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Two cyclists leave towns 295 kilometers apart at the same time and travel toward each other. One cyclist travels 3 km/h slower than the other. If they meet in 5 hours, what is the rate of each cyclist? Please help ASAP GIVING 50 POINTS!!!!!!

Respuesta :

ncsftw

Answer:

28 km/h and 31 km/h

Step-by-step explanation:

let s be the speed of both cyclists.

let A be the speed of the first one, and B be the speed of the second one.

we know that A = B - 3 (he is going 3 km slower)

They close 295 km in only 5 hours, which means that their speed in km/h was 295/5 = 59km/h

thus, A + B = 59

plug in for A -> B - 3 + B = 59

2B = 62

B = 31 km/h

A = 28 km/h

mhanifa

Answer:

  • 28 km/h, 31 km/h

Step-by-step explanation:

Given:

  • Distance 295 km
  • Time 5 hours
  • Speed difference 3 km/h
  • Let the speeds of the lower cyclist is x

Then we have below distance equation:

  • (x + x + 3)*5 = 295
  • 2x + 3 = 295/5
  • 2x + 3 = 59
  • 2x = 56
  • x = 28

The speed of the faster cyclist is:

  • 28 + 3 = 31 km/h

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