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Difficulty:
85%
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Question Stats:
52% (01:52) correct
48%
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wrong
based on 2033
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\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)
A. 1
B. 1.7
C. 2
D. 2.4
E. 3
M11-18
_________________
A. 1
B. 1.7
C. 2
D. 2.4
E. 3
M11-18
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Official Solution:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)
A. 1
B. 1.7
C. 2
D. 2.4
E. 3
First, let's simplify the expression: \(\sqrt{7 + \sqrt{48}} - \sqrt{3} = \sqrt{7 + \sqrt{16*3}} - \sqrt{3} = \sqrt{7 + 4\sqrt{3}} - \sqrt{3}\). Observe that the answer choices are all rational numbers, so irrational parts in the expression must somehow cancel each other out. This observation should be a hint that we might be able to rewrite \(7 + 4\sqrt{3}\) in a way that makes it the square of something, which would allow us to extract the square root and simplify the expression. Bearing this hint in mind, we rewrite the expression as \(4 + 4\sqrt{3} + 3\). Upon examining the expression \(4 + 4\sqrt{3} + 3\), we notice that \(4 + 4\sqrt{3} + 3 = (2 + \sqrt{3})^2\).
Therefore:
Answer: C
_________________
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)
A. 1
B. 1.7
C. 2
D. 2.4
E. 3
First, let's simplify the expression: \(\sqrt{7 + \sqrt{48}} - \sqrt{3} = \sqrt{7 + \sqrt{16*3}} - \sqrt{3} = \sqrt{7 + 4\sqrt{3}} - \sqrt{3}\). Observe that the answer choices are all rational numbers, so irrational parts in the expression must somehow cancel each other out. This observation should be a hint that we might be able to rewrite \(7 + 4\sqrt{3}\) in a way that makes it the square of something, which would allow us to extract the square root and simplify the expression. Bearing this hint in mind, we rewrite the expression as \(4 + 4\sqrt{3} + 3\). Upon examining the expression \(4 + 4\sqrt{3} + 3\), we notice that \(4 + 4\sqrt{3} + 3 = (2 + \sqrt{3})^2\).
Therefore:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = \)
\(=\sqrt{7 + \sqrt{16*3}} - \sqrt{3}=\)
\(=\sqrt{7 + 4\sqrt{3}} - \sqrt{3}=\)
\(=\sqrt{4 + 4\sqrt{3} + 3} - \sqrt{3}=\)
\(=\sqrt{(2 + \sqrt{3})^2} - \sqrt{3} = \)
\(=(2 + \sqrt{3}) - \sqrt{3} = 2\).
\(=\sqrt{7 + \sqrt{16*3}} - \sqrt{3}=\)
\(=\sqrt{7 + 4\sqrt{3}} - \sqrt{3}=\)
\(=\sqrt{4 + 4\sqrt{3} + 3} - \sqrt{3}=\)
\(=\sqrt{(2 + \sqrt{3})^2} - \sqrt{3} = \)
\(=(2 + \sqrt{3}) - \sqrt{3} = 2\).
Answer: C
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If some1 is really fast in calculation then they might try this method as well.
\(\sqrt{ 7+ \sqrt{48}} - \sqrt{3}\) = \(\sqrt{ 14} - \sqrt{3}\) ( approximating 48 as 49)
= 3.7xxx - 1.732
which is close to 2.
but looking at the question it is better to do it the way Bunuel did it as GMAT doesn't want us to be good at calculations but at observation.
\(\sqrt{ 7+ \sqrt{48}} - \sqrt{3}\) = \(\sqrt{ 14} - \sqrt{3}\) ( approximating 48 as 49)
= 3.7xxx - 1.732
which is close to 2.
but looking at the question it is better to do it the way Bunuel did it as GMAT doesn't want us to be good at calculations but at observation.
General Discussion
PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
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You may solve in this way:
Let \(\sqrt{ 7+ \sqrt{48}} - \sqrt{3} = x\)
\(\sqrt{ 7+ \sqrt{48}} = x + \sqrt{3}\)
Squaring both sides
\(7 + \sqrt{48} = 3 + 2\sqrt{3} x + x^2\)
\(4 + \sqrt{48} = 2\sqrt{3}x + x^2\)
\(x^2 + 2\sqrt{3}x - 43 - 4 = 0\)
\(x^2 - 4 + 2\sqrt{3}(x-2) = 0\)
\((x+2) (x-2) + 2\sqrt{3} (x-2) = 0\)
\((x-2) (x + 2 + 2\sqrt{3}) = 0\)
From the above equation, either x-2 = 0 or \(x + 2 + 2\sqrt{3} = 0\)
x = 2 OR \(x = -(2+2\sqrt{3})\)
From the options, x has a positive value, hence Answer = 2 = C
This is a long method; I would prefer Bunuel's method
Let \(\sqrt{ 7+ \sqrt{48}} - \sqrt{3} = x\)
\(\sqrt{ 7+ \sqrt{48}} = x + \sqrt{3}\)
Squaring both sides
\(7 + \sqrt{48} = 3 + 2\sqrt{3} x + x^2\)
\(4 + \sqrt{48} = 2\sqrt{3}x + x^2\)
\(x^2 + 2\sqrt{3}x - 43 - 4 = 0\)
\(x^2 - 4 + 2\sqrt{3}(x-2) = 0\)
\((x+2) (x-2) + 2\sqrt{3} (x-2) = 0\)
\((x-2) (x + 2 + 2\sqrt{3}) = 0\)
From the above equation, either x-2 = 0 or \(x + 2 + 2\sqrt{3} = 0\)
x = 2 OR \(x = -(2+2\sqrt{3})\)
From the options, x has a positive value, hence Answer = 2 = C
This is a long method; I would prefer Bunuel's method
Bunuel
can anyone show me how we reduced( 2+sqrt3)^2-sqrt3 to become just (2+sqrt3)-sqrt3
23a2012
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} =\sqrt{(2+\sqrt{3})^2} - \sqrt{3}= (2+\sqrt{3})-\sqrt{3}=2\)
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gmat6nplus1
Joined: 04 Oct 2013
Last visit: 09 Jan 2019
Posts: 141
Concentration: Finance, Leadership
GMAT 1: 590 Q40 V30
GMAT 2: 730 Q49 V40
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GMAT 2: 730 Q49 V40
Posts: 141
faiint
\(sqrt{48}\) is approximately 7; \(\sqrt{7 + \sqrt{48}\) falls between 3 and 4, say 3,7d \(sqrt{3}\) is approximately 1,7; result is very close to 2. A & B too small D & E too big C = correct.
