Last visit was: 19 Jul 2025, 18:03 It is currently 19 Jul 2025, 18:03
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,627
Kudos: 742,795
 [145]
11
Kudos
Add Kudos
134
Bookmarks
Bookmark this Post
Most Helpful Reply
avatar
shreyast
Joined: 17 Feb 2015
Last visit: 03 Nov 2015
Posts: 23
Own Kudos:
107
 [56]
Given Kudos: 13
GPA: 3
Products:
Posts: 23
Kudos: 107
 [56]
34
Kudos
Add Kudos
21
Bookmarks
Bookmark this Post
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,755
Own Kudos:
34,133
 [14]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,755
Kudos: 34,133
 [14]
11
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
User avatar
AVRonaldo
Joined: 24 Jan 2015
Last visit: 22 Mar 2022
Posts: 56
Own Kudos:
638
 [7]
Given Kudos: 9
GPA: 4
WE:Consulting (Pharmaceuticals and Biotech)
Products:
Posts: 56
Kudos: 638
 [7]
7
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given , using RTD table,

Mary's rate --> R * 5 = 1000 ==> \(\frac{1000}{5}\)
Kate's rate --> R * 6 = 1000 ==> \(\frac{1000}{6}\)

Since they both are running clockwise (same direction), subtract the rates ( Relative rate concept)
\(\frac{1000}{5}\) - \(\frac{1000}{6}\) = \(\frac{1000}{30}\) ==> \(\frac{100}{3}\)

Since Mary and Kate start opposite to each other in a circular track, they are 250m apart.
Distance ran by the Mary to pass Kate and catch her again is 250 + 500 = 750m

\(\frac{100}{3}\) * t = 750

==> t = \(\frac{3}{100}\) * 750 = \(\frac{225}{10}\) ==> 22.5

Answer is 22.5 (B)
User avatar
DmitryFarber
User avatar
Manhattan Prep Instructor
Joined: 22 Mar 2011
Last visit: 16 Jul 2025
Posts: 2,950
Own Kudos:
8,403
 [14]
Given Kudos: 57
GMAT 2: 780  Q50  V50
Expert
Expert reply
GMAT Focus 1: 745 Q86 V90 DI85
Posts: 2,950
Kudos: 8,403
 [14]
10
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
The above solutions work perfectly. Here's an alternative:

First, eliminate D and E. We've already been given the circumference (not the diameter) and it's an integer, so pi is not going to make an appearance in this problem.

Next, get Mary's and Kate's rates in similar terms. If Kate takes 6 minutes to run 1000 meters, how far will Mary go in the same time? In an extra minute she'll go 200 meters, for a total of 1200. So this means that every 6 minutes, Mary gains 200 meters on Kate. Now let's look at the answers:

A) In 7.5 minutes, Mary will only gain 200 meters 1 1/4 times--that's 250 meters (one half circle). She needs to gain 3 half circles, and that leads us to . . .

B) 22.5 minutes. Even if we just approximate, we can see that Mary will gain something in between 600 meters (18 minutes) and 800 meters (24 minutes).

C) 750 minutes. That is a lot of running! :) We don't need to calculate this--we can just recognize that she will gain many thousands of meters in this time.

So why bother with an approach like this? Well, it can be very helpful to have a sense of what kind of answer is reasonable on a given problem. I find that we can fairly often eliminate several answers early on, and sometimes we can even knock out all four wrong answers! However, any estimate that narrows our range will help us to control for error by keeping our answers reasonable. Notice that this approximation technique would help you even if you miscalculated the required distance. For instance, let's say you misapplied 500 as the length of the semicircle and came up with a total distance of 1500. B would still be the only answer choice that looked at all close!
avatar
tomlui2010
Joined: 13 Sep 2014
Last visit: 14 May 2019
Posts: 5
Own Kudos:
8
 [5]
Posts: 5
Kudos: 8
 [5]
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
" Initial separation / relative speed " will give us the time taken to meet for the first time.
In this case, relative speed = 1000/5 - 1000/6
initial separation = 250
Hence, time = 7.5
As of now, they are at the same point and M needs to catch up with K again. Therefore, we can consider that the separation between M and K is 500.
Applying the same method as above, we get that M will catch up after 15.
in total, 7.5 + 15 = 22.5
avatar
PareshGmat
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,538
Own Kudos:
7,900
 [4]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,538
Kudos: 7,900
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Answer = B = 22.5

Refer diagram below

Attachment:
cross.png
cross.png [ 4.83 KiB | Viewed 33853 times ]

Starting points are shown, lets say they meet at a distance "x" from where Kate started

Speed of Mary \(= \frac{1000}{5}\)

Speed of Kate \(= \frac{1000}{6}\)

Setting up the time equation

\(\frac{250+x}{\frac{1000}{5}} = \frac{x}{\frac{1000}{6}}\)

x = 1250

Distance travelled by Mary for first catch = 1250+250 = 1500

Time required by Mary \(= \frac{1500}{\frac{1000}{5}} = 7.5\) ................ (1)

