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# If n is positive, is root(n) > 100 ?

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If n is positive, is root(n) > 100 ? [#permalink]

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11 Jan 2011, 17:57
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

Data Sufficiency
Question: 51
Page: 156
Difficulty: 650

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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If n is positive, is root(n) > 100 ? [#permalink]

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11 Jan 2011, 18:08
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If n is positive, is $$\sqrt {n} > 100$$?

Is $$\sqrt {n} > 100$$? --> is $$n>100^2$$?

(1) $$\sqrt {n-1} > 99$$ --> $$n-1>99^2$$ --> $$n>99^2+1$$: $$99^2+1$$ is less than $$100^2$$ (as $$100^2=(99+1)^2=99^2+2*99+1$$), so $$n$$ may or may not be more than this value. Not sufficient.

(2) $$\sqrt {n+1} > 101$$ --> $$n+1>101^2$$ --> $$n>101^2-1=(101-1)(101+1)=100*102$$, so $$n>100*102>100^2$$. Sufficient.

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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11 Jan 2011, 19:30
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tonebeeze wrote:
If n is positive, is $$\sqrt {n} > 100$$?

1. $$\sqrt {n-1} > 99$$

2. $$\sqrt {n+1} > 101$$

Can someone write out the algebra on this one, I just want to double check work. Thanks.

Even though you asked for algebra, let me point out that you don't really need algebra to solve this.

When you are considering big numbers, square roots of consecutive numbers differ by very little. e.g. $$\sqrt {10000} = 100$$ and $$\sqrt {9999} = 99.995$$... Square roots of even small positive numbers differ by less than 1 e.g. $$\sqrt {1} = 1$$ and $$\sqrt {2} = 1.414$$ - Difference of just 0.414 $$\sqrt {3} = 1.732$$ - Difference of just 0.318. The difference just keeps getting smaller and smaller.

So if $$\sqrt {n-1} > 99$$, $$\sqrt {n}$$ will be very close to $$\sqrt {n-1}$$ and will also be greater than 99. But will it be greater than 100, we cannot say. So not sufficient.

If $$\sqrt {n+1} > 101$$, then $$\sqrt {n}$$ may not be greater than 101, but it will definitely be greater than 100 since between two consecutive integers, the square root difference will not reach 1 (as shown above).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17815 [6], given: 235 Manager Joined: 15 Apr 2011 Posts: 70 Kudos [?]: 20 [0], given: 45 Re: If n is positive, is root(n) > 100 ? [#permalink] ### Show Tags 09 Apr 2012, 12:07 I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks. _________________ http://mymbadreamz.blogspot.com Kudos [?]: 20 [0], given: 45 Math Expert Joined: 02 Sep 2009 Posts: 42264 Kudos [?]: 132782 [2], given: 12372 Re: If n is positive, is root(n) > 100 ? [#permalink] ### Show Tags 09 Apr 2012, 12:19 2 This post received KUDOS Expert's post mymbadreamz wrote: I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks. Question asks whether $$n>100^2$$. From (1) we have that $$n>99^2+1$$. Now, since $$100^2>99^2+1$$, then it's possible that $$n>100^2>99^2+1$$, which would mean that the answer is YES, as well as that $$100^2>n>99^2+1$$, which would mean that the answer is NO. Two different answers, hence not sufficient. Hope it's clear. _________________ Kudos [?]: 132782 [2], given: 12372 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7738 Kudos [?]: 17815 [2], given: 235 Location: Pune, India Re: If n is positive, is root(n) > 100 ? [#permalink] ### Show Tags 09 Apr 2012, 22:05 2 This post received KUDOS Expert's post mymbadreamz wrote: I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks. Stmnt 1 just tells you that $$\sqrt{n-1} > 99$$ Think of 2 diff cases: $$\sqrt{n-1} = 99.2$$ $$\sqrt{n} = 99.205$$ or $$\sqrt{n-1} = 112$$ $$\sqrt{n} = 112.015$$ Can you say whether $$\sqrt{n}$$ is greater than 100? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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23 Jan 2014, 03:23
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

Data Sufficiency
Question: 51
Page: 156
Difficulty: 650

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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23 Jan 2014, 03:24
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SOLUTION

If n is positive, is $$\sqrt {n} > 100$$?

