Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 3008:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates!
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 1211:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
Circumference 5.50 m: $900
Circumference 7.85 m: $1,800
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
74% (03:13) correct
26%
(03:02)
wrong
based on 1158
sessions
History
Date
Time
Result
Not Attempted Yet
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 |
ShowHide Answer
Official Answer
Circumference 5.50 m: $900
Circumference 7.85 m: $1,800
ankitraj88
Joined: 04 Jun 2012
Last visit: 11 Mar 2017
Posts: 4
Given Kudos: 8
Schools: Booth '19 ISB '18 Sloan '19 Wharton '19 CBS '19 (S)
GMAT 1: 740 Q51 V38
20 Sep 2015, 04:04
Kudos
Bookmarks
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
09 Jan 2015, 19:26
Kudos
Bookmarks
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
