Here are five practice questions. Remember: no calculator.
1) A car, moving at a constant speed, covers 100 m in 5 seconds. What is the car’s speed in km/hr? (1 km = 1000 m)

 (A) 20 km/hr

 (B) 54 km/hr

 (C) 60 km/hr

 (D) 72 km/hr
 (E) 90 km/hr
2) Frank and Georgia started traveling from A to B at the same time. Georgia’s constant speed was 1.5 times Frank’s constant speed. When Georgia arrived at B, she turned right around and returned by the same route. She cross paths with Frank, who was coming toward B, when they were 60 miles away from B. How far away are A and B?

 (A) 72 mi

 (B) 120 mi

 (C) 144 mi

 (D) 240 mi
 (E) 300 mi
3) Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around with pause, and returned to A at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. What is the distance between A and B?

 (A) 275 mi

 (B) 300 mi

 (C) 320 mi

 (D) 350 mi
 (E) 390 mi
4) Car A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?

 (A) 60 mph

 (B) 75 mph

 (C) 80 mph

 (D) 96 mph
 (E) 100 mph
5) Cars P & Q are approaching each other on the same highway. Car P is moving at 49 mph northbound and Car Q is moving at 61 mph southbound. At 2:00 pm, they are approaching each other and 120 mi apart. Eventually they pass each other. At what clock time are they moving away from each other and 45 miles apart?