Is $$\sqrt {n} > 100$$? --> is $$n>100^2$$?

(1) $$\sqrt {n-1} > 99$$ --> $$n-1>99^2$$ --> $$n>99^2+1$$: $$99^2+1$$ is less than $$100^2$$ (as $$100^2=(99+1)^2=99^2+2*99+1$$), so $$n$$ may or may not be more than this value. Not sufficient.

(2) $$\sqrt {n+1} > 101$$ --> $$n+1>101^2$$ --> $$n>101^2-1=(101-1)(101+1)=100*102$$, so $$n>100*102>100^2$$. Sufficient.

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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23 Jan 2014, 04:33
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Bunuel wrote:
If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

Sol: We need to find whether $$\sqrt {n} > 100$$------> n>10000 (We can square as both sides are positive

St 1 says $$\sqrt {n-1} > 99$$------> Squaring we get n-1 > (99)^2 or n > (99)^2 +1 So n can be greater than or less than 10000
A and D ruled out

St 2 says $$\sqrt {n+1} > 101$$-----> Squaring we get n+1>(101)^2 -----> n > (101)^2-1. Now (101)^2 -1 is greater than (100)^2

Hence Ans B

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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24 Jan 2014, 00:41
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Bunuel wrote:
If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

[/textarea]

for $$\sqrt{n}>100$$, n would have to be greater than 10000 as $$\sqrt{10000}=100$$or $$\sqrt{100 *100}=100$$

so we need to know if n>100*100 or not.

1)on solving we get n>9802 , so it may or may not be greater than 10,000 , insufficient.
2)On solving we get n>10,200, so n is definitely more than 10,000 sufficient

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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24 Jan 2014, 01:42
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If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2)$$\sqrt {n+1} > 101$$

To prove: $$\sqrt {n} > 100$$
Given that, n is positive, hence squaring both sides, the inequality that is required to be determined $$n > 10000$$

Statement 1

$$\sqrt {n-1} > 99$$
Or, $$n-1 > 99^2$$
Or,$$n-1 > 9801$$
Or, $$n > 9802$$

Since, n can be any number greater than 9802, it is not possible to evaluate $$n > 10000$$; answer is 'no' for values of n in the range 9802< n < 10000 and answer is 'yes' for values of n in the range n > = 10000

Statement (1) is not sufficient.................................(A)(D))

Statement 2

$$\sqrt {n+1} > 101$$
Or, $$n +1 > 101^2$$

Or, $$n +1 > 10201$$
Or,$$n > 10200$$
=> $$n > 10000$$
Or, $$\sqrt {n} > 100$$

Statement (2) alone is sufficient.................................(C)(E))

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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29 Jul 2014, 03:45
First, restate the question: Is $$\sqrt {n} > 100$$ ? --> Is $$n > 10,000$$ ?

(1) $$\sqrt {n-1} > 99$$ --> $$n-1 > 9801$$ Clearly IS.
(2) $$\sqrt {n+1} > 101$$ --> $$n+1 > 10,201$$ --> $$n > 10,200$$ SUFF.

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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08 Sep 2015, 12:49
Bunuel wrote:
SOLUTION

If n is positive, is $$\sqrt {n} > 100$$?

Is $$\sqrt {n} > 100$$? --> is $$n>100^2$$?

(1) $$\sqrt {n-1} > 99$$ --> $$n-1>99^2$$ --> $$n>99^2+1$$: $$99^2+1$$ is less than $$100^2$$ (as $$100^2=(99+1)^2=99^2+2*99+1$$), so $$n$$ may or may not be more than this value. Not sufficient.

(2) $$\sqrt {n+1} > 101$$ --> $$n+1>101^2$$ --> $$n>101^2-1=(101-1)(101+1)=100*102$$, so $$n>100*102>100^2$$. Sufficient.

Can someone please explain why we are considering $$\sqrt{n}$$ positive ?
Suppose if n is 9, then $$\sqrt{n}$$ can be +3 and -3.
Now, according to the rule, if sign on both sides of inequality is unknown, then we cannot square on both sides.
So, in this case $$\sqrt{n}$$ can be +ve or -ve. So we do not know the sign for sure and thus we cannot square on both sides as explained in the solution.
Please correct my concepts and explain what am i doing wrong.

