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

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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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.

Hope it's clear.
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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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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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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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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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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 =

RELATED VIDEO

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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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.

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.

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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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.