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Integrated Reasoning : Two part analysis

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Integrated Reasoning : Two part analysis [#permalink] New post 09 Sep 2012, 15:35
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“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 4πr2, 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.”

Circumference 5.50 m Circumference 7.85 m

Finishing cost
$900

$1,200

$1,800

$2,800

$3,200

$4,500

[Reveal] Spoiler:
5.5 m = $900, 7.85 m = 1800
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Re: Integrated Reasoning : Two part analysis [#permalink] New post 09 Sep 2012, 18:33
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1. 2πr=5.50
=> r = 5.5/2π = 5.5/2x3.14 = .875
So surface area = 4πr^2= 4x3.14x .875^2 = 9.61 m^2
Hence cost = 9.61x92 = 885 ~ $900
2. 2πr=7.85
=> r = 7.85/2π = 7.85/2x3.14 = 1.25
So surface area = 4πr^2= 4x3.14x 1.25^2 = 19.625 m^2
Hence cost = 19.625x92 = 1805 ~ $1800
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Re: Integrated Reasoning : Two part analysis [#permalink] New post 03 Oct 2012, 23:45
Circumference = 2πr
Surface Area of Sphere = 4π(r^2) = (Circumference^2)/π
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Re: Integrated Reasoning : Two part analysis [#permalink] New post 09 Jan 2015, 13:44
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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Re: Integrated Reasoning : Two part analysis [#permalink] New post 09 Jan 2015, 18:26
Expert's post
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:
[Reveal] Spoiler:
900; 1800


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Re: Integrated Reasoning : Two part analysis [#permalink] New post 09 Jan 2015, 18:52
I know this wasn't a short explanation, but thank you!!

EMPOWERgmatRichC wrote:
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:
[Reveal] Spoiler:
900; 1800


GMAT assassins aren't born, they're made,
Rich
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Re: Integrated Reasoning : Two part analysis [#permalink] New post 23 Feb 2015, 07:47
You can always use the the calculator in IR :wink:
jhoop2002 wrote:
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 ???
Re: Integrated Reasoning : Two part analysis   [#permalink] 23 Feb 2015, 07:47
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