Mary travels 500 meters in 2.5 minutes & Kate travels the same in 3 minutes

LCM of 2.5 & 3 = 15

So, after there first catch, they will meet again after 15 minutes .............. (2)

Total time required my Mary = (1) + (2) = 15+7.5 = 22.5
User avatar
MV23
Joined: 10 May 2015
Last visit: 30 Mar 2021
Posts: 296
Own Kudos:
152
 [1]
Given Kudos: 502
Status:Winter is coming
Location: United States
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V40
GMAT 2: 760 Q50 V42
GPA: 3.9
WE:Engineering (Computer Software)
Products:
GMAT 2: 760 Q50 V42
Posts: 296
Kudos: 152
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I am sorry if it is inappropriate to post this here, but,
Could you guys point me to some resources where I can read more about using relative speeds to solve rate problems?
Thank You!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
742,795
 [2]
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,627
Kudos: 742,795
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
MV23
I am sorry if it is inappropriate to post this here, but,
Could you guys point me to some resources where I can read more about using relative speeds to solve rate problems?
Thank You!

Theory on Distance/Rate Problems: distance-speed-time-word-problems-made-easy-87481.html

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
All PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64

For more on relative speed, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/07 ... elatively/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/08 ... -speeding/

Hope it helps.
User avatar
ashutoshsh
Joined: 07 Mar 2016
Last visit: 07 Feb 2017
Posts: 53
Own Kudos:
Given Kudos: 163
Posts: 53
Kudos: 174
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AVRonaldo
Given , using RTD table,

Mary's rate --> R * 5 = 1000 ==> \(\frac{1000}{5}\)
Kate's rate --> R * 6 = 1000 ==> \(\frac{1000}{6}\)

Since they both are running clockwise (same direction), subtract the rates ( Relative rate concept)
\(\frac{1000}{5}\) - \(\frac{1000}{6}\) = \(\frac{1000}{30}\) ==> \(\frac{100}{3}\)

Since Mary and Kate start opposite to each other in a circular track, they are 250m apart.
Distance ran by the Mary to pass Kate and catch her again is 250 + 500 = 750m

\(\frac{100}{3}\) * t = 750

==> t = \(\frac{3}{100}\) * 750 = \(\frac{225}{10}\) ==> 22.5

Answer is 22.5 (B)

Bunuel
Hi! Thanks for the explanation. i understood the method mentioned but i am not getting the question. What i understand is that these two girls are running in opposite direction and we have to calculate the time when they will meet each other. Though the question is saying that first they were running in same direction, but i dont understand how does that matter.
also, you have mentioned that they started in opposite direction when they were 250m apart. from where did we get that.
Thanks.
User avatar
CuriosStud
Joined: 06 Jan 2015
Last visit: 21 Nov 2023
Posts: 52
Own Kudos:
28
 [3]
Given Kudos: 24
Location: India
GMAT 1: 720 Q49 V40
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
ashutoshsh
AVRonaldo
Given , using RTD table,

Mary's rate --> R * 5 = 1000 ==> \(\frac{1000}{5}\)
Kate's rate --> R * 6 = 1000 ==> \(\frac{1000}{6}\)

Since they both are running clockwise (same direction), subtract the rates ( Relative rate concept)
\(\frac{1000}{5}\) - \(\frac{1000}{6}\) = \(\frac{1000}{30}\) ==> \(\frac{100}{3}\)

Since Mary and Kate start opposite to each other in a circular track, they are 250m apart.
Distance ran by the Mary to pass Kate and catch her again is 250 + 500 = 750m

\(\frac{100}{3}\) * t = 750

==> t = \(\frac{3}{100}\) * 750 = \(\frac{225}{10}\) ==> 22.5

Answer is 22.5 (B)

Bunuel
Hi! Thanks for the explanation. i understood the method mentioned but i am not getting the question. What i understand is that these two girls are running in opposite direction and we have to calculate the time when they will meet each other. Though the question is saying that first they were running in same direction, but i dont understand how does that matter.
also, you have mentioned that they started in opposite direction when they were 250m apart. from where did we get that.
Thanks.

Well you have analysed the question wrongly and have been mislead due to its tricky language. The question stem says that the two girls start in opposite direction of the circle and run clockwise. Which means they start at the opposite end of the diameter of the circle and both run clockwise. We are also told that the circumference of the circle is 500. What the question asks is, When will Mary pass Kate once and meet here again.

Now, are given their respective speed and from which we can find their relative speed. Since Mary and Kate both are at the opposite end of the circle, they form a semi circle with a circumference 250 (1/2*500). so for Mary to overtake Kate once and meet her again she will have to cover 750 miles.

