Panter
To answer your simple question simply, yes. Well, more or less.
If you're ignoring friction, air resistance and all other losses, then the only factors to consider are mass (m), gravity (g), height of hill (h) and time (t), which are related to power (P) thus:
P = mgh/t
Transposing gives
t = mgh/P
Since the height of the hill and gravity are both constant*, and assuming no change in power, then time is proportional to mass. Drop the mass by 10% and you'll drop the time by 10%. As jimbo pointed out, it's the total "vehicle" mass which matters, not just your mass.
So, is this simplistic view valid, i.e. do the losses we ingored really matter? Well, I'm afraid they do. If they didn't, then, when riding on the flat, it would take no effort to achieve any speed wished. Which is absurd.
On a well maintained bike, most of your energy is spent battling air resistance. Calculating this is rather trickier, especially as it depends upon your drag coefficient, which is hard to gauge without a wind tunnel!
Luckily, you can make a rough approxiation without this info. Assuming that you're wearing the same gear and on the same bike, then all you need to know is that air resistance is proportional to the square of the your speed.
That means that your 10% mass saving will only be worth about 5% in terms of speed. In fact, unless you start losing ridiculous amounts of weight, the %-age reduction in speed will be about half the %-age reduction in mass.
And disregard all of the above unless the wind is constant across all your rides.
All very approximate, but at least it's a direct answer to the question you actually asked.
Cheers
weevil
* For the pedants, yes, I know gravity varies minutely around the world, but the hill in question is undoubtedly static.
