why these questions are asked in interviews. In real life, people seldom face these problems.
Even in Google (and Microsoft, Oracle, Apple etc.), most people perform very mundane tasks, not much different from any other companies.
I understand these questions might encourage candidates to think differently but more often than not, the employer is looking for someone who will just obey the orders without much counter questions.
PS:
>> Answer - all my own work - Honest.
wiki.answers.com/Q/How_many_times_do_a_clock%27s_hands_overlap_in_a_day
Last edited by: movilogo on Tue 1 May 12 at 10:53
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>>I wonder why these questions are asked in interviews. In real life, people seldom face these problems.
Because you aren't interested in whether the candidate gets the right answer - in fact a candidate getting the right answer tells you less about a candidate than listening to their thought processes as they explain a partial answer to you.
These types of question allow you to explore if a candidate can think rationally - do they get flustered if they can't provide a complete answer - do they become angry or defensive? - do they say anything they can't back up with evidence?
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>> "1/ Answer - all my own work - Honest. "
>>
>> Hmmmm!
>>
>> wiki.answers.com/Q/How_many_times_do_a_clock's_hands_overlap_in_a_day
Damn! I have been plagiarised, That was quick.
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OK. I'll bite. All my own work.
1) Twice
2) One way is run 7m hourglass and 4m hourglass together for 14minutes (2x7m) turning over each as soon as they run out. As the 7m hourglass finishes for the second time stop the 4m hour glass. It will have 2minutes in each half. You are them ready to measure nine minutes with 2 minutes from the 4m hourglass plus another 7minutes.
3)Approximately 50/50 - the same as with each birth.
4) ((3+1)/3)x6
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1) I am not convinced it overlaps only twice in a day!
Whenever hour and minute hand overlap, the second hand will overlap too. In theory, it is assumed that second hand only jumps at one full gap and the width of hands are infinitely small.
In reality, the width of hands have finite thickness and in many watches, second hand moves progressively (without jump).
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OK movi I'll accept I got that wrong. I read it as how often do they line up perfectly. Should have read the question more carefully.
I don't really want the job anyway!
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>> I read it as how often do
>> they line up perfectly.
>>
If it doesn't mean that then it is a meaningless question. What does "overlap" mean? All 3 overlapping each other, or H overlapping M and M overlapping S but H not necessarily overlapping S ?
How thick are the hands?
What style of hands are they - those big swirly ones with elaborate cut out sections?
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>> 1) I am not convinced it overlaps only twice in a day!
I thought that initially too. But thinking about it more - the period of time for which the hour and minute hands overlap is very small, and the second hand at that point in time is somewhere else (apart from at 12am/pm).
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>> >- the period of time
>> for which the hour and minute hands overlap is very small, and the second hand
>> at that point in time is somewhere else (apart from at 12am/pm).
>>
How do you know it is somewhere else?
Set the time at 1 oclock. Then very slowly move the minute hand round until it catches up with the hour hand, at some time just after 5 past 1. Work out what that time is. Now work out where the second hand will be at that time.
I agree it is very unlikely to happen to be at precisely that point too, but how do you calculate it?
Perhaps that was what Zero doing, and he is really very clever after all.
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>> I agree it is very unlikely to happen to be at precisely that point too,
>> but how do you calculate it?
Good question :)
x = angle of minute hand past 12
y = angle of hour hand past 12
Looking for when x = y
For the case just after 1am/pm, at m minutes past the hour:
x = m / 60 (units are fractions of a whole circle)
y = m / 60 / 12 (ie. hour hand is 12 times slower than minute hand)
+ 5 / 60 (because it started at 1, not 12)
Solve:
m / 60 = m / 60 / 12 + 5 / 60
m = m / 12 + 5
m - m / 12 = 5
m = 60 / 11
= 5.454545... minutes
So the second hand will be at 0.454545... x 60 = 27 secs approx, which isn't between 1 and 2.
Of course that's only one of the 11 cases - feel free to solve the rest :)
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I started to try to work it out but then decided...
A - I couldn't
B - I didn't care that I couldn't
Sorry !
:-)
Last edited by: Humph D'Bout on Tue 1 May 12 at 12:09
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>> 3) In the land of Nevernever, all parents only want sons. Every family keeps having
>> children until they get a boy, then they stop. What's the proportion of girls to
>> boys in Nevernever land?
There's no way of working it out. The first child of some parents will be a boy, and then they'll stop. Other parents will have a number of girls before they have a boy. So ......... all parents will have just one boy, but a proportion of those parents will also have an unknown number of girls. Under normal circumstances there will be approximately 985 girls for every 1000 boys. But, if the number of boys is artificially limited then the proportion of girls to boys is indeterminate.
Last edited by: L'escargot on Tue 1 May 12 at 14:54
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>> There's no way of working it out.
