GRE Speed, Distance Problems

Our chapter on GRE Speed, distance problems will make solving a question on GRE Speed, distance an easy task for you. We have just focused on how to use simple formula Speed × Time = Distance. You are supposed to go through our chapter and solve all the problems on GRE Speed, distance problems.

The mathematical model that describes motion has three variables, speed, time & distance. The interrelationship between these three is also the most important formula for this chapter, namely:

Speed × Time = Distance

Problem 1) A car moves for 2 hours at a speed of 25 km/h & another car moves for 3 hours at the same speed. Find the ratio of distance covered by the two cars?

Sol.  Distance covered by first car = 2 × 25 = 50 km

Distance covered by second car = 3 × 25 = 75 km

Ratio of distance covered by the two cars =

\(=\quad (\frac { 50 }{ 75 } )\quad =\quad { (\frac { 2 }{ 3 } ) }^{ Ans }\)

Conversion between km/h to m/sec

1km/h = 1000m/h = \(\quad (\frac { 1000 }{ 3600 } )\) m/sec = \(\quad (\frac { 1000 }{ 3600 } )\) m/sec

Hence, to convert y km/h into m/sec multiply by \(\quad (\frac { 5 }{ 18 } )\)

Thus, y km/h = \(\quad (\frac { 5y }{ 18 } )\) m/sec

And vice-versa: y m/sec = \(\quad (\frac { 18y }{ 5 } )\) km/h.

To convert from m/sec to km/h, multiply by \(\quad (\frac { 18 }{ 5 } )\).

Relative movement, therefore, can be viewed as the movement of one body relative to another moving body.

  • When two bodies are moving in opposite direction at speeds
\({ S }_{ 1 }\quad\) and \({ S }_{ 2 }\quad\) respectively.

The relative speed is defined as \({ S }_{ 1 }\quad\) + \({ S }_{ 2 }\quad\).

  • When two bodies are moving in the same direction.

Relative speed is \(|{ S }_{ 1 }-{ S }_{ 2 }|\).

Suppose a car goes from A to B at an average speed of \({ S }_{ 1 }\)  & then comes back from B to A at an average speed of \({ S }_{ 2 }\). If you had to find out the average speed of the whole journey, what would you do?

Let distance from A to B is d.

Time taken by car to travel from A to B = \(\frac { d }{ { S }_{ 1 } }\)

Time taken by car to travel from B to A = \(\frac { d }{ { S }_{ 2 } }\)

Average speed of whole journey =

\(=\frac { 2d }{ \frac { d }{ { S }_{ 1 } } +\frac { d }{ { S }_{ 2 } } } \\ =\quad \frac { Total\quad distance }{ Total\quad time } \\ =\quad \frac { 2{ S }_{ 1 }{ S }_{ 2 } }{ { S }_{ 1 }+{ S }_{ 2 } }\)

Average speed =

\(\frac { 2{ S }_{ 1 }{ S }_{ 2 } }{ { S }_{ 1 }+{ S }_{ 2 } }\)

The following thing needs to be kept in mind before solving questions on trains:

  • When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. All the rules for relative speed will apply for calculating the relative speed.
  • The distance to be covered when crossing an object whenever a train crosses an object will be equal to:

Length of train + Length of object.

 

Problem 2) Davis drone from Amityville to Betel town at 50 miles per hour & returned by the same rate at 60 miles per hour.

Quantity A:         Davis’s average speed for the round trip, in miles/hour.

Quantity B:         55

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  Let’s use formula for the average speed

Average speed = \(\frac { 2{ S }_{ 1 }{ S }_{ 2 } }{ { S }_{ 1 }+{ S }_{ 2 } } =\quad \frac { 2(50)(60) }{ 50+60 } \\ =\quad 54.54\quad miles/hour\)

Quantity A = 54.54miles/hour

So, Quantity B is greater.

 

Problem 3) Brenda walked a 12-mile scenic loop in 3 hours. If she then reduced her walking speed by half, how many hours would it take Brenda to walk the same scenic loop two more times?

Sol.  Brenda walked a 12-mile scenic loop in 3 hours.

Speed =

\(\frac { Distance }{ Speed } =\frac { 12 }{ 3 } =\quad 4\quad miles/hour\)

She reduced her peed to 2miles/hour.

Time taken by Brenda to walk the same scenic loop two more times =

\(\frac { Total\quad distance }{ Speed } \\ =\quad \frac { 24 }{ 2 } \quad =\quad 12\quad hour\)

 

Problem 4) if a turtle traveled \(\frac { 1 }{ 30 }\) of a mile in 5 minutes, what was its speed in miles per hour?