√of 48 =6.9 so we make it 49. √7+7 -√3 = √14 -√3= 3.7-1.5= 2 (C)
Kinshook
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faiint
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)
\(\sqrt{4 +3 + 2 \sqrt{4*3}} - \sqrt{3} = \sqrt{4} + \sqrt{3} -\sqrt{3} =\sqrt{4} =2\)
IMO C
webarebears
Joined: 22 Jul 2024
Last visit: 26 Feb 2026
Posts: 25
Location: Indonesia
Schools: BU'26 (II)
GMAT Focus 1: 625 Q80 V81 DI82
GPA: 3.81
\(\sqrt{a + 2\sqrt{ab}+ b}\) =\( \sqrt{a}+\sqrt{b}\)
\(\sqrt{7 + \sqrt{48}} - \sqrt{3}\)
=\(\sqrt{7 + 2\sqrt{12}} - \sqrt{3}\)
=\(\sqrt{4}+\sqrt{3}-\sqrt{3}\)
=\(\sqrt{4}\)
= 2
\(\sqrt{7 + \sqrt{48}} - \sqrt{3}\)
=\(\sqrt{7 + 2\sqrt{12}} - \sqrt{3}\)
=\(\sqrt{4}+\sqrt{3}-\sqrt{3}\)
=\(\sqrt{4}\)
= 2
↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} \) =
Since we have nested roots so idea will be to express the values inside the bigger square root as square of some number
\(7 + \sqrt{48}\) => Lets express this as \((a+b)^2 = a^2 + b^2 + 2ab\)
\(\sqrt{48} \) = 2ab
=> \(\sqrt{4 * 3 * 4}\) = 2 * \(\sqrt{3} * \sqrt{4} \) = 2ab
=> a = √3, b = √4 = 2
=> \(7 + \sqrt{48}\) = 3 + 4 + 2 * √3 * 2 = \((√3)^2 + 2 * √3 * 2 + (2)^2\) = \((√3 + 2)^2\)
=> \(\sqrt{7 + \sqrt{48}} - \sqrt{3} \) = \(\sqrt{ (√3 + 2)^2 } - \sqrt{3} \) = √3 + 2 - √3 = 2
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Roots
_________________
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} \) =
Since we have nested roots so idea will be to express the values inside the bigger square root as square of some number
\(7 + \sqrt{48}\) => Lets express this as \((a+b)^2 = a^2 + b^2 + 2ab\)
\(\sqrt{48} \) = 2ab
=> \(\sqrt{4 * 3 * 4}\) = 2 * \(\sqrt{3} * \sqrt{4} \) = 2ab
=> a = √3, b = √4 = 2
=> \(7 + \sqrt{48}\) = 3 + 4 + 2 * √3 * 2 = \((√3)^2 + 2 * √3 * 2 + (2)^2\) = \((√3 + 2)^2\)
=> \(\sqrt{7 + \sqrt{48}} - \sqrt{3} \) = \(\sqrt{ (√3 + 2)^2 } - \sqrt{3} \) = √3 + 2 - √3 = 2
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Roots
_________________
How to Solve :
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GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
naikabhishekjanu
Joined: 13 Jan 2021
Last visit: 09 Aug 2026
Posts: 9
Location: India
Concentration: Technology, Finance
GMAT 1: 660 Q48 V34
GMAT 1: 660 Q48 V34
Posts: 9
Hi,
How can I easily understand if the first square root before 7 is only for 7, or it continues till 48 or it goes on till 3.
How can I easily understand if the first square root before 7 is only for 7, or it continues till 48 or it goes on till 3.
naikabhishekjanu
The first square root covers both 7 and √48, so both are under the root. It does not extend over √3, so that part is separate.
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arushi118
Joined: 21 Jul 2024
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Location: India
Concentration: Leadership, General Management
Schools: INSEAD Aug '26 LBS '26 Kellogg '26
GPA: 8.2/10
Products:
Schools: INSEAD Aug '26 LBS '26 Kellogg '26
Posts: 280
For question which involve a root inside a root, try to write the expression inside the root as a square of some expression. That should usually do the trick to solving these kinds of questions.
can be written as sqrt{4+3+4 sqrt(3)}-sqrt(3)
sqrt{2^2+sqrt(3)^2+2*2*sqrt(3)}-sqrt(3)
sqr{(2+sqrt(3))^2}-sqrt(3)
2+sqrt(3)-sqrt(3)=2
C
sqrt{2^2+sqrt(3)^2+2*2*sqrt(3)}-sqrt(3)
sqr{(2+sqrt(3))^2}-sqrt(3)
2+sqrt(3)-sqrt(3)=2
C
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