|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
31 Aug 2022, 04:22
Kudos
Bookmarks
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:
22% (02:04) correct
78%
(02:43)
wrong
based on 55
sessions
History
Date
Time
Result
Not Attempted Yet
Every morning, a school bus driver starts from point A and picks up children from stop S and returns to point A through a different route. The driver takes the shortest but different route every day. If there is a roadblock at point C, then for how many days can the driver take different routes that do not pass through C?
Screenshot 2022-08-30 103628.png [ 184.02 KiB | Viewed 4061 times ]
A. 7
B. 42
C. 49
D. 62
E. 70
Attachment:
Screenshot 2022-08-30 103628.png [ 184.02 KiB | Viewed 4061 times ]
A. 7
B. 42
C. 49
D. 62
E. 70
31 Aug 2022, 06:27
Kudos
Bookmarks
Total ways from A to S = EEEENN = \(\frac{6!}{4!2!}\) = 15 ways
Point C is roadblock, so reduce the ways thru point C.
A to C = EEEN = \(\frac{4!}{3!}\) = 4 ways
C to S = EN = 2! = 2 ways
So, A to S thru C = 4*2 = 8 ways
Total allowed ways = 15-8 = 7 ways
Return leg- constraint is a different route. If 7 ways from A to S, then 7-1=6 ways from S to A.
Hence, answer = 7*6 = 42 ways i.e. B.
Point C is roadblock, so reduce the ways thru point C.
A to C = EEEN = \(\frac{4!}{3!}\) = 4 ways
C to S = EN = 2! = 2 ways
So, A to S thru C = 4*2 = 8 ways
Total allowed ways = 15-8 = 7 ways
Return leg- constraint is a different route. If 7 ways from A to S, then 7-1=6 ways from S to A.
Hence, answer = 7*6 = 42 ways i.e. B.
01 Sep 2022, 00:41
Kudos
Bookmarks
Given: Every morning, a school bus driver starts from point A and picks up children from stop S and returns to point A through a different route. The driver takes the shortest but different route every day.
Asked: If there is a roadblock at point C, then for how many days can the driver take different routes that do not pass through C?
Ways from A to C = EEEENN = 6!/4!2! = 15
Ways from A to S = EEEN = 4!/3! = 4
Ways from S to C = EN = 2
Ways from A to C without S = 15 - 4*2 = 7
Return ways from A to C without S = 6
Total ways = 7*6 = 42
IMO B
Asked: If there is a roadblock at point C, then for how many days can the driver take different routes that do not pass through C?
Attachment:
Screenshot 2022-08-30 103628.png
Ways from A to C = EEEENN = 6!/4!2! = 15
Ways from A to S = EEEN = 4!/3! = 4
Ways from S to C = EN = 2
Ways from A to C without S = 15 - 4*2 = 7
Return ways from A to C without S = 6
Total ways = 7*6 = 42
IMO B
