Welcome back to our series on GMAT quant rate problems. In the last two articles, we saw how GMAT quant problems employ the concepts of average speed and instantaneous speed. This article will introduce methods for solving problems where the speed of a vehicle changes. These problems can be complex, but like so many other speed/distance/time problems, they can be solved by leaning on an understanding of the direct and inverse relationships between these variables. 

Here’s a relatively straightforward one to get us started:

Due to construction, the speed limit along an 8-mile section of highway is reduced from 55 miles per hour to 35 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit?

(A) 5

(B) 8

(C) 10

(D) 15

(E) 24

If you’ve been reading through this series of articles, you should be a big fan of “factors of change” for the variables of speed, distance, and time. A good way to start this problem is by considering the factor of change from the original speed of 55 mph to the new speed of 35 miles per hour. Both numbers are recognizable as multiples of 5 (5 * 11 and 5 * 7), so the new speed of 35 mph is 7/11ths of the original speed of 55 mph.

Since speed and time over a given distance are inversely related, the time to cover the 8 miles at the new speed will be 11/7ths (the reciprocal of 7/11) of the time at the original speed. The next logical step is to determine how long it takes to cover the 8 mile distance at 55 mph. 

  • Time = distance / speed
  • Time = 8 miles / 55 mph
  • Time = 8/55 hours

This is a bit annoying because it is in hours, but the problem asks us for a time increase in minutes. We can convert the time to minutes by multiplying by 60.

  • Time in minutes = 60 * 8/55
  • Time in minutes = 8 * 60/55

The 60 and the 55 will almost cancel each other, giving us a time of just over 8 minutes.

Now we can think about the 11/7 factor of multiplication to obtain the time at the new speed of 35 mph. Let’s put this in conversation with the answer choices.

(A) 5

(B) 8

(C) 10

(D) 15

(E) 24

Answer B already represents a near doubling of the “just over 8 minutes” time. A multiplication by 11/7 is significantly less than a doubling. In order for the trip to take twice as long, we’d have to be driving only half as fast! But 35 is substantially more than half of 55. The correct answer can only be A: it takes about 5 minutes longer to cover 8 miles at 35 mph than at 55 mph.

Here’s another official GMAT problem

A car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 75 kilometers per hour. At what speed, in kilometers per hour, is the car traveling?

(A) 71.5

(B) 72

(C) 72.5

(D) 73

(E) 73.5

This time, instead of being given a change of speed, we are asked to find the change in speed given a change in time. We can use the inverse relationship of speed and time to figure this out. Whatever the factor of change for the time, the factor of change for the speed will be the reciprocal. So let’s find out what factor of change in time this 2 seconds represents:

75 kilometers per hour means 75 / 60 kilometers per minute. This reduces to 5/4. 5/4 kilometers per minute means 4/5 minutes per kilometer. We can multiply by 60 to obtain seconds per kilometer. 4/5 * 60 = 48, so at 75 kph, it takes 48 seconds to cover 1 kilometer.

Now let’s consider the 2-second increase to this “per kilometer” time of 48 seconds. 2 / 48 = 1 / 24, so we are looking at a 1 / 24th increase in time. Another way to say this is a multiplication by a factor of 25 / 24. So since time and speed are inversely related, the speed responsible for this 2-second increase should be 24 / 25ths (the reciprocal of 25 / 24) of the given speed of 75 mph. (24 / 25) * 75 = 24 * 3 = 72. The correct answer is B.

Let’s try a data sufficiency change of speed problem

A motorboat, which is set to travel at k kilometers per hour in still water, travels directly up and down the center of a straight river so that the change in the boat’s speed relative to the shore depends only on the speed and direction of the current. What is the value of k?

  1. It takes the same amount of time for the boat to travel 4 kilometers directly downstream as it takes for it to travel 3 kilometers upstream.
  2. The current flows directly downstream at a constant rate of 2.5 kilometers per hour.

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are not sufficient.

This problem can be solved simply by applying two skills: algebraic modeling and speed/distance/time fundamentals. After reading the question stem, we can model the speed of the boat upstream and downstream, where k is the speed of the boat in still water and c is the speed of the current:

  • Upstream speed = (k c)
  • Downstream speed = (k + c)

Statement 1 provides the distances that can be traveled downstream and upstream in a given amount of time: 4 kilometers and 3 kilometers, respectively. Since distance and speed are directly related, the ratio of downstream and upstream speeds is equal to the ratio of distances covered traveling downstream or upstream in a given time. So we can set up the following equation using the speeds we modeled:

(k + c) / (kc) = 4 / 3

The upstream speed is 3/4 of the downstream speed, and the downstream speed is 4/3 of the upstream speed. This is useful but insufficient on its own.

Statement 2 supplies the speed of the current as 2.5 kilometers per hour, so the value of the variable c is 2.5. On its own, this can’t be used to determine the value of k. But if we plug this value into the equation derived from statement 1, k will be the only variable remaining and we can simply solve for it. So both statements together are sufficient, and the correct answer is C. 

Here’s a final change of speed problem

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How more miles would he have covered than he actually did if he had driven 2 hours longer at an average rate of 10 miles per hour faster on that day?

(A) 100

(B) 120

(C) 140

(D) 150

(E) 160

This is a complex problem because it involves changes of all three variables: speed, distance, and time! Trying to solve such a problem in a purely conceptual way might get us in a tangle, and algebraic modeling will help.

We are told that a speed increase of 5 miles per hour and a time increase of 1 hour together resulted in a distance increase of 70 miles per hour. Initially, we might model the following equations:

  • D = s * t
  • D + 70 = (s + 5)(t + 1)
  • D + x = (s + 10)(t + 2)

x represents the answer to the question, the distance increase corresponding to a speed increase of 10 kilometers per hour and a time increase of 2 hours. Let’s work with the second equation, with the supplied distance increase of 70 kilometers, and try “foiling” the right side:

  • D + 70 = (s + 5)(t + 1)
  • D + 70 = st + s + 5t + 5

Since distance = speed * time, we can simultaneously subtract D and the product st from the equation:

  • D + 70 = st + s + 5t + 5
  • 70 = s + 5t + 5
  • 65 = s + 5t

We still don’t know the values of s and t, but applying the same expansion to the third equation we modeled shows us how the value of (s + 5t) can be useful:

  • D + x = (s + 10)(t + 2)
  • D + x = st + 2s + 10t + 20

As before, let’s subtract D and the product st:

  • D + x = st + 2s + 10t + 20
  • x = 2s + 10t + 20
  • Now we can factor out a 2 from (2s + 10t):
  • x = 2s + 10t + 20
  • x = 2(s + 5t) + 20

This was the key step, because we know from our other equation that (s + 5t) = 65:

  • x = 2(s + 5t) + 20
  • x = 2(65) + 20
  • x = 130 + 20 = 150

And the correct answer is C.

This concludes our study of change of speed problems. As always, lean on the fundamental relationships between speed, distance, and time. In same cases, this will solve problems on its own. In others, it will be a key ingredient to algebraic solving.

In the next article – the last one in this series – we’ll learn about rate problems with algebraic answers.

Are there any questions that you still have about change of speed problems on the GMAT? Register now for a free 30-minute free consultation with one of our top tutors.

Contributor: Elijah Mize (Apex GMAT Instructor)