1. A container has 100 small balls, of which $36 \%$ are red balls, and the rest are blue balls. Then, after removing ( ) blue balls, the red balls account for $72 \%$ of the remaining balls.
(A) 28
(B) 32
(C) 36
(D) 50
(E) 64
1. D.

Let $x$ be the number of blue balls taken away. Then
$$
\frac{100 \times 36 \%}{100-x} \times 100 \% = 72 \% \Rightarrow x=50 \text {. }
$$