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Is a an integer? 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)!

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85% (hard)

Question Stats:

36% (01:40) correct 64% (01:12) wrong based on 160 sessions

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Is a an integer?

(1) a^2 is an integer.

(2) 4.5a is an integer.
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C
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Is a an integer? 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
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Is a an integer?
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(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
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Is a an integer? 24 Aug 2021, 12:47
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Is a an integer?

(1) a^2 is an integer.

(2) 4.5a is an integer.
Show more

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
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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
\(a^2=x.....a=\sqrt{x}\)
Also \(\frac{9a}{2}\) is an integer say y => \(9a=2y\)
\(9\sqrt{x}=2y=integer\)
Since x is an integer, \(\sqrt{x}\) cannot be a fraction, and hence :-
\(9\sqrt{x}=integer\) tells us that \(\sqrt{x}\) or a is an integer.
Sufficient


C
Show more

Hi chetan2u,
Can you please explain the step -
"Since x is an integer, √x cannot be a fraction, and hence :-
9√x=integer 9x=integer tells us that √x or a is an integer. "

It seems that we started considering √x as integer and then we proved the same.
Pardon , if i misunderstood this concept.
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Is a an integer? 28 Aug 2021, 08:56
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omagra
chetan2u
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
\(a^2=x.....a=\sqrt{x}\)
Also \(\frac{9a}{2}\) is an integer say y => \(9a=2y\)
\(9\sqrt{x}=2y=integer\)
Since x is an integer, \(\sqrt{x}\) cannot be a fraction, and hence :-
\(9\sqrt{x}=integer\) tells us that \(\sqrt{x}\) or a is an integer.
Sufficient


C

Hi chetan2u,
Can you please explain the step -
"Since x is an integer, √x cannot be a fraction, and hence :-
9√x=integer 9x=integer tells us that √x or a is an integer. "

It seems that we started considering √x as integer and then we proved the same.
Pardon , if i misunderstood this concept.
Show more


Hi,

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 \(\sqrt{x}\) has to be an integer.

So we started with x as an integer and deduced that even \(\sqrt{x}\) is an integer, making a as an integer.
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chetan2u
omagra
chetan2u
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
\(a^2=x.....a=\sqrt{x}\)
Also \(\frac{9a}{2}\) is an integer say y => \(9a=2y\)
\(9\sqrt{x}=2y=integer\)
Since x is an integer, \(\sqrt{x}\) cannot be a fraction, and hence :-
\(9\sqrt{x}=integer\) tells us that \(\sqrt{x}\) or a is an integer.
Sufficient


C

Hi chetan2u,
Can you please explain the step -
"Since x is an integer, √x cannot be a fraction, and hence :-
9√x=integer 9x=integer tells us that √x or a is an integer. "

It seems that we started considering √x as integer and then we proved the same.
Pardon , if i misunderstood this concept.


Hi,

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 \(\sqrt{x}\) has to be an integer.

So we started with x as an integer and deduced that even \(\sqrt{x}\) is an integer, making a as an integer.
Show more

Thanks chetan2u.
Now I got it.
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