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# M05-15

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:25
2
6
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Difficulty:

65% (hard)

Question Stats:

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

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If $$x$$ is a positive integer, is $$\sqrt{x} \lt 2.5x - 5$$

(1) $$x \lt 3$$

(2) $$x$$ is a prime number

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Joined: 02 Sep 2009
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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.

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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.

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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Joined: 02 Sep 2009
Posts: 51261

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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.

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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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

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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
Math Expert
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Posts: 51261

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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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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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Math Expert
Joined: 02 Sep 2009
Posts: 51261

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

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