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

Redeem

geometry path question

Expert replies
by hemant_rajput » Wed Feb 27, 2013 9:43 am
four cities are connected by a road network as shown below in the figures. In how many ways can you start from one city and come back to it without traveling on the same road more than one?
Image
a.14
b.12
c.15
d.10
e.16

PS: It is not an official GMAT question, also I know how to solve it, however, I was curious about the approaches to solve this problem with min. time.
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Wed Feb 27, 2013 10:17 am
hemant_rajput wrote:four cities are connected by a road network as shown below in the figures. In how many ways can you start from one city and come back to it without traveling on the same road more than one?
Image
a.14
b.12
c.15
d.10
e.16

PS: It is not an official GMAT question, also I know how to solve it, however, I was curious about the approaches to solve this problem with min. time.
Start at B.
Number of options for the first city visited = 3. (A, D or C).
Number of options for the second city visited = 2. (Either of the 2 cities not yet visited.)
Number of ways to return to B = 2. (Either directly back to B or to the one city not yet visited and then directly back to B.)
To combine these options, we multiply:
3*2*2 = 12.

The correct answer is B.
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

by sonalibhangay » Sat Mar 16, 2013 10:48 am
I thought of solving this in same way as we do for the Problems with Blocks travelled from A to B ones.

There are 4 paths to be travelled to complete visit to each city and back.
This comes to 4! = 24

However, the paths will overlap as we change the start city, so the total number should be divided 4! / 2! . (Similar 2 paths in each arrangement).

Therefore Answer should be 12.

Correct me if this is wrong.

Thanks, Sonali.
Join the discussion

by sonalibhangay » Sat Mar 16, 2013 10:48 am
I thought of solving this in same way as we do for the Problems with Blocks travelled from A to B ones.

There are 4 paths to be travelled to complete visit to each city and back.
This comes to 4! = 24

However, the paths will overlap as we change the start city, so the total number should be divided 4! / 2! . (Similar 2 paths in each arrangement).

Therefore Answer should be 12.

Correct me if this is wrong.

Thanks, Sonali.
Last edited by sonalibhangay on Sat Mar 16, 2013 10:50 am, edited 1 time in total.
Join the discussion

by sonalibhangay » Sat Mar 16, 2013 10:49 am
I thought of solving this in same way as we do for the Problems with Blocks travelled from A to B ones.

There are 4 paths to be travelled to complete visit to each city and back.
This comes to 4! = 24

However, the paths will overlap as we change the start city, so the total number should be divided 4! / 2! . (Similar 2 paths in each arrangement).

Therefore Answer should be 12.

Correct me if this is wrong.

Thanks, Sonali.
Join the discussion