Logan has $3.15 worth of dimes and quarters. He has twice as many dimes as quarters. Determine the number of dimes and the number of quarters that Logan has.

Let's denote the number of quarters Logan has as $q$. According to the problem, Logan has twice as many dimes as quarters, so the number of dimes can be represented as $2q$.

Next, we need to express the total value of the dimes and quarters in terms of $q$. Each quarter is worth $0.25, and each dime is worth$0.10. Therefore, the total value of the quarters is $0.25q$ dollars, and the total value of the dimes is $0.10 \times 2q = 0.20q$ dollars.

The total value of all the coins is given as $3.15. We can set up the following equation to represent this:

$0.25q + 0.20q = 3.15$

Combining the terms on the left side, we get:

$0.45q = 3.15$

To find the value of $q$, we divide both sides of the equation by 0.45:

$q = \frac{3.15}{0.45}$

$q = 7$

So, Logan has 7 quarters. Since he has twice as many dimes as quarters, the number of dimes is:

$2q = 2 \times 7 = 14$

Therefore, Logan has 7 quarters and 14 dimes.