I think there is if you assume a fixed probability of having a boy/girl, although I couldn't tell you what it is :) However, it's quite easy to simulate in a few lines of C code, and that shows that you end up with equal numbers of both (as mentioned earlier(?)).
Last edited by: Focus on Tue 1 May 12 at 15:14
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>>There's no way of working it out.
Assuming equal probability of a boy or a girl in any birth, the population will tend towards a 50/50 proportion of boys/girls.
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That's what I said - I hoped you'd give us an equation :)
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Start with n couples, and assume 50/50 probability of boys/girls.
n/2 have boys and stop.
n/2 have girls and go on to have another try.
So far n/2 boys, n/2 girls.
Then the results the n/2 couples who have another go:
n/4 boys, n/4 girls.
Then the n/4 couples who have third go:
n/8 boys, n/8 girls
and so on.
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Ah yes - thanks JH; neat.
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>> Ah yes - thanks JH; neat.
...although, as I think Crocks had already worked out (EDIT: and possibly JH as well?), if the chances at every birth are 50/50, then it doesn't matter what 'rules' you lay down, the end result is always going to be 50/50. Light dawns...
I won't bother going for that job at Google then :)
Last edited by: Focus on Tue 1 May 12 at 16:09
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>> 2) You have a 4 minute hourglass and a 7 minute hourglass. How do you
>> measure exactly 9 minutes?
Start the two hourglasses simultaneously. When each runs out turn it over and let it start again. The 9 minutes timed period starts when the 4-minute hourglass has completed 12 minutes and ends when the 7-minute hourglass has completed 21 minutes.
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It does say you also have a watch...
:-)
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what happens if some have mixed sex multiple births?
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oh ive just fell in....nevernever land is only occupied by children.......no babies will be born at all
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3.
Could the answer be two boys for every girl?
If the probability is equal of having either sex.
Couple one have a boy
Couple two have a girl, then go on to have a boy by the law of averages (which I know don't exist, but toss a coin enough times and it appears to).
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>> If the probability is equal of having either sex.
But that's all you need to know - doesn't matter what sequence the births follow, if the probability is equal for all births, you will end up with equal numbers of boys and girls.
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If everyone is spending their time watching clocks and fiddling with hourglasses there will be no sex and no girls or boys.
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An interesting question is why since it only takes one male to fertilise many females are all animal populations split 50/50 male and female since on the face of it it seems wasteful to produce so many males.
Clue - consider the grandchildren.
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>> Couple two have a girl, then go on to have a boy by the law
>> of averages (which I know don't exist, but toss a coin enough times and it
>> appears to).
>>
>> Could the answer be two boys for every girl?
>>
No.
See my earlier post.
Last edited by: John H on Tue 1 May 12 at 17:10
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>> If the probability is equal of having either sex.
But it isn't. There are approximately 985 girls born to every 1000 boys. I assume Crankcase wants an exact answer not an approximation.
Last edited by: L'escargot on Tue 1 May 12 at 19:01
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>> >> If the probability is equal of having either sex.
>>
>> But it isn't. There are approximately 985 girls born to every 1000 boys. I assume
>> Crankcase wants an exact answer not an approximation.
The best answer you can give is that the result will tend towards the ratio of boy/girl births ie. a ratio of 985 girls to 1000 boys assuming your figures are correct (although I suspect they've been rounded).
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Crankcase doesn't want an answer at all other than what you want to give, L'es. It's just a bit of fun. Won't be any prizes he's afraid.
I can just about cope with the old chestnut about the man who lives on the 20th floor and every day takes the lift all the way down in the morning but at night only goes to the 15th floor and walks up the rest -why? That's about my standard.
Luckily for me I have a job I enjoy more than working at Google!
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3) Use of 4 min and 7 nin hourglass to time 9 minutes:
At minute 0, start both hourglasses.
At minute 4, turn the empty 4 min hourglass over to start again.
At minute 7, turn the empty 7 min hourglass over to start again. The 4 min hourglass has 1 minute remaining.
At minute 8, when the 4 min hourglass has run out again and the 7 min hourglass has been running for 1 minute, turn the 7 min hourglass back the other way, allowiing the 1 minute of sand to run back through it.
Minute 9 is reached when the 7 min hourglass runs out. Simples.
If the solution has already been linked to then I've missed it on my phone, this was my own work.
The three clock hands will strictly coincide only twice IMO.
Babies, who knows but I like the n/2, n/4, n/8 answer.
((3+1)/3)x6=8
BODMAS...
Essentially (three plus one equals four), four thirds times six equals eight.
Last edited by: Dave_TDCi on Tue 1 May 12 at 22:30
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>> Babies, who knows but I like the n/2, n/4, n/8 answer.
Nice, but over-complicated :)
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I'm far too tired for this. But thinking about it further...
Say 800 couples had babies. 400 boys and 400 girls.
The boys' parents stop, the girls' parents have a second child: 200 boys and 200 girls. So now we have 600 boys and 600 girls.
