Question

The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units. The area increases by 67 square units. Find the dimensions of the rectangle.
A) $l=17,b=9$
B) $l=12,b=7$
C) $l=19,b=7$
D) $l=21,b=7$

Answer 1

Hint:
Here we have to calculate the dimensions of the rectangle. First we will assume the length and breadth to be any variable. Then we will find the area. We will reduce the length by 2 units and increase the breadth by 3 units, we will find its area. We will reduce the original area by 9 square units and then we will equate these areas. Similarly, we will increase the length by 3 units and increase the breadth by 2 units, we will find its area. We will increase the original area by 67 square units and then we will equate these areas. We will solve these two equations by elimination method and we will get the required values.

Complete step by step solution:
Let the length of the rectangle be $L$ and breadth be $B$
Then area of the rectangle $=L\times B$
According to the question, we will reduce the length by 5 units and increase the breadth by 3 units.
Therefore, length of the rectangle becomes $L-5$ and breadth becomes $B+3$
Then area of the rectangle $=\left( L-5 \right)\times \left( B+3 \right)$
Also, the area gets reduced by 9 square units.
Then area of the rectangle $=L\times B-9$
We will equate both areas now as these are equal.
$\Rightarrow L\times B-9=\left( L-5 \right)\times \left( B+3 \right)$
Simplifying the equation further, we get
$\Rightarrow L.B-9=L.B-5B+3L-15$
Simplifying the like terms, we get
$\Rightarrow 3L-5B=6$ ……………………$\left( 1 \right)$
According to the question, we will increase the length by 3 units and increase the breadth by 2 units.
Therefore, length of the rectangle becomes $L+3$ and breadth becomes $B+2$
Then area of the rectangle $=\left( L+3 \right)\times \left( B+2 \right)$
Also, the area gets increased by 67 square units.
Then area of the rectangle $=L\times B+67$
We will equate both areas now as these are equal.
$\Rightarrow L\times B+67=\left( L+3 \right)\times \left( B+2 \right)$
Simplifying the equation further, we get
$\Rightarrow L.B+67=L.B+3B+2L+6$
 Simplifying the like terms, we get
$\Rightarrow 2L+3B=61$ ……………………$\left( 2 \right)$
Now, we will multiply equation $\left( 1 \right)$ with 2 and equation $\left( 2 \right)$ with 3 and then we will subtract equation $\left( 1 \right)$from equation $\left( 2 \right)$
    $\begin{align}
  & \Rightarrow 6L-10B-6L-9B=12-183 \\
 & \Rightarrow -19B=-171 \\
 & \Rightarrow B=9 \\
\end{align}$
 We will put the value of B in equation $\left( 1 \right)$
$\Rightarrow 3L-5\times 9=6$
Simplifying it further, we get
$\begin{align}
  &\Rightarrow 3L=51 \\
 & \Rightarrow L=17 \\
\end{align}$

Therefore, the length of the rectangle is 17 units and the breadth of the rectangle is 9.
Thus, the correct option is A.

Note:
Since we have used elimination methods to solve the equations. In elimination method, we change the equation in one variable by either subtracting or adding two equations. If the coefficients of one variable are opposite then we add the two equations and if the coefficients of one variable are same then we subtract the two equations to convert the equations in one variable.