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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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.

" 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

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!

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.

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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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

\(?\,\,\,:\,\,\,{\rm{minutes}}\,\,{\rm{for}}\,\,\left( {{\rm{catch}}\,\, + 1\,\,{\rm{lap}}\,\,{\rm{ahead}}} \right)\)

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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To get a better idea of what's happening, let's sketch the set-up:


Mary is faster than Kate, we can say that Kate has a 250 meter head start. In other words, when Mary closes that 250-meter 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/time

Mary's speed = 1000 meters/5 minutes = 200 meters per minute
Kate's speed = 1000 meters/6 minutes ≈ 167 meters per minute
200 meters per minute - 167 meters per minute = 33 meters per minute

So, the distance gap between Mary and Kate CLOSES at a rate of 33 meters per minute
We want to reduce the gap by 750 meters
Time = distance/rate
So, time = 750/33

STOP!! 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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