Sol. Turtle traveled \(\frac { 1 }{ 30 }\)  of a mile in 5 minutes.

Distance traveled by turtle in 1 minute = \((\frac { 1 }{ 30\times 5 } )\) miles

Speed of turtle = \((\frac { 1 }{ 150 } )\) miles/minute

1 minute = \((\frac { 1 }{ 60 } )\) hour

Therefore, speed of turtle (in miles/hour)

\(=\quad (\frac { 1 }{ 150 } \times 60)\\ =\quad (\frac { 2 }{ 5 } )\\ =\quad 0.4\quad miles/hour\)

 

Problem 5) Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles/hour & Rob ran at a constant rate of 6 miles/hour. If they began running at the same point on the loop. How many hours later did Sue complete exactly 1 more lap than Rob?

Sol.  Sue ran at a constant rate of 2 miles/hour

Rob ran at a constant rate of 6miles/hour

Relative speed = 2 miles/hour

Time taken by Sue to complete exactly 1 more loop than Rob

\(=\quad (\frac { Relative\quad distance }{ Relative\quad speed } )\\ =\quad (\frac { 10\quad mile }{ 2\quad miles/hour } )\\ =\quad 5\quad hour\)

 

Problem 6) Svetlana ran the first 5 kilometers of a 10 kilometer race at a constant rate of 12 kilometer per hour, if she completed the entire 10 kilometer race in 55 minutes, at which constant rate did she run the last 5 kilometers of the race, in Kilometers per hour?

Sol.  Svetlana ran the first 5 kilometers of a 10-km race at a constant rate of 12 kilometers per hour.

Time taken by Svetlana to complete first 5 km

\(=\quad \frac { Distance }{ Speed } \\ =\quad \frac { 5 }{ 12 } hour\\ =\quad \frac { 5 }{ 120 } hour\)

Time taken by Svetlana to complete entire race \(=\quad \frac { 55 }{ 60 } hour\)

Time taken by Svetlana to complete last 5 km

\(=\quad (\frac { 55 }{ 60 } -\frac { 25 }{ 60 } )\quad hour\\ =\quad (\frac { 30 }{ 60 } )\quad hour\)

Speed at which she ran last 5 km

\(=\quad (\frac { Distance }{ Time } )\quad =\quad (\frac { 5 }{ \frac { 30 }{ 60 } } )\\ =\quad 10\quad km/hour\)

 

Problem 7) Alina took a non-stop car trip that encompassed three different sections of roadways. Each of three sections covered the same distance. Alina averaged 45 miles/hour over the first section, 60 miles/hour over the second section, & 54 miles/hour for the entire trip. What was her average speed for the third section to the nearest miles/hour?

Sol.  Let distance covered by one section be ‘d’ miles.

Let average speed for the third section be V miles/hour.

Total distance = 3d

Average speed in first section = 45 miles/hour

Time taken to complete first section

\(=\quad (\frac { d }{ 45 } )\quad hour\)

Time taken to complete second section

\(=\quad (\frac { d }{ 60 } )\quad hour\)

Time taken to complete third section

\(=\quad (\frac { d }{ V } )\quad hour\)

Average speed for entire trip =

\(=\quad \frac { Total\quad distance }{ Total\quad time } \\ 54\quad =\quad (\frac { 3d }{ \frac { d }{ 45 } +\frac { d }{ 60 } +\frac { d }{ V } } )\quad =\quad (\frac { 3 }{ \frac { 1 }{ 45 } +\frac { 1 }{ 60 } +\frac { 1 }{ V } } )\\ \frac { 1 }{ 45 } +\frac { 1 }{ 60 } +\frac { 1 }{ V } \quad =\quad \frac { 3 }{ 54 } \quad =\quad \frac { 1 }{ 18 } \\ \frac { 1 }{ V } \quad =\quad \frac { 1 }{ 18 } -\frac { 1 }{ 60 } -\frac { 1 }{ 45 } =\quad \frac { 1 }{ 3 } (\frac { 1 }{ 6 } -\frac { 1 }{ 20 } -\frac { 1 }{ 15 } )\quad =\quad \frac { 1 }{ 3 } (\frac { 10-3-4 }{ 60 } )\quad =\quad \frac { 1 }{ 3 } (\frac { 3 }{ 60 } )\\ V\quad =\quad { (60\quad miles/hour) }^{ Ans }\)

 

DistanceSpeedTime
Section 1d45 m/hourd/45
Section 2d60 m/hourd/60
Section 3dV = ?d/V
Total3d54 m/hour3d/54
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