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23 Aug 2021, 21:14
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
36% (01:41) correct
64%
(01:12)
wrong
based on 163
sessions
History
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Time
Result
Not Attempted Yet
Is a an integer?
(1) a^2 is an integer.
(2) 4.5a is an integer.
(1) a^2 is an integer.
(2) 4.5a is an integer.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
23 Aug 2021, 22:09
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Is a an integer?
(1) \(a^2\) is an integer.
a could be a square root => \(\sqrt{3}\) or an integer => 2, 3 and so on
Insufficient
(2) 4.5a is an integer.
\(4.5a=\frac{9a}{2}\) is an integer.
So a could be an even number => 2, 4 etc OR
a could be a fraction with numerator as multiple of 2 and denominator, a multiple of 3 or 9 in its simplest form => \(\frac{2}{3}; \frac{8}{9}\) and so on.
Insufficient
Combined
Either we can straight way deduce that a is square root of an integer or integer, but statement II says a can be a fraction or an integer, we can say combine we get a as integer or PROVE it algebraically as below.
a^2 is an integer, but we are looking for a.
Let a^2=x=integer.
a=\(\sqrt{x}\). We know from statement II that a can be a fraction or integer.
But can a square root of an integer x be a fraction. No, it will be an irrational number.
So \(\sqrt{x}\) or a has to be an integer.
Sufficient
C
(1) \(a^2\) is an integer.
a could be a square root => \(\sqrt{3}\) or an integer => 2, 3 and so on
Insufficient
(2) 4.5a is an integer.
\(4.5a=\frac{9a}{2}\) is an integer.
So a could be an even number => 2, 4 etc OR
a could be a fraction with numerator as multiple of 2 and denominator, a multiple of 3 or 9 in its simplest form => \(\frac{2}{3}; \frac{8}{9}\) and so on.
Insufficient
Combined
Either we can straight way deduce that a is square root of an integer or integer, but statement II says a can be a fraction or an integer, we can say combine we get a as integer or PROVE it algebraically as below.
a^2 is an integer, but we are looking for a.
Let a^2=x=integer.
a=\(\sqrt{x}\). We know from statement II that a can be a fraction or integer.
But can a square root of an integer x be a fraction. No, it will be an irrational number.
So \(\sqrt{x}\) or a has to be an integer.
Sufficient
C
23 Aug 2021, 23:00
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ajaanrkhan
(1) a^2 is an integer.
it can be root 2 or 2 itself
Clearly insufficient
(2) 4.5a is an integer.
it could be 2 or 2/9
Clearly insufficient
However when combined it cannot be a fraction since if it's squared the denominator will be present
and it cannot take root values since then 4.5 *a cannot be an integer
Therefore the only possibility being an integer
Hence IMO C
24 Aug 2021, 12:47
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ajaanrkhan
Statement 1:
Let \(i = 0, 1, 2, 3 ...\). We are given \(a^2 = i\) and \(a = 0, 1, \sqrt{2}, \sqrt{3}...\). Therefore \(a\) is not always an integer. Insufficient.
Statement 2:
Using the method above, we have \(a = 0, \frac{1}{4.5}, \frac{2}{4.5}, \frac{3}{4.5} ...., \frac{9}{4.5} = 2, ...\). Then \(a\) is not always an integer. Insufficient.
Combined:
We can see fractions and irrationals will not have any intersections. The only intersection between these two sets are the integers, thus \(a\) can only be an integer. Sufficient.
Ans: C
