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Anyone know a formula for rate problems when its like

John walked for 5 hours to work at 3km/hr there and 2km/hr back how far was the distance from home to work.

By Emmittsmith (Emmittsmith) on Friday, October 08, 2004 - 07:16 pm: Edit

I don't use formulas for these problems -- I use the units as hints.

So he walked for 5 hours at 3 km/hr.
that's 3 km/hr * 5 hr = 15 km

You see, the hr/hr cancel out, leaving km. If you had a distance and were looking for a time, the obvious thing to do would be to divide the rate into it, so that you'll end up with hr in the numerator.

Now, it appears that the 2 km/hr is extraneous because you already have the distance to there. Unless I am misreading something or you mistyped something. In any case, 5 hours is an ordinate amount of time to spend walking to work.

Hope this helps.

By Neelesh (Neelesh) on Friday, October 08, 2004 - 07:49 pm: Edit

first, i believe emmittsmith is incorect

use the formula d=rt, or distance = rate * time

you need to do this twice.

first, for his way to work, its d=3*x, where x= the time in hours to walk to work

then for his way back, its d=2*(5-x), because tis 5 hours minus the time it took him to walk to work. and because the distance is the same. you set the 2 equations equal to each other, so

3x=2(5-x)

3x=10-2x

x=10 km

By Swizzbeatz905 (Swizzbeatz905) on Friday, October 08, 2004 - 11:02 pm: Edit

thanks alot i knew physics was the answer for this.

By Thermodude (Thermodude) on Saturday, October 09, 2004 - 12:36 am: Edit

Going back on what Neelesh said,

distance = 3 * x

as well as distance = 2 * (5 - x)

However, note that x is in time, or hours.

3x = 2(5 - x)
3x = 10 - 2x
5x = 10
x = 2 hours

Plug 2 hours back into d = 3 * x...and you get 6 km.

This also works if you plug 2 into the other equation:

d = 2 * (5 - x)...which'll also get you 6 km.