I cannot find a specific definition for "within $x$% of $y$". If I want a number within $10$% of $100$, am I looking for numbers in the set $[90,110]$? Or does it mean percentage difference? Or percentage change?

  • $\begingroup$ The set $[90,110]$ is the correct range for interpreting this quantity. $\endgroup$
    – abiessu
    Apr 4, 2016 at 23:03

3 Answers 3


To say that a number $u$ is "within $x \%$ of $y$" means that $|u - y|/|y| \leq x/100$ (assuming $y \neq 0$).

Equivalently, $u \in [ y - \frac{x}{100}|y|, y + \frac{x}{100}|y|]$.

  • 1
    $\begingroup$ This is a lot of formalism for someone probably studying algebra. $\endgroup$ Apr 4, 2016 at 23:08
  • 3
    $\begingroup$ @AlfredYerger Well, OP's history has some questions about differential equations and linear algebra. $\endgroup$
    – littleO
    Apr 4, 2016 at 23:13

Percentage difference and percentage change sound the same to me - maybe there's something I'm just missing.

To compute the upper and lower bounds, you figure out what $x%$ of $y$ is using a proportion, and then add and subtract that number from $y$. In your example, $10%$ of $100$ is $10$, so as observed in the comments, we add and subtract $10$ from $100$.

Subtracting $10$ gives $90$, and adding it gives $110$, so you are within $10%$ of $100$ if you are at least $90$ and less than $110$. In interval notation, you are in the interval $[90, 110]$.


If I want a value, let's call it $a$, that lies within $x$% of $y$,$^\dagger$ then I want

$$y\left(1-\frac{x}{100}\right) \le a \le y\left(1+\frac{x}{100}\right)$$

Following your (correct) example, if I want a value $a$ that lies within $10$% of $100$, then $$90 \le a \le 110$$

In plain(er) English, if you see this statement in the context of a measurement, it means that our measurement's accuracy is bounded by some percentage of the desired value. This is to say we may have some uncertainty within the measurement. In another context, we might feel comfortable with such uncertainty. Then this range indicates our accepted tolerance compared to some ideal value.

$^\dagger$Note: I'm making the assumption that $y>0$. littleO posted a more complete answer.


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