Question 1119273
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Presuming your story occurred sometime after 1971 such that £1 = 100p, proceed as follows:


Let *[tex \Large x] represent the number of children who received 25p.  Then the number of children who received 5p must be *[tex \Large 20\ -\ x].  Then, the total amount given to *[tex \Large x] children must be *[tex \Large 25x] pence, and the total amount given to the *[tex \Large 20\ -\ x] children must be *[tex \Large 5(20\ -\ x)] pence.  These two totals must sum to £2 = 200p.  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  25x\ +\ 5(20\ -\ x)\ =\ 200]


Solve for *[tex \Large x]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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