The problem asks: By what percentage does the area of a rectangle change when its length increases by 20% and its width decreases by 50%?

1. Problem Description

The problem asks: By what percentage does the area of a rectangle change when its length increases by 20% and its width decreases by 50%?

2. Solution Steps

Let $L$ be the original length of the rectangle and $W$ be the original width of the rectangle.

The original area, $A_1$, of the rectangle is given by:

$A_1 = L \times W$

The new length, $L'$, is the original length increased by 20%, so:

$L' = L + 0.20L = 1.20L$

The new width, $W'$, is the original width decreased by 50%, so:

$W' = W - 0.50W = 0.50W$

The new area, $A_2$, of the rectangle is given by:

$A_2 = L' \times W' = (1.20L)(0.50W) = 0.60LW$

The percentage change in area is calculated as:

Percentage Change = $\frac{A_2 - A_1}{A_1} \times 100$

Percentage Change = $\frac{0.60LW - LW}{LW} \times 100$

Percentage Change = $\frac{-0.40LW}{LW} \times 100$

Percentage Change = $-0.40 \times 100$

Percentage Change = $-40$%

Since the percentage change is negative, the area decreases by 40%.

However, the possible answers in the image do not contain -40%. Let's review the question again.

The question asks: by what percentage DOES the area vary. Thus, it asks by what percentage has the area changed, which is 40%.

Since 40% is not among the options, let's examine the information provided again. Length increases by 20% and width decreases by 50%. $L' = 1.2L$ and $W' = 0.5W$. $A_2 = 1.2L * 0.5W = 0.6LW$. The difference is $0.6LW - LW = -0.4LW$. The change relative to the initial area is $-0.4LW/LW = -0.4 = -40\%$. Thus, the new area is $60\%$ of the old one. Thus, the area has changed by $40\%$.

None of the answers given in the image are correct. Let's find the closest answer.

Since the area *decreases* by 40%, none of the options provided are correct. However, if the width increased by 50% instead of decreased, we would have: $W'=1.5W$, and $A_2 = 1.2L * 1.5W = 1.8LW$.

Change = $1.8LW - LW = 0.8LW$. % Change = $0.8LW/LW * 100 = 80\%$. This leads to option A. However, the problem states a *decrease* in the width. Thus, none of the options are correct.

The correct answer is that the area decreases by 40%.

3. Final Answer

None of the options provided are correct. The area decreases by 40%.