Calculating Percentage Error in the Ratio when there is an Error in Numerator or Denominator or both ? I have serached google a bit, but Since Google searches for keywords and also I find it hard to put in a statement for which It can give me useful searches, I am asking it here. So, This is Question of Error Calculation in Ratios
So, Let's say we have to calculate some ratio for precise calculations and terms being involved in the ratio be x and y. Accordingly, $$ Ratio = \frac{x}{y}$$
While calculating , some error was noticed, that can be expressed as $ \Delta x$ and $\Delta y$ respectively for error in measuring Numerator and Denominator values.
Now we need to find the error change in ratio and % Error Change in Ratio. So, the problem can be broken down into these subproblems :


*

*When Only Numerator changes by $\Delta x $ that is $ Ratio = \frac {x+\Delta x}{y}$

*When Only Denominator changes by $\Delta y$ that is $ Ratio = \frac {x}{y+\Delta y}$

*When both Numerator & Denominator changes simultaneously by $\Delta x$ and $\Delta y$ respectively that is $ Ratio = \frac {x+\Delta x}{y+\Delta y}$


Consider both increase and decrease scenario for the calculation. 
 A: Let $r=\frac xy$. Then, $$dr = \frac{ydx-xdy}{y^2}$$
This is the error change in $r$, given the values $x,y$ and the error changes (with +ve sign for increase and -ve sign for decrease) in both of them.
The maximum fractional error is simply:
$$\frac{dr}{r}=\frac{\frac{ydx-xdy}{y^2}}{\frac xy}=\frac{ydx-xdy}{xy}$$
Can you proceed now?

EDIT: For case 1, denominator does not change. So in that case $dy=0$. Similarly, in case 2, numerator does not change. So in that case $dx=0$.
Example problem: if $x=5\pm0.05$ and $y=6$, what would you expect $dr$ and $dr/r$ to be?
Here, $x = 5, y = 6$. Also, $y$ is exactly known, so $dy=0$ but $dx=0.05\neq0$. Now you can put and solve.
A: Gaurang's answer above says that for a fixed error dx in x, you can improve your uncertainty on r by increasing your uncertainty in y. That can't be correct.
The taylor expansion is correct, but then you have to add the variances, and the variance of (-xdy) = +x^2\sigma_y. This yields the well known sum of quadratures formula:
(dr/r)^2 = (dx/x)^2 + (dy/y)^2,
Same as above but definitely no minus!
