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Integrated Reasoning : Two part analysis

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Senior Manager
Joined: 28 Jun 2009
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Location: United States (MA)
Integrated Reasoning : Two part analysis  [#permalink]

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09 Sep 2012, 16:35
5
22
“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

5.5 m = \$900, 7.85 m = 1800
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Re: Integrated Reasoning : Two part analysis  [#permalink]

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09 Sep 2012, 19:33
6
10
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]

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04 Oct 2012, 00:45
4
Circumference = 2πr
Surface Area of Sphere = 4π(r^2) = (Circumference^2)/π
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Re: Integrated Reasoning : Two part analysis  [#permalink]

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09 Jan 2015, 14:44
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]

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09 Jan 2015, 19:26
15
8
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
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

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)

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

900; 1800

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Joined: 09 Jan 2015
Posts: 2
Re: Integrated Reasoning : Two part analysis  [#permalink]

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09 Jan 2015, 19:52
1
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
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

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)

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

900; 1800

GMAT assassins aren't born, they're made,
Rich
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Joined: 01 Feb 2015
Posts: 1
Re: Integrated Reasoning : Two part analysis  [#permalink]

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23 Feb 2015, 08:47
You can always use the the calculator in IR
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 ???
Intern
Joined: 04 Jun 2012
Posts: 5
GMAT 1: 740 Q51 V38
Re: Integrated Reasoning : Two part analysis  [#permalink]

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20 Sep 2015, 04:04
3
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
Intern
Joined: 04 Jun 2012
Posts: 5
GMAT 1: 740 Q51 V38
Re: Integrated Reasoning : Two part analysis  [#permalink]

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20 Sep 2015, 04:07
1
1
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 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 ???
Manager
Joined: 26 Mar 2017
Posts: 114
Re: Integrated Reasoning : Two part analysis  [#permalink]

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06 Jun 2017, 06:37
SOURH7WK wrote:
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

yes thats it, thanks =)
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Re: Integrated Reasoning : Two part analysis  [#permalink]

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23 Jul 2017, 07:58
I was asked this question in the Quant section is GMAT PREP test 2.
Re: Integrated Reasoning : Two part analysis   [#permalink] 23 Jul 2017, 07:58
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