A purse contains a collection of quarters, dimes, nickels, and pennies. The average value of the coins in the purse is 17 cents. If a penny is removed from the purse, the average value of the coins becomes 18 cents. How many nickels are in the purse?
(A) 2
(B) 5
(C) 0
(D) 1
(E) 8
Suppose that there are $n$ coins in the purse to begin with.

Since the average value of the coins is 17 cents, then the total value of the coins is $17 n$.

When one coin is removed, there are $n-1$ coins.

Since the new average value of the coins is 18 cents, then the new total value of the coins is $18(n-1)$.

Since 1 penny was removed, the total value was decreased by 1 cent, or $17 n-1=18(n-1)$ or $17 n-1=18 n-18$ or $n=17$.

Therefore, the original collection of coins has 17 coins worth 289 cents.

Since the value of each type of coin except the penny is divisible by 5 , then there must be at least 4 pennies.

Removing these pennies, we have 13 coins worth 285 cents. These coins may include pennies, nickels, dimes and quarters.

How many quarters can there be in this collection of 13 coins?

12 quarters are worth $12 \times 25=300$ cents, so the number of quarters is fewer than 12 .

Could there be as few as 10 quarters?

If there were 10 quarters, then the value of the quarters is 250 cents, so the remaining 3 coins (which are pennies, nickels and dimes) are worth 35 cents. But these three coins can be worth no more than 30 cents, so this is impossible.

In a similar way, we can see that there cannot be fewer than 10 quarters.

Therefore, there are 11 quarters in the collection, which are worth 275 cents.

Thus, the remaining 2 coins are worth 10 cents, so must both be nickels.

Therefore, the original collection of coins consisted of 11 quarters, 2 nickels and 4 pennies.

ANSWER: (A)