Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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 563 reviewsTarget Test Prep

time speed dist

Expert replies
Post new topic Post Reply
Subscribe topic
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
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
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
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