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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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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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I approached this problem in slightly objective manner.


Since circumference is provided, we can check the ratio of the 2 circumferences and ultimately the ratio of the two radius.

C = 2(pi)(R)

C1/C2= R1/R2 = 5.5/7.85 = 1/√2

Cost is also proportional to the area, thus

A1/A2=Cost1/Cost2= [(R1)/(R2)]^2 = 1/2

The only available combination with that ratio is 900 and 1800



jhoop2002
I was reviewing this question and was curious what tips you have to complete this problem without a calculator. Am I really supposed to manually perform 5.5 / 6.28 ???
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Hi jhoop2002,

This question is essentially a giant multi-step "estimation" question, BUT you have to use the answer choices to your advantage and the work that you do to answer the first question will actually HELP you to answer the second. Here's how:

Using the given formulas for circumference and surface area:

C = 2(pi)(R)
SA = 4(pi)(R^2)

The first sphere has circumference = 5.5m
It's radius is...
5.5 = 2(pi)(R)
5.5/(2pi) = R ....don't do anything more to this....

It's surface area is....
SA = 4pi(5.5/2pi)^2
SA = 4pi[5.5^2/4pi^2)
SA = 5.5^2/pi

5.5 is between 5 and 6, so 5.5^2 is between 25 and 36
We need a rough estimate for....
(25 to 36)/pi

If we say pi = 3 (Note: we ALL know that this isn't super-accurate, but it works in this question. You'll see why this is helpful in a moment...)
(25 to 36)/pi = between 8 and 12
Let's say it's about 10....

With a Surface Area of 10 meters^3 and a cost of $92 per meter^3, we have about....
10(92) = $920
The ONLY answer that's even close is $900.
Lock in THAT value.

Using the same logic, we now deal with the sphere with a circumference of 7.85...and probably work faster (@since we just have to plug in the newer radius into the final calculation)

It's radius is...
7.85 = 2(pi)(R)
7.85/(2pi) = R ....don't do anything more to this....

It's surface area is....
SA = 4pi(7.85/2pi)^2
SA = 4pi[7.85^2/4pi^2)
SA = 7.85^2/pi

7.85 is between 7 and 8, so 7.85^2 is between 49 and 64
We need a rough estimate for....
(49 to 64)/pi

(49 to 64)/pi = about 16 to 21

REMEMBER the work we did on the smaller sphere!!! We "said" its surface area was about 10. The surface area of the larger sphere can't be much more than about 20, which is TWICE the SA, so the cost to paint it must be ABOUT TWICE the cost of painting the smaller sphere....

2(900) = 1800

Final Answer:
900; 1800

GMAT assassins aren't born, they're made,
Rich




Isn't regular surface area = pi(r^2)? is this question changed it to 4pi(r^2), how can we assume circumference equation = 2pi(r) still holds true?
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Hi yanyanjen,

The Area of a CIRCLE is (pi)(R^2), but we're dealing with SPHERES in this question - and the Surface Area of a sphere 4(pi)(R^2)... keep in mind that this formula is information that the prompt GAVE us. The circumference of a sphere is the longest path 'around' the sphere (and looking at that shape 2-dimensionally, that shape would be a circle - meaning that we can use the Circumference formula for a circle here).

GMAT assassins aren't born, they're made,
Rich
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Video solution from Quant Reasoning starts at 32:20
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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url_gal

Given: An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to ​4 pi r squared​, where r is the radius of the sphere.

Asked: In the table, select the value that is closest to the cost of finishing a sphere with a 5.50-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference. Make only two selections, one in each column.

Circumference = \(2*\pi*r \)
Area = \(4\pi r^2\)
Cost of finishing = Area in sq m* $92

Finishing Cost for Circumference 5.50:

\(2\pi*r = 5.5\)
r = .875 m
Cost of finishing =Area in sq m* $92 = $885.85 = approx $900

IMO A

Finishing Cost for Circumference 7.85:

\(2\pi*r = 7.85\)
r = 1.25 m
Cost of finishing = Area in sq m* $92 = $1804.58 = approx $1800

IMO C
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Bunuel bb , this is kind of a geometry question, do you think we will see such questions in the focus edition DI
harikris
An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to \(4πr^2\), where \(r\) is the radius of the sphere.

In the table, select the value that is closest to the cost of finishing a sphere with a 5.50-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference. Make only two selections, one in each column.
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chini
Bunuel bb , this is kind of a geometry question, do you think we will see such questions in the focus edition DI
harikris
An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to \(4πr^2\), where \(r\) is the radius of the sphere.

In the table, select the value that is closest to the cost of finishing a sphere with a 5.50-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference. Make only two selections, one in each column.

Even though hardcore geometry is no longer tested on the GMAT, this question can’t really be considered a geometry question. The formula is given, so there’s no need to recall or apply any geometry concepts. You just plug in the values and use a calculator to get the answer.
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