Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 28 Mar 2017, 05:32

# INSEAD R3 Decisions:

Join Chat Room7 for Latest Decision Updates

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In city A, the streets are aligned in a grid (see attachment

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Apr 2010
Posts: 162
Followers: 1

Kudos [?]: 183 [0], given: 31

In city A, the streets are aligned in a grid (see attachment [#permalink]

### Show Tags

21 Mar 2011, 20:13
3
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

20% (00:00) correct 80% (02:09) wrong based on 10 sessions

### HideShow timer Statistics

In city A, the streets are aligned in a grid (see attachment), where the east-west roads are called 1st Rd, 2nd Rd, 3rd Rd, etc, increasing in number as one moves northward. The north-south roads are called 1st Ave, 2nd Ave, 3rd Ave, etc, increasing in number as one moves eastward. There is a park that runs from 5th Ave to 7th Ave and from 3rd Rd to 5th Rd, as pictured. If Bill needs to walk from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave in the shortest possible time without walking through the park, how many different routes could he take?

1. 45
2. 54
3. 66
4. 98
5. 19

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-city-a-the-streets-are-aligned-in-a-grid-where-the-east-136949.html
[Reveal] Spoiler: OA

Attachments

Doc1.docx [19.17 KiB]
Downloaded 162 times

 To download please login or register as a user

_________________

Consider me giving KUDOS, if you find my post helpful.
If at first you don't succeed, you're running about average. ~Anonymous

Manager
Joined: 03 Mar 2011
Posts: 90
Location: United States
Schools: Erasmus (S)
GMAT 1: 730 Q51 V37
GPA: 3.9
Followers: 2

Kudos [?]: 141 [1] , given: 12

Re: Grid problems [#permalink]

### Show Tags

22 Mar 2011, 11:29
1
This post received
KUDOS
Hello)
This is not an easy question and I hope everyone not to have it on the exam)
Actually this problem is connected with wide and very interesting field of mathematics - graph theory(trees, cycles, 7 bridges of Konigsberg etc.). Probably you have heard about it. If no and you are interested - see here.
http://en.wikipedia.org/wiki/Graph_theory
However, I will try to explain it to you without any extra knowledge.

Obviously, the shortest way is to go to the right and up. I think it is easy to understand. So, the total lenght of the trip is 9 steps. Since you go from the LeftDown to RightUp corner, then you always will go through one of the "key" points, as I call them. See, i circle them in red in the picture.

Now let calculate, how many routes are there to the point 1. Obviously, there is 1 such route. I show it in yellow.

Now let calculate how many routes are there to the point 2. There are 4 such routes. You could easily understand these looking at the picture. I show these routes in yellow, green, blue and red.

How many routes are there from the origin to the 3 point? 6! See the picture.

How many routes to the 4 point? There are 4. Everything is on the picture.

And finally there is only one route to the 5 point. I hope you understand this easily.

And now let do the second part of work. Let calculate, how many options you have going from each of these 5 points to the final point - RightUp corner!
For the 1 point there is only 1 such route - going to the right.
For the 2 point there are 5 routes. See the picture, please.

From the 3 and 4 points there are 4 routes, They are quite similar, so I will show the picture only for one case.

Finally, from the point 5 there are 5 routes to the end, it is like the point 2, so I do not add the picture.

Now we are close to the answer!
What we need is to calculate how many routes, going through each of the 5 "key" points, is there. As you guess we should just multiply the number of routes from the first and second parts.
So, there are 1*1=1 routes going through point 1.
4*5=20 routes going through point 2
6*4=24 routes through point 3
4*4=16 routes through point 4
1*5=5 routes through point 5

The final answer is 1+20+24+16+5=66!

Obviously, this is not the best or shortest way to solve this, but I hope it will help to realise better the essence of the
problem. I sorry for the "quality" of pictures, but I have not much time to draw them)
_________________

If my post is useful for you not be ashamed to KUDO me!
Let kudo each other!

Last edited by bagrettin on 22 Mar 2011, 21:13, edited 1 time in total.
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2019
Followers: 162

Kudos [?]: 1769 [0], given: 376

Re: Grid problems [#permalink]

### Show Tags

22 Mar 2011, 13:34
bhandariavi wrote:
In city A, the streets are aligned in a grid (see attachment), where the east-west roads are called 1st Rd, 2nd Rd, 3rd Rd, etc, increasing in number as one moves northward. The north-south roads are called 1st Ave, 2nd Ave, 3rd Ave, etc, increasing in number as one moves eastward. There is a park that runs from 5th Ave to 7th Ave and from 3rd Rd to 5th Rd, as pictured. If Bill needs to walk from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave in the shortest possible time without walking through the park, how many different routes could he take?
1. 45
2. 54
3. 66
4. 98
5. 19

There must be some short cut ways to solve this problem with combination factorial formula.
Can anyone shed light on this?

Good effort bagrettin. thanks and kudos.

After struggling a lot, this is my way of approaching this problem;

I am going to represent Avenues with suffix A and Roads with suffix R
e.g. (3A,2R) is the intersection of 3rd Avenue and 2nd Road.

Total number of shortest ways from (3A,2R) to (8A,6R) will be $$C^{9}_{5} = \frac{9!}{5!4!} = 126$$. Think it like a string "EEEEENNNN" 5 blocks east and 4 blocks north and a total of 9 blocks.

