Question

When the length of rectangle is reduced by 6 feet and breadth is increase by 4 feet, then the area of a rectangle gets reduced by 10 $f{t}^2$. If we increase the length by 5 feet and breadth by 4 feet, the area is increased by 60 $f{t}^2$. Find the length and breadth of the rectangle.

• A. Length $=\frac{155}{22}\text{ft}$ ; Breadth $=\frac{26}{11}\text{ft}$
• B. Length $=\frac{26}{11}\text{ft}$ ; Breadth $=\frac{155}{22}\text{ft}$
• C. Length $=\frac{18}{11}\text{ft}$ ; Breadth $=\frac{185}{11}\text{ft}$
• D. Length $=\frac{180}{11}\text{ft}$ ; Breadth $=\frac{40}{11}\text{ft}$

Answer 1

The correct option is A Length $=\frac{155}{22}\text{ft}$ ; Breadth $=\frac{26}{11}\text{ft}$

Let the Length of rectangle be $x$ feet.
And, Breadth of the rectangle be $y$ feet.
The area of a rectangle $=x\times y$.

Now, according to first situation the length is reduced by $6$ feet and breadth is increased by $4$ feet which result in area reduction of rectangle by $10f{t}^2$.

$\Rightarrow (x–6)(y+4)=(xy–10)f{t}^2$
On simplifying we get,
$\Rightarrow xy+4x–6y–24=xy–10$
$\Rightarrow xy–xy+4x-6y=24–10$
$\Rightarrow 4x–6y=14$
$\Rightarrow 4x=6y+14$ $......\left(1\right)$

Now,
If we increase the length by $5$ feet and breadth by $4$ feet, the area is increased by $60f{t}^2$.
$\Rightarrow (x+5)(y+4)=(xy+60)$
$\Rightarrow xy+5y+4x+20=xy+60$
$\Rightarrow xy-xy+5y+4x=60–20$
$\Rightarrow 5y+4x=40$

Substituting value of $4x$ from equation $\left(1\right)$ we get,
$\Rightarrow 5y+6y+14=40\Rightarrow 11y=40-14\Rightarrow y=\frac{26}{11}\text{ft}$

Substitute the value of $y=\frac{26}{11}$ in equation $\left(1\right)$ we get,
$\Rightarrow 4x=6\times \frac{26}{11}+14$

$\Rightarrow x=\frac{6\times 26}{11\times 4}+\frac{14}{4}$

$\Rightarrow x=\frac{39}{11}+\frac{7}{2}$

$\Rightarrow x=\frac{78+77}{22}$

$\Rightarrow x=\frac{155}{22}\text{ft}$