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Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Re: Is (7x)^1/2 an integer? [#permalink]
13 Jan 2013, 07:58

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Is \sqrt{7x} an integer?

Notice that we are not told that x is an integer.

(1) \sqrt{\frac{x}{7}} is an integer. Given that \sqrt{\frac{x}{7}}=integer --> square it: \frac{x}{7}=integer^2 --> x=7*integer^2. So, \sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer. Sufficient.

(2) \sqrt{28x} is an integer. If x=\frac{1}{28}, then \sqrt{7x}=\frac{1}{2}\neq{integer} BUT if x=0, then \sqrt{7x}=0=integer. Not sufficient.

Re: Is (7x)^1/2 an integer? [#permalink]
13 Jan 2013, 08:16

Bunuel wrote:

Is \sqrt{7x} an integer?

Notice that we are not told that x is an integer.

(1) \sqrt{\frac{x}{7}} is an integer. Given that \sqrt{\frac{x}{7}}=integer --> square it: \frac{x}{7}=integer^2 --> x=7*integer^2. So, \sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer. Sufficient.

(2) \sqrt{28x} is an integer. If x=\frac{1}{28}, then \sqrt{7x}=\frac{1}{2}\neq{integer} BUT if x=0, then \sqrt{7x}=0=integer. Not sufficient.

Answer: A.

Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Re: Is (7x)^1/2 an integer? [#permalink]
13 Jan 2013, 08:29

2

This post received KUDOS

Expert's post

alexpavlos wrote:

Bunuel wrote:

Is \sqrt{7x} an integer?

Notice that we are not told that x is an integer.

(1) \sqrt{\frac{x}{7}} is an integer. Given that \sqrt{\frac{x}{7}}=integer --> square it: \frac{x}{7}=integer^2 --> x=7*integer^2. So, \sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer. Sufficient.

(2) \sqrt{28x} is an integer. If x=\frac{1}{28}, then \sqrt{7x}=\frac{1}{2}\neq{integer} BUT if x=0, then \sqrt{7x}=0=integer. Not sufficient.

Answer: A.

Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Thanks again! Alex

If x is NOT an integer you cannot make its prime factorization.

If we were told that x IS an integer, then \sqrt{28x}=2\sqrt{7x}=integer. Thus x must be of the form 7^{odd}*integer^2, so in this case \sqrt{7x}=\sqrt{7*(7^{odd}*integer^2)}=\sqrt{7^{even}*integer^2}=integer.

Re: Is (7x)^1/2 an integer? [#permalink]
26 May 2013, 20:17

alex1233 wrote:

Is \sqrt{7x} an integer?

(1) \sqrt{\frac{x}{7}} is an integer

(2) \sqrt{28x} is an integer

Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks! Alex

Quote:

\sqrt{7x} = integer ? or 7x= int^2 ?

(1) \sqrt{\frac{x}{7}} is an integer ---- lets say b x/7 = b^2 --> where b is an integer x= 7 *b^2 7x = (7*b) ^2 hence 7x=int^2--- sufficient

(2) \sqrt{28x} is an integer ---- lets say c

28x= c^2 7x=c^2/4 = (c/2)^2 If C is odd , then ans is No and if C is even then yes . ( If you knew that x was an integer , then this would have sufficed) Insufficient. _________________

Re: Is (7x)^1/2 an integer? [#permalink]
05 Jun 2013, 22:53

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Expert's post

alex1233 wrote:

Is \sqrt{7x} an integer?

(1) \sqrt{\frac{x}{7}} is an integer

(2) \sqrt{28x} is an integer

From F.S 1 , we know that \sqrt{\frac{x}{7}} = Integer(I)--> \sqrt{\frac{x*7}{7*7}} = \sqrt{7x}*\frac{1}{7} = I--> \sqrt{7x} = 7*I. Thus, an integer. Sufficient.

From F.S 2, we know that \sqrt{28x} = Integer(I) --> \sqrt{7*4x} = 2*\sqrt{7x} = I. Thus, \sqrt{7x} =\frac{I}{2}. Depending on I being odd/even, the given expression will be a non-integer/integer. Insufficient.