Last visit was: 12 Jul 2024, 16:13 It is currently 12 Jul 2024, 16:13
Close
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
Your Progress

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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94302
Own Kudos [?]: 640177 [43]
Given Kudos: 84576
Send PM
Most Helpful Reply
Intern
Intern
Joined: 12 Jul 2023
Posts: 17
Own Kudos [?]: 16 [5]
Given Kudos: 9
GMAT 1: 650 Q49 V30
GMAT 2: 690 Q48 V37
Send PM
General Discussion
GMAT Club Legend
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)
Send PM
Intern
Intern
Joined: 25 Jun 2023
Posts: 24
Own Kudos [?]: 12 [4]
Given Kudos: 3
Location: India
Concentration: Leadership, Entrepreneurship
GMAT 1: 690 Q49 V34
GMAT 2: 760 Q50 V42
GRE 1: Q169 V155
GPA: 3.74
Send PM
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
Manager
Joined: 07 May 2023
Posts: 199
Own Kudos [?]: 243 [4]
Given Kudos: 47
Location: India
Send PM
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
Manager
Joined: 28 Dec 2020
Posts: 106
Own Kudos [?]: 136 [3]
Given Kudos: 523
Send PM
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
Send PM
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

Answer C
GMAT Club Bot
Re: Around the World in 80 Questions (Day 1): The first cyclist from point [#permalink]
Moderator:
Math Expert
94302 posts