### Solution:

Given the area of rectangle gets reduced by 80 sq. units if its length is decreased by 5 units and breadth is increased by 2 units.If there is an increment in the length by 10 units and decrement in the breadth by 5 units, then area is increased by 50 square units.

Let the length and breadth of rectangle be x and y respectively.

We know the area of a rectangle is given by,

$A=\mathit{length}\times \mathit{breadth}$

So, here we have,

Area = $x\times y=\mathit{xy}$

According to the question, we have,

$\n \n \n \n \n \n xy\u2212(x\u22125)(y+2)\n \n \n \n =80\n \n \n \n \n \n \u21d2xy\u2212(xy+2x\u22125y\u221210)\n \n \n \n =80\n \n \n \n \n \n \u21d22x\u22125y+70\n \n \n \n =0\u2026\u2026.(1)\n \n \n \n \n $

Similarly, from the second condition,

$\n \n \n \n \n \n (x+10)(y\u22125)\u2212xy\n \n \n \n =50\n \n \n \n \n \n \u21d2(xy\u22125x+10y\u221250)\u2212xy\n \n \n \n =50\n \n \n \n \n \n \u21d25x\u221210y+100\n \n \n \n =0\n \n \n \n \n \n \u21d2x\u22122y+20\n \n \n \n =0\u2026\u2026(2)\n \n \n \n \n $

The solution of the equations ${a}_{1}x+{b}_{1}y+{c}_{1}=0\mathit{and}{a}_{2}x+{b}_{2}y+{c}_{1}=0$by cross multiplication method is given by the formula,

$\frac{x}{{b}_{1}{c}_{2}-{b}_{2}{c}_{1}}=\frac{y}{{a}_{2}{c}_{1}-{a}_{1}{c}_{2}}=\frac{1}{{a}_{1}{b}_{2}-{a}_{2}{b}_{1}}$

Here, ${a}_{1}=2,{b}_{1}=-5,{c}_{1}=70,{a}_{2}=1,{b}_{2}=-2,{c}_{2}=20.$

Using the cross-multiplication method to find the value of x and y,

$\Rightarrow \frac{x}{-5\left(20\right)-70\left(-2\right)}=\frac{y}{70\left(1\right)-\left(20\right)2}=\frac{1}{2(-2)-\left(1\right)(-5)}$

$\Rightarrow \frac{x}{40}=\frac{y}{30}=\frac{1}{1}$

$\Rightarrow x=\frac{40}{1}=40$

$\Rightarrow y=\frac{30}{1}=30$

Therefore, the length and breadth of the rectangle are 40 units and 30 units respectively.

Hence, option (1) is correct.