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

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Bunuel
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If n is positive, is root(n) > 100 ? [#permalink] New post   11 Jan 2011, 17:57
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If n is positive, is \(\sqrt {n} > 100\)?

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

(2) \(\sqrt {n+1} > 101\)
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If n is positive, is root(n) > 100 ? [#permalink] New post   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) \(\sqrt {n-1} > 99\):

    \(n-1>99^2\);

    \(n>99^2+1\)

Now, \(99^2+1\) 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) \(\sqrt {n+1} > 101\):

    \(n+1>101^2\);

    \(n>101^2-1\);

    \(n>(101-1)(101+1)\);

    \(n>100*102\);

    \(n>100*102>100^2\).


Sufficient.

Answer: B.
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   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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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   09 Apr 2012, 12:07
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] New post   09 Apr 2012, 12:19
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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.
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   09 Apr 2012, 22:05
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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?
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   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

Difficulty level 650 is okay
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   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

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

Answer: B.


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] New post   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.

Answer: B.


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] New post   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

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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   11 Sep 2016, 06:10
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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.

Answer: B
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   08 Jun 2021, 04:31
Bunuel wrote:
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.

Answer: B.


Hey In statement 1, how can you square both sides as it is not necessary that n-1 is positive?

Gagan
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   08 Jun 2021, 06:23
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gagan0303 wrote:
Bunuel wrote:
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.

Answer: B.


Hey In statement 1, how can you square both sides as it is not necessary that n-1 is positive?

Gagan


If n - 1 is negative \(\sqrt {n-1}\) cannot not be greater than 99. Also, if n - 1 were negative, then we'd have even root from negative number and even roots from negative numbers are not defined on the GMAT: \(\sqrt[even]{negative}\) is undefined.
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If n is positive, is root(n) > 100 ? [#permalink] New post   03 Dec 2021, 18:54
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I can not stress this enough! DO THE ALGEBRA! Especially when there are not many variations of algebra you can do on this problem.

If n is positive, is n√>100n>100?

Square both sides and we are trying to find n>10,000

(1) n−1−−−−−√>99n−1>99

Translates to n-1>99^2

Just do 99*100 which is 9900 and then subtract by 99 so we get n - 1 > 9801 which turns into n > 9802

UH OH N can be less than or greater than 10,000! Not Suff.


(2) n+1−−−−−√>101

Same thing, square both sides

N + 1 = > 101*101

10100 + 100 = 10200

N + 1 > 10200

N > 10199 So YES N is greater than 10,000 we are good to go. SUFF.
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   01 Apr 2022, 12:27
The question if sqrt(n) > 100 can be rephrased by squaring both sides, which yields: is n > 10000?

1) We can do the same thing to this statement. n - 1 > 9801, add 1 to both sides and we get that n > 9802. Insufficient.

2) Once again, square both sides. N + 1 > 10201, deduct 1, n > 10200. Sufficient to answer the original (rephrased) question if n > 10000.

B.
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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   02 Apr 2022, 05:12
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Bunuel wrote:
If n is positive, is \(\sqrt {n} > 100\)?

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

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


Question: Is √n > 100?
Question: Is \(n > 100^2\)?
Question: Is \(n > 10^4\)?

Statement 1: \(\sqrt {n-1} > 99\)

i.e. \({n-1} > 99^2\)
i.e. \({n} > (100-1)^2+1\)
i.e. \({n} > 100^2 + 1^2 - 2*1*100+1\)
i.e. \({n} > 10^4 - 198\)

i.e. n may or may not be greater than \(10^4\) hence

NOT SUFFICIENT

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

i.e. \({n+1} > 101^2\)
i.e. \({n} > (100+1)^2-1\)
i.e. \({n} > 100^2 + 1^2 + 2*1*100-1\)
i.e. \({n} > 10^4 + 200\)

i.e. n is definitely greater than \(10^4\) hence

SUFFICIENT

Answer: Option B

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Re: If n is positive, is root(n) > 100 ? [#permalink] New post   29 Jun 2023, 08:06
KarishmaB wrote:
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?


I got it wrong too. So, I think the key is n is not declared as an "integer". Therefore, option 1 has multiple possibilities. Hence, insufficient.
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Re: If n is positive, is root(n) > 100 ? [#permalink]
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