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Find the area of the rectangle ABCD, if the length and bread

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Find the area of the rectangle ABCD, if the length and bread

by BTGmoderatorDC » Thu Jun 06, 2019 5:46 am

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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

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Source: — Data Sufficiency |

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by Ian Stewart » Thu Jun 06, 2019 6:07 am

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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.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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