# What does “X% faster” mean?

I was reading something today that was talking in terms of 10%, 100% and 1000% faster. I assumed that 10% faster means it takes 10% less time (60 seconds down to 54 seconds).

If that is correct wouldn't 100% faster mean 0 time and 1000% mean traveling back in time?

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You should leave a comment on the site to this effect. –  user02138 Mar 15 '11 at 15:42
I think he actually explains those numbers in the post. He is using "100% faster" instead of "twice as fast" and "1000% faster" instead of "ten times as fast". –  Brian Mar 15 '11 at 15:52
So if you are talking about execution time (like he is in the article), 1000% would be 1/10 of the original execution time? Would 100% faster be the same execution time? –  Abe Miessler Mar 15 '11 at 19:10

I second Peláez's answer. In addition, I want to explain this mistake of yours: “I assumed that 10% faster means it takes 10% less time (60 seconds down to 54 seconds).” 10% normalized = 0.1 , oldTime=60 , and the correct result is not 54:

$newTime = oldTime/(1+0.1) = 54.54545454545454\dots$ $\neq 54 = oldTime\cdot(1-0.1)$.

If $X\to 0$, then the difference between formulas $\to 0$, so in calculations $\cdot(1-X)$ is often used.

But the following statement is correct and precise: 10% faster means that something moves 10% further.

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As beroal suggests, the difference is between $1+x$ and $\frac{1}{1-x}$, which are close for $x\lt\lt 1$ but diverge as $x$ gets close to $1$. –  Ross Millikan Mar 16 '11 at 2:36
Faster here is referring to the speed, it means that if you normalize $X$ as a number between $0$ and $1$, you will have that $$newSpeed = oldSpeed(1+X)$$ But faster doesn't mean that the time taken for the activity (this case is data processing or something like that) to complete is 0, or even negative, because the time is always some ratio of the form $t = \frac{d}{s}$, where $d$ would be the number of activities or data to process, an $s$ is the speed to complete one activity or process one data, so $t$ is always positive. But of course as X increases, the time tends to 0.