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Mary and Kate are running clockwise around a circular track with a cir
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27 Mar 2015, 06:00
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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. \(45\pi\) Kudos for a correct solution.
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Re: Mary and Kate are running clockwise around a circular track with a cir
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30 Mar 2015, 00:17
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 timesthat'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 thiswe 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
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27 Mar 2015, 06:32
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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Mary and Kate are running clockwise around a circular track with a cir
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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. \(45\pi\)
Kudos for a correct solution. 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
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28 Mar 2015, 00:33
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
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30 Mar 2015, 21:54
" 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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Re: Mary and Kate are running clockwise around a circular track with a cir
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30 Mar 2015, 23:25
Answer = B = 22.5 Refer diagram below Attachment:
cross.png [ 4.83 KiB  Viewed 9080 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
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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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Re: Mary and Kate are running clockwise around a circular track with a cir
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28 May 2015, 04:06
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! 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
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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) BunuelHi! 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
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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) BunuelHi! 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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Re: Mary and Kate are running clockwise around a circular track with a cir
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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. chetan2uPosted from my mobile device



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Re: Mary and Kate are running clockwise around a circular track with a cir
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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. chetan2uPosted from my mobile device 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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Re: Mary and Kate are running clockwise around a circular track with a cir
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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. chetan2uPosted 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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Re: Mary and Kate are running clockwise around a circular track with a cir
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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
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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



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Re: Mary and Kate are running clockwise around a circular track with a cir
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03 Feb 2018, 08:03
Mary's speed = \(\frac{1000}{5}\) Kate's speed = \(\frac{1000}{6}\) Relative speed = \(\frac{1000}{5}  \frac{1000}{6} = \frac{100}{3}\) Mary and Kate start at opposite end on circular track. So at the beginning they are 250 m apart. This is same as giving Kate a head start of 250 m. In this case how long Mary will take to catch Kate? To catch Kate again, Mary will have to run 500 m more. This is same as giving Kate a head start of 500 m. In this case how long Mary will take to catch Kate? (See how 500 m comes. Let, Kate and Mary start from a point of A of the circumference track. Mary crosses Kate the moment they both started from A. So, to catch Kate, Mary will have to run full circumference plus the distance Kate passed. Therefore, Mary has to run 500 m more during the same time Kate runs)
The answer would be the total time to cover both 250 m and 500 m, i.e, 750 m. We know for relative speed of bodies moving in same direction, the equation is (Sx  Sy) X t = Distance
Here, total distance to be covered is 250+500=750 m \(\frac{100}{3}*t = 750\) \(t=22.5\)
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Re: Mary and Kate are running clockwise around a circular track with a cir
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23 Jan 2019, 05:03
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. \(45\pi\)
\(?\,\,\,:\,\,\,{\rm{minutes}}\,\,{\rm{for}}\,\,\left( {{\rm{catch}}\,\, + 1\,\,{\rm{lap}}\,\,{\rm{ahead}}} \right)\) 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.
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Re: Mary and Kate are running clockwise around a circular track with a cir
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24 Jan 2019, 08:54
Speed of Mary is x = 200m/min i.e 600/3 m/min Speed of Kate is y = 500/3 m/min since they are running in same direction their relative speed will be xy i.e 100/3 m/min They are already 250 m apart, and the requirement is to cover those 250 m plus one more round of 500 m to pass her again with the relative speed of 100/3 m/min Hence time taken will be 750*3/100 = 22.5 min



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Re: Mary and Kate are running clockwise around a circular track with a cir
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24 Jan 2019, 11:57
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. To get a better idea of what's happening, let's sketch the setup: Mary is faster than Kate, we can say that Kate has a 250 meter head start. In other words, when Mary closes that 250meter gap, she will pass Kate for the FIRST time. However, since we want to find the time for Mary to pass Kate for the SECOND time, we can say that Kate has a 750 meter head start. So we want to determine the time it takes Mary to close the gap from 750 meters to 0 meters. Mary runs 1000 meters every five minutes and Kate runs 1000 meters every six minutes. Speed = distance/timeMary's speed = 1000 meters/5 minutes = 200 meters per minuteKate's speed = 1000 meters/6 minutes ≈ 167 meters per minute200 meters per minute  167 meters per minute = 33 meters per minuteSo, the distance gap between Mary and Kate CLOSES at a rate of 33 meters per minuteWe want to reduce the gap by 750 meters Time = distance/rateSo, time = 750/ 33STOP!! Before we perform this somewhat tedious calculation, we should scan the answer choices. When we do so, we see that the answer choices are quite spread apart, which means we can be quite aggressive with our estimation. So, rather than divide by 33, let's divide by 30 That is, 750/ 33 ≈ 750/ 30 = 25 So, the correct answer is close to 25 Answer: B Cheers, Brent
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