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Deconstructing the Question
We only care about the packed dirt and grass segments.
On packed dirt, the athlete runs at 201 meters per minute.
On grass, she runs at 100.5 meters per minute.
The grass distance is 4 times the packed dirt distance, and the total time on these two terrains is 21 minutes.
The key shortcut is this: the grass speed is exactly half the packed dirt speed. So if the grass distance is 4 times as large, then the time on grass must be 8 times the time on packed dirt.
Step-by-step
Let the time on packed dirt be \(t\).
Then the time on grass is \(8t\).
So:
\(t+8t=21\)
\(9t=21\)
\(t=\frac{21}{9}=\frac{7}{3}\)
Now find the packed dirt distance:
\(201\cdot \frac{7}{3}\)
Since \(201\div 3=67\), this becomes:
\(67\cdot 7=469\)
So the packed dirt distance is 469.
The grass distance is 4 times that:
\(4\cdot 469=1876\)
So the grass distance is 1,876.
Answer: Packed dirt = 469, Grass = 1,876
We only care about the packed dirt and grass segments.
On packed dirt, the athlete runs at 201 meters per minute.
On grass, she runs at 100.5 meters per minute.
The grass distance is 4 times the packed dirt distance, and the total time on these two terrains is 21 minutes.
The key shortcut is this: the grass speed is exactly half the packed dirt speed. So if the grass distance is 4 times as large, then the time on grass must be 8 times the time on packed dirt.
Step-by-step
Let the time on packed dirt be \(t\).
Then the time on grass is \(8t\).
So:
\(t+8t=21\)
\(9t=21\)
\(t=\frac{21}{9}=\frac{7}{3}\)
Now find the packed dirt distance:
\(201\cdot \frac{7}{3}\)
Since \(201\div 3=67\), this becomes:
\(67\cdot 7=469\)
So the packed dirt distance is 469.
The grass distance is 4 times that:
\(4\cdot 469=1876\)
So the grass distance is 1,876.
Answer: Packed dirt = 469, Grass = 1,876
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