How to Analyze an IR Two-Part Question

by on August 11th, 2012

This is the latest in a series of “How To Analyze” articles that began with the general “How To Analyze A Practice Problem” article (click on the link to read the original article).

This week, we’re going to analyze a specific IR question from the Two-Part prompt category. First, give yourself up to 2.5 minutes to try the below GMATPrep® problem.

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to 4πr^2, where r is the radius of the sphere.

In the above table, select the value that is closest to the cost of finishing a sphere with a 5.50-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference. Make only two selections, one in each column.”

After trying the problem, checking the answer, and reading the given solution (if any), I then try to answer the questions listed below. First, I’ll give you what I’ll call the “standard” solution (that is, one we might see in an Official Guide book if this were an official guide problem – a correct solution but not necessarily one that shows us the easiest way to do the problem). Then we’ll get into the analysis.

“Standard” solution: The formula for circumference is C = 2πr. We can use this to calculate the radii of the two spheres (note that the problem asks us to find the “closest” values, so we can estimate):

5.50 m circumference sphere: 5.5 = 2πr. (5.5/2π) = r. (Use calculator) r = 0.876 (approx.).
7.85 m circumference sphere: 7.85 = 2πr. (7.85/2π) = r. (Use calculator) r = 1.25 (approx.).

Next, the problem tells us that the formula for surface area is SA = 4πr^2. We can plug in to calculate the surface area of each sphere:

5.50 m circumference sphere: r = 0.876. SA = 4 π(0.876)^2. (Use calculator) SA = 9.63
7.85 m circumference sphere: r = 1.25. SA = 4 π(1.25)^2. (Use calculator) SA = 19.625

Finally, we can multiply each surface area by $92 per square meter to find the cost required to finish each sphere:

5.50 m circumference sphere: cost = (SA) × (cost per square meter) = (9.63) × (92) = $886
7.85 m circumference sphere = (19.625) × (92) = $1,805

The correct answers are $900 for the first column and $1,800 for the second column.

1. Did I know WHAT they were trying to test?

- Was I able to CATEGORIZE this question by topic and subtopic? By process / technique? If I had to look something up in my books, would I know exactly where to go?

→ The question is an IR Two-Part prompt. The question prompt is pretty distinctive: the answer choices are presented in table form and I have to select two of them. This problem is wordy, but it’s not a verbal problem – it definitely falls on the quant side of the fence.

- Did I COMPREHEND the symbols, text, questions, statements, and answer choices? Can I comprehend it all now, when I have lots of time to think about it? What do I need to do to make sure that I do comprehend everything here? How am I going to remember whatever I’ve just learned for future?

→ I got the question right but I spent way too much time doing it – nearly 4 minutes. The math was tedious and I had to use the calculator repeatedly. Reading the “standard” solution really didn’t help much – that is pretty much what I did, but it’s just so much work! I can already tell that I’m going to have to try to figure out a more efficient way to do this one (see section 2, below).

- Did I understand the actual CONTENT (facts, knowledge) being tested?

→ This part was fine I think. I knew the formula for circumference. They gave me the surface area formula, and I did put everything together okay. It just took me too long.

2. How well did I HANDLE what they were trying to test?

- Did I choose the best APPROACH? Or is there a better way to do the problem? (There’s almost always a better way!) What is that better way? How am I going to remember this better approach the next time I see a similar problem?

→ When I was solving, I remember thinking one thing was annoying: at one point, I divided by π and then a few steps later I had to multiply by π again. Maybe there’s a way I could’ve cut out that step altogether. I’m going to go play around with just the formulas and see what happens.

→ Let’s see, first I had C = 2πr and the point here was that I had to solve for the radius: C/2π = r. My next step was to plug that r into the surface area formula:

SA = 4πr^2

SA = 4π(C/2pi)^2

SA={4pi C^2}/{4pi^2}

→ Oh, wow, check it out! I’m going to be able to cancel out some stuff and make this way easier!

