Sasha has $\$3.20$ in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could have?
Suppose Sasha has $q$ quarters.  Then she also has $q$ nickels, and the total value of her quarters and nickels is $.25q + .05q = .30q$.  Since $3.20/0.30 = 10\frac{2}{3}$, this means that she has at most $\boxed{10}$ quarters.  (This amount is obtainable; for example we can let the rest of her coins be pennies.)
10
Level 4
Prealgebra
Problem: Sasha has $\$3.20$ in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could have?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to define variables for the number of quarters and nickels. Let \( q \) represent the number of quarters and \( n \) represent the number of nickels. According to the problem, \( q = n \).

Next, we calculate the total value of the coins. The value of quarters is \( 0.25q \) and the value of nickels is \( 0.05n \). Since \( n = q \), we can substitute \( n \) with \( q \) in the equation. Thus, the total value of the coins can be expressed as:

\[
0.25q + 0.05q = 0.30q
\]

We know that the total amount of money Sasha has is $3.20. Therefore, we set up the equation:

\[
0.30q = 3.20
\]

To find \( q \), we solve for \( q \) by dividing both sides of the equation by \( 0.30 \):

\[
q = \frac{3.20}{0.30}
\]

Calculating the right side gives:

\[
q = \frac{3.20 \times 100}{30} = \frac{320}{30} = \frac{32}{3} \approx 10.67
\]

Since \( q \) must be a whole number (as you cannot have a fraction of a coin), we take the greatest integer less than or equal to \( 10.67 \), which is \( 10 \).

Thus, the greatest number of quarters Sasha could have is \( 10 \).<|END_OF_LATENT|> The final answer is $10$.