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

geomatry

Expert replies
by vipulgoyal » Wed Jan 22, 2014 2:26 am
In the fig above angle BAC and angle ADC is a right angle, If CD = 3 , AD = ??
please refer attachment

D
You do not have the required permissions to view the files attached to this post.
Join the discussion
Source: — Data Sufficiency |

by dimochka » Wed Jan 22, 2014 8:59 am
I'm trying to put together the actual solution to find AD just for reference, but regardless the below question can be solved without doing that (and you should aim to solve similar questions without finding the actual answer, but rather just find out if it can be solved)

We have a triangle with essentially 6 segments of various lengths: AB,AC,BC (main triangle), AD (the line through it which is also the height),BD,CD (that make up BC).

Now realistically, we only really have four variables: AB,AC,AD,BD
This is because CD = 3 and BC = BD + CD = BD + 3, so we can substitute for those two.

In order to solve any equation with four variables we need four equations (and in general - to solve an equation with X variables, we'd need X dissimilar equations). We currently have three, because we have three right triangles:
1. AC^2 + AB^2 = BC^2 --> you can replace that with AC^2 + AB^2 = (BD+3)^2
2. AD^2 + 3^2 = AC^2
3. AD^2 + BD^2 = AB^2 (we know angle ADB is a right angle because angle ADC is a right angle)

Therefore, all we need is one more equation, or one value for either of the variables in our equations. Since either choice gives us a value for one of these variables, the answer is indeed D
Please feel free to private message me with any questions; I don't check the forums regularly, whereas PMs go straight to my email.
Join the discussion

by theCodeToGMAT » Wed Jan 22, 2014 10:17 am
[spoiler]{D}[/spoiler]
You do not have the required permissions to view the files attached to this post.
R A H U L
Join the discussion

by GMATGuruNY » Wed Jan 22, 2014 10:36 am
Image

A height drawn through the right angle of a triangle forms 3 SIMILAR TRIANGLES.
Image
Proof:
Let ∠CAB = x and ∠BAD = y.
Since ∠BAC is a right angle, x+y=90.
The result is that all 3 triangles -- ∆ACD, ∆BAD, and ∆ABC -- have the same combinations of angles:
x-y-90.
Triangles that have the same combination of angles are SIMILAR.

The legs of similar triangles are in the SAME RATIO.
Thus, in all 3 triangles:
(leg opposite x) : (leg opposite y) = (leg opposite x) : (leg opposite y).
In ∆ACD, (leg opposite x) : (leg opposite y) = CD : AD.
In ∆BAD, (leg opposite x) : (leg opposite y) = AD : BD.
Since the two ratios are equal, we get:
CD / AD = AD / BD
(AD)² = (BD)(CD).

Statement 2: BD = 16/3
Since BD = 16/3, CD=3, and (AD)² = (BD)(CD), we get:
(AD)² = (16/3)(3)
(AD)² = 16
AD=4.
SUFFICIENT.

Statement 1: AB = 20/3
Since the 2 statements cannot contradict each other, it must be possible that BD=16/3 and AD=4 in statement 1.
To confirm that these values imply that AD=20/3, apply the Pythagorean theorem to ∆BAD:
BD² + AD² = AB²
(16/3)² + 4² = (20/3)²
256/9 + 144/9 = 400/9
400/9 = 400/9.
This works.

Moreover, NO VALUES OTHER than BD=16/3 and AD=4 will yield that AB=20/3.
As shown above, (AD)² = (BD)(CD).
Thus:
If the value of BD decreases, so will the value of AD, with the result that AB<20/3.
If the value of BD increases, so will the value of AD, with the result that AB>20/3.
Thus, in order that AB=20/3, it must be true that BD=16/3 and AD=4, as indicated in Statement 1.
SUFFICIENT.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion