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m11 Q18

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08 Jul 2010, 13:13
1
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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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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

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

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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08 Jul 2010, 16:51
Ahh... that explains...

Thanks lagomez =p
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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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06 Jul 2012, 00:34
1
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Expert's post
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?

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

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

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

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

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

P.S. It's not a good idea to use approximation for this question.
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02 Jun 2013, 09:03
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?

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

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

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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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?

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

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

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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26 May 2014, 08:39
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?

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

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

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