A bowl contained $10\%$ blue candies and $25\%$ red candies.  A bag containing three quarters red candies and one quarter blue candies was added to the bowl.  Now the bowl is $16\%$ blue candies.  What percentage of the candies in the bowl are now red?
1. Let the number of candies in the bowl originally be \( x \) and the number of candies in the bag be \( y \).

2. The number of blue candies in the bowl originally is \( 0.1x \) and the number of red candies in the bowl originally is \( 0.25x \).

3. The number of blue candies in the bag is \( 0.25y \) and the number of red candies in the bag is \( 0.75y \).

4. After adding the bag to the bowl, the total number of blue candies in the bowl is \( 0.1x + 0.25y \) and the total number of red candies in the bowl is \( 0.25x + 0.75y \).

5. The total number of candies in the bowl after adding the bag is \( x + y \).

6. We are given that the percentage of blue candies in the bowl is now \( 16\% \). Therefore, we have the equation:
   \[
   \frac{0.1x + 0.25y}{x + y} = 0.16
   \]

7. Solving for \( x \) in terms of \( y \):
   \[
   0.1x + 0.25y = 0.16(x + y)
   \]
   \[
   0.1x + 0.25y = 0.16x + 0.16y
   \]
   \[
   0.25y - 0.16y = 0.16x - 0.1x
   \]
   \[
   0.09y = 0.06x
   \]
   \[
   x = \frac{0.09y}{0.06} = 1.5y
   \]

8. We want to find the percentage of red candies in the bowl now:
   \[
   \frac{0.25x + 0.75y}{x + y}
   \]

9. Substitute \( x = 1.5y \) into the expression:
   \[
   \frac{0.25(1.5y) + 0.75y}{1.5y + y}
   \]
   \[
   \frac{0.375y + 0.75y}{2.5y}
   \]
   \[
   \frac{1.125y}{2.5y}
   \]
   \[
   \frac{1.125}{2.5} = 0.45
   \]

10. Convert the fraction to a percentage:
    \[
    0.45 \times 100\% = 45\%
    \]

The final answer is \(\boxed{45\%}\)