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M05-15

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M05-15  [#permalink]

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New post 15 Sep 2014, 23:25
2
6
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A
B
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D
E

Difficulty:

  65% (hard)

Question Stats:

62% (01:10) correct 38% (01:09) wrong based on 172 sessions

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Re M05-15  [#permalink]

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New post 15 Sep 2014, 23:25
1
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, 06:48
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, 22:34
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, 12:30
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, 00:49
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, 04:52
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, 09:50
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, 10:11
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M05-15  [#permalink]

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New post 22 Aug 2018, 14:09
How can you do this quickly? Without testing cases?
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Re: M05-15 &nbs [#permalink] 22 Aug 2018, 14:09
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