percent problem

[saorabh]

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??

[nimish.tiwari]

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 %.

[nigerian.horse]

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

[RonPurewal]

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.

[devneeetbajaj]

Hey Ron, which numbers would you pick for this?

[esledge]

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.

[NewSc2]

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.

[tim]

cool..