Percentage Decrease
Percentage decrease measures how much a quantity has fallen relative to its original value, expressed as a percent.
Formula
\text{percentage decrease} = \left(\frac{\text{original} - \text{new}}{\text{original}}\right) \times 100\%
Definition
Percentage decrease tells you how much something went down compared to where it started: find the amount it decreased, divide by the original amount, and multiply by $100$. Formally, percentage decrease $= ((\text{original} - \text{new})/\text{original}) \times 100\%$, or new value $= \text{original} \times (1 - p/100)$; a $100\%$ decrease means the value reaches zero, and a decrease greater than $100\%$ is not defined in most real-world settings. For exponential decay $f(t) = P e^{-rt}$, the percentage decrease over interval $T$ is $(1 - e^{-rT}) \times 100$, and the half-life $T_{1/2} = \ln(2)/r$, the time for a $50\%$ decrease, is fundamental in radioactive decay and pharmacokinetics.
Example
A jacket that was $\$80$ and is now $\$60$ decreased by $\$20$, and $20/80 \times 100 = 25\%$. A car depreciating from $\$24{,}000$ to $\$18{,}000$ is also a $25\%$ decrease, confirmed by the multiplier $\$24{,}000 \times 0.75 = \$18{,}000$; two successive $20\%$ decreases give $0.80 \times 0.80 = 0.64$, a total $36\%$ decrease, not $40\%$. Carbon-14 has a half-life of $5730$ years, so after $11{,}460$ years (two half-lives) the amount has decreased by $75\%$, leaving only $25\%$ remaining.
Key Insight
Just like percentage increase, percentage decrease is always compared to the original (starting) value: losing $\$20$ from $\$80$ is a $25\%$ decrease, but the same $\$20$ loss from $\$200$ would only be a $10\%$ decrease. Percentage decrease and increase are not symmetrical either: a $50\%$ decrease followed by a $50\%$ increase does not return to the original, $\$100 \times 0.5 = \$50$, then $\$50 \times 1.5 = \$75$, since the increase is applied to the smaller value. This asymmetry reflects the multiplicative structure of ratios, a $p\%$ decrease followed by a $p\%$ increase yields a factor of $(1-p/100)(1+p/100) = 1 - (p/100)^2 < 1$, related to the AM-GM inequality, with the surprising consequence that any sequence of percent ups and downs including even one round trip results in a net loss.