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

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

by nchaswal » Tue May 31, 2016 8:23 am

How can this question be even solved without doing intricate mathematics?
It is GMAT. So what?

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by [email protected] » 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