Lesson Selected Travel and Distance problems from the archive

Selected Travel and Distance problems from the archive

Problem 1

Let u = first train' speed, in mph.
Then the second train' speed is u+12 mph, according to the condition.

The first train traveled 9 hours before the second train overtakes the first one.
During this time the first train covered the distance of 9*u miles.

The second train traveled 7 hours before overtakes the first one.
During this time the second train covered the distance of 7*(u+12) miles.

These distances are the same. It gives you an equation

9u = 7*(u+12),   or 

9u = 7u + 84  --->  2u = 84  --->  u =  = 42 mph.

Then u+12 = 54 mph.

The problem is solved.

Answer. The first train' speed is 42 mph. The second train' speed is 54 mph.

Problem 2

Your guiding equation is 

9*u = 7*(u+14),

where u is Andy' speed on the way to the rock concert, in mph.  
Each side of this equation is the distance between home and concert.

Simplify and solve it:

9u = 7u + 98,

2u = 98,

u = .

The solution is u = 49 mph. It is Andy' speed on the way to concert.

So, the Andy' speed on the way home is 49+14 = 63 mph.

Problem 3

Let D be the distance under the question.
Then your equation is

 = .

The first  addend in the left side is the time of driving at 32 mph.
The second addend in the left side is the time of driving at 48 mph.
The right side is the total traveling time, according to the condition.

To solve the equation, multiply both sides by 96, which is LCM of the denominators. You will get

3D + 2D = 120,

5D = 120,

D =  = 24.

Answer. The distance under the question is 24 miles.

Problem 4

Let D be the distance under the question.

Then your equation is 

 -  = 2.   (1)

The first term in the left side is the time cycling at 11 km/h.
The second term in the left is is the time driving at 33 km/h.
The number "2" on the right is the time difference, according to the condition.

To solve (1), multiply both sides by 33. You will get

3D - D = 2*33,  or

2D = 66,  or

D = 33.

Answer. The distance is 33 km.

Problem 5

Your equation is 

 = .

The first addend in the left side is the time spent to cover 98 miles.
The second addend is the time spent to cover 133 miles.
Here "u" is he ship's rate for the first 98 mi, which is under the question.

To solve it, multiply both sides by u*(u+5). You will get this quadratic equation:

98*(u+5) + 133u = 14u*(u+5).

Simplify and solve it for "u":

 = ,  or

 = ,

 =  =  = .

Only the positive root u = 14 makes sense.

Answer. The speed under the question is 14 miles per hour.

Problem 6

45 minutes =  of an hour.

48 minutes =  of an hour.

Let "u" be the speed of the train in the first scenario. in .
Then the speed in the second scenario is (u-5) .

Then you have this equation

 = ,

which says that the distance is the same.

Simplify it:

 = ,

 = 4,

 = 4,

 = 4,

u = 80.

Thus the train speed is 80 km/h. 
Hence, the distance is  = 60 km.

Answer. The distance is 60 km.