mals24 wrote:Does rectangle A have a greater perimeter than rectangle B?
1. The length of a side of rectangle A is twice the length of a side of rectangle B.
2. The area of rectangle A is twice the area of rectangle B.
OA after a few discussions.
As always, let's start by jotting down the relevant formulas on our scratch paper:
Area = l * w
perimiter = 2l + 2w
(1) we know that 1 of the sides of A is twice one of the sides of B.
However, we know nothing about the other sides, so (1) is insufficient.
(2) Knowing the relationships between the areas does NOT help us determine which has a bigger perimiter.
For example, A could be a 6*5 rectangle and B could be a 3*5 rectangle, in which case A would have a bigger perimiter.
On the other hand, A could be a 6*5 rectangle and B could be a (.01)*(1500) rectangle, in which case B would have a bigger perimiter.
Therefore, (2) is also insufficient.
For combining, let's define some variables to make the discussion simpler:
Let p and q be the sides of Rectangle A and let x and y be the sides of Rectangle B.
From (2), we know that pq = 2(xy).
From (1), we know that one of p or q is 2 times one of x or y. We really don't care which is which, so let's say that p = 2x.
Subbing into (2), we now get:
2xq = 2xy
q = y
So, we know that one side of Rectangle A is bigger than a side of Rectangle B and the other two sides of the rectangles are equal. Therefore, Rectangle A must have a greater perimiter than Rectangle B. Together, we get a definite "yes" answer: choose (C).
