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555-605 (Medium)|   Roots|      
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Bunuel
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 13 Mar 2015, 07:46
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D
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25% (medium)

Question Stats:

78% (01:32) correct 22% (01:30) wrong based on 462 sessions

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If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

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


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D
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 13 Mar 2015, 08:48
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\((\sqrt{12}+\sqrt{108}) (12[/square_root]+\sqrt{108})=N\)
120 + 2×3×3×4 = N
N=192

Answer: D
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 13 Mar 2015, 08:57
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Bunuel
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

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


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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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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 13 Mar 2015, 09:35
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Bunuel
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

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


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

N=192
ANSWER D
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 13 Mar 2015, 21:34
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12=2*2*3
108=2*2*3*3*3

we have

2*sqrt3+6*sqrt3=8*sqrt3

8*sqrt3=8^2*3=192

D
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 13 Mar 2015, 22:11
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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....

Final Answer:
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D

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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 14 Mar 2015, 09:02
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Bunuel
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

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


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+1 for D.

2 Root 3+6Root 3=8 Root 3= Root N
Square Both sides,
64*3=N=192
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 15 Mar 2015, 22:58
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Bunuel
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

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


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MAGOOSH OFFICIAL SOLUTION:

First of all, the mistake: we CANNOT add through
Attachment:
cgpwe_img20.png
cgpwe_img20.png [ 1.32 KiB | Viewed 7177 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 7202 times ]

Answer = (D).
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 17 Mar 2015, 03:14
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\(\sqrt{12}+\sqrt{108}=\sqrt{N}\)

Squaring both sides

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

Answer = D
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 14 Sep 2016, 07:51
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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 = 20 Mar 2018, 09:55
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Bunuel
If \(\sqrt{12}+\sqrt{108}=\sqrt{N}\), then N =

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


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

\(\sqrt{12}\) 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 = 29 Oct 2020, 02:26
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Sqrt of 12 is 3.x and sqrt of 108 is 10.x

Adding both will lead us to a sqrt of close to 14. Therefore, N should be 192.

Answer D
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If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 18 Jan 2026, 07:15
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