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M05-15

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M05-15  [#permalink]

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New post 16 Sep 2014, 00:25
3
8
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A
B
C
D
E

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  55% (hard)

Question Stats:

59% (01:41) correct 42% (01:48) wrong based on 200 sessions

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Re M05-15  [#permalink]

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New post 16 Sep 2014, 00:25
2
Official Solution:


First of all notice that since \(x\) is a positive number then \(\sqrt{x} \gt 0\).

(1) \(x \lt 3\). Since \(x\) is a positive integer then \(x=1\) or \(x=2\). For both those values, the right hand side (\(2.5x - 5\)) is less than or equal to zero, so it cannot be more than the left hand side (\(\sqrt{x}\)) which is positive. Hence the answer to the question is NO. Sufficient.

(2) \(x\) is a prime number. If \(x=2\) then the answer is NO but if \(x=11\) then the answer is YES. Not sufficient.


Answer: A
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M05-15  [#permalink]

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New post 03 Sep 2015, 07:48
Bunuel wrote:
Official Solution:


First of all notice that since \(x\) is a positive number then \(\sqrt{x} \gt 0\).

(1) \(x \lt 3\). Since \(x\) is a positive integer then \(x=1\) or \(x=2\). For both those values, the right hand side (\(2.5x - 5\)) is less than or equal to zero, so it cannot be more than the left hand side (\(\sqrt{x}\)) which is positive. Hence the answer to the question is NO. Sufficient.

(2) \(x\) is a prime number. If \(x=2\) then the answer is NO but if \(x=11\) then the answer is YES. Not sufficient.


Answer: A


Hey Bunuel,

Thanks for your help on this forum. Maybe I am wrong, but doesn't x>0 only tell me that the square root has a real value?
So, as far as (1) why don't you consider negative value of \sqrt{x}? For example \sqrt{2}=+ or - 1.4 and this could be < or > 0.

Thanks!
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Re: M05-15  [#permalink]

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New post 03 Sep 2015, 23:34
gmatminator wrote:
Bunuel wrote:
Official Solution:


First of all notice that since \(x\) is a positive number then \(\sqrt{x} \gt 0\).

(1) \(x \lt 3\). Since \(x\) is a positive integer then \(x=1\) or \(x=2\). For both those values, the right hand side (\(2.5x - 5\)) is less than or equal to zero, so it cannot be more than the left hand side (\(\sqrt{x}\)) which is positive. Hence the answer to the question is NO. Sufficient.

(2) \(x\) is a prime number. If \(x=2\) then the answer is NO but if \(x=11\) then the answer is YES. Not sufficient.


Answer: A


Hey Bunuel,

Thanks for your help on this forum. Maybe I am wrong, but doesn't x>0 only tell me that the square root has a real value?
So, as far as (1) why don't you consider negative value of \sqrt{x}? For example \sqrt{2}=+ or - 1.4 and this could be < or > 0.

Thanks!


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice though, that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Re: M05-15  [#permalink]

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New post 04 Aug 2016, 13:30
Can we solve this question as below:-

Question stem asks if √x <2.5x−5
This means, we need to prove if 2.5x - 5>0 => if x > 2? (we don't have equality sign because question mentions that x is a positive integer. This means >0).

Statement 1. x <3
As x is a positive integer, possible values of x are 1 and 2 and none of these is larger than 2. So answer to the question asked is NO. - Sufficient

Statement 2. x is a prime number.
x can be any prime number including 2 so we don't know if x>2 or not. - Insufficient

Answer - A
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Re: M05-15  [#permalink]

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New post 23 Jul 2017, 01:49
hi brunel

Question stem asks if √x <2.5x−5
we know x is positive , so squaring both sides gives x< 6.25x^2 + 25 - 25x i.e 26x< 6.25x^2 + 25

so if put 1 answer is yes. for x = 2 answer is no.How can answer be A
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Re: M05-15  [#permalink]

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New post 23 Jul 2017, 05:52
sidagar wrote:
hi brunel

Question stem asks if √x <2.5x−5
we know x is positive , so squaring both sides gives x< 6.25x^2 + 25 - 25x i.e 26x< 6.25x^2 + 25

so if put 1 answer is yes. for x = 2 answer is no.How can answer be A


Why are you squaring? We can only square inequality when we are certain that both sides are non-negative. We don't know that.

