The length and breadth of a rectangle are in the ratio 4:3, respectively. When the length of the rectangle is increased by 30% and the breadth remains the same, the area of the rectangle increases by 360 sq m. Find the original length (in meters) of the rectangle.

The correct option is 40.

Let the original length and breadth of the rectangle be $l$ and $b$, respectively. Since they are in the ratio 4:3, we can represent them as:
Original length, $l=4x$
Original breadth, $b=3x$
where $x$ is a positive scaling factor.

The original area of the rectangle ($A_1$) is given by:
$A_1=l\times b=4x\times 3x=12x^2$

According to the question, the length of the rectangle is increased by 30% and the breadth remains the same. Let's find the new dimensions:
New length, $l^{\text{'}}=4x+30\%\text{of}4x=4x+1.2x=5.2x$
New breadth, $b^{\text{'}}=3x$

The new area of the rectangle ($A_2$) is:
$A_2=l^{\text{'}}\times b^{\text{'}}=5.2x\times 3x=15.6x^2$

We are given that the area of the rectangle increases by 360 sq m. Therefore:
$A_2-A_1=360$

Substitute the expressions for $A_1$ and $A_2$ into the equation:
$15.6x^2-12x^2=360$
$3.6x^2=360$
$x^2=\frac{360}{3.6}$
$x^2=100$
$x=10$ (since $x$ must be positive)

Now, we can find the original length of the rectangle:
Original length = $4x=4\times 10=40\text{meters}$.