if x is a positive integer, is sqrt{x} < 2.5x - 5 ?
1. x < 3
2. x is a prime number
help!
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- eagleeye
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Let's rephrase the question as Is sqrt(x) + 5 - 2.5x < 0mehaksal wrote:if x is a positive integer, is sqrt{x} < 2.5x - 5 ?
1. x < 3
2. x is a prime number
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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Hey eagleeye,eagleeye wrote:Let's rephrase the question as Is sqrt(x) + 5 - 2.5x < 0mehaksal wrote:if x is a positive integer, is sqrt{x} < 2.5x - 5 ?
1. x < 3
2. x is a prime number
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
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.
- anuprajan5
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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.
Regards
Anup
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.
Regards
Anup
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- Brent@GMATPrepNow
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Absolutely.anuprajan5 wrote:My assumption was that on the GMAT sqrt of a certain number is always positive. We do not take the negative number.
Example: sqrt(49) = 7 (not -7)
However, the equation x^2 = 49 has two solutions (x=7, -7)
Cheers,
Brent
- eagleeye
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Please read this post. It might help. sqrt(x) is never negative for a real number.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.
https://www.beatthegmat.com/squares-squa ... tml#476551