if R=V+I and V=(100+_5)and I= (10+_2). what is the percentage error ​

Dear Student,
Here in this question voltage V = $\left(100\pm 5\right)V$ and current , I = $\left(10\pm 2\right) A$
Resistance R = $\frac{V}{I}=\frac{100}{10}=10 \Omega$
Now , $\frac{∆R}{R}=\pm \left(\frac{∆V}{V}+\frac{∆I}{I}\right)=\pm \left(\frac{5}{100}+\frac{2}{10}\right)=\pm 0·25$
Therefore , $∆R =\pm 0·25\times R=\pm 0·25\times 10=\pm 2·5\Omega$
Resistance with error limit , R =$\left(10\pm 2·5\right)\Omega$
Percentage error in R = $\frac{∆R}{R}\times 100\% =\frac{\pm 2·5}{10}\times 100 =\pm 25\%$
Regards