Find the ratio of shaded to unshaded area. The picture is attached. I saw that the shaded area was described in two different forms, and each depicted a different number (3 and 4).
I multiplied 3:8 by 4 to get 12:32. Then I multiplied 4:9 by 3 to get 12:27.
I added the unshaded areas (32 and 27) to get 59. The answer has to be D (12:59), but the answer booklet says 12:35. How come?

 A: The unshaded area is $$\frac{5}{8}A_X + \frac{5}{9}A_Y$$
where $A_X$ and $A_Y$ are areas of squares $X$ and $Y$.
For the shaded area $A_s$ it is given that $A_s=\frac{3}{8}A_X = \frac{4}{9}A_Y$ so
$$\frac{5}{8}A_X = \frac{5}{3}A_s$$
$$\frac{5}{9}A_Y = \frac{5}{4}A_s$$
then the ratio shaded/unshaded is $$\frac{A_s}{\frac{5}{3}A_s+\frac{5}{4}A_s} = \frac{12}{35}$$

Maybe this is too much, you can follow your method using $12:(32-12)$ and $12:(27-12)$ so that you use the unshaded areas instead of total areas
A: Pictographically, what you did was to find

The shaded area is $ \ 12 \ $ parts in $ \ 32 \ $ of square X and $ \ 12 \ $ parts in $ \ 27 \ $ of square Y .  (Making a "common numerator" was not a bad idea...)  So the shaded area is $ \ 12 \ $ parts in a total area of $ \ 20 + 12 + 15 \ \ . $  The question asks for the ratio of the shaded area to the unshaded area, which is $ \ 12 \ $ to $ \ 20 + 15 \ \ , $ choice (C) .
[The total area then is $ \ 47 \ $ units.  You came up with $ \ 59 \ $ because simply adding $ \ 32  + 27 \ $ "double-counts" the shaded area.]
