All but $$10$$ of the coins are $$1\text{p}$$ coins, which tells us that the number of $$2\text{p}$$ and $$5\text{p}$$ coins adds up to $$10$$. Similarly, the total of the $$1\text{p}$$ and $$5\text{p}$$ coins is $$10$$ and the total of the $$1\text{p}$$ and $$2\text{p}$$ coins is $$10$$. He therefore has $$5$$ of each coin, giving a total of $$(5 \times1)+ (5 \times2)+(5\times5)=40\text{p}$$. 
Karl Lecter has been collecting $$1\text{p}$$, $$2\text{p}$$ and $$5\text{p}$$ coins in a jar. All but $$10$$ of his coins are $$1\text{p}$$ coins, all but $$10$$ are $$2\text{p}$$ coins, and all but $$10$$ are $$5\text{p}$$ coins. How much money does he have? 
Options:
a. $$8\text{p}$$ 
b. $$10\text{p}$$ 
c. $$25\text{p}$$ 
d. $$40\text{p}$$ 
e. $$80\text{p}$$ 