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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: 1. Please provide your solutions to the questions; 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.

(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.

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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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.

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: 1. Please provide your solutions to the questions; 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.

(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.

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

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))

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.

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.

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.

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.

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

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]

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