Time, Speed and distance 2

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Time, Speed and distance 2

by datonman » Wed Feb 04, 2015 11:49 am
The ratio between the speeds of Mary and Jerry is 7:8 while covering a distance. If Mary takes 20 mins more than Jerry, calculate the time taken by Jerry to cover the distance.

Is there truly a way to solve this by inverting the ratio in regards to time, hence, 7/8 times 8/7?
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by GMATGuruNY » Wed Feb 04, 2015 12:12 pm
datonman wrote:The ratio between the speeds of Mary and Jerry is 7:8 while covering a distance. If Mary takes 20 mins more than Jerry, calculate the time taken by Jerry to cover the distance.

Is there truly a way to solve this by inverting the ratio in regards to time, hence, 7/8 times 8/7?
The RATE RATIO for Mary and jerry is 7:8.
Since rate and time are RECIPROCALS, the TIME RATIO for Mary and Jerry is 8:7.
Implication:
For every 8 minutes that Mary travels, Jerry travels 7 minutes.
Since 8:7 = 80:70 = 160:140, Mary travels for 160 minutes, while Jerry travels for 140 minutes, for a difference of 20 minutes.

Final Answer: 140.
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by GMATinsight » Thu Feb 05, 2015 9:37 am
datonman wrote:The ratio between the speeds of Mary and Jerry is 7:8 while covering a distance. If Mary takes 20 mins more than Jerry, calculate the time taken by Jerry to cover the distance.

Is there truly a way to solve this by inverting the ratio in regards to time, hence, 7/8 times 8/7?
In the given two cases the constant element is DISTANCE

Distance = Speed x Time

For Mary, Distance = 7a x (t+20) [t is in mins]
For Jerry, Distance = 8a x t

therefore 7a(t+20) = 8at
7t + 140 = 8t
t = 140 mins (Time of Jerry)
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by GMATinsight » Thu Feb 05, 2015 9:44 am
Another Method:

In the given two cases the constant element is DISTANCE

Distance = Speed x Time

For Jerry, Distance = 8a x t
For Mary, Distance = 7a x (8t/7) [Since distance is constt. therefore if speed becomes 7/8 of previous speed then time will become 8/7 of previous time

and the difference of these times is goven 20 mins

i.e. (8t/7) - t = 20
i.e. t/7 = 20
i.e. t = 140 Mins
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by Matt@VeritasPrep » Mon Feb 09, 2015 12:04 am
The crucial insight here is that Mary's Distance = Jerry's Distance. (They both covered "a distance", so it's the same for each of them.) Since Mary's rate and time are given in terms of Jerry's rate and time, we'll assign variables to Jerry, then set Mary's stats in terms of his.

Jerry's R = j
Jerry's T = t

Mary's R = (7/8)j
Mary's T = t + 20

Since their distances are equal, j*t = (7/8)j * (t + 20). Dividing by j, we have t = (7/8)(t + 20), (1/8)t = (140/8), or t = 140.