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
_________________

I approached this problem in slightly objective manner.


Since circumference is provided, we can check the ratio of the 2 circumferences and ultimately the ratio of the two radius.

C = 2(pi)(R)

C1/C2= R1/R2 = 5.5/7.85 = 1/√2

Cost is also proportional to the area, thus

A1/A2=Cost1/Cost2= [(R1)/(R2)]^2 = 1/2

The only available combination with that ratio is 900 and 1800

2pi R = 5.50

Therefore, R = 5.50/2*7/22 = 0.875 Find R^2 = 0.7656

So, Surface area = 4 PI R^2 = 9.625 Sq M.

Cost of finishing a sphere = 92* 4 PI R^2 = 92 * 9.625 = 885 So Approx. $900

Find the same way for other circumference of 7.85M

On the Integrated Reasoning section, you have access to a calculator.

So just solve the formula for radius of the sphere first.

If you know circumference as 5.50 meters, then pi * d = 5.50

d = 5.50 / 3.14
d = 1.75

Since radius is half the diameter, radius is 0.876.

Now we have to solve for surface area then multiply by cost / square meter.

Surface area is 4pi*r^2

4 * 3.14 * 0.876 * 0.876 = 9.636

$92 / square meter

OK so $92 / sq meter * 9.636 square meters = $886.5

After some rounding that is $900 so that is the best answer for the first column.

I approached this problem in slightly objective manner.


Since circumference is provided, we can check the ratio of the 2 circumferences and ultimately the ratio of the two radius.

C = 2(pi)(R)

C1/C2= R1/R2 = 5.5/7.85 = 1/√2

Cost is also proportional to the area, thus

A1/A2=Cost1/Cost2= [(R1)/(R2)]^2 = 1/2

The only available combination with that ratio is 900 and 1800

this question is easy. I come up with a super easy solution.

Find the ratio between the 2 spheres, later / former = (7.85 / 5.5) ^ 2 = approximately 1.4^ 2 = 1.96 = approximately 2
then, look at the options => 900 and 1800

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?

Hi yanyanjen,

The Area of a CIRCLE is (pi)(R^2), but we're dealing with SPHERES in this question - and the Surface Area of a sphere 4(pi)(R^2)... keep in mind that this formula is information that the prompt GAVE us. The circumference of a sphere is the longest path 'around' the sphere (and looking at that shape 2-dimensionally, that shape would be a circle - meaning that we can use the Circumference formula for a circle here).

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