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E
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Difficulty:
95%
(hard)
Question Stats:
17% (01:39) correct
83%
(02:02)
wrong
based on 52
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\(a\) and \(b\) are integers. \([x]\) is an integer less than or equal to \(x\). Is \([\frac{a}{b}] \geq {1}\)?
(1) \(ab = 64\)
(2) \(a=b^2\)
(1) \(ab = 64\)
(2) \(a=b^2\)
03 Sep 2018, 05:01
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Official Solution:
We’ll go for LOGICAL because there is a logic to understanding the operator.
Since \([\frac{a}{b}]\) is defined as an integer either equal to or less than \(\frac{a}{b}\), no possible value of \((a,b)\) can ever make \([\frac{a}{b}]\) necessarily greater than any integer. This means a definitive answer to the question stem, if there is sufficient information, can only be ‘NO!’ – when \([\frac{a}{b}] < {1}\). This would be the case if \(b\) is greater than \(a\). (1) gives us no such information, and (2) gives us the opposite: the greater integer \(b\) is, integer \(a\) becomes even greater. Thus, combining the two we’ll still get \(a > b\). Therefore, (E) is correct.
Answer: E
We’ll go for LOGICAL because there is a logic to understanding the operator.
Since \([\frac{a}{b}]\) is defined as an integer either equal to or less than \(\frac{a}{b}\), no possible value of \((a,b)\) can ever make \([\frac{a}{b}]\) necessarily greater than any integer. This means a definitive answer to the question stem, if there is sufficient information, can only be ‘NO!’ – when \([\frac{a}{b}] < {1}\). This would be the case if \(b\) is greater than \(a\). (1) gives us no such information, and (2) gives us the opposite: the greater integer \(b\) is, integer \(a\) becomes even greater. Thus, combining the two we’ll still get \(a > b\). Therefore, (E) is correct.
Answer: E
