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Register now and save up to $200 Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code ## IR - Architect & spheres (CAT1 ) This topic has 2 expert replies and 3 member replies saralys Newbie | Next Rank: 10 Posts Joined 07 Feb 2015 Posted: 3 messages #### IR - Architect & spheres (CAT1 ) Wed Feb 25, 2015 2:45 am Hi everyone, Please let me know in case this question has already been answered. I am reviewing my answers to the GMAT Prep CAT1 and I am stuck with this question (IR). Could anyone help? Thanks a lot. Sara saralys Newbie | Next Rank: 10 Posts Joined 07 Feb 2015 Posted: 3 messages Mon Mar 09, 2015 12:26 pm Thanks again! I went again through this question and now I realize I had simply not been able to recompose the circumference formula. I'm starting to understand better when you can estimate and when you should use the calculator. Sara Marty Murray Legendary Member Joined 03 Feb 2014 Posted: 2050 messages Followed by: 131 members Upvotes: 955 GMAT Score: 800 Sat Mar 07, 2015 3:15 pm Sara, when judiciously used, the calculator can be a good tool for speeding up calculation on IR. Of course, if you use it when it doesn't make sense to use it, doing so can actually slow you down. Another tool that you can use to can really speed things up in IR is the sorting function included in many of the tables. When they ask about one category of data, you can sort for that. Then for the next question you sort for another, or even one after another. By doing this sorting one can make the work much easier and one can get to answers much faster. _________________ Marty Murray GMAT Coach m.w.murray@hotmail.com http://infinitemindprep.com/ In Person in the New York Area and Online Worldwide ### GMAT/MBA Expert DavidG@VeritasPrep Legendary Member Joined 14 Jan 2015 Posted: 2613 messages Followed by: 118 members Upvotes: 1153 GMAT Score: 770 Wed Feb 25, 2015 8:53 am Note that you can use an on-screen calculator for IR questions. (I used a calculator here, but you could also estimate.) First, solve for the radius If C = 5.5 2 Pi * r = 5.5 2*3.14 *r = 5.5 6.28 * r = 5.5 r = 5.5/6.28 r = .876 Next, solve for surface area 4 Pi * r^2 = 4 * 3.14 * .876^2 = 9.64 (Surface area) Last, calculate cost 9.64 * 92 = 887 Closest to 900 If C = 7.85 2 Pi * r = 7.85 2*3.14 *r = 7.85 6.28 * r = 7.85 r = 7.85/6.28 r = 1.25 4 * Pi * r^2 = 4 * 3.14 * 1.25^2 = 19.625 (surface area) 19.625 * 92 = 1805 Closest to 1800 _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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saralys Newbie | Next Rank: 10 Posts
Joined
07 Feb 2015
Posted:
3 messages
Wed Feb 25, 2015 12:53 pm
Thanks for your prompt answer. I got so anxious that I even forgot about the calculator... Tells you much about my chances for an honorable GMAT score... To be continued!

Sara

### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
Joined
23 Jun 2013
Posted:
9184 messages
Followed by:
472 members
2867
GMAT Score:
800
Wed Feb 25, 2015 1:07 pm
Hi Sara,

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

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

_________________
Contact Rich at Rich.C@empowergmat.com

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