 (A) 3:06 pm

 (B) 3:30 pm

 (C) 3:54 pm

 (D) 5:21 pm
 (E) 6:15 pm
Solutions will be given at the end of this article.
Motion problems
The test loves motion problems. See this introductory article about rate questions for the basics. In particular, that article talks about the tricky issue of “average speed” or “average velocity,” and has some practice problems involve those topics.
Gaps between vehicles
Some problems, such as the ones above, specify distances between two cars — how far apart they are. Just as D = RT is true for each individual moving item, it is also true for a gap, and often, thinking about the D = RT of the gap is an enormous shortcut in a problem.
Fact: when two cars are moving in opposite directions, either approaching or receding, we add the individual speeds to get the rate for the gap. When two cars are moving the same direction, we subtract the individual speeds to get the rate for the gap.
Keep in mind, at any time, the gap may be expanding or shrinking, and so the rate may be how fast the gap is getting bigger or how fast the gap is getting smaller.
Here are four cases to keep in mind.
Case I: cars in opposite directions, approach
Here, the gap is getting smaller, and the sum of the two individual speeds of the cars is the rate at which the gap is shrinking.
Case II: cars in opposite directions, receding
Here, the gap is getting larger, and the sum of the two individual speeds of the cars is the rate at which the gap is expanding.
Case III: cars in same directions, faster car behind slower car
Here, the gap is getting smaller, and the difference of the two individual speeds of the cars is the rate at which the gap is shrinking.
Case IV: cars in same directions, faster car ahead of slower car
Here, the gap is getting larger, and the difference of the two individual speeds of the cars is the rate at which the gap is expanding.
Using these four cases, if we figure out the rate at which the gap is getting larger or smaller, then it would take a simple sum or difference to related one speed to the other. Even if both individual speeds are unknown, and we need to set up simultaneous equations, and the “gap” equation can be one of those two equations.
Summary
If you had any insights while reading this, you may want to revisit the practice problem above. If you have any insights about these problems you want to share, please let us know in the comments section below.
Practice problem solutions
1) This question is more about pure unit conversion than about motion per se. In an hour, there are 60 minute, each with 60 seconds, so 1 hr = 60*60 = 3600 seconds.
The car’s speed is (100 m)/(5 s) = 20 m/s, so the question is: what is 20 m/s in km per hour.
Well, in one hour (i.e. 3600 second), a car moving 20 m/s would cover
D = RT = (20 m/s)(3600 s) = 72,000 m
That’s 72 km, or 72 km/hr. Answer = (D).
2) Let F = Frank’s speed, and G = Georgia’s speed. We know G = 1.5*F. Let T be the time from start until they pass one another, and let D be the distance between the two locations. At time T, Frank has not gotten to destination B yet, and so has traveled a distance of (D – 60); meanwhile, by time T, Georgia has gone the entire distance of D and 60 miles beyond that, for a total distance of (D + 60).
Thus, our D = RT equations for the two cars are:
Frank: D – 60 = F*T
Georgia: D + 60 = G*T = 1.5F*T
Since the first equation, Frank’s equation, gives us an expression for F*T, and we don’t need the values for either of those variables, substitute that in to the second equation:
D + 60 = 1.5(D – 60)
D + 60 = 1.5D – 90
60 = 0.5D – 90
150 = 0.5D
300 = D
Answer = (E)
3) In the last 15 miles of his approach to B, Kevin was traveling at 60 mph, so he traveled that distance in ¼ hr, or 15 minutes. That means, when he arrived at B, 15 minutes had elapsed, and he took (4 hr) – (15 min) = 3.75 hr to drive the distance D at 80 mph. It will be easier to leave that time in the form (4 hr) – (15 min).
D = RT = (80 mph)[ (4 hr) – (15 min)] = 320 mi – 20 mi = 300 mi
Answer = (B)
4) Let A = speed of Car A, and B = speed of Car B. We know A = 1.25*B. Let D be the distance from where the cars pass each other at 1:30 pm to Town Y. Car A covers the distance D in 1 hr 45 min, or 1.75 hr; in that same time period, B covers (D – 35).
Formula for A: D = A*1.75
Formula for B: D – 35 = B*1.75
Substitute the former into the latter, to eliminate D.
A*1.75 – 35 = B*1.75
7A – 140 = 7B
A – 20 = B
Now, substitute A = 1.25*B (it’s easier to solve for B, then find A).
1.25*B – 20 = B
0.25*B – 20 = 0
B = 80 mph
A = 1.25*80 = 100 mph
Answer = (E)
5) Rather than focus on the motion of the individual cars, it will be much easier in this problem to think about the gap between the cards, as discussed above. When the cars are approaching, the rate at which the gap is shrinking is the sum of the two velocities, 110 mph, and that time, the time until meeting, would be the T = D/R = 120 mi/110 mph. Don’t calculate yet: just hold that thought. Once the cars pass, the gap is expanding, and the rate at which it expands is again the sum of the velocities, 110 mph, so we would again divide that distance by 110 mph.
Once we have those two fractions for the time, we will add them, and of course, they have a common denominator, so we can simply combine them by adding the numerators. In essence, if we think about the gap, there is a total distance of (120 mi + 45 mi) = 165 mi that is traversed at 110 mph. That would take a time of:
T = (165 mi)/(110 mph) = 15/10 = 1.5 hours
Therefore, the cars would be 45 mile apart and receding 1.5 hours later, at 3:30 pm
Answer = (B)
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In question number 4, can you please help me to understand how you got 175 in formula for B
Formula for B: D – 35 = B*175
What I understand is Distance= Rate* Time
D is the distance from Y to the point where the 2 cars met subtract the distance B needs to travel = rate of the car B multiple by time. ?
Thanks
Turkish
Turkish: You’ve got it exactly! We have the variable for the full length of D; that would be the distance car A traveled, because car A got all the way to the destination. This is why Distance = the speed of A times 1.75. The speed of A allowed it to get all the way from the passing point to its destination in 1.75 hours (or one hour 45 minutes).
On the other hand, B did not make it all the way to the destination in 1:45. Instead, car B was still 35 miles short of the destination after 1.75 hours and was thus also 35 miles behind car A after 1.75 hours. In other words, car B travels at a rate of speed that gets it a distance of the complete D minus 35 miles in 1.75 hours.
So if D = A*1.75, then it must also be true that D35 = B*1.75. D is the distance from Y to the point where the two cards met. Subtract the distance D needs to travel to get to Y (35 miles), and you get something equivalent to the rate of speed for car B times 1.75 hours, as you say. 🙂
Hi,
I’m confused about #5 because it says that Car P is moving at 49mph North and Car P is moving at 61mph South. I think you meant to put Car Q for one of them, otherwise it’s impossible because Car P can’t be moving in two different directions at the same time. Which one is moving North and which one is moving South? Or can the problem still be solved even if you don’t know which car is which? Thanks!
Dear Carly,
Good eye, my friend! 🙂 Yes, that was a typo, and I just changed it. Actually, it doesn’t matter at all for the purposes of this problem which one is moving N or S, and which one is moving at which given speed, since the question purely concerns their separation. We could swap around which one was moving N or S, and which one had which given speed, as long as both speeds and both directions were assigned to one car or the other, and the calculation & answer would be exactly the same in every possible combination. That’s deeply important to appreciate about the nature of this problem, and problems of this sort.
The details of who is moving how fast or in what direction are not necessary at all to the mathematics, but the GMAT is always good about giving a “pretty story” to a problem of this type, so I neatened up the “story” aspect by fixing the typo. Again, this “prettiness” is purely cosmetic and wholly irrelevant to the underlying mathematical arrangement.
Does all this make sense?
Mike 🙂
HI all,
For #3. Are you saying that he reached B and turned around or are you saying he left A turned around at some random point and then headed back to B?
Thank you
Dear Michael,
I’m sorry. I see how the original phrasing of that question might have been ambiguous. I just changed the wording, to clarify this question. In your opinion, is everything clear now?
Mike 🙂