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27 Apr 2020, 03:22
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
51% (01:22) correct
49%
(01:14)
wrong
based on 121
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x > 0\), is \(\sqrt{x}\) a positive integer?
(1) \(x^2\) is a positive integer
(2) \(x = m*b\) where \(m = b\)
Are You Up For the Challenge: 700 Level Questions
(1) \(x^2\) is a positive integer
(2) \(x = m*b\) where \(m = b\)
Are You Up For the Challenge: 700 Level Questions
27 Apr 2020, 03:46
Kudos
Bookmarks
Bunuel
(Statement1): \(x^2\) is a positive integer
\(x^{2}\) could be \(81\)
--> \(x=9\) --> \(√x= √9= 3\) (YES)
\(x^{2}\) could be \(9\)
--> \(x=3\) --> \(√x = √3\) (NO)
Insufficient
(Statement2): \(x = m*b\) where \(m = b\)
Nothing tells us about what m and b are.
x could be \(m*b= 3*3=9\)
--> \(√x =√9= 3\) (YES)
\(x\) could be \(m*b= √3*√3=3 \)
--> \(√x = √3\) (NO)
Insufficient
Taken together 1&2,
\(x\) could be \(m*b= 3*3=9\)
--> \(√x =√9= 3\) (YES)
\(x\) could be \(m*b= √3*√3=3\)
--> \(√x = √3\) (NO)
Insufficient
Answer (E)
27 Apr 2020, 04:38
Kudos
Bookmarks
Bunuel
Solution
Step 1: Analyse Question Stem
- • x > 0
- • For \(\sqrt{x}\) be an integer, x must be a perfect square.
- o We need to find whether x is a perfect square of an integer or not
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: \(x^2\) is a positive integer.
- • Case I: If x = 2, then \(x^2\) is positive integer and x is an integer
• Case II: If x = \(\sqrt{2}\), then \(x^2\) = 2 is an intger but x is not
Statement 2: x = m*b, where m = b
- • Since x is positive, the sign of m and b will same i.e., either both are positive or both are negative.
• \(x =b^2=m^2\).
- o \(\sqrt{x} = b = m\)
- o Thus, we cannot find the answer
Step 3: Analyse Statements by combining.
From statement 1: \(x ^2\) is a positive integer.
From statement 2: x = \(b^2 = m^2\)
- • We cannot find whether x is an integer or not
