**Sometimes, the best choice from among two options is neither of them! **

## Approaches to the GMAT Quantitative Section

Many quantitative problems contain numbers and demand numerical calculations. Others contain variables, and one can use either an algebraic approach or a numeric approach. See this blog post for more on that choice. Folks are often aware of these two basic approaches, algebraic and numeric, for attacking GMAT Quant problems. What folks often forget is that there’s third approach, at least as powerful, which magically unlocks otherwise intractable problems. First, a sample difficult DS question.

## Sample Problem

1) A, B, P, and Q are four positive numbers. Does AQ = BP?

Statement #2: In the x-y plane, the line through (A, B) and (P, Q) goes through the origin.

You may want to wrestle with this a bit before reading on.

## I See!

Of course, that third perspective is the geometric/visual perspective. A gigantic chunk of our cerebral cortex — basically the entire occipital lobe — is devoted to processing visual information. We homo sapiens are very essentially visual creatures. Yet, in the mad rush of GMAT Quant, we forget this powerful capacity of ours. Part of the problem is that Geometry often gets short shrift, both in school and in GMAT review, so we underappreciate what humble branch of math offers. Algebra we have to calculate. Geometry we often see immediately.

Of course, in any problem with a diagram, or in any problem that mentioned a geometric shape, you probably will use visual/geometric thinking. This problem is subtle: only statement #2 even mentions the x-y plane, i.e. coordinate geometry. As a general rule, when the problem gives even the slightest glimmer that there might be something visual/geometric about it, that is a strong clue that the visual/geometric will yield fast insights. We will use a visual/geometric approach in solving the above problem.

## Sample Problem Solution

**Big geometry idea** **#1**: every point in the x-y plane is the vertex of a unique little right triangle with legs parallel to the axes, with a second vertex on the x-axis and the third vertex at the origin. The length of the vertical leg is the y-coordinate, and the length of the horizontal leg is the x-coordinate.

That one fact has several enormous implications. Notice, the distance from the point (x,y) to the origin is the hypotenuse, so we could find that distance with the Pythagorean Theorem.

This should look familiar. That’s exactly the form we have for points (A, B) and (P, Q) in statement #1. The statement:

is entirely equivalent to the statement that (A, B) and (P, Q) are equal distances from the origin. In other words, if we made a circle with center at the origin, we know both points would lie on the same circle.

Well, how does this help us answer the question: Does AQ = BP? At least in the diagram as it is currently drawn, A and Q are the “big” legs and P and B are the “small” legs, and (big)*(big) ≠ (small)*(small). For this example, clearly AQ ≠ BP — a “no” answer to the prompt. But, if A = P and B = Q, the two points would be at the same place. Statement #1 would still be satisfied, and the answer to the prompt would then be “yes.” Statement #1 is consistent with both a “yes” answer and “no” answer to the prompt, so by itself it is **insufficient**.

Now, forget about Statement #1 and focus exclusively on Statement #2. We are told the line through (A, B) and (P, Q) goes through the origin, like this:

**Big geometry idea** **#2**: Two triangles of the same shape and same angles but different sizes are called *similar*. **Similar triangles have proportional sides**!! That is one of the deepest and most powerful ideas in all of geometry, and it has staggering implications for problem-solving on the GMAT.

Because the two right triangles are obvious similar, we can set up a proportion among the four sides of interest:

Now, cross-multiply:

AQ = PB

Voila! It turns out: the prompt question is a proportionality question in disguise. (will you remember that trick on the GMAT?) Statement #2 allows us to answer the prompt with a definitive “Yes!”, so it is **sufficient**. Answer =** B.**

## Epilogue

The poet Edna St. Vincent Millay (1892 – 1950) wrote: “Euclid alone has looked on beauty bare,” her tribute to the inherent aesthetic appeal of geometry, especially as formulated by its founder, Euclid (c. 300 bce). Some would argue that the cogent elegance that geometry brings to problem-solving is part and parcel of its aesthetic appeal. Whatever aesthetic satisfaction you find in geometry, its profound power as a problem-solving tool is undeniable. Master this perspective, especially the ideas related to similar triangles, and a whole new dimension of problem-solving possibilities will open to you.

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Hi Mike,

the way I thought about staement 2 was “if they both go through the origin, then the x-coefficent of each point on the plane is zero, therefore when you cross multiply, you’ll have the same result”. Is that right? I realize I made an assumption that the origin is (0,0), but that isn’t always the case. Any thoughts?

Thank you!

Hi Toby 🙂

You’re correct that when we refer to the origin in the x-y plane, we are referring to the point (0,0). However, I’m not quite sure what you mean by the x-coefficient of each point on the plane. (A, B) and (P, Q) are not necessarily the origin. In fact, if both point were, they would be the same point. Rather, we can draw one line that passes through the origin (both the x and y intercepts are 0) and this line will also pass through (A, B) and (P, Q). The two points are on the same line but this line could have any number of slopes.

Hope that helps 🙂

Hi Mike,

For the second statement, how can we assume both the lines have the same slope ?

It plainly states that the lines goes through the origin, which could also imply that (A,B) and (P,Q) have different slopes.

Hey Mike,

Thanks for the response; sorry if I came across as narrow minded!

Sammy, quite alright, sir. It’s always a challenge for each one of us to see beyond one’s own self. Best of luck to you.

With respect, Mike 🙂

For debunking the first statement, you can simply use the algebraic rule of ‘x^2 + b^2 = x^2 + 2xy + y^2’ (after squaring both sides) and see that the first statement will be insufficient.

I feel that the geometric approach might be a tad time consuming.

As for the second statement, since both points lie on the same line (same slope) – you can write out each set of point in slop-intercept form. You will get Q = mP & B = mA. Solve for m and you should get AQ = PB.

Sammy, first of all, I’m not sure I understand your the first equation you cite. Clearly, (x + y)^2 = x^2 + 2xy + y^2, but I’m not sure of the relationship of b to the terms involving y. If we square both sides of Statement #1, we get A^2 + B^2 = P^2 + Q^2, which I’m not sure helps us. Even after squaring both sides, I don’t know that one can readily solve Statement #1 with algebra alone. For statement #2, clearly you could use slope-intercept form and get to the answer with algebra, but with all due respect, I think you are missing the larger point. You appear to be an individual with a facility with algebra, which is fantastic, but understand: thereby, you are not representative of the general population. For folks for whom algebra is a chore, a visual/geometric approach can be lightning fast by comparison. For someone like you, who already is lightning fast with algebra, the possible additional advantage of a geometric/visual approach is not nearly as great. Thus, an approach that seems “a tad time consuming” for you might be a windfall blessing for another user. It’s good to be aware of the vantage from which you speak. Does that make sense, my talented friend? With respect, Mike 🙂