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solving look around you's math problems

some may say that the math problems featured in [look around you][1] are
impossible, but i am here to attempt to calculate them using the power of
maths.

### problem one


	
*question:**

> jean is shorter than brutus, but taller than imhotep. imhotep is taller than
  jean but shorter than lord scotland. lord scotland is twice the height of
  jean and brutus combined, but only a tenth of the height of millsy. millsy is
  at a constant height of x-y. if jean stands exactly 1 nautical mile away from
  lord scotland, how tall is imhotep?


	
*official answer:** imhotep is invisible.


the joke for this problem is that it's obviously contradictory and impossible,
however, the problem never specifies that there can't be two people with the
same name.

imhotep < jean 1 < brutus  
jean 2 < imhotep < lord scotland  
lord scotland = (jean 1 + brutus) * 2  
millsy = lord scotland / 10  
millsy = x-y  
jean 2 one nautical mile away from lord scotland  
imhotep = ?

i don't know where to go from here

### problem two


	
*question:**

> eight ladies go into eight shops at eight o'clock in the morning. each lady
  wants to buy eight spiders. for each spider, eight spider shoes must be
  bought.  but, they only have eight pounds between them. with each spider
  costing eightpence, and each spider shoe costing an eighthpence each, will
  the ladies have enough change for the bus home, a journey costing eightpence
  per stop and made up of eight stops.


	
*official answer:** the ladies were eightpence short.


this one is fairly simple, it's just a matter of multiplication.

8 ladies * (8 spiders * 0.08 pounds + 8 spider shoes * 0.008 pounds + 8 stops * .08 pounds) =  

	
*10.752 pounds**


the ladies were 2.76 pounds short.

### problem three


	
*question:**

> it's the future, and queen elizabeth iii and queen elizabeth iv are
  going to party held by queen elizabeth v, and are keen to make the right
  impression, so it's important that they choose their outfits carefully. queen
  elizabeth iii has forty dresses to choose from, whereas queen elizabeth iv
  has four thousand. queen elizabeth v has just one dress, but it has the
  ability to transform itself into the shape of any dress. the night before the
  party, queen elizabeth iv's handmaiden steals the patterns to queen elizabeth
  iii's dresses, and working through till dawn, makes forty exact replicas. can
  you calculate the probability that all three queens will be wearing the same
  dress at the party, and how many times can queen elizabeth v's dress change
  before it overheats?


	
*official answer:** the party was cancelled


assuming that every dress in elizabeth iii's and elizabeth iv's wardrobes are
equally likely to be chosen, then each dress in elizabeth iii's collection has
a 1/40 chance of being warn, while elizabeth iv's dresses have a 1/4040 chance
after the duplicates are made. however, since queen elizabeth v dress can
change into *any* dress, the probably of her wearing any specific dress
approaches zero. therefore, it can be assumed that the probability of all three
wearing the same dress is zero. to answer the second part, the video shows that
elizabeth the v's dress can change fourteen times before overheating.


[1]: https://en.wikipedia.org/wiki/Look_Around_You

tags: miscellaneous