Repeat the process, now 700 boys and 700 girls.
And again, now 750 boys and 750 girls.
And so on, and so on.
So the answer is, the ratio of boys to girls will be 50/50, but half the boys will be older, only children whereas all the girls will have siblings.
Can I go to bed now? :)
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>> So the answer is, the ratio of boys to girls will be 50/50
Yes. As I said previously I suspect the answer they're looking for is that it doesn't matter what shenanigans the parents go through to achieve the offspring they want - it doesn't affect the overall ratio of boys to girls (whatever that is - 50/50, 859/1000, ..). Anything else is a red herring.
>> but half the boys will be older, only children whereas all the girls will have siblings.
You'd probably get extra points for that :)
Last edited by: Focus on Wed 2 May 12 at 06:26
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>> 3) Use of 4 min and 7 nin hourglass to time 9 minutes:
>>
>> At minute 0, start both hourglasses.
>> At minute 4, turn the empty 4 min hourglass over to start again.
>> At minute 7, turn the empty 7 min hourglass over to start again. The 4
>> min hourglass has 1 minute remaining.
>> At minute 8, when the 4 min hourglass has run out again and the 7
>> min hourglass has been running for 1 minute, turn the 7 min hourglass back the
>> other way, allowiing the 1 minute of sand to run back through it.
>> Minute 9 is reached when the 7 min hourglass runs out. Simples.
There is clearly more than one solution to this question. I maintain that my solution (see my post Tue 1 May 12 16:03) is just valid as yours.
Last edited by: L'escargot on Wed 2 May 12 at 06:36
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>> There is clearly more than one solution to this question. I maintain that my solution
>> (see my post Tue 1 May 12 16:03) is just valid as yours.
As was Crocks' (10:53); in fact it looks like he answered all 3 correctly. He should go for the job :)
Last edited by: Focus on Wed 2 May 12 at 08:53
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Thanks, Focus but I'm not sure I want it.
I think Dave_TDCi should have it.
I just gave the answers, he gave a lot more explanation which is sure to impress Number Cruncher. :-)
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>> ie. a ratio of 985 girls to 1000 boys assuming your figures
>> are correct (although I suspect they've been rounded).
I did say approximately 985 girls to every 1000 boys. The 985 figure varies approximately +/- 1 for various parts of the world.
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But why, given that only a few males are needed to maintain a species do all species have a 50/50 distribution of males and females. Why does evolution perpetuate this apparently wasteful distribution?
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Evolution creates males not just to produce offsprings but also to protect the race from external attacks. Granted in modern life females don't need much protection but in the grand schema of universe, this modern life of last 100 years is just a blink.
There is one theory of dinosaurs' extinction which claimed this happened because they were either producing too few males or females (but not both).
Evolution is much more powerful than we think. There are many unisex animals (usually at very lower level of animal hierarchy). So nature thinks that is the way forward it will happen.
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>> But why, given that only a few males are needed to maintain a species do
>> all species have a 50/50 distribution of males and females. Why does evolution perpetuate this
>> apparently wasteful distribution?
Because you need males to hunt and gather for the females.
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Also, not all males are born equal. In animal world, often males have to fight for female. Only the strongest wins and can mate. This ensures that only the best genes are carried forward.
Further to that, the male semen determines the gender of the child. If there are too few males, the chance of having boys will be even lower. If this goes on for a long time, the male sex will be wiped off (and thus entire race will disappear). So, 50:50 ratio keeps it happening in a proper way.
Last edited by: movilogo on Wed 2 May 12 at 09:54
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"Because you need males to hunt and gather for the females."
Not the answer - you think in terms of evolutionary success and grandchildren.
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in evolution terms we are still in the hunter gatherer phase. You need to think in much longer timescales.
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Forget humans. All animals have a 50/50 split of sexes. The question is why should this be.
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One answer: www.askabiologist.org.uk/answers/viewtopic.php?id=5227
EDIT: sorry not read it yet so not sure if covers other species
Last edited by: Focus on Wed 2 May 12 at 11:31
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Yes it applies for all species. A genetic tendency to produce mainly female offspring would initially give an individual a huge advantage in perpetuating its genes. However as a the number of females in the population increase the advantage would be lost and a male producing tendency would then be rewarded. Over time this oscillation would finally stabilise at a 50/50 split
Last edited by: CGNorwich on Wed 2 May 12 at 19:54
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Bees are the obvious oddities, but there are others.
This paper, for example, says that whilst the proportion of female to male bees varies over the year in this species, the overall proportion is 38% male.
beheco.oxfordjournals.org/content/10/4/401.full
And then there's the Barramundi fish, which actually changes sex as it gets older. As Les Hiddens says, a funny sort of arrangement.
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I think my missus acquired Barramundi genes the day after we married! :-(
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Mildly diverting brain teasers - Iffy
