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m11 Q18

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m11 Q18 [#permalink] New post 08 Jul 2010, 13:13
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\sqrt{7+\sqrt{48}}-\sqrt{3}=?

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

Last edited by Bunuel on 06 Jul 2012, 00:35, edited 1 time in total.
Edited the question and the OA.
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Re: m11 Q18 [#permalink] New post 08 Jul 2010, 14:36
honkergirl wrote:
\sqrt{7}+\sqrt{48}-\sqrt{3}

Choices:
a. 1.0
b. 1.7
c. 2.0
d. 2.4
e. 3.0

The explanation for this qtn tells us to simplify
\sqrt{7}+\sqrt{48}=4+4\sqrt{3}+3
This leads to answer choice 2.

Butif I use normal algebra somehow I get a different answer:
\sqrt{7} = 2.646
\sqrt{48}=4\sqrt{3}
\sqrt{3}=1.73

Answer:7.84

Can someone please point out where I'm going wrong?

Thanks,

HG


Yes you need to take the square root of the square root of 48. You have the problem written incorrectly.
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Re: m11 Q18 [#permalink] New post 08 Jul 2010, 16:51
Ahh... that explains...

Thanks lagomez =p
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Re: m11 Q18 [#permalink] New post 05 Jul 2012, 20:44
How can I use estimation to solve this? I tried using estimation ...but always get trapped in some wrong choice....what to do avoid getting a wrong answer when using estimation in a question that has answer choices that are too close?
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Re: m11 Q18 [#permalink] New post 06 Jul 2012, 00:34
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teal wrote:
How can I use estimation to solve this? I tried using estimation ...but always get trapped in some wrong choice....what to do avoid getting a wrong answer when using estimation in a question that has answer choices that are too close?


The question should read.

\sqrt{7+\sqrt{48}}-\sqrt{3}=?

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}.

So, \sqrt{7+\sqrt{48}}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2.

Answer: C.

P.S. It's not a good idea to use approximation for this question.
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Re: m11 Q18 [#permalink] New post 02 Jun 2013, 06:12
Bunuel - what led you to think of breaking up the 7 into 4+3? I understand everything in terms of the math, but I don't think I would of thought of breaking up the 7 into two parts and then factoring?

Is it just something that will become intuitive with practice or was there something that led you to think that?

Bunuel wrote:
teal wrote:
How can I use estimation to solve this? I tried using estimation ...but always get trapped in some wrong choice....what to do avoid getting a wrong answer when using estimation in a question that has answer choices that are too close?


The question should read.

\sqrt{7+\sqrt{48}}-\sqrt{3}=?

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}.

So, \sqrt{7+\sqrt{48}}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2.

Answer: C.

P.S. It's not a good idea to use approximation for this question.
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Re: m11 Q18 [#permalink] New post 02 Jun 2013, 09:03
Expert's post
vaj18psu wrote:
Bunuel - what led you to think of breaking up the 7 into 4+3? I understand everything in terms of the math, but I don't think I would of thought of breaking up the 7 into two parts and then factoring?

Is it just something that will become intuitive with practice or was there something that led you to think that?

Bunuel wrote:
teal wrote:
How can I use estimation to solve this? I tried using estimation ...but always get trapped in some wrong choice....what to do avoid getting a wrong answer when using estimation in a question that has answer choices that are too close?


The question should read.

\sqrt{7+\sqrt{48}}-\sqrt{3}=?

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}.

So, \sqrt{7+\sqrt{48}}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2.

Answer: C.

P.S. It's not a good idea to use approximation for this question.


I think such kind of tricks should come with practice.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: m11 Q18 [#permalink] New post 26 May 2014, 07:31
Bunuel wrote:
teal wrote:
How can I use estimation to solve this? I tried using estimation ...but always get trapped in some wrong choice....what to do avoid getting a wrong answer when using estimation in a question that has answer choices that are too close?


The question should read.

\sqrt{7+\sqrt{48}}-\sqrt{3}=?

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}.

So, \sqrt{7+\sqrt{48}}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2.

Answer: C.

P.S. It's not a good idea to use approximation for this question.


Bunuel, what is wrong with using approximation for this question, because aside from using the technique that you just used, it is the only method left?
I did the following and answered the question correctly.

sqrt(48) ~ sqrt(49) = 7
sqrt(7+sqrt(48)) - sqrt(3)=
sqrt(14) - sqrt(3)

sqrt(9) < sqrt(14) < sqrt(16)
3 < sqrt(14) < 4
We know that sqrt(14) is closer to 4 than it is to 3, so let's try some numbers.
Given that the answer choices are numbers with only a tenths place, we can estimate sqrt(14) to a number to just the tenths place.
sqrt(14) ~ 3.8
3.8^2 = 14.44 | Close enough
3.8-sqrt(3) = 3.8 - 1.7 = 2.1
The closest answer to this is 2.

C
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Re: m11 Q18 [#permalink] New post 26 May 2014, 08:39
Expert's post
TooLong150 wrote:
Bunuel wrote:
teal wrote:
How can I use estimation to solve this? I tried using estimation ...but always get trapped in some wrong choice....what to do avoid getting a wrong answer when using estimation in a question that has answer choices that are too close?


The question should read.

\sqrt{7+\sqrt{48}}-\sqrt{3}=?

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

\sqrt{7+\sqrt{48}}=\sqrt{7+4\sqrt{3}}=\sqrt{4+4\sqrt{3}+3}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}.

So, \sqrt{7+\sqrt{48}}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2.

Answer: C.

P.S. It's not a good idea to use approximation for this question.


Bunuel, what is wrong with using approximation for this question, because aside from using the technique that you just used, it is the only method left?
I did the following and answered the question correctly.

sqrt(48) ~ sqrt(49) = 7
sqrt(7+sqrt(48)) - sqrt(3)=
sqrt(14) - sqrt(3)

sqrt(9) < sqrt(14) < sqrt(16)
3 < sqrt(14) < 4
We know that sqrt(14) is closer to 4 than it is to 3, so let's try some numbers.
Given that the answer choices are numbers with only a tenths place, we can estimate sqrt(14) to a number to just the tenths place.
sqrt(14) ~ 3.8
3.8^2 = 14.44 | Close enough
3.8-sqrt(3) = 3.8 - 1.7 = 2.1
The closest answer to this is 2.

C


Nothing wrong. Good approach.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

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Re: m11 Q18   [#permalink] 26 May 2014, 08:39
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