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
Isn't regular surface area = pi(r^2)? is this question changed it to 4pi(r^2), how can we assume circumference equation = 2pi(r) still holds true?