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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08 Sep 2015, 13:53
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kunal555 wrote:
Bunuel wrote:
SOLUTION

If n is positive, is $$\sqrt {n} > 100$$?

Is $$\sqrt {n} > 100$$? --> is $$n>100^2$$?

(1) $$\sqrt {n-1} > 99$$ --> $$n-1>99^2$$ --> $$n>99^2+1$$: $$99^2+1$$ is less than $$100^2$$ (as $$100^2=(99+1)^2=99^2+2*99+1$$), so $$n$$ may or may not be more than this value. Not sufficient.

(2) $$\sqrt {n+1} > 101$$ --> $$n+1>101^2$$ --> $$n>101^2-1=(101-1)(101+1)=100*102$$, so $$n>100*102>100^2$$. Sufficient.

Can someone please explain why we are considering $$\sqrt{n}$$ positive ?
Suppose if n is 9, then $$\sqrt{n}$$ can be +3 and -3.
Now, according to the rule, if sign on both sides of inequality is unknown, then we cannot square on both sides.
So, in this case $$\sqrt{n}$$ can be +ve or -ve. So we do not know the sign for sure and thus we cannot square on both sides as explained in the solution.
Please correct my concepts and explain what am i doing wrong.

This excerpt is from GMATCLUB math book: "When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}$$=5, NOT +5 or -5. In contrast, the equation $$x^2$$=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT"

Additionally, $$n^2 = 25$$ is a 2nd degree equation in 'n' and hence MUST have 2 solutions. But $$\sqrt{n^2}$$ no longer is a 2nd degree equation and hence should only have 1 solution.

As such, per GMAT's stance, $$\sqrt{n^2} = +n$$ ONLY.

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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24 Dec 2015, 04:34
Important consideration: n is not an integer, not stated explicitly; assuming the same would given an incorrect conclusions

Statement 1 gives many values which could be greater than 99 but not greater than 100, or greater than 100. Insufficient

Statement 2 gives a sure conclusion that it is greater. Hence, sufficient
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Re: If n is positive, is root(n) > 100 ? [#permalink]

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22 Aug 2016, 15:00
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tonebeeze wrote:
If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

Target question: Is √n > 100?

This is a good candidate for REPHRASING the target question.
Take √n > 100 and square both sides to get n > 10,000
So, we get: REPHRASED target question: Is n > 10,000?

Statement 1: √(n - 1) > 99
Square both sides to get n - 1 > 99²
Evaluate: n - 1 > 9801
Add 1 to both sides to get: n > 9802
So, x COULD equal 9803, in which case n < 10,000
Conversely, x COULD equal 10,001, in which case n > 10,000
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: √(n + 1) > 101
Square both sides to get n + 1 > 101²
Evaluate: n + 1 > 10,201
Subtract 1 from both sides to get: n > 10,200
If x is greater than 10,200, then we can be certain that x > 10,000
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

[Reveal] Spoiler:
B

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Re: If n is positive, is root(n) > 100 ? [#permalink]

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11 Sep 2016, 06:10
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If n is positive, is $$\sqrt {n} > 100$$?

(1) $$\sqrt {n-1} > 99$$

(2) $$\sqrt {n+1} > 101$$

We need to determine if √n > 100. We can square both sides of the inequality and obtain n > 10,000. That is, if we can determine that n > 10,000, then √n > 100.

Statement One Alone:

√(n – 1) > 99

Again, let’s square both sides of the inequality; we obtain n – 1 > 9,801. We add 1 to both sides we obtain n > 9,802. However, we cannot determine whether n > 10,000. Statement one alone is not sufficient. Eliminate answer choices A and D.

Statement Two Alone:

√(n + 1) > 101

Once again, let’s square both sides of the inequality; we obtain n + 1 > 10,201. We subtract 1 from both sides and we have n > 10,200. Thus, n > 10,000; statement two alone is sufficient.

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Re: If n is positive, is root(n) > 100 ?   [#permalink] 22 Aug 2017, 18:29
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