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help!

by mehaksal » Sun Aug 12, 2012 11:19 am
if x is a positive integer, is sqrt{x} < 2.5x - 5 ?

1. x < 3
2. x is a prime number

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by mehaksal » Sun Aug 12, 2012 11:21 am
m hell confused a to why many ppl r suggesting C???
is B a trap??

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by eagleeye » Sun Aug 12, 2012 4:03 pm
mehaksal wrote:if x is a positive integer, is sqrt{x} < 2.5x - 5 ?

1. x < 3
2. x is a prime number
Let's rephrase the question as Is sqrt(x) + 5 - 2.5x < 0

Given: x is a positive integer.

1. x < 3
Since x is a positive integer, x can either be 1 or 2.
For x=1, sqrt(1) + 5 -2.5 > 0
For x=2, sqrt(2) + 5 -2.5*2 > 0.
In both cases, the expression is > 0, sufficient.

2. x is a prime number.
We already tested for 2 above and got >0, let's test a larger number.

For x=3, sqrt(3) + 5 -2.5*3 = 1.732 + 5 - 7.5 <0. Hence we get opposite sign for x = 3. Insufficient.

A is correct.

Let me know if this helps :)

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by mehulsayani » Tue Aug 14, 2012 4:09 am
eagleeye wrote:
mehaksal wrote:if x is a positive integer, is sqrt{x} < 2.5x - 5 ?

1. x < 3
2. x is a prime number
Let's rephrase the question as Is sqrt(x) + 5 - 2.5x < 0

Given: x is a positive integer.

1. x < 3
Since x is a positive integer, x can either be 1 or 2.
For x=1, sqrt(1) + 5 -2.5 > 0
For x=2, sqrt(2) + 5 -2.5*2 > 0.
In both cases, the expression is > 0, sufficient.

2. x is a prime number.
We already tested for 2 above and got >0, let's test a larger number.

For x=3, sqrt(3) + 5 -2.5*3 = 1.732 + 5 - 7.5 <0. Hence we get opposite sign for x = 3. Insufficient.

A is correct.

Let me know if this helps :)
Hey eagleeye,
Here, we are only given that X is a positivenumber. How can you say that Squareroot(x) is also positive??

Since x is a positive integer, x can either be 1 or 2.
For x=1, sqrt(1) + 5 -2.5 => -1 +2.5 or 1+ 2.5 ==> always>0
For x=2, sqrt(2) + 5 -2.5*2 , -1.71 + 0 or 1.71 + 0, So its not always 0.
In both cases, the expression is > 0, is NOT sufficient.

2. even two is not sufficient

So, i feel the answer is E.

Correct me if I m wrong.

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by anuprajan5 » Tue Aug 14, 2012 4:30 am
My assumption was that on the GMAT sqrt of a certain number is always positive. We do not take the negative number.

My rationale:

x^2 = 16,

if we were to square root it, then sqrt (x^2) = mod x = 4, -4

If we were saying x = sqrt (16), then the number is only positive 4


Hence A should be sufficient.



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by Brent@GMATPrepNow » Tue Aug 14, 2012 7:23 am
anuprajan5 wrote:My assumption was that on the GMAT sqrt of a certain number is always positive. We do not take the negative number.
Absolutely.
Example: sqrt(49) = 7 (not -7)

However, the equation x^2 = 49 has two solutions (x=7, -7)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by eagleeye » Tue Aug 14, 2012 9:45 am
mehulsayani wrote: Hey eagleeye,
Here, we are only given that X is a positivenumber. How can you say that Squareroot(x) is also positive??

Correct me if I m wrong.
Please read this post. It might help. sqrt(x) is never negative for a real number.

https://www.beatthegmat.com/squares-squa ... tml#476551