How to Logically Solve Algebra Problems on the GMAT

Just as all roads lead to Rome (well, all roads in Europe anyways), there are many ways to solve math questions on the GMAT. Any question can conceivably be solved in a variety of ways, but they must always be logical. No method is inherently superior to any other, so often it’s a question of which method will solve this particular problem in the most efficient way possible.

Algebra questions in particular lend themselves to multiple solutions. Solving the question using only algebra will always work in theory; however this strategy can turn messy in practice. Since the GMAT is a multiple choice exam, you can always backsolve by using the answer choices, but this will rarely yield the correct answer directly. Any question with variables can be made less abstract by picking values for the variables X and Y (Z, λ and ♥ as well if needed). Sometimes you can solve the question without a single pen stroke (or keystroke) just by using the concept.

If all of these concepts work to solve problems on the GMAT, then the question becomes: Which should I use? The answer is simply this: Any one of them that works for you and that makes sense and can be solved in 2-3 minutes or less. If you have all of these strategies in your back pocket, you can solve the question in a variety of ways (hopefully all of which give the same answer).

Let’s look at a tricky algebra question and review various different ways to solve it. Hopefully at least one of these methods will make sense to you and then that method can be your default strategy for similar GMAT questions.

On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

(A)   1/4
(B)   4/5
(C)   1/5
(D)   1/6
(E)    1/7

Let’s look at this logically. We know that s must be an integer, and somehow between s and s+1, Derek is averaging a brisk 2.8 miles/hour. Logically, we now know that s must be 2 and s+1 must be 3. If you don’t recognize this requirement, think about any other option. If s had been 1 and s+1 had been 2, Derek could have never averaged 2.8 miles/hour. Ditto for s=3 or greater, so we’re sandwiched in our values of s. If Derek had averaged 2.5 miles/hour, we would have known that he spent as much time at 2 mph as at 3 mph. Since the average is 2.8, we know that he spent more time at the higher speed, specifically 4/5 of his time at the speed and 1/5 of his time at the lower speed.

Now that we have a 360 degree view of the information presented in the question, let’s examine various ways to get the correct answer.

Algebraically, we know Derek spent some portion of his time walking at 2 MPH (k) and the rest of the time (1-k) at 3 MPH. Set the total time to be d and set up d1 for the slow walk and d2 for the fast walk, all we have to do is solve for variable k:

 

 

This can be rewritten as:

 

 

Putting the right side on a common denominator to (eventually) isolate K:

 

 

Multiplying both sides by 6:

 

 

Simplifying:

 

 

Subtracting 2 from both sides:

 

 

The algebraic solution yields answer choice E: 1/7 of the time is spent at the slow speed. However, this kind of algebra can lead to mistakes if you’re not careful, so alternative solutions can be just as valid.

Picking numbers, the speeds are in a relation of 1:4. For every minute Derek spent at 2 mph, he spent 4 at the higher speed. We can translate this relationship of time into a relationship of distance using the units of Miles Per Hour (MPH, not to be confused with Neil Patrick Harris). If Derek spent 1 hour at 2 MPH, he’d cover 2 miles. He’d also have to spend 4 hours at 3 MPH, covering an additional 12 miles. On total, he would have covered 2 miles out of a total of 14 miles, which is 1/7 of the distance, or answer choice E.

Looking at the answer choices, once you’ve found that he spent only 1/5 of his time at the slower speed, it becomes immediately clear that he must have covered less than 1/5 of the distance at the slow speed. This leaves either 1/6 or 1/7 as possible answer choices. Let’s try 1/6 and see what that gives! If he spent 1/6 of the time at 2 MPH, set the total distance to 6 miles, 1 of which at the slow speed and the other 5 at the higher speed. 1 mile at 2 MPH means 0.5 hours at speed s. This leaves 5 miles at 3 MPH, for a total of 1.67 hours spent walking in the sun. We need these numbers to be a ratio of 1:4, so it doesn’t work. E backsolves elegantly into 1 mile at the slow speed and 6 miles at 3 MPH, a perfect 1:4 ratio.

As you can see, even a question that’s all about algebra can have multiple elegant solutions. The math here isn’t trivial, so if you can find an alternative route to the solution, then so much the better. There are other solutions to this problem, so I encourage you to experiment and try and find your own solution. Needless to say, if you have three different methods that you’re comfortable with, including the algebra, you’ll be able to navigate the thick (#Thicke) and thin of the GMAT and solve any question asked of you.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And, be sure to find us on Facebook and Google+, and follow us on Twitter!

Ron Awad is a GMAT instructor for Veritas Prep based in Montreal, bringing you weekly advice for success on your exam.  After graduating from McGill and receiving his MBA from Concordia, Ron started teaching GMAT prep and his Veritas Prep students have given him rave reviews ever since.

Leave a Reply

Spam protection by WP Captcha-Free