GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Nov 2019, 22:53

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59095
If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

11 Jan 2011, 17:57
4
61
00:00

Difficulty:

45% (medium)

Question Stats:

66% (01:42) correct 34% (01:46) wrong based on 1548 sessions

### HideShow timer Statistics

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.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 59095
If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

11 Jan 2011, 18:08
14
19
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.

_________________
Director
Joined: 25 Apr 2012
Posts: 654
Location: India
GPA: 3.21
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

23 Jan 2014, 04:33
5
1
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

Difficulty level 650 is okay
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
##### General Discussion
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9784
Location: Pune, India
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

11 Jan 2011, 19:30
9
2
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).
_________________
Karishma
Veritas Prep GMAT Instructor

Manager
Joined: 15 Apr 2011
Posts: 59
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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 59095
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

09 Apr 2012, 12:19
3
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.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9784
Location: Pune, India
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

09 Apr 2012, 22:05
2
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

Math Expert
Joined: 02 Sep 2009
Posts: 59095
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

23 Jan 2014, 03:23
1
14
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.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 59095
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

23 Jan 2014, 03:24
3
3
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.

_________________
Director
Joined: 27 May 2012
Posts: 936
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

24 Jan 2014, 00:41
1
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

_________________
- Stne
Manager
Joined: 04 Oct 2013
Posts: 150
Location: India
GMAT Date: 05-23-2015
GPA: 3.45
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

24 Jan 2014, 01:42
1
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))

Manager
Joined: 21 Oct 2013
Posts: 177
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

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.

Manager
Joined: 29 Jul 2015
Posts: 154
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

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.
CEO
Joined: 20 Mar 2014
Posts: 2570
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

08 Sep 2015, 13:53
1
1
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.
Manager
Joined: 17 Jun 2015
Posts: 191
GMAT 1: 540 Q39 V26
GMAT 2: 680 Q46 V37
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

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
_________________
Fais de ta vie un rêve et d'un rêve une réalité
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4064
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

22 Aug 2016, 15:00
8
Top Contributor
1
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

RELATED VIDEO

_________________
Test confidently with gmatprepnow.com
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2812
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

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.

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Director
Joined: 14 Dec 2017
Posts: 510
Location: India
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

12 Jun 2018, 00:09
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) $$\sqrt {n+1} > 101$$

Given $$n >0$$ , we are asked is $$\sqrt {n} > 100$$?

or rephrased as $$n > 10^4$$?

Statement 1:

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

Squaring both sides we get,

$$(n-1) > 99^2$$

$$(n-1) >{(100 - 1)}^2$$

$$n > 10^4 - 200 + 2$$

$$n > 9802$$

Hence $$n > 10^4$$ or $$9802 < n < 10^4$$

Statement 1 alone is Not Sufficient.

Statement 2:

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

Squaring both sides, we get

$$(n + 1) > 101^2$$

$$(n + 1) > (100 + 1)^2$$

$$n > 10^4 + 200 + 2$$

$$n > 10^4 + 202$$

Statement 2 alone is Sufficient.

Thanks,
GyM
_________________
Director
Joined: 19 Oct 2013
Posts: 511
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

06 Nov 2018, 10:15
The way I attempted this question is by squaring both sides

is n > 100^2? from the question stem and 100^2 = 99^2 + 99 + 100

Statement 1) n - 1 > 99^2

n > 99^2 + 1 so not sufficient.

Statement 2)

n + 1 > 101^2

101^2 = 100^2 + 100 + 101 =

n > 101^2 - 1 = 100^2 + 100 + 101 - 1 = 100^2 + 200

it is larger than 100^2 Sufficient.

I found that I somehow forgot that the easier way for statement 2 is to try difference of a square (101-1) (101+1)

Manager
Joined: 03 Aug 2017
Posts: 79
Re: If n is positive, is root(n) > 100 ?  [#permalink]

### Show Tags

22 Oct 2019, 22:49
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) $$\sqrt {n+1} > 101$$

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.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!

This is how i solved

Target question : We need to find whether n√>100n>100------ or if n>10000 (We can square as both sides are positive )

St 1 says n−1−−−−−√>99n−1>99------> Squaring we get n-1 > (99)^2 or n > (99)^2 +1 = 9801+1 = 9802
So n can be 9803 which is lesser than 10000 or more than 10000 lets say 10003 as n > 9803

So n can be greater than or less than 10000
A and D ruled out and statment 1 alone is not suffcient

St 2 says n+1−−−−−√>101n+1>101-----> Squaring we get n+1>(101)^2 -----> n > (101)^2-1. Now (101)^2 -1 is greater than (100)^2
Which basically means N > 10100 if so n is definately greater than 10000 so it is sufficient

Hence Ans B
Re: If n is positive, is root(n) > 100 ?   [#permalink] 22 Oct 2019, 22:49
Display posts from previous: Sort by