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# Which of the following is the best approximation?

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Intern
Joined: 09 Oct 2012
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Which of the following is the best approximation? [#permalink]

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11 Oct 2012, 12:56
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$$(\sqrt{2}+\sqrt{5})^2$$ which of the following is the best approximation?
a. 7
b. 10
c. 13
d. 15
e. 17

Is there somebody who could explain how to make this calculate quickly?
I find this question in the Gmat prep software, what is the level of the question?
[Reveal] Spoiler: OA
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Re: which of the following is the best approximation? [#permalink]

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11 Oct 2012, 13:03
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Expert's post
IanSolo wrote:
$$(\sqrt{2}+\sqrt{5})^2$$ which of the following is the best approximation?
a. 7
b. 10
c. 13
d. 15
e. 17

The right answer is c.
Is there somebody who could explain how to make this calculate quickly?
I find this question in the Gmat prep software, what is the level of the question?

$$(a+b)^2=a^2+2ab+b^2$$, thus $$(\sqrt{2}+\sqrt{5})^2=2+2*\sqrt{2}*\sqrt{5}+5=7+2\sqrt{10}$$.

Now, $$3^2=9$$, so $$\sqrt{10}$$ is a little bit greater than 3, which means that $$7+2\sqrt{10}\approx{7+2*3}=13$$.

P.S. I'd say it's ~600 level question.
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Re: Which of the following is the best approximation? [#permalink]

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10 Dec 2012, 23:36
$$(\sqrt{2}+\sqrt{5})(\sqrt{2}+\sqrt{5})$$
$$2 + \sqrt{10 + [square_root]10} + 5[/square_root]$$
$$7 + 2\sqrt{10}$$

\sqrt{9} = 3
\sqrt{16} = 4
[fraction]10[/fraction] - a little more than 3

$$7 + 2*3 = 7 + 6 ~ 13$$

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Re: Which of the following is the best approximation? [#permalink]

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04 Aug 2014, 11:03
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Re: Which of the following is the best approximation? [#permalink]

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04 Aug 2014, 20:51
Its (a+b)^2 formula..

(\sqrt{2}+\sqrt{5})^2 = 2 + 5 + 2*\sqrt{2}*\sqrt{5}

7+ 2*\sqrt{10} = 7 + (~6) ~=13.
Re: Which of the following is the best approximation?   [#permalink] 04 Aug 2014, 20:51
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# Which of the following is the best approximation?

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