25 rectangles

The sum of the Xrectangles' perimeters is equal to the big rectangle's perimeter, since any two of them is in a different row and in a different column.

The sum of the Yrectangles' perimeters is equal to the big rectangle's perimeter, since any two of them is in a different row and in a different column.

So the perimeter of the big rectangle is $46+47+45+49+50=237$ .

From that the perimeter of the pink rectangle is: $237-(45+48+51+46)=\boxed{47}$

Let the width of $n$ th column (from left to right) be $w_n$ and the height of $n$ th row (from top to bottom) be $h_n$ . Then the perimeter of the pink rectangle is $p=2(w_5+h_5)$ .

Consider the orange and yellow rectangle of the 3rd column, we have:

$\begin{cases} 2(w_3+h_4) = 45 & \implies w_3+h_4 = 22.5 & ...(1) \\ 2(w_3+h_5) = 46 & \implies w_3+h_4 = 23 & ...(2) \end{cases}$

$(2)-(1): \implies h_5-h_4 = 0.5 \implies h_4 = h_5 - 0.5$ . Similarly, column 1: $h_3 = h_4+0.5 = h_5$ , column 4: $h_1 = h_3 + 0.5 = h_5+0.5$ . Similar for row 2: $w_2 = w_5 + 1$

We note thatf for red ractangle

$\begin{aligned}2(w_2+h_1) & = 50 \\ 2(w_5+1+h_5+0.5) & = 50 \\ 2(w_5+h_5)+3 & = 50 \\ \implies 2(w_5+h_5) & = \boxed{47} \end{aligned}$