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amp0201
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If x is a positive integer, is x^1/2 an integer (1) 28 Mar 2011, 11:34
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If x is a positive integer, is x^1/2 an integer

(1) (36x)^1/2 is an integer

(2) (3x + 4) ^ 1/2 is an integer

1. Statement 1 ALONE is sufficient to answer the question, but statement 2 alone is NOT sufficient.
2. Statement 2 ALONE is sufficient to answer the question, but statement 1 alone is NOT sufficient.
3. BOTH statements 1 and 2 TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
4. Each statement ALONE is sufficient to answer the question.
5. Statement 1 and 2 TOGETHER are NOT sufficient to answer the question.



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A
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fluke
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If x is a positive integer, is x^1/2 an integer (1) 28 Mar 2011, 13:02
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amp0201
If x is a positive integer, is x^1/2 an integer

(1) (36x)^1/2 is an integer

(2) (3x + 4) ^ 1/2 is an integer

1. Statement 1 ALONE is sufficient to answer the question, but statement 2 alone is NOT sufficient.
2. Statement 2 ALONE is sufficient to answer the question, but statement 1 alone is NOT sufficient.
3. BOTH statements 1 and 2 TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
4. Each statement ALONE is sufficient to answer the question.
5. Statement 1 and 2 TOGETHER are NOT sufficient to answer the question.
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Q: Is \(sqrt{x}\) an integer?

1.
\(sqrt{36x}=6sqrt{x}=Integer\)
6 is an integer. Integer*Integer=Integer; 6*Integer=Integer
Thus,
\(sqrt{x}\) must be an integer.
Sufficient.

2.
\(sqrt{3x + 4}=Integer\)
x may or may not be an integer.
For x=4; \(sqrt{3*4 + 4}=sqrt{16}=4=Integer\); \(sqrt{x}=sqrt{4}=2=Integer\)
For x=7; \(sqrt{3*7 + 4}=sqrt{25}=5=Integer\); But, \(sqrt{x}=sqrt{7}=Non Integer\)
Not Sufficient.

Ans: "A"
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If x is a positive integer, is x^1/2 an integer (1) 28 Mar 2011, 14:55
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I have one question over here in option A -

1. 6 * [x^1/2] = integer.

now 6 * 1.5 = 9 (integer)

so how can x^1/2 still be an integer.
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If x is a positive integer, is x^1/2 an integer (1) 28 Mar 2011, 15:06
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amp0201
I have one question over here in option A -

1. 6 * [x^1/2] = integer.

now 6 * 1.5 = 9 (integer)

so how can x^1/2 still be an integer.
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Because; x is a +ve integer. (1.5)^2=2.25 is not an integer.
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If x is a positive integer, is x^1/2 an integer (1) 04 Apr 2011, 21:37
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For me, instead of guessing possible values where "(3x + 4) ^ 1/2 is an integer" is true/false -
Let (3x + 4) ^ 1/2 =Y
=>3x+4=Y^2
=>x=(Y^2-4)/3
=>x=(Y-2)*(Y+2)/3
Now, its easy to fine two values of Y where either of Y+2 or Y-2 is a multiple of 3 so that x is an integer.
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If x is a positive integer, is x^1/2 an integer (1) 05 Apr 2011, 03:40
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fluke


1.
\(sqrt{36x}=6sqrt{x}=Integer\)
6 is an integer. Integer*Integer=Integer; 6*Integer=Integer
Thus,
\(sqrt{x}\) must be an integer.
Sufficient.
Show more

from the question S1 it can be deduced that 6*x^1/2=interger
but to say integer*interger=integer,hence x^1/2 is also integer is not the correct way to put it.
integer*noninteger can also be = integer,so x^1/2 can be noninteger as well.

but for x^1/2 to be non integer, the denominator of the fraction represented by x^1/2 should be 2/3or6.(since 6* x^1/2 = integer)
it is given that x is integer,so above case it not possible.
so it can be concluded that x^1/2 is also integer
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If x is a positive integer, is x^1/2 an integer (1) 05 Apr 2011, 07:33
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1. sufficient
36x is a perfect square
=> x is a perfect square as 36 is perfect square

2. Not sufficient

3x+4 is a perfect square, but that doesnt mean x is a square.

Answer A

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