el1981 wrote:
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100
(B) 120
(C) 140
(D) 150
(E) 160
Plug in a value for the RATE and solve for the TIME.
Let the actual rate = 10 mph and the actual time = t hours
Actual Distance = (actual rate)(actual time) = 10t
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually didSince the rate here is 5 mph greater than the actual rate of 10 mph and the time 1 hour longer than the actual time of t hours, we get:
Distance 2 = (rate 2)(time 2) = (10+5)(t+1) = 15(t+1) = 15t+15
Since Distance 2 is 70 miles greater than the Actual Distance, we get:
Distance 2 - Actual Distance = 70
(15t+15) - 10t = 70
5t = 55
t = 11
In this case:
Actual Distance = (actual rate)(actual time) = 10*11 = 110 miles
Distance 2 = (rate 2)(time 2) = 15*12 = 180 miles --> 70 miles longer than the actual 110-mile distance
How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?Since the rate here is 10 mph greater than the actual rate of 10 mph and the time 2 hours longer than the actual time of 11 hours, we get:
Distance 3 = (rate 3)(time 3) = (10+10)(11+2) = 20*13 = 260 miles ----> 150 miles longer than the actual 110-mile distance
Note:
The actual rate, time and distance can be different values.
Let the actual rate = 5 mph and the actual time = t hours
Actual Distance = (actual rate)(actual time) = 5t
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually didSince the rate here is 5 mph greater than the actual rate of 5 mph and the time 1 hour longer than the actual time of t hours, we get:
Distance 2 = (rate 2)(time 2) = (5+5)(t+1) = 10(t+1) = 10t+10
Since Distance 2 is 70 miles greater than the Actual Distance, we get:
Distance 2 - Actual Distance = 70
(10t+10) - 5t = 70
5t = 60
t = 12
In this case:
Actual Distance = (actual rate)(actual time) = 5*12 = 60 miles
Distance 2 = (rate 2)(time 2) = 10*13 = 130 miles --> 70 miles longer than the actual 60-mile distance
How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?Since the rate here is 10 mph greater than the actual rate of 5 mph and the time 2 hours longer than the actual time of 12 hours, we get:
Distance 3 = (rate 3)(time 3) = (5+10)(12+2) = 15*14 = 210 miles --> 150 miles longer than the actual 60-mile distance
Another case:
Let the actual rate = 30 mph and the actual time = t hours
Actual Distance = (actual rate)(actual time) = 30t
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually didSince the rate here is 5 mph greater than the actual rate of 30 mph and the time 1 hour longer than the actual time of t hours, we get:
Distance 2 = (rate 2)(time 2) = (30+5)(t+1) = 35(t+1) = 35t+35
Since Distance 2 is 70 miles greater than the Actual Distance, we get:
Distance 2 - Actual Distance = 70
(35t+35) - 30t = 70
5t = 35
t = 7
In this case:
Actual Distance = (actual rate)(actual time) = 30*7 = 210 miles
Distance 2 = (rate 2)(time 2) = 35*8 = 280 miles --> 70 miles longer than the actual 210-mile distance
How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?Since the rate here is 10 mph greater than the actual rate of 30 mph and the time 2 hours longer than the actual time of 7 hours, we get:
Distance 3 = (rate 3)(time 3) = (30+10)(7+2) = 40*9 = 360 miles --> 150 miles longer than the actual 210-mile distance
As the cases above illustrate:
The actual rate, time and distance cannot be determined.
In the first case, actual rate = 10 mph, actual time = 11 hours, actual distance = 110 miles.
In the second case, actual rate = 5 mph, actual time = 12 hours, actual distance = 60 miles.
In the third case, actual rate = 30 mph, actual time = 7 hours, actual distance = 210 miles.