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Could someone provide a solution to this GMATPrep IR Questio

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Could someone provide a solution to this GMATPrep IR Questio [#permalink]

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New post 25 Oct 2012, 07:24
Hi Guys,

PFA the GMAT PREP IR Question as a pic file. Could someone please provide a solution to this GMATPrep IR Question?




Thanks,
harikris
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Re: Could someone provide a solution to this GMATPrep IR Questio [#permalink]

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New post 30 Oct 2012, 17:15
2pi R = 5.50

Therefore, R = 5.50/2*7/22 = 0.875 Find R^2 = 0.7656

So, Surface area = 4 PI R^2 = 9.625 Sq M.

Cost of finishing a sphere = 92* 4 PI R^2 = 92 * 9.625 = 885 So Approx. $900

Find the same way for other circumference of 7.85M

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Re: Could someone provide a solution to this GMATPrep IR Questio [#permalink]

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New post 06 Nov 2012, 13:16
On the Integrated Reasoning section, you have access to a calculator.

So just solve the formula for radius of the sphere first.

If you know circumference as 5.50 meters, then pi * d = 5.50

d = 5.50 / 3.14
d = 1.75

Since radius is half the diameter, radius is 0.876.

Now we have to solve for surface area then multiply by cost / square meter.

Surface area is 4pi*r^2

4 * 3.14 * 0.876 * 0.876 = 9.636

$92 / square meter

OK so $92 / sq meter * 9.636 square meters = $886.5

After some rounding that is $900 so that is the best answer for the first column.

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Re: Could someone provide a solution to this GMATPrep IR Questio [#permalink]

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New post 11 Oct 2017, 20:37
this question is easy. I come up with a super easy solution.

Find the ratio between the 2 spheres, later / former = (7.85 / 5.5) ^ 2 = approximately 1.4^ 2 = 1.96 = approximately 2
then, look at the options => 900 and 1800

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Re: Could someone provide a solution to this GMATPrep IR Questio   [#permalink] 11 Oct 2017, 20:37
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