Also, you really do not need to square put x = 1 and x = 2 into √x <2.5x −5. What do you get?
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Re: M05-15  [#permalink]

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New post 02 Sep 2017, 10:50
Bunuel wrote:
sidagar wrote:
hi brunel

Question stem asks if √x <2.5x−5
we know x is positive , so squaring both sides gives x< 6.25x^2 + 25 - 25x i.e 26x< 6.25x^2 + 25

so if put 1 answer is yes. for x = 2 answer is no.How can answer be A


Why are you squaring? We can only square inequality when we are certain that both sides are non-negative. We don't know that.

Also, you really do not need to square put x = 1 and x = 2 into √x <2.5x −5. What do you get?



Hey Bunuel- I know there is no need to square here as the stem is rather simple. But just to correct you, you can actually square in this case. LHS is a square root, which is definitely positive. RHS is given to be greater than LHS, so obviously RHS is greater than a positive number, so there is no problem with squaring this inequality. Thanks
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Re: M05-15  [#permalink]

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New post 02 Sep 2017, 11:11
bkpolymers1617 wrote:
Bunuel wrote:
sidagar wrote:
hi brunel

Question stem asks if √x <2.5x−5
we know x is positive , so squaring both sides gives x< 6.25x^2 + 25 - 25x i.e 26x< 6.25x^2 + 25

so if put 1 answer is yes. for x = 2 answer is no.How can answer be A


Why are you squaring? We can only square inequality when we are certain that both sides are non-negative. We don't know that.

Also, you really do not need to square put x = 1 and x = 2 into √x <2.5x −5. What do you get?



Hey Bunuel- I know there is no need to square here as the stem is rather simple. But just to correct you, you can actually square in this case. LHS is a square root, which is definitely positive. RHS is given to be greater than LHS, so obviously RHS is greater than a positive number, so there is no problem with squaring this inequality. Thanks


No, you are not right. If you tried and solve with and without squaring you'd get different ranges.

Simple example:
\(\sqrt{x}<x\) --> x>1

But if you square you'll get: \(x<x^2\) --> x<0 or x>1.
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Re: M05-15  [#permalink]

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New post 22 Aug 2018, 15:09
How can you do this quickly? Without testing cases?
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Re: M05-15  [#permalink]

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New post 01 Sep 2019, 11:11
sarimzahid2395 wrote:
How can you do this quickly? Without testing cases?


sarimzahid2395 - The point here is that testing cases is a time-efficient method of solving the question. I like to say that being a good reader is what helps boost your Quant scores, and this problem is a case in point. Remember, it is analytical reasoning that lies at the heart of the test, not necessarily mathematical prowess. Here, the keywords about the unknown are positive and integer. Statement (1) restricts those infinite possible positive integers to just two: 1 or 2, since 0 is not considered positive. The 1 breaks the square root down without performing any complicated analysis. The problem then reads, "is 1 < 2.5(1) - 5?" Reminder: whenever the square root symbol appears on the GMAT™, as Bunuel pointed out above in another response, only the positive root is considered. Getting back to the reformed question, it is clear that 1 is not, in fact, less than -2.5, but we have a definitive answer. The only way to overturn (A) as an answer is if 2 leads to a different scenario. In the question, "is √2 < 2.5(2) - 5," the answer will again come to no, since √2 is greater than 0. A "no" response to both 1 and 2, the only possible x values, provides a consistent picture, so the answer for now is that (A), Statement (1), is SUFFICIENT.

Looking at Statement (2), our prime positive integers, combining the information in the statement with that of the problem, will be 2, 3, 5, 7, 11, and so on, once again stretching to infinity. We have already tested 2, so we can test any other valid number in the list of primes and see whether we also get a consistent picture:

is √3 < 2.5(3) - 5?

Just by focusing on the right-hand side, we can quickly determine that 2.5(3) - 5 = 2.5, and the question thus becomes,

is √3 < 2.5?

You do not have to be a math genius to see that √3 must be less than 2.5, since √4 would give us 2. Hence, we now have a yes answer to the same question we had a no to before, with 2, from this very statement. We can say goodbye to (D), then, and choose (A) as the answer.

I am no Quant maven, even if I aspire to be one. But with a little number sense, I cracked this one in 1:16 without writing down a thing. The concepts and vocabulary that are brought to bear in the question are simple enough. It is a mistake to think that you have to figure out everything all the time when you just need to answer the question that is being asked.

Good luck in your studies.

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Re: M05-15   [#permalink] 01 Sep 2019, 11:11
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