IR - Architect & spheres (CAT1 )

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IR - Architect & spheres (CAT1 )

by saralys » Wed Feb 25, 2015 2:45 am
Hi everyone,

Please let me know in case this question has already been answered.
I am reviewing my answers to the GMAT Prep CAT1 and I am stuck with this question (IR).
Could anyone help?

Thanks a lot.
Sara

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by [email protected] » Wed Feb 25, 2015 8:53 am
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
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by saralys » Wed Feb 25, 2015 12:53 pm
Thanks for your prompt answer. I got so anxious that I even forgot about the calculator... Tells you much about my chances for an honorable GMAT score... To be continued!

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by [email protected] » Wed Feb 25, 2015 1:07 pm
Hi Sara,

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: 900, 1800

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by Marty Murray » Sat Mar 07, 2015 3:15 pm
Sara, when judiciously used, the calculator can be a good tool for speeding up calculation on IR.

Of course, if you use it when it doesn't make sense to use it, doing so can actually slow you down.

Another tool that you can use to can really speed things up in IR is the sorting function included in many of the tables. When they ask about one category of data, you can sort for that. Then for the next question you sort for another, or even one after another. By doing this sorting one can make the work much easier and one can get to answers much faster.
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by saralys » Mon Mar 09, 2015 12:26 pm
Thanks again! I went again through this question and now I realize I had simply not been able to recompose the circumference formula. I'm starting to understand better when you can estimate and when you should use the calculator.

Sara