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If 12^(1/2) + 108^(1/2) = N^(1/2), then N =

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If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 13 Mar 2015, 07:46
1
2
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

77% (01:30) correct 23% (01:24) wrong based on 225 sessions

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If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 13 Mar 2015, 08:48
\((\sqrt{12}+\sqrt{108}) (12[/square_root]+\sqrt{108})=N\)
120 + 2×3×3×4 = N
N=192

Answer: D
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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 13 Mar 2015, 08:57
1
1
Bunuel wrote:
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

(A) 96
(B) 120
(C) 144
(D) 192
(E) 288


Kudos for a correct solution.



there are two ways ,

1) \(\sqrt{12}+\sqrt{108}=\sqrt{N}\)....
N=12+108+2\(\sqrt{12}*\sqrt{108}\)=120+2*12*3=192..

2)\(\sqrt{12}\)=nearly 3.5(between 9 &16)..
\(\sqrt{108}\) =nearly 10.4(between 100&121)..
so 3.5+10.4=13.9..
ans should be between 13^2 and 14^2.. only D is possible

ans D
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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 13 Mar 2015, 09:35
3
1
Bunuel wrote:
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

(A) 96
(B) 120
(C) 144
(D) 192
(E) 288


Kudos for a correct solution.


\(\sqrt{12}+\sqrt{108}=\sqrt{N}\)
\(\sqrt{12}+\sqrt{12*3*3}=\sqrt{N}\)
\(\sqrt{12}+3 \sqrt{12}=\sqrt{N}\)
\(4\sqrt{12}=\sqrt{N}\)

N=192
ANSWER D
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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 13 Mar 2015, 21:34
1
12=2*2*3
108=2*2*3*3*3

we have

2*sqrt3+6*sqrt3=8*sqrt3

8*sqrt3=8^2*3=192

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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 13 Mar 2015, 22:11
1
Hi All,

The answer choices to this question are sufficiently 'spread out' that we can avoid most of the 'technical math' and use estimation to get to the correct answer.

Root(12) is between 3 and 4
Root(108) is between 10 and 11

The correct answer has to be BETWEEN 10+3=13 and 4+11=15

13^2 = 169
15^2 = 225

So we're looking for an answer between 169 and 225....There's only one that fits....

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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 14 Mar 2015, 09:02
1
Bunuel wrote:
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

(A) 96
(B) 120
(C) 144
(D) 192
(E) 288


Kudos for a correct solution.


+1 for D.

2 Root 3+6Root 3=8 Root 3= Root N
Square Both sides,
64*3=N=192
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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 15 Mar 2015, 22:58
1
Bunuel wrote:
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

(A) 96
(B) 120
(C) 144
(D) 192
(E) 288


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

First of all, the mistake: we CANNOT add through
Attachment:
cgpwe_img20.png
cgpwe_img20.png [ 1.32 KiB | Viewed 2390 times ]


That incorrect thinking would lead to the trap answer of (B). Instead, we have to simplify each square root on the left.
Attachment:
cgpwe_img21.png
cgpwe_img21.png [ 5.29 KiB | Viewed 2391 times ]


Answer = (D).
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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 17 Mar 2015, 03:14
\(\sqrt{12}+\sqrt{108}=\sqrt{N}\)

Squaring both sides

\(n = 12 + 108 + 2\sqrt{12*108} = 120 + 72 = 192\)

Answer = D
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Re: If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 14 Sep 2016, 07:51
Really like lucky's and bunuels approach ...way to go!!!
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N =  [#permalink]

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New post 20 Mar 2018, 09:55
Bunuel wrote:
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

(A) 96
(B) 120
(C) 144
(D) 192
(E) 288


Kudos for a correct solution.


\(\sqrt{12}+\sqrt{108}=\sqrt{N}\)

\(\sqrt{12}\) is approx 3.5

\(\sqrt{108}\)=\(\sqrt{N}\) is approx 10.5

Adding both we get, 3.5 + 10.5 = 14

Therefore 14 =\(\sqrt{N}\)

\(N = (14)^2\)

\(N = 196\)

(D) is the nearest
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = &nbs [#permalink] 20 Mar 2018, 09:55
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