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## 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
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#### Is this question even solvable? It is too hard I feel

Tue May 31, 2016 8:23 am

How can this question be even solved without doing intricate mathematics?

_________________
It is GMAT. So what?

### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
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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
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) 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 _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert DavidG@VeritasPrep Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 120 members Upvotes: 1153 GMAT Score: 770 Tue May 31, 2016 12:33 pm nchaswal wrote: 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.) 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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