# math

posted by
**lea** on
.

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