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17 Jul 2024, 06:59
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Distance covered on packed dirt (meters): 469
Distance covered on grass (meters): 1,876
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
78% (03:18) correct
22%
(03:34)
wrong
based on 860
sessions
History
Date
Time
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An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.
|
| Distance covered on packed dirt (meters) | Distance covered on grass (meters) | |
| 316.5 | ||
| 469 | ||
| 1,266 | ||
| 1,876 | ||
| 5,064 | ||
| 7,504 |
ShowHide Answer
Official Answer
Distance covered on packed dirt (meters): 469
Distance covered on grass (meters): 1,876
17 Jul 2024, 07:04
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Bunuel
Experts' Global Explanation:
Let the speed of the athlete in the packed dirt and the grass be \(S_P\) and \(S_G\), respectively.
Let the distance covered on packed dirt and the grass be \(D_P\) and \(D_G\), respectively.
Let the time taken to cover packed dirt and the grass be \(T_P\) and \(T_G\), respectively.
\(S_P= 210\) meters per minute
\(S_G= 100.5\) meters per minute
Since the distance covered on grass was four times the distance covered on packed dirt: \(D_G= 4D_P\).
Since it took 21 minutes for the athlete to cover both the packed dirt and the grass: \(T_P+ T_G= 21\) minutes.
\(T_P+ T_G = (\frac{D_P}{ S_P}) + (\frac{D_G}{ S_G})\)
\(21 = (\frac{D_P}{ 201}) + (\frac{D_G}{ 100.5})\)
\(21 = (\frac{D_P}{ 201}) + (\frac{2D_G }{ 201})\)
\(21 = \frac{(D_P+ 2D_G)}{201}\)
\(21 =\frac{ (D_P+ 2 * 4D_P)}{201}\)
\(21 = \frac{(D_P+ 8D_P)}{201}\)
\(21 = \frac{(9D_P)}{201}\)
\(D_P = 469\) meters
\(D_G = 4D_P = 4 * 469 = 1,876\) meters.
17 Jul 2024, 07:29
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An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.
Distance covered on packed dirt (meters): 469
Distance covered on grass (meters): 1876
x/200,1 + 4x/100,5 = 21 minutes => x = 469 meters => 4x = 1876 meters
Distance covered on packed dirt (meters): 469
Distance covered on grass (meters): 1876
x/200,1 + 4x/100,5 = 21 minutes => x = 469 meters => 4x = 1876 meters
