math
posted by lea .
it took cindy 2h to bike from abbott to benson at a constant speed. the return trip took only 1.5h because she increased her speed by 6km/h. how far apart are abbott and benson?
distance = rate*time
a to b. d= r*2 hrs
b to a d=(r+6 km/hr)*1.5 hrs.
Set distances equal.
r*2=(r+6)*1.5 and solve for r which is the rate to travel in the 2 hours.
Distance = 2 hours x rate = ??
OR distance = 1.5 hours x (rate + 6 km/hr). I found r = 18 km/hr
so 18 km/hr x 2 hrs = 36 km.
OR (18+6) km/hr x 1.5 hours = 36 km.
Let L be the unknown separation distance. Let V be the speed for the frist trip.
L = (2 hours) * V
L = (1.5 hours) * (V + 6)
You have two equations in two unknowns and can sor both V and L.
2V = 1.5 V + 9
0.5 V = 18 V = 36 km/hr
Use that V to solve for L.
I made a mistake in the last equation of my previous answer. V = 18 km/hr
L = 18 x 2 = 36 km
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