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

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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

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

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

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

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I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks.
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Re: If n is positive, is root(n) > 100 ?  [#permalink]

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

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

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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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2. Please vote for the best solutions by pressing Kudos button;
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Re: If n is positive, is root(n) > 100 ?  [#permalink]

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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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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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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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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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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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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 $$\sqrt{x}$$, 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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GMAT 1: 540 Q39 V26 GMAT 2: 680 Q46 V37 Re: If n is positive, is root(n) > 100 ?  [#permalink]

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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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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

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

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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]

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

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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)

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

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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
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