Lesson Had a car move faster it would arrive sooner

Had a car move faster it would arrive sooner

Problem 1

Let r" be the actual speed of the car, in mph.

Then the hypothetical speed is (r+13) mph.


Moving at the speed of "r" miles per hour, the car spends  hours.

Traveling with the hypothetical speed, the car would spend  hours.

The condition says

 = 5.


It is your "time" equation.  You must solve it for "r". 
To do it, multiply both sides by the common denominator r*(r+13). You will get

1950*(r+13) - 1950*r = 5r*(r+13),

 = ,

 = 0.

Factor it:

(r+78)*(r-65) = 0.


The only positive root is r = 65.


Answer. The actual speed of the car is 65 mph.

Problem 2

Let d be the distance from his house to work, in miles.

Then the time he spends normally is .
The time he spends when the weather is bad is .

The condition says: normal speed minus slower speed is 6 mph.
It gives an equation

 -  = 6.

Multiply equation by 36 (both sides) to get rid of the denominators. You will get

36* - 36* = 36*6.

Cancel the denominators

9d - 8d = 216

d = 216 miles to his house.

Check.  The actual speeds are   = 54 mph  and   = 48 mph.  The difference is  6 mph.

Problem 3

Let L be the travel distance (now unknown).

Then the passenger train spends  hours to cover this distance, while
the freight train spends  hours.

According to the condition, the difference of these time intervals is 1.5 hour. It gives you an equation

 -  = 1.5.

Multiply both sides by 40*60 to get off denominators. You will get

60*L - 40*L = 40*60*1.5,   or

20*L = 3600.

Hence, L =  = 180 km.

Then the passenger train spends  = 3 hours, while the greight train spends  = 4.5 hours.

The problem is solved.