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# Could someone provide a solution to this GMATPrep IR Questio

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Intern
Joined: 12 Jul 2012
Posts: 29
Could someone provide a solution to this GMATPrep IR Questio  [#permalink]

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25 Oct 2012, 08:24
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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Q2.jpg [ 123.01 KiB | Viewed 1712 times ]

Intern
Joined: 11 Oct 2012
Posts: 2
Re: Could someone provide a solution to this GMATPrep IR Questio  [#permalink]

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30 Oct 2012, 18:15
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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Joined: 14 Apr 2009
Posts: 2264
Location: New York, NY
Re: Could someone provide a solution to this GMATPrep IR Questio  [#permalink]

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06 Nov 2012, 14:16

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

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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Joined: 12 Dec 2016
Posts: 1709
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
Re: Could someone provide a solution to this GMATPrep IR Questio  [#permalink]

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11 Oct 2017, 21:37
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
Re: Could someone provide a solution to this GMATPrep IR Questio &nbs [#permalink] 11 Oct 2017, 21:37
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# Could someone provide a solution to this GMATPrep IR Questio

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