If b is an integer, is

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Jan 01, 2020 1:55 am

If b is an integer, is $\sqrt {a^2+b^2}$ an integer?

(1) a^2 + b^2 is an integer.
(2) a^2 - 3b^2 = 0

OA B

Source: Official Guide

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Source: Beat The GMAT — Data Sufficiency | Wed Jan 01, 2020 1:55 am

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Wed Jan 01, 2020 10:36 pm

BTGmoderatorDC wrote:If b is an integer, is $\sqrt {a^2+b^2}$ an integer?

(1) a^2 + b^2 is an integer.
(2) a^2 - 3b^2 = 0

OA B

Source: Official Guide

Let's take each statement one by one.

(1) a^2 + b^2 is an integer.

Case 1: Say a = âˆš3 and b = 1, then $\sqrt {a^2+b^2}= \sqrt {3+1}=2$, an integer. The answer is yes.
Case 2: Say a = 3 and b = 1, then $\sqrt {a^2+b^2}= \sqrt {9+1}=âˆš10$, not an integer. The answer is no.

No unique answer. Insufficient.

(2) a^2 - 3b^2 = 0

a^2 = 3b^2

Thus, $\sqrt {a^2+b^2}= \sqrt {4b^2}=2b$, an integer. The answer is yes. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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