distance

grandh01 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

Let the actual distance traveled by the motorist be d
and the actual speed be s
and the actual time be t
Then d = s * t

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, implies speed = s + 5, time = t + 1, then distance traveled = d + 70
So, d + 70 = (s + 5)(t + 1)
d + 70 = st + 5t + s + 5
d + 65 = d + 5t + s (since d = st)
5t + s = 65 ... Equation (1)

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? implies speed = s + 10, time = t + 2
Actual distance covered = d.
So let the additional distance covered be x
Hence, total distance covered = d + x

Then d + x = (s + 10)(t + 2)
d + x = st + 10t + 2s + 20
x = 10t + 2s + 20
x = 2(5t + s) + 20 (From equation 1, 5t + s = 65)
x = (2 * 65) + 20 = 150

The correct answer is D.