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Register now and save up to $200 Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code ## Is this question even solvable? It is too hard I feel This topic has 2 expert replies and 0 member replies nchaswal Senior | Next Rank: 100 Posts Joined 17 Apr 2015 Posted: 44 messages Followed by: 2 members Thanked: 6 times Target GMAT Score: 760 GMAT Score: 740 #### Is this question even solvable? It is too hard I feel Tue May 31, 2016 8:23 am ttp://postimg.org/image/vfgflw263/" target="_blank"> How can this question be even solved without doing intricate mathematics? _________________ It is GMAT. So what? Thanked by: jijosei Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums! ### GMAT/MBA Expert Rich.C@EMPOWERgmat.com Elite Legendary Member Joined 23 Jun 2013 Posted: 8820 messages Followed by: 463 members Thanked: 2815 times GMAT Score: 800 Tue May 31, 2016 9:14 am Hi nchaswal, 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)

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

Thanked by: lpierce10

### GMAT/MBA Expert

DavidG@VeritasPrep Legendary Member
Joined
14 Jan 2015
Posted:
2368 messages
Followed by:
115 members
Thanked:
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GMAT Score:
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Tue May 31, 2016 12:33 pm
nchaswal wrote:
ttp://postimg.org/image/vfgflw263/" target="_blank">

How can this question be even solved without doing intricate mathematics?
Note that you can use an on-screen calculator for IR questions. (I used a calculator here, but you could also estimate.)

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

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