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Mary and Kate are running clockwise around a circular track

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by swerve » Tue Jan 22, 2019 12:56 pm
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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. 7.5Ï€
E. 45Ï€

The OA is B

Source: Magoosh
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by GMATGuruNY » Tue Jan 22, 2019 2:21 pm
swerve 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. 7.5Ï€
E. 45Ï€
Since Mary is behind Kate by 250 meters -- half the circumference of the track -- she will catch up to Mary when she has traveled 250 meters more than Kate.
To catch up to Kate a second time, Mary must then travel one more lap than Kate -- in other words, 500 meters more than Kate.
Implication:
To catch up to Kate TWICE, Mary must travel a total of 750 meters more than Kate.

When people compete, SUBTRACT THEIR RATES.
Mary's rate = (1000 meters)/(5 minutes) = 200 meters per minute.
Kate's rate = (1000 meters)/(6 minutes) = 500/3 meters per minute.
Rate difference = 200 - 500/3 = 600/3 - 500/3 = 100/3 meters per minute.
Implication:
Every minute, Mary travels 100/3 meters more than Kate.

Since Mary must travel 750 meters more than Kate -- and every minute she travels 100/3 meters more than Kate -- the time for Mary to catch up to Kate twice = 750/(100/3) = (750*3)/100 = 7.5 * 3 = 22.5 minutes.

The correct answer is B.
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Mary and Kate are running clockwise around a circular track

by fskilnik@GMATH » Wed Jan 23, 2019 3:48 am
swerve 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. 7.5π
E. 45π
Source: Magoosh
Let´s use immediately RELATIVE VELOCITY (speed) and UNITS CONTROL, two powerful tools covered in our course!

$$?\,\,\,:\,\,\,{\text{minutes}}\,\,{\text{for}}\,\,\left( {{\text{catch}}\,\, + 1\,\,{\text{lap}}\,\,{\text{ahead}}} \right)$$
$$\left. \matrix{
{V_M} = {{1000\,\,{\rm{m}}} \over {5\,\,\min }}\,\,\,\, \hfill \cr
{V_K} = {{1000\,\,{\rm{m}}} \over {6\,\,\min }} \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{V_{M \to K}}\,\, = \,\,1000\,\,\underbrace {\left( {{1 \over 5} - {1 \over 6}} \right)}_{ = \,\,{1 \over {30}}}\,\,\,\, = \,\,\,\,{{100\,\,{\rm{m}}} \over {3\,\,\min }}$$
$$\left( {{\text{relative}}} \right)\,\,{\text{distance}}\,\,\,\,\,{\text{ = }}\,\,\left( {\frac{1}{2} + 1} \right) \cdot 500\,\,{\text{m}}$$
$${\text{?}}\,\,\,{\text{ = }}\,\,\,\frac{{3 \cdot 500}}{2}\,\,{\text{m}}\,\,\, \cdot \,\,\,\left( {\frac{{3\,\,\min }}{{100\,\,{\text{m}}}}} \right)\,\,\,\, = \,\,\,\,\frac{{9 \cdot 5}}{2}\,\,\min \,\,\, = \,\,\,22.5\,\,\min $$

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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Brent@GMATPrepNow » Thu Jan 24, 2019 10:46 am
swerve 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. 7.5π
E. 45π

The OA is B

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

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.
Image

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.
Image

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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by Scott@TargetTestPrep » Sun Jan 27, 2019 6:14 pm
swerve 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. 7.5π
E. 45π
Since Mary and Kate start opposite each other, in order for Mary to pass Kate and catch her again, she must run 1½ laps, or 750 meters more than Kate. Also, since Mary’s rate is 1000/5 = 200, and Kate’s rate is 1000/6 = 500/3, the difference in rates is 200 - 500/3. Thus the time, in minutes, Mary will pass Kate and catch her again is

750/(200 - 500/3)

2250/(600 - 500)

2250/100

22.5

Answer: B

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