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

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Math Expert
Joined: 02 Sep 2009
Posts: 58410
If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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17 Jun 2019, 23:58
00:00

Difficulty:

15% (low)

Question Stats:

81% (01:24) correct 19% (01:53) wrong based on 48 sessions

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If $$8<\sqrt{(n+6)(n+1)}<9$$, then n could equal

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

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Joined: 31 Oct 2013
Posts: 1467
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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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.

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

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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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Joined: 16 Jan 2019
Posts: 489
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Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal  [#permalink]

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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. $$\sqrt{150}$$

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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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
Re: If 8 < [(n+6)(n+1)]^(1/2) < 9, then n could equal   [#permalink] 18 Jun 2019, 05:00
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