Question

If the length and breadth of the room are increased by 1 m each, its area would be increased by $31{\text{m}}^2$. If the length is increased by 1 m and breadth is decreased by 1 m, the area would decrease by $9{\text{m}}^2$ . Find the area of the floor of the room in ${\text{m}}^2$.

• A. $200$
• B. $209$
• C. $250$
• D. $150$

Answer 1

The correct option is B

$209$

Step 1: Use given information to obtain equations.

Let $x,y$ and $z$ be the length, breadth, and area of a room respectively.

Then the area of a room is given by $\Rightarrow \left(\text{length}\times \text{breadth}\right)=\left(x\times y\right)=z$.

Now, if the length and breadth of the room are increased by 1 m each, its area would be increased by $31{\text{m}}^2$.

$\begin{array}{rcl}& \Rightarrow & \left[\left(x+1\right)\times \left(y+1\right)\right]=z+31\\ & \Rightarrow & xy+x+y+1=z+31\\ & \Rightarrow & xy+x+y=z+30...\left(1\right)\end{array}$

Also, if the length is increased by 1 m and the breadth is decreased by 1 m, the area would decrease by $9{\text{m}}^2$.

$\begin{array}{rcl}& \Rightarrow & \left[\left(x+1\right)\times \left(y-1\right)\right]=z-9\\ & \Rightarrow & xy-x+y-1=z-9\\ & \Rightarrow & xy-x+y=z-8...\left(2\right)\end{array}$

Step 2: Use the elimination method.

By subtracting equation (2) from (1) to get:

$\begin{array}{rcl}\left(xy+x+y\right)-\left(xy-x+y\right)& =& z+30-z+8\\ & \Rightarrow & 2x=38\\ & \Rightarrow & x=19\end{array}$

Thus, the length of the room is $19\text{m}$.

Now, $z=x\times y$$\Rightarrow z=19y$

Put the value of $z$ in equation (1).

$\begin{array}{rcl}& \Rightarrow & 19y+19+y=19y+30\\ & \Rightarrow & y=30-19\\ & \Rightarrow & y=11\end{array}$

Thus, the breadth of the room is $11\text{m}$.

Now, the area of the room is:

$\begin{array}{rcl}z& =& x\times y\\ & =& 19\times 11\\ & =& 209\end{array}$.

Thus, the area of the room is $209{\text{m}}^2$.

Hence option (b) is correct.