Kate runs 1000 miles every six minutes and Mary runs 1000 miles every 5 minutes. So every six minutes Mary runs 1200 miles. hence she covers 200 miles every 6 minutes. So to cover 750 miles it takes 750*6/200 = 22.5 mins
avatar
rahulkashyap
Joined: 09 Oct 2015
Last visit: 24 Feb 2019
Posts: 170
Own Kudos:
Given Kudos: 28
Posts: 170
Kudos: 72
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi
I don't understand how it's plus 500? Could someone pls explain why it's plus 500. The 250 part is quite clear.

chetan2u

Posted from my mobile device
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 19 Jul 2025
Posts: 11,294
Own Kudos:
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,294
Kudos: 41,843
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rahulkashyap
Hi
I don't understand how it's plus 500? Could someone pls explain why it's plus 500. The 250 part is quite clear.

chetan2u

Posted from my mobile device

Hi,
They both are starting opposite to each other, so they are 500/2 m away...
So first time M has to travel 250 more than K to catch up with her...
But we have to find the second time she catches her and this involves another complete circle that is 500, so total 250+500
User avatar
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 885
Own Kudos:
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 885
Kudos: 1,703
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Mary and Kate are running clockwise around a circular track with a circumference of 500 meters, each at her own constant speed. Mary runs 1000 meters every five minutes and Kate runs 1000 meters every six minutes. If Mary and Kate start opposite one another on the circular track, how many minutes must Mary run in order to pass Kate and catch her again?

A. 7.5
B. 22.5
C. 750
D. \(7.5\pi\)
E. \(45\pi\)
\(?\,\,\,:\,\,\,{\rm{minutes}}\,\,{\rm{for}}\,\,\left( {{\rm{catch}}\,\, + 1\,\,{\rm{lap}}\,\,{\rm{ahead}}} \right)\)

Let´s use immediately RELATIVE VELOCITY (speed) and UNITS CONTROL, two powerful tools covered in our course!

\(\left. \matrix{\\
{V_M} = {{1000\,\,{\rm{m}}} \over {5\,\,\min }}\,\,\,\, \hfill \cr \\
{V_K} = {{1000\,\,{\rm{m}}} \over {6\,\,\min }} \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{V_{M \to K}}\,\, = \,\,1000\,\,\underbrace {\left( {{1 \over 5} - {1 \over 6}} \right)}_{ = \,\,{1 \over {30}}}\,\,\,\, = \,\,\,\,{{100\,\,{\rm{m}}} \over {3\,\,\min }}\)
\(\left( {{\rm{relative}}} \right)\,\,{\rm{distance}}\,\,\,\,\,{\rm{ = }}\,\,\left( {{1 \over 2} + 1} \right) \cdot 500\,\,{\rm{m}}\)

\({\rm{?}}\,\,\,{\rm{ = }}\,\,\,{{3 \cdot 500} \over 2}\,\,{\rm{m}}\,\,\, \cdot \,\,\,\left( {{{3\,\,\min } \over {100\,\,{\rm{m}}}}} \right)\,\,\,\, = \,\,\,\,{{9 \cdot 5} \over 2}\,\,\min \,\,\, = \,\,\,22.5\,\,\min\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
User avatar
GmatPoint
Joined: 02 Jan 2022
Last visit: 13 Oct 2022
Posts: 247
Own Kudos:
131
 [1]
Given Kudos: 3
GMAT 1: 760 Q50 V42
GMAT 1: 760 Q50 V42
Posts: 247
Kudos: 131
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Given the total circumference of the circle is 500 meters.
Since both of them start on the opposite sides. They have a distance gap of 250 meters in between them.
The rate of speed of Mary is \(\frac{1000}{5}\)
The rate of speed of Jane is \(\frac{1000}{6}\)
Since they both travel in the same direction the relative speed is :
\(\left(\frac{1000}{5}-\ \frac{1000}{6}\right)\cdot t\ =\ 250\)
t = 7.5 minutes. This is the first time mary reaches Jane.
Mary in order to meet Jane for the second time :
The distance Mary needs to travel now is equal to the circumference.
Hence the time required is :
\(\left(\frac{1000}{5}-\ \frac{1000}{6}\right)\cdot t\ =\ 500\)
t = 15 minutes.
The total time required is 15+7.5 minutes = 22.5 minutes
User avatar
Dereno
Joined: 22 May 2020
Last visit: 19 Jul 2025
Posts: 458
Own Kudos:
Given Kudos: 364
Products:
Posts: 458
Kudos: 423
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Speed = distance / time. Here two persons are involved in a race.so it becomes relative speed = distance apart/ time .

Two persons running in
same direction relative speed needs to be subracted
opposite direction relative speed needs to be added..

Here it’s same direction. Even though there are exactly opposite to each other. Which is 180 degrees apart. Circumference is 500 meters. Distance apart = 500/2 = 250 metres., which is the initial separation between two runners.

Relative speed = 1000/5 - 1000/6 = 100/3 ( as given in question Mary runs 1000 meters every five minutes and Kate runs 1000 meters every six minutes ).

Time = separation/ relative speed.
= 250/ (100/3)
= 7.5 mins.

At time 7.5 mins, both Mary and Kate are at the same point. Mary needs to catch up Kate again. Now the separation between them is one full circumference which is equal to 500 metres.

Time = 500/ (100/3) = 15 mins.

Hence the total time = 7.5 +15 = 22.5 mins.
Moderators:
Math Expert
102627 posts
PS Forum Moderator
698 posts