# Carlos and his Bike

• Sep 9th 2009, 06:34 PM
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!(Thinking)
• Sep 9th 2009, 09:01 PM
Keilan
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.
• Sep 10th 2009, 05:16 AM