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A rectangle is defined to be "silver" if and only

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by BTGmoderatorDC » Sun Sep 16, 2018 12:29 am

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Answers

A

B

C

D

E

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Difficulty—

A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is rectangle R silver?

(1) R has the same area as S
(2) The ratio of one side of R to one side of S is 2 to 1.

OA E

Source: Veritas Prep
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by Jay@ManhattanReview » Sun Sep 16, 2018 3:58 am
BTGmoderatorDC wrote:A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is a rectangle R silver?

(1) R has the same area as S
(2) The ratio of one side of R to one side of S is 2 to 1.

OA E

Source: Veritas Prep
Given: Rectangle S is silver, i.e., the ratio of its length to its width is 2 to 1

Question: Is a rectangle R silver?

Question rephrased: Does rectangle R has the ratio of its length to its width is 2 to 1?

Let's take each statement one by one.

(1) R has the same area as S.

Case 1: Say the length and the width of the rectangle S and R are 2 and 1, respective. The answer is yes.
Case 2: Say the length and the width of the rectangle S are 2 and 1, respective. However, the length and the width of the rectangle R are 1 and 2, respective. The answer is no.

Insufficient.

(2) The ratio of one side of R to one side of S is 2 to 1.

Cases discussed above are applicable here too. Insufficient.

(1) and (2) together

Cases discussed above are applicable here too. Insufficient.

The correct answer: E

Hope this helps!

-Jay
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by fskilnik@GMATH » Sun Sep 16, 2018 11:08 am
BTGmoderatorDC wrote:A rectangle is defined to be "silver" if and only if the ratio of its length to its width is 2 to 1. If rectangle S is silver, is rectangle R silver?

(1) R has the same area as S
(2) The ratio of one side of R to one side of S is 2 to 1.

Source: Veritas Prep
Image

The correct answer is therefore [spoiler]__(E)______[/spoiler] .


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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