As the basketball playoffs get underway, the LA Lakers are present once again. While they’re missing their star player, let’s take a moment to reflect on how he helped get them there – and how it can help you on the GMAT. In the entire history of basketball, perhaps no one has been a better example of focus and perseverance than Kobe Bryant. And if there’s one thing you’ll need to be successful on the GMAT, it’s a good dose of Kobe-esque grit and determination. What lessons can we take from the legend to use in our own preparation and game day execution?

As if he hadn’t already given us plenty to look to in his long and storied career, his finest example may have come just days ago as he faced his first serious setback. In making a routine push to the basket, he ruptured his Achilles tendon. The injury could not be more fitting; like the Ancient Greek hero, Kobe has seemed virtually invincible from the first day he stepped on the professional court. And, like Achilles, the injury came after pushing his ‘army’ as far as he could, giving a superhuman effort to win and get his team to the playoffs.

What did he do in the face of such a critical setback? Under NBA rules, another player could have been substituted in. But not with Kobe. He stood up, walked to the free throw line, and sunk two free throws to tie the game with only minutes left.

On the GMAT, this kind of perseverance will be rewarded. You will almost without a doubt get to a question – or several – that will appear too difficult to solve, at least not in the few minutes you can dedicate to it.

When that happens, channel your inner Mamba (Kobe’s serpentine nickname). Recognize two things:

- The same focus and perseverance that has gotten you this far into the exam—not only through the hours of study and past questions that seem relatively manageable, but all the way to the toughest stuff the GMAT can throw at you—will help keep you calm and focused at that moment
- There is always a solution that is manageable in the short time you have to think.

Take the following data sufficiency problem as an example of how perseverance can help:

*If integers A and B form an equation such that √( A ^{3} – A^{2} – B ) = 7, what is the value of A and B?*

If I see this on the test, I might rather my Achilles was torn. This looks difficult! But let’s get some grit and focus – all we need to do is get enough information to figure out A and B. They’re integers, and since we understand the solution is manageable, they’re probably in the single sigits to very low double digits at worst. Not only that, we can simplify already to ( A^{3} – A^{2} – B ) = 49. That already feels better. Now we just have to make our free throws.

**Free Throw 1) A ^{2} – A = 12**

Alright, we can figure out how to break this down. Astute test takers instantly see this as a quadratic equation that can be factored into (A – 4)(A + 3) = 0, but even if you don’t catch the shortcut on a question, you can always make educated guesses to quickly test and check. A can equal 4 or -3… probably not sufficient – plug in each value to make sure either could be plausible, and you see they both can work. If you’re curious, you could determine that either 48 – B = 49 or -36 – B = 49… so B is either -1 or -85. Certainly -1 is more likely, but we don’t know for sure.

**Free Throw 2) B ^{2} – B = 2.**

No problem here, the same logic applies. (B – 2)(B + 1) = 0 does the trick, and B is either 2 or -1. Presto! A value of -1 matches one of our options, and we have a winner. Statement 2 completes the puzzle and together they work – so answer choice C seems obvious. But wait. That ended up being a bit simpler than two free throws should be when we’ve encountered a problem that looks as bad as a ruptured Achilles. Watch Kobe’s free throws one more time. There is no way he could make those shots with out extra focus and determination, and we may not get the correct answer without the same. The GMAT is aiming right for our own Achilles with these kinds of questions.

Let’s make sure now – If B is either 2 or -1, then when we plug it back into ( A^{3} – A^{2} – B ) = 49, we will see that A^{3} – A^{2} = either 51 or 48. If we think about that, it might seem that only one possibility really exists, since we know A has to be an integer. We can try to figure out the exact value of A (it’s 4), but it doesn’t really matter; the second statement is all we really need to determine that there is in fact only one possible value for A and B, and answer Choice B is correct. That kind of answer only comes by truly focusing on the question and pushing ourselves to carefully consider all the possibilities of what’s knowable with the given information.

**The correct answer is (B). **

Kobe Bryant has built a career on his grit and determination. He’s had ups and downs and persevered to do amazing things: 5 championships and 15 all-star games, a sometimes strained relationship with former teammate and basketball great Shaquille O’Neal, an 81 point game. And now the Achilles, a testament to his perseverance. He managed all this through consistent mental focus and a lot of hard work, and if you can build the same habits of perseverance during your own GMAT preparation, I promise you will find yourself outperforming on routine questions (like Kobe’s 81 pt game) and gritting it out for the win on the most tough-to-tackle problems, even those you might normally consider your own Achilles’ heel.

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*Joseph Dise has been teaching GMAT preparation for Veritas Prep for the last 4 years in Paris and New York City. Check out another one of his basketball related posts here. *

I do not understand this part: “If we think about that, it might seem that only one possibility really exists, since we know A has to be an integer.”

How can you tell?

Thank you!

Hi Steve,

Since we’re left with (A^3 – A^2) = either 51 or 48, we could use this to test the most likely integer values of A. There are a couple approaches we can take. The most obvious is to ‘plug and play’ using the values of A (4, -3) we deduced from the prior statement, which return 48 and -36 respectively — a good start as 48 is an expected return.

If that doesn’t strike you, start with a small, plausible and manageable number — say 2 — and build from there. 2 returns 4, so we need to go higher. 3 returns 18, 4 returns 48 (an expected number), and 5 returns 100. The least efficient approach is to try to factor out A or A^2 in some fashion; it mostly complicates and confuses the algebra.

The real efficiency — and the answer to your question — is in being able to assess that given the proximity of 48 to 51, and how the function (A^3 – A^2) changes as integer A changes, there can only be one integer A that works. Because 2,3,4,5 returns 4,18,48,100 respectively, 51 is not actually a possibility, as it would require a non-integer value of A. Since 48 as a return for (A^3 – A^2) narrows A to only one possible integer, there is now only one integer value for both A and B.