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Which of the following is the best approximation of

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Which of the following is the best approximation of  [#permalink]

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New post 01 Jul 2010, 13:33
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Which of the following is the best approximation of \(\sqrt{3.8*10^{25}}\)?

A. 1.9*10^5

B. 6.2*10^5

C. 1.9*10^12

D. 6.2*10^12

E. 1.9*10^25
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Re: Quantitative Roots problem tough  [#permalink]

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New post 01 Jul 2010, 13:44
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ksharma12 wrote:
2. Which of the following is the best approximation of \sqrt{3.8x10^25}?


a) 1.9x10^5

b) 6.2x10^5

c) 1.9x10^12

d) 6.2x10^12

e) 1.9x10^25


Sometimes its best to turn a decimal into a whole number. If you change it to 38 x 10^24 you can bring of the ten to the 24 as 10^12.

the square root of 38 is going to be a little bit bigger than 6. since 6 x 6 = 36.

so the closest answer here is D
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Re: Which of the following is the best approximation of  [#permalink]

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New post 05 Feb 2014, 01:27
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1
ksharma12 wrote:
Which of the following is the best approximation of \(\sqrt{3.8*10^{25}}\)?

A. 1.9*10^5

B. 6.2*10^5

C. 1.9*10^12

D. 6.2*10^12

E. 1.9*10^25


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Which of the following is the best approximation of  [#permalink]

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New post 04 Jul 2016, 12:19
ksharma12 wrote:
Which of the following is the best approximation of \(\sqrt{3.8*10^{25}}\)?

A. 1.9*10^5

B. 6.2*10^5

C. 1.9*10^12

D. 6.2*10^12

E. 1.9*10^25


Lets' denote the inside of the root with 'original tag ('original'=3.8*10^{25})
notice that:
- 3.8*10^{25}=38*10^{24}
- 38*10^{24}<49*10^{24}


\sqrt{36*10^{24}}<\sqrt{'original'}<\sqrt{49*10^{24}}

so we get

6*10^{24}<'original'<7*10^{24}

and from here we can see the answer
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Re: Which of the following is the best approximation of  [#permalink]

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New post 23 Sep 2016, 06:49
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ksharma12 wrote:
Which of the following is the best approximation of \(\sqrt{3.8*10^{25}}\)?

A. 1.9*10^5

B. 6.2*10^5

C. 1.9*10^12

D. 6.2*10^12

E. 1.9*10^25


first of all, sqrt(x^25) is not x^5. it can be rewritten as: x^(25/2)
A, B, and E can be eliminated right away.

now..take one factor of 10 and multiply by 3.8. we got 38*10^24.
10^24 under radical is 10^2.
38 squared is slightly more than 6.

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

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New post 12 Nov 2017, 08:24
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ksharma12 wrote:
Which of the following is the best approximation of \(\sqrt{3.8*10^{25}}\)?

A. 1.9*10^5

B. 6.2*10^5

C. 1.9*10^12

D. 6.2*10^12

E. 1.9*10^25


We want the exponent of the 10 to be an even number, because taking the square root is easier if the exponent is even. Let’s simplify the given equation:

√(3.8 x 10^25) = √(38 x 10^24) = √38 x √10^24 = √38 x 10^12 ≈ 6.2 x 10^12

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

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New post 11 Jan 2019, 07:39
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Re: Which of the following is the best approximation of   [#permalink] 11 Jan 2019, 07:39
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