SA = C^2/pi

→ That’s amazing. All I have to do is square the circumference figure and divide by π. Let’s see, first I’m going to square 5.5. That’s about 30 (since 5^2 = 25 and 6^2 = 36). Also, I can estimate not only because the problem told me I could (by asking for the “closest” number) but also because of those answer choices. Look how far apart they are! Okay, now divide by 3 (close enough – we’re estimating!) and the surface area is 10. The cost per square meter is $92, so the approximate cost is $920. Only one answer is close. Wow – I just did that whole thing without having to use my calculator once!

→ Let’s do the other one. I need to square 7.85… which is almost 8. 8^2 = 64, so let’s say 7.85^2 = 60. Divide by 3 to get 20. Then, 20 × $92 is about $1,800 (because 2 × 9 = 18).

→ So a little work up front made my life immensely easier in the end. How would I know to do this next time? Well certainly, first, I’m going to write down the two formulas (circumference and surface area) side-by-side. That will let me notice that they both have r variables and pi symbols. Next, the problem literally tells me that I can approximate (“closest to”), plus the answers are really far apart. But when I look at the formulas separately, there doesn’t seem to be any reasonable way to approximate – for instance, to find the radius for the first sphere, I’d have to divide 5.5 by 2 and by 3, or 5.5 / 6 – which is smaller than 1 – should I estimate to 1 or would I have to use 0.9? Hmm, this seems messy – maybe there’s some algebra manipulation I can do first… yes, yes, there is.

→ That’s how I’d get myself to realize that I should combine those two equations up front. That might take about 45 to 60 seconds, then I’d use another 45 to 60 seconds to do the algebra, and now I’ve got 30 to 60 seconds left to do the last bit of math, which has now become so easy that I don’t even need the calculator.

- Did I have the SKILLS to follow through? Or did I fall short on anything?

→ I had the math skills, yes, but not the “test taker savvy” skills to realize that I should’ve done some algebra simplification before I started to plug and chug. I need to retrain myself to “look before I leap” – in other words, to see how I can simplify things before I just dive right in.

- Did I make any careless mistakes? If so, WHY did I make each mistake? What habits could I make or break to minimize the chances of repeating that careless mistake in future?

→ I didn’t on this one, although I could certainly imagine that it would have been easy to do so!

- Am I comfortable with OTHER STRATEGIES that would have worked, at least partially? How should I have made an educated guess?

→ Certainly, the smaller sphere should cost less, so I would’ve kept that in mind if I’d had to guess. Alternatively, I could’ve worked backwards starting from the answers – but I’m thinking that would be too time consuming on this one, because I have 6 answers and have to find values for 2 spheres.

- Do I understand every TRAP & TRICK that the writer built into the question, including wrong answers?

→ Actually, from the way that they wrote this, I do think that they were trying to get me to just start plugging and chugging (and wasting a bunch of time). Next time I have the urge to hop on the calculator immediately, I’m going to make sure to write stuff out first and see whether that’s really my best course of action.

3. How well did I or could I RECOGNIZE what was going on?

- Did I make a CONNECTION to previous experience? If so, what problem(s) did this remind me of and what, precisely, was similar? Or did I have to do it all from scratch? If so, see the next bullet.

- Can I make any CONNECTIONS now, while I’m analyzing the problem? What have I done in the past that is similar to this one? How are they similar? How could that recognition have helped me to do this problem more efficiently or effectively? (This may involve looking up some past problem and making comparisons between the two!)

→ They got me to default to pulling up the calculator and chugging away. Now that I’ve seen this, though, I’m going to be prepared for next time (see below).

- HOW will I recognize similar problems in the future? What can I do now to maximize the chances that I will remember and be able to use lessons learned from this problem the next time I see a new problem that tests something similar?

→ The key thing here, I think, was the two formulas that shared a variable – that, coupled with the fact that they told me I could estimate, yet it didn’t seem reasonable at all (at first) to estimate. Those two things together are a really good clue that perhaps the two formulas can be combined and simplified, and then I will actually be able to estimate (and possibly entirely avoid using the calculator in the first place!).

And that’s it! Note that, of course, the details above are specific to each individual person – such a write-up would be different for every single one of you, depending upon your particular strengths, weaknesses, and mistakes. Hopefully, though, this gives you a better idea of the way to analyze an IR problem. This framework also gives you a valuable way to discuss problems with fellow online students or in study groups – this is the kind of discussion that really helps to maximize scores.

* GMATPrep® question courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

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