(16) Tom took a 1900-mile trip by bus and plane. The bus averaged 60
miles per hour and the plane averaged 700 miles per hour. If the trip
took 5 hours, how many miles did Tom travel by bus? [150 miles]math
If he traveled x miles by bus, then he went 1900-x by plane.
Since time = distance/speed,
x/60 + (1900-x)/700 = 5
To solve this problem, we can use the formula:
Distance = Speed x Time
Let's denote the distance traveled by bus as 'x' and the distance traveled by plane as 'y'.
From the given information, we know that Tom took a 1900-mile trip and the trip took 5 hours. Now let's set up two equations:
1) x + y = 1900 (eq. 1) - representing the total distance traveled
2) x/60 + y/700 = 5 (eq. 2) - representing the total time taken
From equation 1, we can solve for y by substituting the value of 'x' in terms of 'y' in equation 2:
x = 1900 - y
Substituting this back into equation 2:
(1900 - y)/60 + y/700 = 5
Now, we can simplify this equation to get it in terms of 'y'. Multiplying through by 4200 (the least common multiple of 60 and 700) to clear the denominators:
(4200 * (1900 - y))/60 + (4200 * y)/700 = 5 * 4200
(70 * (1900 - y)) + (60 * y) = 21000
133000 - 70y + 60y = 21000
-10y = -112000
y = 112000 / 10 = 11200
Now, substituting the value of 'y' back into equation 1:
x + 11200 = 1900
x = 1900 - 11200 = -9300
Since distance cannot be negative, it means there is no negative distance. Therefore, there is no valid solution to this problem.