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.

500 ds

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Community Manager
Posts: 789
Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member
Location: Silicon valley, California
Thanked: 30 times

by jayhawk2001 » Wed Apr 04, 2007 9:12 pm
We have to use properties of similar triangles here. Draw lines
from B and C to the ground. Lets call these points D and E respectively
on the ground. BD and CE are parallel

We are given that AB = BC.

(2) says that BD = 5. We can use properties of similar triangles now.
BD / CE = AB / AC. We can hence compute CE

So B is the correct answer.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 269
Joined: Sun Apr 01, 2007 5:41 am

by dunkin77 » Thu Apr 05, 2007 7:50 am
Thanks for explanation Jay.

I am still a bit confused cause we only know AB=AC and BD=5,
so, BD / CE = AB / AC (let's say AB=AC=3)

5 / CE = 3 / 3

5/CE=1

CE=5..........?? :cry:

I would appreciate it if you could explain a bit more... thanks again for your help!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member
Location: India
Thanked: 51 times

by gabriel » Thu Apr 05, 2007 10:01 am
dunkin77 wrote:Thanks for explanation Jay.

I am still a bit confused cause we only know AB=AC and BD=5,
so, BD / CE = AB / AC (let's say AB=AC=3)

5 / CE = 3 / 3

5/CE=1

CE=5..........?? :cry:

I would appreciate it if you could explain a bit more... thanks again for your help!!

dunkin u have got evrything right except for the part that AB= AC ... AB is actually = BC ... therefore AC=2AB ... so we have 5/CE = AB/AC .... 5/CE = 1/2 .. therefore CE = 10 .. hence the answer is B
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Tue Mar 27, 2007 2:53 pm

by vk.neni » Thu Apr 05, 2007 10:52 am
Hi,
Couldn't we solve this using information in (i) x=30? With this info the small triangle (ABD) becomes a 30-60-90 and we know the length of one of the sides, so we can figure the other two sides. Once, we've this info, we can use the similar triangle method to solve it. So, the answer would be D.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Sat Mar 10, 2007 9:22 pm

Re: 500 ds

by myprepgmat » Sat Apr 07, 2007 5:06 am
Ans is B

I'll explain...

Let the height of point B is h and the point is H. Let the length of seesaw ie AC is 2d..

Now draw perpendicular lines to ground from point B and C. Let they intersect at ground at D ( from Point B) & E(point C).

Now we will get two similar triangles.. ie ABD and ACE..

we have BD/AB = CE/AC

ie h/d= H/2d given AB=BC we assumed AC=2d

i.e. h=H/2

given h=5ft

therefore H=10ft

case 1 doesnot give any unique value as the length of AC or height of the point B is not given.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member
Location: India
Thanked: 51 times

by gabriel » Sat Apr 07, 2007 9:51 am
vk.neni wrote:Hi,
Couldn't we solve this using information in (i) x=30? With this info the small triangle (ABD) becomes a 30-60-90 and we know the length of one of the sides, so we can figure the other two sides. Once, we've this info, we can use the similar triangle method to solve it. So, the answer would be D.
dude one of the most basic mistake for a DS... carrying information from one statement to the other .. the length of the side is mentioned only in the second statement.. so cant use it for the first ..
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1