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Mary and Kate are running clockwise around a circular track with a cir

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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. \(45\pi\)

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Re: Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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Mary's speed is \(\frac{1000}{5}\) and Kate's speed is \(\frac{1000}{6}\)
Relative speed is \(\frac{1000}{5} - \frac{1000}{6} = \frac{100}{3}\)

Mary and Kate start at opposite end on circular track, so initially they are 250 m apart. So Mary will have to cover that distance first and then to pass her again she'll have to cover another 500 m.

To cover the total 750 m, time taken will be\(\frac{750}{(100/3)} = 22.5\)

Hence option B.

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New post 27 Mar 2015, 10:22
Bunuel wrote:
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. \(45\pi\)

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M =\(\frac{1000}{5}\)
K= \(\frac{1000}{6}\)

As they are running in same direction ,
relative speed of Mary = \(\frac{1000}{5} - \frac{1000}{6}\) = \(1000* \frac{1}{30}\)
with this speed Mary will take \(\frac{750*30}{1000}\) mins to pass Kate and catch her again.
Answer B.
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Re: Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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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)

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Re: Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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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!
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Re: Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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" 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

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Answer = B = 22.5

Refer diagram below

Attachment:
cross.png
cross.png [ 4.83 KiB | Viewed 3937 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
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Re: Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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New post 27 May 2015, 21:05
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!

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MV23 wrote:
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: http://www.veritasprep.com/blog/2012/07 ... elatively/
http://www.veritasprep.com/blog/2012/08 ... -speeding/

Hope it helps.
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Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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New post 28 Jul 2016, 11:46
AVRonaldo wrote:
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.

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Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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New post 15 Aug 2016, 06:23
ashutoshsh wrote:
AVRonaldo wrote:
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

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New post 15 Aug 2016, 07:17
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.

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New post 15 Aug 2016, 08:30
rahulkashyap wrote:
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

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That is because Mary has to pass Kate once and meet her again. So first time Mary meets Kate she will travel 250M and to meet her again she will have to take another complete round.

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New post 15 Aug 2016, 10:08
rahulkashyap wrote:
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
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New post 04 Sep 2017, 20:46
Bunuel wrote:
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. \(45\pi\)

Kudos for a correct solution.



Mary would pass Kate initially by crossing half the circle. so the distance is 250. To catch her again she has to traverse the entire track again i.e 500 meters.
So total distance = 250+500= 750 meters.

Relative speed = 1000/5- 1000/6 ( being in the sme direction)

So Time = 750/1000* 30=90/4= 22.5

Ans:B

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Re: Mary and Kate are running clockwise around a circular track with a cir [#permalink]

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New post 03 Nov 2017, 07:55
Bunuel wrote:
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. \(45\pi\)

Kudos for a correct solution.


hi

Since the length of the circular track is 500 meters, the initial distance between Mary and Kate is 250 meters.

so, the time required to meet each other for the first time is

250
______________
1000/5 - 1000/6

=7.5 minutes

time required to meet each other for the second time is

500
______________
1000/5 - 1000/6

= 15 minutes

Total = 7.5 + 15 = 22.5 minutes

thanks
:cool:

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Re: Mary and Kate are running clockwise around a circular track with a cir   [#permalink] 03 Nov 2017, 07:55
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