# Using Algebra vs. Logic on GMAT Quant Questions

In pretty much every class I teach, at some point I’ll get the algebra vs. strategy question. Which is better? How do you know? I sympathize with the students’ confusion, as we’ll use the two approaches in different scenarios, but there doesn’t seem to be any magic formula to determine which is preferable. In many instances, both approaches will work fine, and the choice will mostly be a matter of taste and comfort for the test-taker.

In other cases, the question seems to have been specifically designed to thwart an algebraic approach. While there’s no official litmus test, there are some predictable structural clues that will often indicate that algebra is going to be nothing short of hemorrhage-inducing.

Here’s my personal heuristic; if an algebraic scenario involves hideously complex quadratic equations, I avoid the algebra. If, on the other hand, algebra leaves me with one or two linear equations to solve, it will almost certainly be a viable option. You might not recognize which category the question falls under until you’ve done a bit of leg-work. That’s fine. The key is not to get too invested in one approach and to have the patience and flexibility to alter your strategy midstream, if necessary.

Let’s look at some scenarios with unusually complex algebra. Here’s a GMATPrep® question:

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200
B. 19,600
C. 20,000
D. 20,400
E. 20,800

Simple enough. Let’s say the sides of this rectangular park are a and b. We know that the perimeter is 2a + 2b, so 2a + 2b = 560. Let’s simplify that to a + b = 280.

The diagonal of the park will split the rectangle into two right triangles with sides a and b and a hypotenuse of 200. We can use the Pythagorean theorem here to get: a^2 +b^2 = 200^2.

So now I’ve got two equations. All I have to do is solve the first and substitute into the second. If we solve the first for a, we get a = 280- b. Substitute that into the second to get: (280 – b)^2 + b^2 = 200^2. And then… we enter a world of algebraic pain. We’re probably a minute in at this point, and rather than flail away at that awful quadratic for several minutes, it’s better to take a breath, cleanse the mental palate, and try another approach that can get us to an answer in a minute or so.

Anytime we see a right triangle question on the GMAT, it’s worthwhile to consider the possibility that we’re dealing with one of our classic Pythagorean triples. If I see root 2? Probably dealing with a 45:45:90. If we see a root 3? Probably dealing with a 30:60:90. Here, I see that the hypotenuse is a multiple of 5, so let’s test to see if this is, in fact, a 3x:4x:5x triangle. If it is, then a + b should be 280.

Because 200 is the hypotenuse it corresponds to the 5x. 5x = 200 à x = 40. If x = 40, then 3x = 3*40 = 120 and 4x = 4*40 = 160. If the other two sides of the triangle are 120 and 160, they’ll sum to 280, which is consistent with the equation we assembled earlier.

And we’re basically done. If the sides are 120 and 160, we can just multiply to get 120*160 = 19,200. (And note that as soon as we see that ‘2’ is the first non-zero digit, we know what the answer has to be.)

Here’s one more from the Official Guide:

A store currently charges the same price for each towel that it sells. If the current price of each towel were to be increased by \$1, 10 fewer of the towels could be bought for \$120, excluding sales tax. What is the current price of each towel?

1. \$1
2. \$2
3. \$3
4. \$4
5. \$12

First the algebraic setup. If we want T towels that we buy for D dollars each, and we’re spending \$120, then we’ll have T*D = 120.

If the price were increased by \$1, the new price would be D+1, and if we could buy 10 fewer towels, we could then afford T -10 towels, giving us (T-10)(D+1) = 120.

We could solve the first equation to get T = 120/D. Substituting into the second would give us (120/D – 10)(D + 1) = 120. Another painful quadratic. Cue hemorrhage.

So let’s work with the answers instead. Start with D. If the current price were \$4, we could buy 30 towels for \$120. If the price were increased by \$1, the new price would be \$5, and we could buy 120/5 = 24 towels. But we want there to be 10 fewer towels, not 6 fewer towels so D is out.

So let’s try B. If the initial price had been \$2, we could have bought 60 towels. If the price had been \$1 more, the price would have been \$3, and we would have been able to buy 40 towels. Again, no good, we want it to be the case that we can buy 10 fewer towels, not 20 fewer towels.

Well, if \$4 yields a gap that’s too narrow (difference of 6 towels), and \$2 yields a gap that’s too large (difference of 20 towels), the answer will have to fall between them. Without even testing, I know it’s C, \$3.

This is all to say that it’s a good idea to go into the test knowing that your first approach won’t always work. Be flexible. Sometimes the algebra will be clean and elegant. Sometimes a strategy is better. If the algebra yields a complex quadratic, there’s an easier way to solve. You just have to stay composed enough to find it.