Question

The length of a rectangle is 1 ft less than twice the width. If the width is doubled, and the length is increased by 7 ft, then the perimeter becomes 320 ft. What is the width of the original rectangle?

• A. $\frac{77}{4}\text{ft}$
• B. $\frac{77}{2}\text{ft}$
• C. $39\text{ft}$
• D. $80\text{ft}$

Answer 1

The correct option is B $\frac{77}{2}\text{ft}$

Let the width of the original rectangle be $“w”$.

Length $=2w-1$

New length $=2w-1+7=2w+6$

New width $=2w$

New perimeter $=2[2w+(2w+6\left)\right]$

New perimeter $=320$

$2[2w+(2w+6\left)\right]=320$

$2[4w+6]=320$

$4w+6=160$ [dividing by 2 on both the sides]

$4w+6-6=160-6$ [subtracting 6 from both the sides]

$4w=154$

$w=\frac{154}{4}$ [dividing by 4 on both the sides]

$w=\frac{77}{2}\text{ft}$

Thus, the width of the original rectangle is $\frac{77}{2}\text{ft}$.