Of the following, which is the best approximation to 0.0026^1/2 ? : GMAT Problem Solving (PS)
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# Of the following, which is the best approximation to 0.0026^1/2 ?

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Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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01 Mar 2010, 02:38
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Of the following, which is the best approximation to $$\sqrt{0.0026}$$?

(A) 0.05
(B) 0.06
(C) 0.16
(D) 0.5
(E) 0.6
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Dec 2014, 01:35, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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01 Mar 2010, 03:03
abhi758 wrote:
Of the following, which is the best approximation to $$\sqrt{0.0026}$$?
(A) 0.05
(B) 0.06
(C) 0.16
(D) 0.5
(E) 0.6

Kindly provide the steps to the solution. OA to be posted soon.

SQRT .0026 =
SQRT (26 x 10^-4) =
SQRT 26 x SQRT (10^-4) =
(approx.) 5*10^(-4/2) =
5*10^-2 =
.05

A
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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01 Mar 2010, 05:17
abhi758 wrote:
Of the following, which is the best approximation to $$\sqrt{0.0026}$$?
(A) 0.05
(B) 0.06
(C) 0.16
(D) 0.5
(E) 0.6

Kindly provide the steps to the solution. OA to be posted soon.

Backsolving 0.05^2 = .0025 it is nearest to .0026 hence A.
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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01 Mar 2010, 05:45
Even better bangalorian2000. Now it should take less than 1 min to arrive at the solution.. Thnks
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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01 Mar 2010, 12:49
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When you have square or cubed roots of numbers that are less than 0 you can use the following.
For square roots:

Square root the number at the end of the decimal (so if it were (0.0026)^0.5, you would square root 26) and divide the number of decimal places by 2. In this case, we know that the square root of 26 is a tiny bit more than 5, and dividing the number of decimal places (4) by 2 gives you 2 decimal places. Thus the answer is approximately 0.05

Cube root of a number: same procedure as above, but we cube root the number at the end of the fraction and divide the number of decimal places by 3.
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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02 Mar 2010, 01:13
What is the difficulty level of this question? Has to be less than 600.
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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06 Mar 2010, 09:55
abhi758 wrote:
Of the following, which is the best approximation to $$\sqrt{0.0026}$$?
(A) 0.05
(B) 0.06
(C) 0.16
(D) 0.5
(E) 0.6

Kindly provide the steps to the solution. OA to be posted soon.

Square root of a number that has four digits to the right of decimal would have two digits to the right of decimals in the solution. So by this we rule out option D & E

Now the approx value of (26)^1/2 would be 5. The solution would therefore be A.

Hope this helps.
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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06 Mar 2010, 15:48
this is how I did it....
nearest to 26 is 25..root of this is 5
therefore, btw a) and d)
with 2 zeros before the decimal point, I took .05*.05 = 0.0025

so it is A...interesting one

HTH
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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11 Mar 2010, 06:20
I used scientific notation.

\sqrt{26*10^-4}

\sqrt{26}close to square root of 5 (25) ... 10^-4 becomes 10^-2 and ans is 0.05.

A.

Make sense?
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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15 Mar 2010, 07:41
nickk wrote:
When you have square or cubed roots of numbers that are less than 0 you can use the following.
For square roots:

Square root the number at the end of the decimal (so if it were (0.0026)^0.5, you would square root 26) and divide the number of decimal places by 2. In this case, we know that the square root of 26 is a tiny bit more than 5, and dividing the number of decimal places (4) by 2 gives you 2 decimal places. Thus the answer is approximately 0.05

Cube root of a number: same procedure as above, but we cube root the number at the end of the fraction and divide the number of decimal places by 3.

Can you give an example the same way u did for the square root for the cube root?
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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24 Dec 2014, 18:04
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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25 Dec 2014, 01:37
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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09 Apr 2016, 06:08
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Re: Of the following, which is the best approximation to 0.0026^1/2 ? [#permalink]

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22 Aug 2016, 11:19
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abhi758 wrote:
Of the following, which is the best approximation to $$\sqrt{0.0026}$$?

(A) 0.05
(B) 0.06
(C) 0.16
(D) 0.5
(E) 0.6

√0.0026 = √(26/10000)
= (√26)/(√10000)
≈ 5/100
≈ 0.05

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Re: Of the following, which is the best approximation to 0.0026^1/2 ?   [#permalink] 22 Aug 2016, 11:19
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