Find the area of the rectangle ABCD, if the length and breadth are odd positive integers.
(1) The perimeter of rectangle ABCD = 28 cm
(2) The area of the square whose side length is equal to the length of the rectangle, is 80% greater than the area of the rectangle.
OA C
Source: e-GMAT
Find the area of the rectangle ABCD, if the length and bread
This topic has expert replies
-
- Moderator
- Posts: 7187
- Joined: Thu Sep 07, 2017 4:43 pm
- Followed by:23 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
From Statement 1, 2L + 2W = 28, so L + W = 14. There are a few pairs of positive odd integers that add to 14, say 9 and 5 or 3 and 11, and these will give us different areas, so this is not sufficient.
From Statement 2, the area of an L by L square is 80% bigger than the area of our L by W rectangle. So
L^2 = 1.8 LW
L = 1.8W
L = 9W/5
Since 9W/5 is equal to the integer L, it must be true that W is divisible by 5. It's also an odd integer, but that's really all we know - W could be 5, and L could be 9, or W could be 15 and L could be 27, among other possibilities.
But combining the two statements, W must be 5 and L must be 9, so the answer is C.
From Statement 2, the area of an L by L square is 80% bigger than the area of our L by W rectangle. So
L^2 = 1.8 LW
L = 1.8W
L = 9W/5
Since 9W/5 is equal to the integer L, it must be true that W is divisible by 5. It's also an odd integer, but that's really all we know - W could be 5, and L could be 9, or W could be 15 and L could be 27, among other possibilities.
But combining the two statements, W must be 5 and L must be 9, so the answer is C.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com