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
GMATBootcamp Starts Sep 28
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE BOOTCAMP

Live Online Bootcamp Class with Top GMAT Expert Chris Peckover

15 live classes from Sep 28, 2026

Schedule
Mon to Fri · 7:00 to 10:00 PM ET
Included
Live classes + 6 months of TTP OnDemand
  • Boost your GMAT score in less than one month in a live online class
  • 6 months access to TTP OnDemand video courses included
View bootcamp & 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
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
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.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

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

time speed dist

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 56
Joined: Sun Mar 21, 2010 11:06 am
by mitaliisrani » Mon Aug 16, 2010 9:41 pm
X and Y run between point A and point B which are 6 km apart. X starts at 10 a.m. from A, reaches B, and returns to A. Y starts at 10:30 a.m. from B, reaches A, and comes back to B. Their speeds are constant with Y's speed being twice that of X. While returning to their starting points they meet at a point which is exactly midway between A and B. When do they meet for the first time?

a) 10.33 1/2 am
b) 10.37 1/2 am
c) 10.33 am
d)10.33 2/3 am

OA A

Plz give approach
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 161
Joined: Mon Apr 05, 2010 9:06 am
Location: Mumbai
Thanked: 37 times

by 4GMAT_Mumbai » Mon Aug 16, 2010 10:02 pm
Hi,

Let us see the approach ...

What is the distance ran by X before meeting Y at the mid point ? 6 + 3 = 9 km

What is the distance ran by Y before meeting X at the mid point ? 6 + 3 = 9 km

Time taken by X for running 9km = Time taken by Y for running 9km + 0.5 hours

If a km / h is the speed of X, 2a km / h is the speed of Y

So, (9 / a) = (9 / 2a) + 0.5

Solving this, one should be able to get 'a' to be 9 km / h

Coming to the 2nd part of the question,

By 10:30, X would have run 4.5 kms. The first meeting thus happens a little after 10:30 in the 1.5 km stretch closer to B.

Distance = 1.5 kms; Relative speed = 27 km / hr (as they are running in opposite directions) (9 km / h + 18 km / h)

Time = 1.5 / 27 = 3 / 54 hours

Converting into time,

Time = 3 * 60 / 54 = 10 / 3 minutes = 3 minutes and 20 seconds.

Mmm ... Not getting any of the answer choices ... Wondering what I am missing ... Help please !
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 56
Joined: Sun Mar 21, 2010 11:06 am

by mitaliisrani » Mon Aug 16, 2010 10:18 pm
4GMAT_Mumbai wrote:Hi,

Let us see the approach ...

What is the distance ran by X before meeting Y at the mid point ? 6 + 3 = 9 km

What is the distance ran by Y before meeting X at the mid point ? 6 + 3 = 9 km

Time taken by X for running 9km = Time taken by Y for running 9km + 0.5 hours

If a km / h is the speed of X, 2a km / h is the speed of Y

So, (9 / a) = (9 / 2a) + 0.5

Solving this, one should be able to get 'a' to be 9 km / h

Coming to the 2nd part of the question,

By 10:30, X would have run 4.5 kms. The first meeting thus happens a little after 10:30 in the 1.5 km stretch closer to B.

Distance = 1.5 kms; Relative speed = 27 km / hr (as they are running in opposite directions) (9 km / h + 18 km / h)

Time = 1.5 / 27 = 3 / 54 hours

Converting into time,

Time = 3 * 60 / 54 = 10 / 3 minutes = 3 minutes and 20 seconds.

Mmm ... Not getting any of the answer choices ... Wondering what I am missing ... Help please !
Hey dude your approach is absolutely correct..as you mentioned they meet after 10.30 ...and it takes them 3 minutes 20 seconds to meet..thus,,,..the y meet at 10.33 1/3am...which is option a
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 56
Joined: Sun Mar 21, 2010 11:06 am

by mitaliisrani » Mon Aug 16, 2010 10:35 pm
mitaliisrani wrote:
4GMAT_Mumbai wrote:Hi,

Let us see the approach ...

What is the distance ran by X before meeting Y at the mid point ? 6 + 3 = 9 km

What is the distance ran by Y before meeting X at the mid point ? 6 + 3 = 9 km

Time taken by X for running 9km = Time taken by Y for running 9km + 0.5 hours

If a km / h is the speed of X, 2a km / h is the speed of Y

So, (9 / a) = (9 / 2a) + 0.5

Solving this, one should be able to get 'a' to be 9 km / h

Coming to the 2nd part of the question,

By 10:30, X would have run 4.5 kms. The first meeting thus happens a little after 10:30 in the 1.5 km stretch closer to B.

Distance = 1.5 kms; Relative speed = 27 km / hr (as they are running in opposite directions) (9 km / h + 18 km / h)

Time = 1.5 / 27 = 3 / 54 hours

Converting into time,

Time = 3 * 60 / 54 = 10 / 3 minutes = 3 minutes and 20 seconds.

Mmm ... Not getting any of the answer choices ... Wondering what I am missing ... Help please !
Hey dude your approach is absolutely correct..as you mentioned they meet after 10.30 ...and it takes them 3 minutes 20 seconds to meet..thus,,,..the y meet at 10.33 1/3am...which is option a

Sorry an error in the first part 10/3 minutes is 3 1/3 (1/3rd is 20 seconds since 60 seconds is one minute)..you were correct all through out:)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1