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# GMAT Practice Problems on Motion

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)

1. (A) 20 km/hr
1. (B) 54 km/hr
1. (C) 60 km/hr
1. (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?

1. (A) 72 mi
1. (B) 120 mi
1. (C) 144 mi
1. (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?

1. (A) 275 mi
1. (B) 300 mi
1. (C) 320 mi
1. (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?

1. (A) 60 mph
1. (B) 75 mph
1. (C) 80 mph
1. (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?

1. (A) 3:06 pm
1. (B) 3:30 pm
1. (C) 3:54 pm
1. (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

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

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

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 