Since Blaxican is such a buzz-kill I'm going to give it a shot.
Here's a tabular for the relations.
Since we know that there's always a constant number of days in-between them we can construct (and solve) the following matrix.
Meaning that the relation is such that the sister is 4/3rds as old as her brother.
So the thread is now about checking if that's accurate.
Originally posted by AstnerThe future sister would be 2z.
Since Blaxican is such a buzz-kill I'm going to give it a shot.Here's a tabular for the relations.
Since we know that there's always a constant number of days in-between them we can construct (and solve) the following matrix.
Meaning that the relation is such that the sister is 4/3rds as old as her brother.
So the thread is now about checking if that's accurate.
Read it again.
Re: Funny riddle
(x+y)/2 - z = x - y = 2z - x = c (constant, which is difference between ages)
Gonna do present n future first.
2z - y - c = 0
Oh shit, I gotta differentiate both y and z and find out the constant, then do the same logic for the past part of the sum.
I gave up on Maths at A Level, and it's 20 past 4. **** that, I'm going to bed, but I think I layed down some of the foundation stuff.
Re: Re: Funny riddle
Originally posted by lord xyz
(x+y)/2 - z = x - y = 2z - x = c (constant, which is difference between ages)
Gonna do present n future first.
2z - y - c = 0
Oh shit, I gotta differentiate both y and z and find out the constant, then do the same logic for the past part of the sum.I gave up on Maths at A Level, and it's 20 past 4. **** that, I'm going to bed, but I think I layed down some of the foundation stuff.
I think you "laid" down a gross misspelling.