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Mathematics 7 Online
OpenStudy (lollylau):

(Very Long Question) A takes his dog for a walk. B is 1 kilometre away. They are walking towards each other at 3km/hr. The dog sees B and runs to B at 9km/hr. When he reaches B, it immediatly turns back to A, and runs back to B, etc. A and B are still walking at 3km/hr. When A and B are 1/5 km apart, the dog sees a kitty cat and goes to try and catch it at 12km/hr. When A and B meet, how many km has the dog ran?

OpenStudy (phi):

The trick to this question is to ignore the back and forth of the dog, and just figure out how long the dog is running (we assume he changes speed instantaneously!) If A and B both walk at 3 km/hr, they are "closing" at 6 km/hr. Using rate*time= distance 6*t= 4/5 km, or t= 4/30 during which time the dog has run d1= 4/30*9= 36/30 km A and B take an additional (1/5) km / 6 km/hr = 1/30 hr to meet. During this time, the dog travels an additional d2= 12*1/30= 12/30 km total distance is (12+36)/30 = 48/30 = 16/10 = 1.6 km

OpenStudy (lollylau):

thank you very much!!!!

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