Executive Assessment: Quant Strategies for Faster Solutions – Part 4

by on November 13th, 2017

math2Welcome to the fourth installment of Quant Strategies for the EA!

If you’re just joining us now, head back to the first part and work your way back here.

We’re going to try another Integrated Reasoning (IR) question from the official free practice set—but this time, we’ll do a 2-Part question (this is labeled #1 in the 2-Part set on the EA website, as of November 2017). Give yourself about 3 minutes to do the question. Good luck!

“The Quasi JX is a new car model. Under ideal driving conditions, the Quasi JX’s fuel economy is E kilometers per liter (E{km/L}) when its driving speed is constant at S kilometers per hour (S{km/h}).

“In terms of the variables S and E, select the expression that represents the number of liters of fuel used in 1 hour of driving under ideal driving conditions at a constant speed S, and select the expression that represents the number of liters of fuel used in a 60 km drive under ideal driving conditions at a constant speed S. Make only two selections, one in each column.”

This question has a lot going on. It’s going to be important to think carefully about how to organize this on paper and what solution path to choose.

Glance: Wall of Text. Math-focused (glance at those answer choices!). Hmm, and the answers all have variables in them. Could we use Smart Numbers to solve? Think about that while …

… Reading the question. I’m focused initially on defining the variables—and, actually, I’m really glad that they put what they did in the parentheses. This is defining the variables for me. Let’s Jot this down.

 

I’m not a fan of rate problems, so right now, I’d stop and tell myself what this means in “real world” terms.

E represents how many kilometers I can go if I put 1 liter of fuel in my car.

S represents how many kilometers I can go in 1 hour.

Both of these are under “ideal driving conditions,” whatever that is.

Let’s keep reading. First, they want me to find this:

“ … select the expression that represents the number of liters of fuel used in 1 hour of driving under ideal driving conditions at a constant speed S

Oh, I see. The “ideal driving conditions” thing is just a way of saying that I don’t need to worry about the rate of things changing. Everything’s constant. So I can ignore that part.

The question is how much fuel I’m going to use if I drive for an hour. Here’s where I want to use some real numbers—Smart Numbers!—to make sure I get this right. This is a strategy I can use whenever the problem has a bunch of variables and never gives me real numbers for those things. I get to just choose my own numbers and turn the problem from an algebra problem into an arithmetic problem.

Here’s how it works. Let’s say that I can go 2 kilometers for each liter of fuel (E = 2 km/L). And let’s say that my speed is 10 kilometers in 1 hour (S = 10 km/h).

Now, figure out how the problem works assuming that there are no variables. Everywhere the problem says E, it now says 2. And everywhere the problem says S, it now says 10.

How much fuel would I use in an hour? Well, in one hour, I would be able to drive 10 km.

I can go 2 kilometers on 1 liter of fuel. So if I need to go 10 kilometers, then I need to put 5 liters of fuel in my car.

Boom! I use 5 liters of fuel. Let’s go plug E = 2 and S = 10 into the answers to find the one that equals 5.

The very first one works. I’m going to run my eye down the rest, just to make sure. Nope, (B) doesn’t work. (C) and (D) don’t either. And (E) and (F) are fractions—too small.

The correct answer for the first column is (A).

Here’s the question for the second part:

“ … select the expression that represents the number of liters of fuel used in a 60 km drive under ideal driving conditions at a constant speed S.”

I’m still using E = 2 km/L and S = 10 km/h.

This time, they tell me I have to drive 60 km. Well, if I’m driving at a rate of 10 km/h, then how long will it take me to drive 60 km? Right, 6 hours.

Also, for every 2 km I drive, I need 1 liter of fuel, so if I’m driving 60 km, I need 60 / 2 = 30 liters of fuel.

Oh … and just that last part is my answer. I didn’t actually have to figure out how long it would take me to drive 60 km—I only needed the E = 2 km/L figure for this part of the problem.

That’s a big hint, actually. When I go to the answers, I should have to plug in for E, not for S, so I only need to check the answers that have E in them. That means I can knock out 4 answers right now! (If you’re not sure about that, you can still test those other answers.)

The two answers that contain the variable E are reciprocals, either 60/E or E/60. The first one works, so the second one can’t be the same number—we’re done!

The correct answer for the second column is (C).

And one more thing: Remember how I did some math that I didn’t have to do? I found the driving time for the second part when I didn’t need to know that. That’s a good reminder that I need to think about what I’m solving for—not just plow ahead and calculate something because I can. Don’t do more than you have to on the EA!

Key Takeaways for EA Smart Numbers:

(1) If the problem gives you variables, including in the answer choices, you can likely choose your own numbers to work through the problem. Just make sure that the problem never gives you a real value for that variable anywhere. (If it does, you have to use that real value.)

(2) If you can use smart numbers, do! Think about what will make the problem easy—don’t think about picking “real world” numbers necessarily. 10 km/h is really slow! And 2 km/L is pretty bad fuel economy. :) But that doesn’t matter; what matters is that those are nice, small numbers to use to solve.

(3) Turn that knowledge into Know the Code flash cards:

* Executive Assessment questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

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