Asif123 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
In this question, we're comparing a HYPOTHETICAL trip to an ACTUAL trip.
We're told that the hypothetical trip would have been 70 miles longer than the actual trip.
So, we can start with the following word equation:
(distance of hypothetical trip) - (distance of actual trip) = 70Let r = the speed during the ACTUAL trip
Let t = the total time of the ACTUAL tripSo, r + 5 = the speed during the HYPOTHETICAL trip
And t + 1 = the total time of the HYPOTHETICAL tripSince distance = (rate)(time), we can substitute our values into the original word equation:
((r + 5)(t + 1)) - (rt) = 70Expand to get:
rt + r + 5t + 5 - rt = 70Simplify:
r + 5t + 5 = 70Subtract 5 from both sides:
r + 5t = 65 [we'll use this information shortly]The question: 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?So, we want to determine the following:
(distance travelled during this new hypothetical trip) - (distance travelled during the actual trip)In this case, r + 10 = the rate, and t + 2 = the travel time
Plug these values into the above expression to get:
(r + 10)(t + 2) - (r)(t)Expand:
(rt + 2r + 10t + 20 - rtSimplify:
2r + 10t + 20Rewrite as follows:
2(r + 5t) + 20Since we already learned that
r + 5t = 65, we can plug this into the above expression to get:
2(65) + 20, which simplifies to
150Answer: D
05 Feb 2020, 07:09 
75% (02:44) correct
