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24 Dec 2017, 01:44
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
47% (01:40) correct
53%
(01:30)
wrong
based on 134
sessions
History
Date
Time
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If z is a positive integer, is \(\sqrt{z}\) an integer?
(1) \(\sqrt{3z}\) is an integer.
(2) \(\sqrt{4z}\) is not an integer.
(1) \(\sqrt{3z}\) is an integer.
(2) \(\sqrt{4z}\) is not an integer.
24 Dec 2017, 02:59
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Bunuel
Statement 1: \(\sqrt{3z}=Integer\)
\(=>\sqrt{z}=\frac{Integer}{\sqrt{3}}=>\frac{Integer}{irrational}\).
This will not be an integer. sufficient
Statement 2: \(\sqrt{4z}=non{ }integer=>2\sqrt{z}=noninteger\)
\(\sqrt{z}=\frac{noninteger}{2}=>\frac{noninteger}{integer}\). This will not be an integer. Sufficient
Option D
24 Dec 2017, 07:05
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Answer is D.
1. says √3z = I (I =Integer)
That is 3z must be square no to become √3z Integer. For example, z = 3, 12 etc. SUFFICIENT
2. √4z is not an Integer
For that, √4z must not be a square no. SUFFICIENT
1. says √3z = I (I =Integer)
That is 3z must be square no to become √3z Integer. For example, z = 3, 12 etc. SUFFICIENT
2. √4z is not an Integer
For that, √4z must not be a square no. SUFFICIENT
