Math Basics – Distance, Rate and Time

The $4^{th}$ of July weekend is upon us, and many of us will be taking road trips, or even taking a plane somewhere. To commemorate this collective movement, let’s learn the most fundamental formula when dealing with movement over time. First, let’s meet Bob…

Q1. Bob drives at an average rate of $50$ mph from Berkeley to Los Angeles, a distance of $350$ miles. How long does it take him to complete the trip?

(A) $4$ hrs

(B) $5$ hrs

(C) $7$ hrs

(D) $10$ hrs

(E) $12$ hrs

When dealing with $distance$, $rate$ and $time$, we always want to remember the nifty little formula, $D = R \times T$, in which $D$ stands for the distance, $R$ stands for the rate (or speed), and $T$ stands for the time.

With the problem above, the $distance$ between Berkeley and Los Angeles is $350$ miles. So $D = 350$. Bob is traveling at $50$ mph, so that is his $rate$. The question is how long will it take him to complete this trip. Therefore, we have to solve for $T$. Let’s set up the equation, plugging in the values for $D$ and $R$:

$350 = 50T$.

Solving for $T$, we get $7$. Because we are dealing with miles per hours, the $7$ corresponds to hours. So $T = 7$.

Answer (C)

Now let’s try another problem:

Q2. Charlie takes $2.5$ hours to fly from Los Angeles to Mexico City, a distance of $1200$ miles. What is the average speed of his plane in miles per hour?

(A) $200$ mph

(B) $240$ mph

(C) $410$ mph

(D) $480$ mph

(E) $533$ mph

Setting up the equation, we get $1200 (Distance) = 2.5 (Time) \times R$; $1200 = 2.5R$. Before solving (assuming this is on the current GRE, in which you do not have a calculator), let’s change $2.5$ to $\dfrac{5}{2}$, as it is much easier to do the math with fractions than with decimals. We get $1200 = \dfrac{5}{2R}$. Solving for $R$, we multiply both sides by the reciprocal, $\dfrac{5}{2}$. This gives us R on the right-hand side of the equation, and $1200 \times \dfrac{5}{2}$ on the left-hand side. $\dfrac{2400}{5} = 480 (D)$.

Note you could have solved this problem back-solving, in which you put the answer choices back into the question. Or, even better, we could just plug-in $500$ mph. Of course that is not one of the answer choices, but using $500$ will make the math very easy. If the number is a little too high when we plug-in $500$, then the answer must be (D). If it is a little low, then the answer must be (E) $533$. Again, you never try to back solve with an ugly number like $533$ — it will take too long.

Using $500$ mph, we get $1200 (Distance) = 500 (Rate) \times 2.5 (Time)$. You can see that multiplying these two numbers gives us $1250$. Meaning, we flew too fast and missed Mexico City by $50$ miles. Therefore, we have to slow down the plane a little — the answer is $480$.

Answer (D)

So, this weekend, whether you are traveling abroad or just making a trip to a friend’s BBQ, you can figure out your average distance, rate, or time. Who said GRE doesn’t relate to the real world?