Dilemma in calculation of percentage error while measuring focal length on an optical bench 
An optical bench has $1.5$ $m$ long scale having four equal divisions in each cm. While measuring
the focal length of a convex lens, the lens is kept at $75$ $cm$ mark of the scale and the object pin
is kept at $45$ $cm$ mark. The image of the object pin on the other side of the lens overlaps with
image pin that is kept at $135$ $cm$ mark. In this experiment, the percentage error in the measurement of the focal length of the lens is ____

This was asked in JEE Advance $2019$ Paper 2 , and this is the solution :
Least count = $0.25$ $cm$
${1\over v}+ {1\over u}={1\over f}$
(sign convention has already been applied , $u$ and $v$ here are magnitude of object distance and image distance)
hence ,
${{dv\over v^2}}+ {{du\over u^2}}={{df\over f^2}}$
Now here is the part I have objections with ,
$|du|=0.25+0.25=0.5$ $cm$
$|dv|=0.25+0.25=0.5$ $cm$
After this if we substitute these values of $|du|$ and $|dv|$ , and do further calculations we'll get the answer as $1.38 \%$
I understand that since we have to measure the values of $u$ and $v$ as $(x_2-x_1)$ , so a maximum error of 2 times the least count can occur , but the problem is that practically it can not be 2 times the least count for both $|u|$ and $|v|$ at the same time.
For clarification:
If the lens is kept at $75.25$ $cm$ mark and object pin is kept at $44.75$ $cm$ mark , then we do get an error of $+0.5$ $cm$ in value of $u$ but now when this has happened there's no way that we can simultaneously get an error of $+0.5$ $cm$ in value of $v$ (as image pin can be kept only in between $134.75$ $cm$ and $135.25$ $cm$ marks).
And if I calculate like this I get a maximum error of about $1.1 \%$
So I think that the above answer $1.38 \%$ is not physically possible and can never happen . But sadly the official answer given is also $1.38 \%$
Am I missing something?
Could someone please tell me where I have gone wrong, or is that the official answer is indeed wrong and the correct answer should be $1.1 \%$ only?
 A: You are observing that the errors in $u$ and $v$ are correlated, while the textbook solution appears to assume that they are uncorrelated.
The argument that independent uncertainties add in quadrature (that is, that $dz = \sqrt{dx^2 + dy^2}$) assumes that the individual uncertainties are independent of each other and normally distributed random errors.  If you have a more sophisticated understanding of your apparatus and its error distributions you can make more accurate estimates of how much scatter to expect in repeated measurements from your experiment, and you can test those estimates by modeling their distribution.  But this is in general a very complicated problem and we hide it from introductory-level students.
If you were planning on using such a setup many times, or to make a measurement which has never been made before, you might have statistics-related reasons to worry whether your measurement uncertainty is 1.1% or 1.4%.  But many successful physicists do not ever need to conduct such a second-order error analysis.
