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If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal

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Math Expert
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If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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New post 17 Jun 2019, 23:58
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

83% (01:25) correct 18% (01:33) wrong based on 40 sessions

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Concentration: Accounting, Finance
GPA: 3.68
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Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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New post 18 Jun 2019, 00:03
Bunuel wrote:
If \(8<\sqrt{(n+6)(n+1)}<9\), then n could equal

A. 5
B. 6
C. 7
D. 8
E. 9


Given,

\(8<\sqrt{(n+6)(n+1)}<9\)

Square all the parts of the inequality.

\(8^2<(\sqrt{(n+6)(n+1)})^2<9^2\)

\(64<(n+6)(n+1)<81\)

Only option A works.

Thus n must be 5.

A is the best answer.
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Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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New post 18 Jun 2019, 00:07
Bunuel wrote:
If \(8<\sqrt{(n+6)(n+1)}<9\), then n could equal

A. 5
B. 6
C. 7
D. 8
E. 9


I approached this question by substitution. (N+6)*(N+1) must be between 64 to 81 (non inclusive) for N to be acceptable,

N=7 means 13*8=104, So N must be less than 7.
N=6 means 12*7=84, So N must be less than 6.

IMO A, Check 11*6=66
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Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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New post 18 Jun 2019, 00:13
Plug in the answer choices to see which one lies within \(\sqrt{64}\) and \(\sqrt{81}\)

A. \(\sqrt{66}\)
B. \(\sqrt{84}\)
C. \(\sqrt{104}\)
D. \(\sqrt{126}\)
E. \(\sqrt{150}\)

Answer is (A)

Hit Kudos if this helped!
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Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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New post 18 Jun 2019, 05:00
Bunuel wrote:
If \(8<\sqrt{(n+6)(n+1)}<9\), then n could equal

A. 5
B. 6
C. 7
D. 8
E. 9


square sides
we get
64<(n+6)(n+1)<81
n=5
sufficient
IMO A
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Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal   [#permalink] 18 Jun 2019, 05:00
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