Let d,n and q be the number of dime,nickel and quarter respectively.

0.05n+0.1d+0.25q=8.35
=> 5n + 10d + 25q = 835
=> n + 2d + 5q = 167.....1)

and

n+d+q=100.....2)

Subtracting 2) from 1

d + 4q = 67

The equation will have max d when d=67 and q=0 (since it is not necessary for the piggy bank to contain all the three type of coins)
The equation will have min value when d=3 and q=16.

Difference between largest and smallest number of dime = 67 - 3=64. (D)

Let n, d, and q be the number of nickels, dimes, and quarters, respectively. We can generate two equations:

n + d + q = 100 (Number of coins equation)

and

5n + 10d + 25q = 835 (Money value equation, in cents)

n + 2d + 5q = 167

Subtracting the first equation from the second, we have:

d + 4q = 67

To maximize the number of dimes, we minimize the number of quarters. If we have no quarters (q = 0), we will have 67 dimes, and this is the maximum number of dimes.

Now, to minimize the number of dimes, we must maximize the number of quarters. So 4q must be as large as possible, and this occurs when q = 16. When q = 16, then d = 3. Thus, the minimum number of dimes is 3.

The difference between the maximum and minimum number of dimes is 67 - 3 = 64.

Answer: D
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