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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point 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.

Speed problem - Is this solution correct ?

Expert replies
by pbanavara » Sun Nov 09, 2008 9:31 pm
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours






OA : B
Join the discussion
Source: — Problem Solving |

by raunekk » Sun Nov 09, 2008 10:20 pm
1st cyclist 2nd cyclist

R 6m/h 10m/h


T t t-3

D d =6t d =10(t-3)


Both distances will be sam when they meet...

thus 6t=10(t-3)

t=7.5

thus time taken by 2nd cyclist will be =t-3 = 7.5 - 3 = 4.5

hence B
Join the discussion

by cramya » Mon Nov 10, 2008 6:04 am
Please see attachment.

Hope this helps. Good luck.
Attachments
cyclist.xls
(14 KiB) Downloaded 101 times
Join the discussion

by jnellaz » Mon Nov 10, 2008 7:30 am
Gents, Thanks for the solutions.
Join the discussion