The length of a rectangle is increased by 60%. By what percent the width have to be decreased to maintain the same area ?

Question

Accepted Answer

To solve the problem of maintaining the same area of a rectangle when its length is increased by 60%, we need to determine by what percentage the width must be decreased.

Let's start with the original dimensions of the rectangle:
- Let the original length be $ L $.
- Let the original width be $ W $.
- The original area $ A $ is given by $ A = L \times W $.

When the length is increased by 60%, the new length becomes:
$$ L_{\text{new}} = L + 0.6L = 1.6L $$

To maintain the same area, the new area must equal the original area:
$$ A_{\text{new}} = L_{\text{new}} \times W_{\text{new}} = A = L \times W $$

Substituting the new length into the equation for the area:
$$ 1.6L \times W_{\text{new}} = L \times W $$

Solving for the new width $ W_{\text{new}} $:
$$ W_{\text{new}} = \frac{L \times W}{1.6L} = \frac{W}{1.6} $$

The percentage decrease in width is calculated as follows:
- The original width is $ W $.
- The new width is $ \frac{W}{1.6} $.

The decrease in width is:
$$ \text{Decrease} = W - \frac{W}{1.6} = W \left(1 - \frac{1}{1.6}\right) = W \left(\frac{0.6}{1.6}\right) = W \times \frac{3}{8} $$

The percentage decrease is:
$$ \text{Percentage decrease} = \left(\frac{3}{8}\right) \times 100\% = 37.5\% $$

Thus, the width must be decreased by $ 37\frac{1}{2}\% $ to maintain the same area, which corresponds to Option 1.

**Explanation for why other options are incorrect:**

- **Option 2 (60%)**: This option suggests that the width should be decreased by 60%. However, a 60% decrease would result in a new width of $ 0.4W $, which would not maintain the original area when the length is increased by 60%.

- **Option 3 (75%)**: A 75% decrease in width would result in a new width of $ 0.25W $. This would significantly reduce the area, not maintaining the original area.

- **Option 4 (120%)**: A decrease of 120% is not feasible, as it implies a negative width, which is not possible in practical terms.

Therefore, the correct answer is Option 1, $ 37\frac{1}{2}\% $, as it accurately maintains the original area of the rectangle when the length is increased by 60%.