(related to Problem: The Puzzling Money-Boxes)
The correct answer to this puzzle is as follows: John put into his money-box two double florins ($8$s.), William a half-sovereign and a florin ($12$s.), Charles a crown ($5$s.), and Thomas a sovereign ($20$s.). There are six coins in all, of a total value of $45$s. If John had $2$s. more, William $2$s. less, Charles twice as much, and Thomas half as much as they really possessed, they would each have had exactly $10$s.
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