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555-605 Level|   Algebra|   Inequalities|   Roots|                              
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I didn't understand why statement 1 is not sufficient. Can someone please explain? thanks.
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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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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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Bunuel
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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Bunuel
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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Bunuel
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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Bunuel
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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tonebeeze
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

Answer =
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Bunuel
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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Bunuel
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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Bunuel
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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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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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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Bunuel
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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KarishmaB
mymbadreamz
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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