9514 1404 393
Answer:
165 km
Explanation:
Gilly's speed at 15 kph is 5 kph more than Snap's speed of 10 kph. That is, Gilly moves away from Snap at 5 kph.
Moving toward Snap, Gilly's closure speed is 10 kph +20 kph = 30 kph.
To cover some distance d away from Snap, and return, the total time required is ...
time = distance/speed
t_away = d/5
t_toward = d/30
The total distance Gilly travels is the product of speed and time.
t_away = (15)(d/5) = 3d
t_toward = (20)(d/30) = 2/3d
Then Gilly's average speed for the round trip is ...
speed = distance/time = (3d +2/3d)/(d/5 +d/30) = (11/3)/(7/30) = 110/7 . . . kph
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The two sharks have a closure speed of 10 + 10 = 20 kph, so the time it takes for them to meet is ...
time = distance/speed = (210 km)/(20 km/h) = 10.5 h
In that time, Gilly travels ...
distance = speed ยท time = (110/7 km/h)(10.5 h) = 165 km
The total distance Gilly swims before the seals meet is 165 km.
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Additional comment
There are several ways this problem can be worked. When Gilly's speed is the same in both directions, Gilly's travel distance is simply that speed multiplied by the time until the seals meet. Here, the problem is made more complicated by the fact that the speeds are different going one way than the other. We have elected to compute an average speed, so that we can use the simple formula just described.
Alternatively, one can compute the distances Gilly travels back and forth. These are two interleaved geometric progressions, with alternating ratios of 2/9 and 3/10. The distances Gilly travels between meetings are ...
126, 28, 8.4, 28/15, 0.56, ... km
This sequence has a sum of 165 km that can be found by considering alternate terms. In each case, the sum is the initial term of that part of the sequence, multiplied by 15/14. Then the overall sum is (126 +28)(15/14) = 165.