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07 Feb 2023, 12:19
Kudos
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Hi,
I am studying for an exam that is the equivalent of the GMAT but in French (TAGE MAGE).
One of the practice questions is the following:
Anaïs completes a 20 km run in 1 hour 45 minutes. Anaïs starts the race quickly. After 2 km, her average speed drops. She takes 30 seconds more per kilometer compared to her average speed over the first two kilometers of the race. Finally, she accelerates over the last 5 km, which took her 11 minutes to complete. What is her approximate average speed from kilometer 2 to kilometer 15?
The solution is the following:
Let v be the average speed of Anaïs between the 2nd kilometer and the 15th kilometer.
T1=time spent on the first two km
T1 = distance/speed= 2/v + 1
T2=time spent between the 2nd km and the 15th km
T2=distance/speed = 13 / v
Anaïs takes 1 h 45 min - 11 min that is 94 minutes to cover the first 15 km, so : T1+T2=94minutes
2/v + 1 + 13/v = 94
=>15v=93
=>v=1593=3∗53∗31=531≈0,16
So Anais is going 0.16 kilometers per minute between the 2nd and 15th kilometers.
Does anyone understand why the distance spent on the first two km is 2 / v + 1 ?
Thank you in advance.
I am studying for an exam that is the equivalent of the GMAT but in French (TAGE MAGE).
One of the practice questions is the following:
Anaïs completes a 20 km run in 1 hour 45 minutes. Anaïs starts the race quickly. After 2 km, her average speed drops. She takes 30 seconds more per kilometer compared to her average speed over the first two kilometers of the race. Finally, she accelerates over the last 5 km, which took her 11 minutes to complete. What is her approximate average speed from kilometer 2 to kilometer 15?
The solution is the following:
Let v be the average speed of Anaïs between the 2nd kilometer and the 15th kilometer.
T1=time spent on the first two km
T1 = distance/speed= 2/v + 1
T2=time spent between the 2nd km and the 15th km
T2=distance/speed = 13 / v
Anaïs takes 1 h 45 min - 11 min that is 94 minutes to cover the first 15 km, so : T1+T2=94minutes
2/v + 1 + 13/v = 94
=>15v=93
=>v=1593=3∗53∗31=531≈0,16
So Anais is going 0.16 kilometers per minute between the 2nd and 15th kilometers.
Does anyone understand why the distance spent on the first two km is 2 / v + 1 ?
Thank you in advance.
Archived Topic
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10 Feb 2023, 02:37
Kudos
Bookmarks
Does anyone understand why the distance spent on the first two km is \(\frac{2 }{ v} + 1\) ?
It should be
\( \frac{2}{v} -1\)
Since we are calculating the time for 2 km distance using v (average speed between the 2nd km and the 15th km) and it's given that Anais takes 30 seconds more per km while running from 2 - 15 km compared to the first 2 kms
Effectively -> \(\frac{2}{v}\) is the time for 2 km at speed v and 1 is twice of 30 sec which has to be subtracted from \(\frac{2}{v} \)
\(\frac{2}{v} -1 + \frac{13}{v} = 94\)
v = 0.157 km/min
Hope your query is resolved
It should be
\( \frac{2}{v} -1\)
Since we are calculating the time for 2 km distance using v (average speed between the 2nd km and the 15th km) and it's given that Anais takes 30 seconds more per km while running from 2 - 15 km compared to the first 2 kms
Effectively -> \(\frac{2}{v}\) is the time for 2 km at speed v and 1 is twice of 30 sec which has to be subtracted from \(\frac{2}{v} \)
\(\frac{2}{v} -1 + \frac{13}{v} = 94\)
v = 0.157 km/min
Hope your query is resolved
12 Feb 2023, 01:05
Kudos
Bookmarks
It should be 2/v - 1, if we consider the speed between the 2nd and 15th kilometer as v.
This is because if 'v' results into 30 seconds more time per km, so calculating the first 2 kms must have accounted for an extra minute (30 seconds twice) if we consider the time as '2/v'.
Therefore, this 1 minute needs to be subtracted.
2/v - 1 + 13/v = 94 is the equation.
This is because if 'v' results into 30 seconds more time per km, so calculating the first 2 kms must have accounted for an extra minute (30 seconds twice) if we consider the time as '2/v'.
Therefore, this 1 minute needs to be subtracted.
2/v - 1 + 13/v = 94 is the equation.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
