# What Makes a Hard GMAT Quant Question: Part 1

Two years ago, GMAC reached out to me with an offer to curate and write explanations for their GMAT Official Advanced Questions. Since the job required selecting the “hardest” of the hard questions, they asked me to explain to them what I thought constituted a difficult GMAT question. I wrote up a pretty lengthy explanation for them, and was ultimately offered the job. Unfortunately, we eventually decided not to pursue it due to contractual issues, so I’m making my write-ups available for the GMAT population. As an independent GMAT tutor, I work with many students who’ve gone through a course or self-study and came to the end of it with scores far below their goal and what they’re capable of. In many cases, the discrepancy is due to misconceptions about what skills the GMAT is actually testing. All too often, students leave a test-prep program with the belief that proficiency on the GMAT is merely a matter of learning and internalizing rules, while all too often neglecting the relevant strategies and reasoning framework required to do well on tougher questions. In this series of blogs, I’m going to use official questions to illustrate what really constitutes difficulty on the GMAT.

Difficulty is, of course, relative, so the best way to illustrate a question’s difficulty is to compare questions that test similar concepts/quantitative abilities but that differ in ways fundamental to what I think constitute difficult questions. Please note that these different characteristics can and often do overlap on one question (indeed, it’s often the overlap of these characteristics that differentiate some of the hardest questions from their counterparts), but, for the purposes of this demonstration, I’ve restricted my analysis of any question to the criterion I consider most relevant.

To make the most of this article, I suggest that you first attempt the questions, then read my analysis below!

Example #1:

Example #2:

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

Question 1 explanation:

Option 1: Algebraic

Let h = the # of stocks closing at a higher price
Let = the # of stocks closing at a lower price

Since there 2,420 total stocks, we know: h + l = 2,420
We’re further told that the # of stocks that closed higher was 20% greater than the # of stocks that closed lower, so we can create a second equation: h = 1.2l. To solve for h, we can now substitute 1.2l in for h in the first equation, solve for l, and then multiply that value by 1.2 (there are other ways of doing this algebra, but this is the approach I’ll take).

1.2l + l = 2,420
2.2l = 2,420
Multiply both sides by 10:
22= 24,200
Divide by 10:
l = 24200/22 = 1,100
h = 1.2l = 1,100 (1.2) = 13,200

Option 2: Use the Choices:

For many test-takers, the algebraic approach on this question can be complicated and time-consuming. Though it might appear that we’re left with no other recourse, we can leverage the dispersion of the choices to our advantage by using a more intuitive approach!

Intuitively, we know two things: a) the total # of stocks is 2,420 and b) there are more stocks that ended up priced higher than priced lower.

Had the # of stocks that ended up higher been the same as the # of stocks that ended up lower, the answer would be 2,420/12 = ~1200. Since more stocks ended up higher than lower, we know the answer must be above ~1,200, so we’re left between D and E. In addressing these two choices, we can use the fact that, proportionately, these numbers are quite far apart. Taking advantage of that fact, we can ask ourselves which choice would make it so that the # of stocks priced higher is 20% greater than the # of stocks priced lower. If we try choice E, we’ll see that if h = ~1600, then l = ~800, yielding a 2:1 ratio. This is greater than the 20% difference we’re looking for, so we know E is too large, leaving us with just choice D.

Question 2 explanation:

Option 1: Algebraic:

Let l = length of the rectangle and w = width of the rectangle.

Since the perimeter is 560, we know that 2l + 2w = 560 –> l + w = 280
The diagonal of the rectangle forms a right triangle with the sides of the rectangle as its legs and the diagonal as the hypotenuse, so we can use the pythagorean theorem to say: l^2 + w^2 = 280^2.

We now have two equations:
a) l + w = 280
b) l^2 + w^2 = 200^2

This is where things get interesting/messy. How will we combine these two to solve for lw? We can of course substitute to solve for each individual variable, but that will be time-consuming and error prone. A faster approach is to use quadratic templates to our advantage:

square both sides of equation 1: (l+w)^2 = 280^2 –> l^2 + 2lw + w^2 = 280^2
Using equation 2, substitute 200^2 for (l^2 + w^2) in equation 1 to arrive at: 2lw + 200^2 = 280^2.
Subtract 200^2 from both sides: 2lw = 280^2 – 200^2
Notice that the right side is a difference of squares that can be re-written as (280 + 200)(280 – 200). Therefore:
2lw = (280 + 200)(280 – 200) = 480(80)
Divide by 2:
lw = 480(40) = 19,200.