Events & Promotions
|
|
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.
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 1308:30 AM PDT
-09:30 AM PDT
In Episode 4 of our GMAT Ninja CR series, we tackle the most intimidating CR question type: Boldface & "Legalese" questions. If you've ever stared at an answer choice that reads, "The first is a consideration introduced to counter a position that...- May 14
08:30 AM PDT
-09:30 AM PDT
Most GMAT test-takers are intimidated by the hardest GMAT Verbal questions. In this session, Target Test Prep GMAT instructor Erika Tyler-John, a 100th percentile GMAT scorer, will show you how top scorers break down challenging Verbal questions..
24 Oct 2007, 17:43
Kudos
Bookmarks
Could I solve this by setting johns time to D-2 and Jacob to D. Also set John's time to T and Jacobs time to t+1 with their rates? It seems to give me the right answer but it is not the way the book solves it so I am not sure if it is just a fluke or it can be answered either way. Also is there a way to solve this using only the difference in thier rates instead of their actual rates?
John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)
Thanks!!
John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)
Thanks!!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
24 Oct 2007, 18:21
Kudos
Bookmarks
I looked at the problem the following way:
John stopped for 1 hour, and then started again. During that time, Jacob continuously rides, at 12 mph, covering a distance of 12 miles. If X is the time to catch up, then John needs to go X PLUS 1 hour @ Jacob's speed:
15(x)=12(x+1)
15x=12x+12
3x=12, x=4 hours
John stopped for 1 hour, and then started again. During that time, Jacob continuously rides, at 12 mph, covering a distance of 12 miles. If X is the time to catch up, then John needs to go X PLUS 1 hour @ Jacob's speed:
15(x)=12(x+1)
15x=12x+12
3x=12, x=4 hours
24 Oct 2007, 22:36
Kudos
Bookmarks
First, you must find the element they have in common (distance), so therefore D = R*T.
Second, you must set both sides of the following equation to equal each other, accounting for (1) the initial 40 mins; and (2) John's 1 hour break:
15x = 12(x+1) - (2/3)(15-12)
15x = 12x + 12 - 2
15x = 12x + 10
3x = 10
x = 10/3
x = 3hrs 20 mins.
Any suggestions?
Second, you must set both sides of the following equation to equal each other, accounting for (1) the initial 40 mins; and (2) John's 1 hour break:
15x = 12(x+1) - (2/3)(15-12)
15x = 12x + 12 - 2
15x = 12x + 10
3x = 10
x = 10/3
x = 3hrs 20 mins.
Any suggestions?
