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.

Distance Problem

Expert replies
by okigbo » Fri Feb 26, 2010 8:08 am
Skier Lindsey Vonn completes a straight 300-meter downhill run in t seconds and at an average speed of (x + 10) meters per second. She then rides a chairlift back up the mountain the same distance at an average speed of (x - 8) meters per second. If the ride up the mountain took 135 seconds longer than her run down the mountain, what was her average speed, in meters per second, during her downhill run?

(A) 10

(B) 15

(C) 20

(D) 25

(E) 30
Join the discussion
Source: — Problem Solving |

by ajith » Fri Feb 26, 2010 8:56 am
okigbo wrote:Skier Lindsey Vonn completes a straight 300-meter downhill run in t seconds and at an average speed of (x + 10) meters per second. She then rides a chairlift back up the mountain the same distance at an average speed of (x - 8) meters per second. If the ride up the mountain took 135 seconds longer than her run down the mountain, what was her average speed, in meters per second, during her downhill run?

(A) 10

(B) 15

(C) 20

(D) 25

(E) 30
Time taken for the ride down = 300/(x+10)
Time taken for the ride up = 300/(x-8)

Now 300/(x+10) +135 = 300/(x-8)
20/(x+10) +9 = 20/(x-8)

x=10 clearly satisfies this

speed for the downhill journey = x+10 = 20 m/sec
Always borrow money from a pessimist, he doesn't expect to be paid back.
Join the discussion

by m&m » Fri Feb 26, 2010 9:30 am
It may take some time to solve the algebra from the post above - though it is absolutely correct and will always lead to the right answer. However, there is only 5 possible answers so try plugging the first:

300/(10+10) = 15

300/(10-8) = 150

150-15 = 135

Luckily this matches on the first try with A

Say we would have started with C (mid-pt of all answers)

300/(20+10) = 10

300/(20-8) = 25

25-10 = 15 so we know that 20 is MUCH too high so must be either A or B

I would then pick A - which is right but even if wrong I would know that B is right from elimination


Hope this helps
Join the discussion

by ajith » Fri Feb 26, 2010 9:35 am
m&m wrote:
Luckily this matches on the first try with A
u matched the try with C

it asks for the speed for downhill journey which is x+10 and not x
Always borrow money from a pessimist, he doesn't expect to be paid back.
Join the discussion

by m&m » Fri Feb 26, 2010 12:14 pm
Ur absolutely right t. Knowing x=10

avg speed is 10+10 = 20 C is correct

Thanks for correcting
Join the discussion