Now we need to subtract the routes that are actually not there because of the park;

We just need to consider 2 intersections for that;
(6A,3R), (5A,4R)

Ways to get to (6A,3R) from (3A,2R) = $$C^{4}_{3}=4$$
"From (6A,3R) to (6A,5R) via (6A,4R)" AND "From (6A,3R) to (7A,4R) via (6A,4R)" are two routes that need to be subtracted.
All we need to do now is to find out the number of ways to reach (8A,6R) from (6A,5R) & (7A,4R)

Ways from (6A,5R) to (8A,6R) = $$C^{3}_{2}=3$$
Ways from (7A,4R) to (8A,6R) = $$C^{3}_{1}=3$$

Number of ways to be subtracted = 4*3+4*3 = 12+12=24

Similarly, we can repeat it for intersection (5A,4R)

Ways to reach (5A,4R) from (3A,2R) = $$C^{4}_{2} = 6$$

"From (5A,4R) to (6A,5R) via (6A,4R)" AND "From (5A,4R) to (7A,4R) via (6A,4R)" are two routes that need to be subtracted.
All we need to do now is to find out the number of ways to reach (8A,6R) from (6A,5R) & (7A,4R)

We already found that;
Ways from (6A,5R) to (8A,6R) = $$C^{3}_{2}=3$$
Ways from (7A,4R) to (8A,6R) = $$C^{3}_{1}=3$$

Number of ways to be subtracted= 6*3+6*3 = 18+18=36

Total number of ways to be subtracted = 24+36=60

Number of possible shortest ways = 126-60 = 66.

Ans: "C"
_________________
Manager
Joined: 04 Apr 2010
Posts: 162
Followers: 1

Kudos [?]: 183 [0], given: 31

Re: Grid problems [#permalink]

### Show Tags

22 Mar 2011, 18:25
Thanks fluke & bagrettin.
_________________

Consider me giving KUDOS, if you find my post helpful.
If at first you don't succeed, you're running about average. ~Anonymous

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7255
Location: Pune, India
Followers: 2205

Kudos [?]: 14359 [2] , given: 222

Re: Grid problems [#permalink]

### Show Tags

22 Mar 2011, 19:05
2
This post received
KUDOS
Expert's post
bhandariavi wrote:
In city A, the streets are aligned in a grid (see attachment), where the east-west roads are called 1st Rd, 2nd Rd, 3rd Rd, etc, increasing in number as one moves northward. The north-south roads are called 1st Ave, 2nd Ave, 3rd Ave, etc, increasing in number as one moves eastward. There is a park that runs from 5th Ave to 7th Ave and from 3rd Rd to 5th Rd, as pictured. If Bill needs to walk from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave in the shortest possible time without walking through the park, how many different routes could he take?
1. 45
2. 54
3. 66
4. 98
5. 19

There must be some short cut ways to solve this problem with combination factorial formula.
Can anyone shed light on this?

You might have seen a simpler problem without the park. You need to take shortest possible time so you cannot retrace any path. You cannot go down or left. You need to go up and right only.
To go from bottom left corner to top right, you will have to go 4 paces Up and 5 paces Right.
You can do this in many ways e.g. UUUURRRRR, UURRRUURR etc
There will be total 9!/(5!*4!) = 126 ways
Attachment:

Ques2.jpg [ 35.49 KiB | Viewed 3558 times ]

From these we need to remove those ways which use the red paths because they pass from the park. Note that you need to reach one of the thick red paths first. Only then can you reach either one of the two thin red paths. So if we just remove the cases using the thick Red paths, we will be done.

In how many ways can you go from bottom left to top right using the thick Up path?
You will need to take 3 Rights and 1 Up (in any order), then the Up thick Red path, then either (Up thin Red path, 2 Rights and 1 Up) or (Right thin Red path, 2 Ups and 1 Right)
You can do this in $$\frac{4!}{(3!)} * 1* 2 * \frac{3!}{2!} = 24$$ ways

In how many ways can you go from bottom left to top right using the thick Right path?
You will need to take 2 Rights and 2 Ups (in any order), then the Right thick Red path, then either (Up thin Red path, 2 Rights and 1 Up) or (Right thin Red path, 2 Ups and 1 Right)
You can do this in $$\frac{4!}{(2!)(2!)} * 1* 2 * \frac{3!}{2!} = 36$$ ways

Now all we need to do is: 126 - 24 - 36 = 66 ways
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Senior Manager
Joined: 05 Jul 2010
Posts: 359
Followers: 15

Kudos [?]: 51 [0], given: 17

Re: Grid problems [#permalink]

### Show Tags

21 Aug 2011, 15:01
Thanks Karishma for this awesome explanation! I was scratching my head on a GMAT problem for the very first time in my GMAT prep. I can use this strategy in many problems Awesome!
Re: Grid problems   [#permalink] 21 Aug 2011, 15:01
Similar topics Replies Last post
Similar
Topics:
20 In city A, the streets are aligned in a grid, where the east 10 07 Aug 2012, 09:20
13 Alicia lives in a town whose streets are on a grid system 11 01 Feb 2012, 11:05
22 If the 4 x 4 grid in the attached picture is filled with the 11 31 Jan 2012, 19:25
1 In city A, the streets are aligned in a grid, where the east 1 19 Jun 2011, 07:56
35 If A is the center of the circle shown above (see attachment 19 07 Sep 2009, 07:19
Display posts from previous: Sort by

# In city A, the streets are aligned in a grid (see attachment

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.