BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Math Problem - Help needed to solve

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Tue Mar 16, 2010 12:53 pm

Math Problem - Help needed to solve

by priyasaibaba » Wed Apr 21, 2010 7:08 am
Can anyone help me solve this problem?

As the figure above shows, two circles lie on a line and tangent to each other. If the radius of two circles are 2r and r, respectively, in terms of r, AB=?
(A) 5r/2
(B) 8r/3
(C) 2*2^1/2*r
(D) 3r
(E) 2*3^1/2*r

Pls find the figure in the attached doc.

Thanks for your help.
Attachments
Doc1.docx
(14.16 KiB) Downloaded 153 times
Last edited by priyasaibaba on Wed Apr 21, 2010 7:44 am, edited 3 times in total.
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Mon Mar 15, 2010 6:15 am
Thanked: 32 times
Followed by:1 members

by eaakbari » Wed Apr 21, 2010 7:33 am
Do attach the figure, you will not be able to paste it. I cannot understand the figure solely through your explanation
Whether you think you can or can't, you're right.
- Henry Ford
Top
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Tue Mar 16, 2010 12:53 pm

by priyasaibaba » Wed Apr 21, 2010 7:46 am
Hi eaakbari, I have attached the figure. Pls let me know, if u find the answer.

Thanks,
Priya
Top
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Tue Mar 16, 2010 12:53 pm

by priyasaibaba » Wed Apr 21, 2010 7:53 am
D is not the answer. C is the answer. Can someone explain the steps to get the answer?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Mon Mar 15, 2010 6:15 am
Thanked: 32 times
Followed by:1 members

by eaakbari » Wed Apr 21, 2010 7:55 am
Answer is C. wait let me explain
Whether you think you can or can't, you're right.
- Henry Ford
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Mon Mar 15, 2010 6:15 am
Thanked: 32 times
Followed by:1 members

by eaakbari » Wed Apr 21, 2010 8:10 am
Image

CD is a line joining the 2 centers
ED is a line parallel and equal to AB which meets line AC at point E

Now CD = 2r + r = 3r
CE = CA-EA = 2r - r = r

Hence Using pythogoras

ED^2 = 9r^2 + r^2
ED = AB = ROOT(8r^2)
= 2root2 * r

HTH
Whether you think you can or can't, you're right.
- Henry Ford
Top
Legendary Member
Posts: 809
Joined: Wed Mar 24, 2010 10:10 pm
Thanked: 50 times
Followed by:4 members

by akhpad » Wed Apr 21, 2010 8:22 am
Correct, C is the answer
Top
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Tue Mar 16, 2010 12:53 pm

by priyasaibaba » Wed Apr 21, 2010 8:22 am
Wow, thanks very much. Appreciate your help.
Top
Master | Next Rank: 500 Posts
Posts: 266
Joined: Mon Oct 19, 2009 9:46 pm
Thanked: 8 times
GMAT Score:690

by pkw209 » Wed Apr 21, 2010 4:56 pm
good problem. Haven't come across one like that yet.
Top
Master | Next Rank: 500 Posts
Posts: 266
Joined: Mon Oct 19, 2009 9:46 pm
Thanked: 8 times
GMAT Score:690

by pkw209 » Wed Apr 21, 2010 4:56 pm
...until now
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Wed Apr 21, 2010 10:08 pm
eaakbari wrote:Image

CD is a line joining the 2 centers
ED is a line parallel and equal to AB which meets line AC at point E

Now CD = 2r + r = 3r
CE = CA-EA = 2r - r = r

Hence Using pythogoras

ED^2 = 9r^2 + r^2
ED = AB = ROOT(8r^2)
= 2root2 * r

HTH
Great job!

You ended up with the correct answer, just want to correct one small typo so no one gets confused along the way:
ED^2 = 9r^2 + r^2
9(r^2) is actually the hypotenuse, not ED^2, so that should be:

ED^2 = 9(r^2) - r^2

You did the math as though it were a subtraction sign, so all is good after that point!
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Post new topic Post Reply
• Page 1 of 1