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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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    If it is referring to speed, the statement "10% faster" makes sense. If it is referring to completion time, it does not. However, advertisers like to misuse percentages in various ways. For another example, the new toilets where I work have a sign that says they "save 30% more water per flush than a regular toilet".2017-02-23

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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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When I see numbers like that I don't believe they are honest numbers-they really mean "faster", "much faster" and "much much faster". But taking them seriously, if you were traveling 60 mph, 10% faster would be 66 mph (taking 9.09% less time), 100% faster would be 120 mph (taking 50% less time) and 1000% faster would be 660 mph (taking 91.91% less time).

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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 $1$.2011-03-16
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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.