Problem Solving

BTGmoderatorDC wrote:A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are

(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3

OA B

Source: Official Guide

We COULD use algebra to solve this question.
However, it's probably faster to just test the answer choices

The answer choices give us the length and width of the rectangle.
So, the sum of two values will equal HALF the perimeter of the rectangle.
Since we want a perimeter of 10, the correct answer choice will have a sum of 5 ( HALF the perimeter)

(A) 3/2 by 7/2
3/2 + 7/2 = 10/2 = 5
Yay!!!
Hold on!
The question tells us the length is TWICE the width, yet 7/2 is NOT twice 3/2
ELIMINATE A

(B) 5/3 by 10/3
5/3 + 10/3 = 15/3 = 5. GREAT!
Also, 10/3 IS TWICE 5/3

Answer: B

Cheers,
Brent

BTGmoderatorDC wrote:A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are

(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3

OA B

Source: Official Guide

Approach #2: Algebra

Let x = the width of the rectangle
So, 2x = the length of the rectangle

If the perimeter is 10, we can write: x + x + 2x + 2x = 10
Simplify: 6x = 10
Solve: x = 10/6 = 5/3
So, the width (x) is 5/3
And the length (2x) is 10/3

Answer: B

Cheers,
Brent