# Cyclists

Cyclist who rides at an average speed 16 km/h travels trip distance 10 min before the cyclist who rides at an average speed 11 km/h.

What is the length of this cyclist trip(distance in km)?

Result

s =  5.87 km

#### Solution:

$16\cdot (t-\dfrac{ 10}{60})= 11\cdot t \ \\ t = 0.53 \ h \ \\ s = 11 \cdot t \ \\ s = 5.87 \ \text{km}$

16*(60t-10)=11*t*60
s = 11 t

300t = 160
s-11t = 0

s = 8815 ≈ 5.866667
t = 815 ≈ 0.533333

Calculated by our linear equations calculator.

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