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11 Jul 2024, 23:48
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We are trying to catch the orange team.
What a crazy leading. But, I can't wait for our comeback.
it's coming.
Let's get started with our explanation for this topic on day four:
First step in Two Parts, before reading the text I understand what is the question and what is it asking for:
Rephrase the Question:
Question: This is kind of Logic question. don't get overwhelm by the story.
the question asks about 2 scenarios: Going towards City Centre and Laurel Lane.
We have some constraints.
first step is DO NOT do over calculation. try to find a simple shortcut.
Rephrase - Reading and Understanding the text:
Given:
- 3 routes of 3 Miles
- 2 routes of 4 Miles
- Everyday we have 3 routes chosen
- we are dealing with 3 days
- Everyday she runs 10 Miles
- She can't run the same route in all 3 days. => only 2 days.
- she never avoid (Trick) 2 days in a row. meaning if she start on route A, she will run it tomorrrow too or the day after tomorrow.
Solve:
We know we must have only 1 of 4 miles each day and 2 of 3 miles each day.
if we start with 1 we must run it tomorrow or the day after tomorrow.
let's call the routes A->E
Let's start with the easy scenario of City Centre (CC):
Day1 - D , A , B - 10 Miles
Day2 - D , A , C - 10 Miles
Day3 - E , B , C - 10 Miles
We can see that we have 9 routes, each of which she runs toward CC so we have a total of 9
Part 2 - Laurel Lane (LL)
the only one who don't have LL in its route is A.
so as you can see, we must take A in 2 days of running because of our constrains:
Day1 - D , A , B - 10 Miles -> 2
Day2 - D , A , C - 10 Miles -> 2
Day3 - E , B , C - 10 Miles -> 3
We can see that we have 7 routes, without LL, each of which she runs toward LL so we have a total of 7.
THE END
I hope you liked the explanation, I have tried my best here.
Let me know if you have any questions about this question or my explanation.
pranjalshah
Joined: 11 Dec 2023
Last visit: 10 Nov 2025
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Location: India
Concentration: Operations, General Management
Schools: ISB '27 (A$$$$)
GMAT Focus 1: 735 Q90 V87 DI82 
12 Jul 2024, 00:31
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Answer: City Center: 9, Laurel Lane: 7
Only way to satisfy the given constraints is the following combo:-
124
135
234
Any other combination will violate the given constraints.
In the above combo, she runs to city center 9 times and Laurel Lane 7 times.
Only way to satisfy the given constraints is the following combo:-
124
135
234
Any other combination will violate the given constraints.
In the above combo, she runs to city center 9 times and Laurel Lane 7 times.
12 Jul 2024, 00:34
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A cross-country runner trains by running three predetermined routes every day. Each day, her total distance travelled is 10 miles. She does not run any of the routes more than two days in a row and never avoids a route two days in a row.
For the City Centre, select the number of times the runner will run towards the city centre over the span of any 3-day period. For Laurel Lane, select the number of times the runner will run towards Laurel Lane over the span of any 3-day period. Make only two selections, one in each column.
To have a total travelling distance of 10 miles the cross-runner needs to take 2 routes with 3 miles distance and 1 route with 4 miles distance. This is the only possible combination. So, we can already answer that she will run towards the city center 9 times over the span of any 3-day period. Every route has a city centre as final destination, and each days she takes 3 rotes, so 3 times 3 gves 9.
The cross-runner does not run any of the routes more than two days in a row and never avoids a route two days in a row.
Route 2, route 3, route 4 and route 5 include the Laurel Lane.
No matter in which order she will choose routes, she will need to run at least twice the route 1.
So, 9-2 = 7 times
The runner will run towards the city centre 9 times over the span of any 3-day period. The runner will run towards Laurel Lane 7 times over the span of any 3-day period.
For the City Centre, select the number of times the runner will run towards the city centre over the span of any 3-day period. For Laurel Lane, select the number of times the runner will run towards Laurel Lane over the span of any 3-day period. Make only two selections, one in each column.
To have a total travelling distance of 10 miles the cross-runner needs to take 2 routes with 3 miles distance and 1 route with 4 miles distance. This is the only possible combination. So, we can already answer that she will run towards the city center 9 times over the span of any 3-day period. Every route has a city centre as final destination, and each days she takes 3 rotes, so 3 times 3 gves 9.
The cross-runner does not run any of the routes more than two days in a row and never avoids a route two days in a row.
Route 2, route 3, route 4 and route 5 include the Laurel Lane.
No matter in which order she will choose routes, she will need to run at least twice the route 1.
So, 9-2 = 7 times
The runner will run towards the city centre 9 times over the span of any 3-day period. The runner will run towards Laurel Lane 7 times over the span of any 3-day period.
