# Thread: Carlos and his Bike

1. ## Carlos and his Bike

Please help with this story problem, I'm still recovering from summer math-deprivation, and I need a refresher!

Carlos cycles for 2 hours at a fast speed, then cycles for three more hours at a speed 10 kph slower. If the entire distance traveled is 90 km, find Carlos' speed for the first two hours of travel.

Thanks!

2. Alright, here's what we know.

He travels 2 hours at a fast speed, lets call that speed f km/h
He then travels 3 hours at (f-10) km/h
In these 5 hours, he travels 90km

So we have the formula velocity = distance/time. If you don't know where that formula comes from, let me know and I'll explain.

So we have:
$\displaystyle v = \frac{d}{t}$
d and t are given to us as 90 and 5 respectively
v is going to be his average speed, so we take 2 times the fast speed plus 3 times the slower speed and divide it by 5. This gives us the average speed he travelled over the 5 hours.
$\displaystyle \frac{2f + 3(f-10)}{5} = \frac{90}{5}$
$\displaystyle 2f + 3f -30 = 90$ by cancelling the 5's in the denominator and expanding $\displaystyle 3(f-10)$
$\displaystyle 5f = 120$
$\displaystyle f = 24 km/h$

So his faster speed is 24 km/h. Let me know if this solution did not make sense to you at any point.