During a trip on expressway,don drove a total of x miles.His average speed on a 5 mile section is 30miles/hr and his average speed of remainder of trip was 60 miles/hr.His travel time was what % greater than it would have been if he had traveled at constant rate of 60 miles /hr.

OA is 500/x%.

Kindly help??

consider that travel time for 5 miles @30mph = t1
travel time for (x-5) miles @ 60mph= t2
and travel time if driven all along by 60 mph is = t3.

Don would take more time by driving at variable speeds of 30 mph and 60mph than he'd take to drive at constant speed of 60mph.

So, reqd % increase = [(t1+t2) - t3]/ t3 * 100

Now, t1 = 5/30 = 1/6
t2 = (x-5)/60
t3 = x/60

Solving, you'll get the % increase to be 500/x %.

During a trip on expressway,don drove a total of x miles.His average speed on a 5 mile section is 30miles/hr and his average speed of remainder of trip was 60 miles/hr.His travel time was what % greater than it would have been if he had traveled at constant rate of 60 miles /hr.

Solution:
x can be broken as: 5 miles and x-5 miles
dert (distance=rate*time)
hence for 5 mile case, t=d/r=5/30
for next journey: t=x-5/60
so total time is: 5/30+(x-5)/60=(x+5)/60

If the journey were only due to 60mph, then t=x/60

now, confusing word problem: His travel time was what % greater than.........60 miles /hr. So this happens to be in the following form..

org travel time-travel time with 60mph *100%
------------------------------------------------------
travel time with 60mph

x/60+5/60-x/60
--------------------- *100%
x/60

5/60*60/x*100%=500%/x.....OA

please follow the forum rules. ALWAYS POST PROBLEMS WITH ALL ANSWER CHOICES.

on this problem especially, the easiest way to solve the problem is by the VIC (plug-in-numbers) method. if you don't have the answer choices, you can't use that method.

Hey Ron, which numbers would you pick for this?

I'd pick easy numbers that give an easy, integer result to any calculations I do.

So you don't have to scroll, here's the question again:
During a trip on expressway,don drove a total of x miles.His average speed on a 5 mile section is 30miles/hr and his average speed of remainder of trip was 60 miles/hr.His travel time was what % greater than it would have been if he had traveled at constant rate of 60 miles /hr?

He went 30 mph for 5 miles ==> that part of the trip took 5/30 hour = 1/6 hour = 10 minutes.

He went 60 mph for the remainder of the trip (x-5 miles), and we also need to think about how long he would have traveled at 60 mph for the whole trip (x miles.) I picked x so that x/60 and (x-5)/60 are clean fractions--that is, travel times are easy, neat fractions of an hour, easily converted into minutes.

Thus, I picked x = 20.

So Don went 60 miles for the remaining 15 miles of the trip==> 1/4 hour = 15 min.
He total actual trip was thus 10 minutes + 15 minutes = 25 minutes.

If he had gone 60mph for the whole 20 miles, he would have traveled 1/3 hour = 20 minutes.

His trip was thus 25% longer (25 min. instead of 20 min.) than it would have been at a constant 60 mph.

We can plug this in to choice (A) to determine that (500/x)% = 25% when x = 20.
BUT AGAIN, THIS METHOD CAN ONLY BE USED IF YOU HAVE ALL FIVE ANSWER CHOICES. PLEASE FOLLOW OUR POSTING GUIDELINES, INCLUDING THE COMPLETE PROBLEM EVERY TIME.

_________________
Emily Sledge
Instructor
ManhattanGMAT

For reference, this was on the GMATPrep #2 CAT.

Answer choices:
A) 8.5%
B) 50%
C) x/12%
D) 60/x%
E) 500/x%

I first plugged in 10 miles, which resulted in a 15 minute journey, and a 50% longer travel time. This fit both B and E (annoying), and I knew B was unlikely, as the longer Don traveled the lower the percentage should be. So I then plugged in 20 miles total to figure out E, but this pushed the problem past 2 minutes total.

cool..

_________________
Tim Sanders
Manhattan GMAT Instructor