Last visit was: 12 Jul 2024, 16:13 It is currently 12 Jul 2024, 16:13
Toolkit
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

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# Around the World in 80 Questions (Day 1): The first cyclist from point

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94302
Own Kudos [?]: 640177 [43]
Given Kudos: 84576
Intern
Joined: 12 Jul 2023
Posts: 17
Own Kudos [?]: 16 [5]
Given Kudos: 9
GMAT 1: 650 Q49 V30
GMAT 2: 690 Q48 V37
General Discussion
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5274
Own Kudos [?]: 4167 [2]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Intern
Joined: 25 Jun 2023
Posts: 24
Own Kudos [?]: 12 [4]
Given Kudos: 3
Location: India
GMAT 1: 690 Q49 V34
GMAT 2: 760 Q50 V42
GRE 1: Q169 V155
GPA: 3.74
Re: Around the World in 80 Questions (Day 1): The first cyclist from point [#permalink]
3
Kudos
1
Bookmarks
Let speed of first cyclist be x and speed of second cyclist by y. Let T be time take by second cyclist to reach after from the meeting point.
Distance = speed x time

36x=0.9yT
36y = 2x(T+6)
y=2x(T+6)/36

36x = 0.9(2x(T+6)/36)T
Simplifying this, equation becomes : T^2 + 6T - 720

Solve for T : 24 min

From A to B, first cyclist took, 36+24+6 mins --> 1 hour 6 mins
Manager
Joined: 07 May 2023
Posts: 199
Own Kudos [?]: 243 [4]
Given Kudos: 47
Location: India
Re: Around the World in 80 Questions (Day 1): The first cyclist from point [#permalink]
3
Kudos
1
Bookmarks
Bunuel wrote:
The first cyclist from point A and the second cyclist from point B simultaneously started moving toward each other, at their respective constant speeds. They met each other after 36 minutes and continued moving without stopping. After the meeting, the first cyclist doubled his speed, and the second cyclist reduced his speed by 10%. The first cyclist reached point B 6 minutes after the second cyclist reached point A. How much time did the first cyclist take to travel from point A to point B ?

A. 1 hour
B. 1 hour and 4 minutes
C. 1 hour and 6 minutes
D. 1 hour and 10 minutes
E. 1 hour and 36 minutes

 This question was provided by GMAT Club for the Around the World in 80 Questions Win over $20,000 in prizes: Courses, Tests & more Speed of first cyclist = a units / min Speed of second cyclist = b units / min Until A and B meet - Distance traveled by A in 36 mins = 36a Distance traveled by B in 36 mins = 36b After A and B meet - Time taken by A = 36b/2a = 18b/a Time taken by B = 36a/0.9b = 40a/b 18b/a - 40a/b = 6 ---(1) Divide by 2 9b/a - 20a/b = 3 Let b/a = x 9x - 20/x = 3 9x^2 - 20 - 3x = 0 x = 5/3 Total time = 36 + 36 * 5/6 = 36 + 30 = 66 IMO C Manager Joined: 28 Dec 2020 Posts: 106 Own Kudos [?]: 136 [3] Given Kudos: 523 Re: Around the World in 80 Questions (Day 1): The first cyclist from point [#permalink] 1 Kudos 2 Bookmarks Let the speed of first cyclist be a m/min (taking miles/min just for sake of mentioning minutes. We can take km too, the working will be same) and speed of second cyclist be b m/min They meet after 36 min. So, distance travelled by first cyclist = 36a distance travelled by second cyclist = 36b After the meet, speed of first cyclist = 2a m/min and speed of second cyclist = 0.9b m/min Now the distance to be travelled by first cyclist = distance already travelled by second cyclist = 36b Similarly the distance to be travelled by second cyclist = distance already travelled by first cyclist = 36a Using the above information - Time taken by first cyclist to travel remaining distance after meeting $$= \frac{36b}{2a} = \frac{18 b}{ a}$$ Similarly, time taken by second cyclist to travel remaining distance after meeting $$= \frac{36a}{0.9b} = \frac{40 a}{b}$$ Total time taken by first cyclist to travel from A to B $$= 36 + \frac{18 b}{ a}$$ Now , we are given that the first cyclist reached point B 6 minutes after the second cyclist reached point A $$\frac{18 b}{ a} = \frac{40 a}{b} + 6\\ \\ Let \frac{b}{ a} = k\\ \\ 18 k = \frac{40}{k }+ 6\\ \\ 9k^2 - 3k - 20 = 0\\ \\ 9k^2 + 12k - 15k - 20 = 0\\ \\ (3k+4) (3k-5)=0$$ Since ratio of two speeds will be positive ,$$k = \frac{5}{3}$$ Now, total time taken by first cyclist to travel from A to B $$= 36 + \frac{18 b}{ a} = 36 + 18k$$ $$= 36 + 18 * \frac{5}{3}\\ \\ =66 min\\ \\ \\ = 1 hour 6 min$$ Tutor Joined: 26 Jun 2014 Status:Mentor & Coach | GMAT Q51 | CAT 99.98 Posts: 450 Own Kudos [?]: 808 [0] Given Kudos: 8 Re: Around the World in 80 Questions (Day 1): The first cyclist from point [#permalink] Expert Reply Bunuel wrote: The first cyclist from point A and the second cyclist from point B simultaneously started moving toward each other, at their respective constant speeds. They met each other after 36 minutes and continued moving without stopping. After the meeting, the first cyclist doubled his speed, and the second cyclist reduced his speed by 10%. The first cyclist reached point B 6 minutes after the second cyclist reached point A. How much time did the first cyclist take to travel from point A to point B ? A. 1 hour B. 1 hour and 4 minutes C. 1 hour and 6 minutes D. 1 hour and 10 minutes E. 1 hour and 36 minutes  This question was provided by GMAT Club for the Around the World in 80 Questions Win over$20,000 in prizes: Courses, Tests & more

Let the meeting point between A and B is P
Also, let the cyclist from A be named X and the one from B be named Y
X covered AP in 36 min, Y covered BP in 36 min
Let A would have covered PB in x min, but at twice his speed, he will take x/2 min
Let B would have covered PA in y min, but at 90% of his speed, he will take y/0.9 = 10y/9 min
Thus: x/2 = 6 + 10y/9 ... (i)
Also, X covered AP in 36 min and BP in x min (usual speed)
Y covered BP in 36 min and AP in y min (usual speed)
=> speed ratio = 36/y = x/36 => y = 1296/x

From (i): x/2 = 6 + 1440/x
=> 1440/x = 6 + x/2

Solving, we get: x = 60
Thus, actual time taken = x/2 = 30 mins to cover PB
Thus, total time = 36 + 30 = 66 min