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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:50) correct
69%
(03:02)
wrong
based on 141
sessions
History
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A) 6
B) 66
C) 72
D) 126
E) 262
Is there any formula to find out directly??
07 Aug 2012, 23:03
Kudos
Bookmarks
SOURH7WK
In city A, the streets are aligned in a grid, 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?
A) 6
B) 66
C) 72
D) 126
E) 262
Is there any formula to find out directly??
A) 6
B) 66
C) 72
D) 126
E) 262
Is there any formula to find out directly??
Let me add a little bit of detail to the solution given above.
You need to go from a point that is to the bottom left to a point that is to the top right. So you should take steps towards right and top. Since Bill wants to take shortest possible time, he should not go left or down because that is the opposite direction. His destination lies towards right and up.
Say, he takes one step to go from one intersection to the next one. He can take various routes e.g.
RRRRRUUUU (R represents one step right and U represents one step up)
RRRRUUUUR
etc
The total number of ways is basically obtained by re-arranging 5 Rs and 4 Us. You can do it in 9!/5!*4! = 126 ways (we divide by 5! and 4! because all Rs and all Us are identical)
Now, what happens due to the park? Everything is the same except that one intersection is not available - 4th Rd, 6th Ave. You cannot include this intersection in your journey. So what do you do? You remove all paths that include this intersection.
Now our question is this: In how many ways can you go from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave when you include the corner of 4th Rd, 6th Ave?
From the corner of 2nd Rd and 3rd Ave to the corner of 4th Rd and 6th Ave - You need to take 3 steps right and 2 steps up - RRRUU etc. No of ways = 5!/3!*2! = 10
From the corner of 4th Rd and 6th Ave to the corner of 6th Rd and 8th Ave - You need to take 2 steps right and 2 steps up - RRUU etc. No of ways = 4!/2!*2! = 6
Total number of ways in which the 4th Rd, 6th Ave is included = 10*6 = 60 ways
Number of paths in which the corner of 4th Rd, 6th Ave is not included = 126 - 60 = 66
07 Aug 2012, 12:55
Kudos
Bookmarks
SOURH7WK
In city A, the streets are aligned in a grid, 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?
A) 6
B) 66
C) 72
D) 126
E) 262
Is there any formula to find out directly??
A) 6
B) 66
C) 72
D) 126
E) 262
Is there any formula to find out directly??
Really don't know of any formula that could help...
Pretty cruel, definitely not a real GMAT test question.
The best I could come up with is as follows:
Each optimal path (of minimal length) would consist of 5 walks to the right (R) and 4 walks up (U).
The total number of such paths is given by \(9C4=\frac{9*8*7*6}{2*3*4}=126.\)
We have to eliminate those optimal paths that go through the center of the park (the intersection between 4th Road and 6th Avenue). He can still walk around the park. To reach the center of the park we need 3R and 2U walks - a total of 5C2=5*4/2=10 possibilities. From the center of the park, to reach the final destination we need 2R and 2U walks - another 4C2=4*3/2=6 possibilities. This would give 10*6=60 paths to eliminate, leaving 126-60=66 possibilities.
Answer B (not C).
Are you sure OA is C?
