Last visit was: 28 Apr 2026, 10:24 It is currently 28 Apr 2026, 10:24
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Events & Promotions
User avatar
Rosy
User avatar
Joined: 08 Oct 2007
Last visit: 08 Oct 2007
Posts: 2
Posts: 2
Kudos: 0
A man had covered 2/3 of distance by cycling and remaining 08 Oct 2007, 02:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
A man had covered 2/3 of distance by cycling and remaining covered on feet. For this he took twice the time to the cycling. Find how much speed is greater for cycling than on feet.
Regards,
Rosy.



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!
User avatar
Fig
User avatar
Joined: 01 May 2006
Last visit: 02 Feb 2025
Posts: 1,031
Own Kudos:
253
Posts: 1,031
Kudos: 253
A man had covered 2/3 of distance by cycling and remaining 08 Oct 2007, 04:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Welcome to GMATClub ! :D :)

Let be:
- v1, d1 and t1 : the speed, distance and time done with the cycle
- v2, d2 and t2 : the same things for walking

t2 = 2 * t1
<=> d2 / v2 = 2 * d1 / v1
<=> v1 = ((2*d1)/d2) * v2

With d1 = 2*d2, we have:
v1 = ((2*d1)/d2) * v2
<=> v1 = ((2* (2*d2))/d2) * v2
<=> v1 = 4 * v2

So, it was 4 time faster cycling than walking.
User avatar
beckee529
User avatar
Joined: 11 Jun 2007
Last visit: 23 Feb 2012
Posts: 393
Own Kudos:
1,977
Posts: 393
Kudos: 1,977
A man had covered 2/3 of distance by cycling and remaining 08 Oct 2007, 07:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rosy
A man had covered 2/3 of distance by cycling and remaining covered on feet. For this he took twice the time to the cycling. Find how much speed is greater for cycling than on feet.
Regards,
Rosy.
Show more


not sure if my logic is correct:

general formula r = d/ t

cycling =
d1 = 2/3
t1 = 2t
r1 = d1 / t1
r1 = (2/3)/ 2t = 1/3

by feet =
d2 = 1/3
t2 = t
r2 = d2 / t2
r2 = (1/3)/ t = 1/3

so the same by cycling or on foot
User avatar
GMATBLACKBELT
User avatar
Joined: 29 Mar 2007
Last visit: 03 Jun 2013
Posts: 1,138
Own Kudos:
1,915
Posts: 1,138
Kudos: 1,915
A man had covered 2/3 of distance by cycling and remaining 08 Oct 2007, 10:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rosy
A man had covered 2/3 of distance by cycling and remaining covered on feet. For this he took twice the time to the cycling. Find how much speed is greater for cycling than on feet.
Regards,
Rosy.
Show more


Question was worded poorly i.e. grammer is bad. but still solveable. :P


anyway. Pick a value for the distance. Something that is divisible by both 2/3 and 1/3 to make solving the problem easier.

D=60 Let x be the rate for cycling.
Let y be the rate for walking.
Let t equal the time for cycling.

So we have:
C W
R x y
T t 2t
D 40 20

Now solve for t:

C: t=40/x W: 2t=20/y Now plug in t from C.

we get 2(40/x)=20/y ---> 80/x=20/y ---> 20x=80y ----> x=4y.

So the cycling rate x is 4 times greater than the walking rate of y.
User avatar
rishi2377
User avatar
Joined: 18 Jun 2007
Last visit: 25 Mar 2009
Posts: 125
Own Kudos:
319
Posts: 125
Kudos: 319
A man had covered 2/3 of distance by cycling and remaining 08 Oct 2007, 13:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
15sec. Just plug in numbers.
take a nuber for total distance (to make it easy take a number whic is a multiple of 3). I took 90.
D covered by cycle = 90*2)/ 3= 60
D coverd on feet= remaining 1/3 of 90 =30

suppose time taken for first 60miles =1hr
then time taken for last 30 miles ( on feet)= 2hr
Speed by cycle= 60m/h
Speed on feet= 30/2=15m/h


ratio= 60/15=4 times.

PS...you need to work on verbal :P btw welcome to club.



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 above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!