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secondary 4 | E Maths
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Part (b)
Thank you☺️
Date Posted: 7 years ago
Views: 502
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This is the answer for part a.
Date Posted:
7 years ago
Thank you, but I still don’t know about part (b) . Hope you can help :D Thanks a lot!!
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Change their speed into the number of seconds that each car will take to go 1 round back the the starting point. This means 300 / 4 = 75 seconds. The slowest car will take 75 seconds to go 1 round. Do the same for all 3 cars.
After that do LCM for these numbers. You will find that when it is 300 seconds, all 3 cars will be back at the starting point together.
After that do LCM for these numbers. You will find that when it is 300 seconds, all 3 cars will be back at the starting point together.
Date Posted:
7 years ago
But in 300 seconds, the fastest car would have travelled 300 x 15m = 4500 m. It has exceed the 3300 m distance. This means the 3 cars didn't meet at the starting point again.
Ok :D
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