BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Rate problem

Expert replies
by garuhape » Tue Feb 15, 2011 6:33 am
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was the average speed for the entire trip.

Official Answer: 12000/(x+200)

I don't understand the solution.

I did it like that:

x*40+(1-x)*60=average speed of total trip.

Let's say that she traveled 30% of the distance at an average speed of 40 and 70% at an average speed of 60. Thus the average speed of the total trip is:

0.3*40+0.7*60=54

:?:
Join the discussion
Source: — Problem Solving |

by Night reader » Tue Feb 15, 2011 6:55 am
OA may skip important details :) see my solution below

D=distance, Speed=S, T=time; xD/100 at average S(1)=40; (100-x)D/100 at average S(2)=60; find D/(T1+T2)-?
T1=xD/100 : 40 and T2=(100-x)D/100 :60; T1+T2=[3xD+2D(100-x)]/(100*120)= (xD+200D)/(100*120)

D: (xD+200D)/(100*120)= 12000/(x+200)
garuhape wrote:During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was the average speed for the entire trip.

Official Answer: 12000/(x+200)

I don't understand the solution.

I did it like that:

x*40+(1-x)*60=average speed of total trip. <--- it says precisely x% but we don't know the value of x, e.g. x=20% you take 1-20 which is -19, but should be x%=x/100 and (100-x)/100

Let's say that she traveled 30% of the distance at an average speed of 40 and 70% at an average speed of 60. Thus the average speed of the total trip is:

0.3*40+0.7*60=54

:?:
Join the discussion

by GMATGuruNY » Tue Feb 15, 2011 7:14 am
GMATGuruNY wrote:During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?

a. (180-X)/(2)
b. (x+60)/(4)
c. (300-x)/(5)
d. (600)/(115-x)
e. (12,000)/(x+200)
Plug in a value for the distance and a value for x.

Let distance = 100 miles.
Let x = 40.
Distance for 40% of the trip = .4*100 = 40 miles.
Since the rate for this portion is 40mph, time = d/r = 40/40 = 1 hour.
Distance for remainder of the trip = 100-40 = 60 miles.
Since the rate for this portion is 60mph, time = d/r = 60/60 = 1 hour.
Total time = 1+1 = 2 hours.
Average speed for the whole trip = (total distance)/(total time) = 100/2 = 50. This is our target.

Now we plug x = 40 into all the answer choices to see which yields our target of 50.

Only answer choice E works:
(12,000)/(x+200) = 12000/(40+200) = 50.

The correct answer is E.[/quote]
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

by garuhape » Tue Feb 15, 2011 7:21 am
Using what you have written in blue

x/100*40 + (100-x)/100*60 = 40x/100+60(100-x)/100 = 40x/100+(6000-60x)/100 = (6000-20x)/100 = (300-x)/5,

which is one of the answer choices but which apparently is wrong. However, if you try it out the formula works perfectly fine.
Join the discussion

by garuhape » Tue Feb 15, 2011 7:27 am
Thanks guys, now I understood it ;)
Join the discussion