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Difficulty:
55%
(hard)
Question Stats:
72% (03:16) correct
28%
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wrong
based on 2411
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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 \(4πr^2\), 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.
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 |
ankitraj88
Joined: 04 Jun 2012
Last visit: 11 Mar 2017
Posts: 4
Schools: Booth '19 ISB '18 Sloan '19 Wharton '19 CBS '19 (S)
GMAT 1: 740 Q51 V38
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
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
EMPOWERgmatRichC
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Last visit: 31 Dec 2023
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Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert reply
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:
GMAT assassins aren't born, they're made,
Rich
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
General Discussion
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
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
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.
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.
ankitraj88
Joined: 04 Jun 2012
Last visit: 11 Mar 2017
Posts: 4
Schools: Booth '19 ISB '18 Sloan '19 Wharton '19 CBS '19 (S)
GMAT 1: 740 Q51 V38
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
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
EMPOWERgmatRichC
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?
EMPOWERgmatRichC
Expert
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,775
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert reply
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
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
avigutman
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Tutor
Joined: 17 Jul 2019
Last visit: 27 Jun 2026
Posts: 1,285
Location: Canada
Schools: NYU Stern (WA)
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Expert reply
Video solution from Quant Reasoning starts at 32:20
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Kinshook
Major Poster
Joined: 03 Jun 2019
Last visit: 11 Sep 2026
Posts: 6,127
Location: India
Schools: HBS '28 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
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
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
chini
Joined: 08 Jun 2021
Last visit: 17 Aug 2026
Posts: 16
Location: India
Concentration: General Management, Sustainability
Schools: INSEAD Aug '26 Stanford '26 HBS '28
GMAT Focus 1: 695 Q85 V86 DI82
WE:Information Technology (Consulting)
chini
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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How do you now 5.5/7.85 is 1/√2 right away?
ankitraj88
but you need to know the formula for circumference, which is not required on the focus, no?
Bunuel
GMATINATOR9000
Check these two topics:
While specific geometry knowledge is not tested on GMAT Focus, not everything involving shapes, volumes, or areas requires specialized geometry knowledge. The area of a square or rectangle, the volume of a cube or rectangular solid, and the Pythagorean theorem are not considered specific geometry knowledge by the GMAT and can still be tested. Moreover, a question can involve shapes but test another area, such as combinations or other topics. There are several questions involving these concepts in the GMAT Prep Focus mocks.
The chapter on coordinate geometry, including planes and slopes, is still present in the recent Official Guides. That said, it is tested to a much smaller extent. For example, you might see some questions involving graphs, either in Problem Solving or in Data Insights graph-based questions. However, those usually fall under the functions category rather than pure coordinate geometry, so they typically won’t involve distance calculations, angles, or similar topics.
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Posts: 9
This question is hard, need 3 mins minimum to solve this.
harikris
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