How can I get the ratio between the areas? I have a statement that says:

If a rectangle increases its width by one fifth of its length and its
  length decreases it in half, then its initial area will:

The answer must be: Decreases $2/5$ of the initial area.
So my development was:
$w_1*L_1 = A_1$
$w_2 = w_1 + L_1/5$
$L_2 = L_1/2$
So, $A_2 = (w_1 + L_1/5)(L_1/2)$
$A_2 = (w_1L_1)/2 + (L_1)^2/2$
Then, $A_2/A_1 = \frac{1}{2} + \frac{L}{5W_1}$
But the answer must be purely numerical, in which I am failing or how do I arrive at the result? Thanks in advance.
 A: Your logic is sound and there is no single number that is the ratio of areas.  You can just take a couple rectangles to show the example.  Let rectangle $1$ be $2 \times 100$ with area $200$.  It changes to $22 \times 50$ with area $1100$ so the ratio is $5.5$.  Let rectangle $2$ be $10 \times 10$ with area $100$.  It changes to $12 \times 5$ with area $60$ so the ratio is $0.6$
A: Given the edit providing the answer, it's very likely that (as in my comment) the question was garbled somewhere, or was just badly written. The statement

If a rectangle increases its width by one fifth of its length and its
  length decreases it in half, then its initial area will:

means

If the width of a rectangle increases by one fifth of itself and the
  length decreases to half of itself then ...

... the width and length of the new rectangle are $1.2W$ and $0.5L$. Multiplying those together (to find the area) shows that the area has changed by a factor 
$1.2 \times 0.5 = 0.6$. So the new area is $60\%$ of the original area. It has decreased by $2/5$ of itself.
A: $A_2 = (w_1 + \frac{L_1}{5})(\frac{L_1}{2})$
$A_2 = \frac{(w_1L_1)}{2} + \frac{(L_1)^2}{10}$
$\frac{A_2}{A_1} = \frac{(w_1L_1)}{2(w_1L_1)} + \frac{(L_1)^2}{10(w_1L_1)}$
$$\frac{A_2}{A_1} = \frac{1}{2} + \frac{L_1}{10w_1}$$
Increasing the width by one fifth of the length negates this from being an increase in area by a constant factor.
To decrease by $\frac{2}{5}$of the original area, the width increasing by one fifth and the length decreasing by one half is one possible solution.
