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

In the figure above, A is the center of the circle, DF is 5,

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members
by BTGmoderatorDC » Wed Oct 03, 2018 10:09 pm
Timer
00:00
Your answer
Difficulty–

right wrong

Pick an answer to see how members did

Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

A. 5
B. 7
C. 8
D. 10
E. 15

OA B

Source: Veritas Prep
Top
Join the discussion
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16206
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 04, 2018 8:36 am
BTGmoderatorDC wrote:Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

A. 5
B. 7
C. 8
D. 10
E. 15
Let x = the length of AF
Image
This means AD = x + 5 = radius of the circle.

This is convenient, because AE is also a radius of the circle.
So, AE must have length x + 5
Image

At this point, we can focus on the RIGHT TRIANGLE below:
Image
When we apply the Pythagorean Theorem we get: (x + 5 )² + x² = 25²
Expand to get: x² + 10x + 25 + x² = 625
Simplify: 2x² + 10x + 25 = 625
Set equal to zero: 2x² + 10x - 600 = 0
Divide both sides by 2 to get: x² + 5x - 300 = 0
Factor: (x + 20)(x - 15) = 0
So, EITHER x = -20 or x - 15
Since x cannot be negative, we know that x = 15

So, let's add this to our diagram.
Image
Also, notice that I added some symbols to represent the 3 angles in ∆EAF

At this point we might recognize that ∆EAF and that ∆ECB are similar triangles
Here's why:
Since EB is the diameter of the circle, we know that ∠C is 90°
Also, both triangles have ∠B
If ∆EAF and ∆ECB share two angles, then the 3rd angles must also be equal.
So, the two triangles must be similar.
Image

Let y = the length of CF
So, side EC has length 25 + y

If two triangles are similar, the ratio of their corresponding sides must be equal.
We can write: 25/40 = 20/(25 + y)
Cross multiply to get: 25(25 + y) = (40)(20)
Expand: 625 + 25y = 800
Then: 25y = 175
So y = 7

In other words, CF = 7

Answer: B
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16206
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 04, 2018 8:39 am
BTGmoderatorDC wrote:Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

A. 5
B. 7
C. 8
D. 10
E. 15
As you can see from my solution above, this question is a time-killer (even if you answer it correctly)
So, if you're running short on time, you can reduce the correct answers to two options, and make your best guess
How so?
IMPORTANT: the diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise.
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16206
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 04, 2018 8:45 am
BTGmoderatorDC wrote:Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

A. 5
B. 7
C. 8
D. 10
E. 15
As you can see from my solution above, this question is a time-killer (even if you answer it correctly!!)
So, if you're running short on time, you can reduce the correct answers to two options, and make your best guess
How so?
IMPORTANT: the diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise.
Image

So, if FD has length 5, how long do you think CF is?
To me, CF looks a little longer than FD (which has length 5). So, I'd say the length of CF is either 7 or 8.
So, the correct answer is either B or C.
Make your best guess and move on!

DOWNSIDE: You just guessed
UPSIDE: You have a 50% chance of guessing correctly, AND you just saved a ton of time.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

In the figure above, A is the center of the circle, DF is 5,

by fskilnik@GMATH » Thu Oct 04, 2018 10:39 am
BTGmoderatorDC wrote:Image

In the figure above, A is the center of the circle, DF is 5, and EF is 25. What is the length of CF?

A. 5
B. 7
C. 8
D. 10
E. 15
Brent´s solution is full of VERY important GMAT properties/details.

On the other hand, the "3k, 4k, 5k" shortcut (GMAT´s scope) and a very-easy Geometry property (out-of-GMAT´s scope) do the job in 15 seconds:
$$? = CF = x$$
$$\Delta AFE\,\, = \left( {R - 5,R,25} \right)\,\,\mathop = \limits^{{\text{works}}\,!} \,\,\left( {3 \cdot 5,4 \cdot 5,5 \cdot 5} \right)\,\,\,\,\, \Rightarrow \,\,\,R = 20$$
$$\underbrace {\left[ {R + (R - 5)} \right]}_{GF}\,\,\, \cdot \,\,\,\underbrace {\,5\,}_{FD}\,\, = \underbrace {\,25\,}_{EF} \cdot \underbrace {\,x\,}_{FC}\,\,\,\,\,\mathop \Rightarrow \limits^{R\, = \,20} \,\,\,\,\,\,? = x = 7$$
Image

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Legendary Member
Posts: 2604
Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Thu Oct 04, 2018 11:30 am
Let Af be x
EA = AB = AD = AF +5 =x +5

Using Pythagoras theorem in triangle EAF.

25^2 = (x+5)^2 + x^2 or
2x^2 +10x - 600 = 0
or x^2 +5x -300 = 0
or x =15 or -20

Since x cannot be negative, x has to be 15.

Using similar triangle property in triangle EBC and EAF, we can say that
EC/EB = EA/EF

or EC/40 = 20/25
or EC =32

EC = EF +FC = 32
or 25+FC =32
or FC = 7

Hence, B is the correct answer. Regards!
Top
Join the discussion
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Sat Oct 06, 2018 12:52 am
Let AF =X
Length = EA = AB = AD = AF, 15 = X + 15
Using Pythagoras theorem in triangle EAF
$$25^2=\ \left(x\ +\ 15^2\right)+x^2$$
$$625=\ \left(x\ +\ 5\right)+\left(x+5\right)\ +\ x^2$$
$$625=\ x^2+5x\ +\ 5x\ +\ 25\ +\ x^2$$
$$2x^2\ +\ 10x\ -\ 600\ =\ 0$$
Dividing through with 2
$$\frac{2x^2}{2}+\ \frac{10x}{2}-\ \frac{600}{2}=0$$
$$x^2+5x\ -\ 300\ =\ 0$$
Factorizing the equation.
(x - 15) (x + 20)= 0
x - 15 = 0 or x + 20 = 0
x = 15 or x = -20
The length AF cannot be negative, hence x = 15
For triangle EBC and EAF they are similar or congruent triangle in which their corresponding sides are all in the same proportion.
Hence, $$\frac{EC}{EB}\ =\ \frac{EA}{EF}$$
$$\frac{EC}{40}\ =\frac{20}{25}$$
$$EC=\ \frac{\left(20\ \cdot\ 40\right)}{25}=32$$
EC = EF + FC = 32
25 +FC = 32
FC = 32 - 25
FC = 7

Option B is CORRECT.
Top
Join the discussion