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 = $0.25$ $cm$

${1over v}+ {1over u}={1over f}$
(sign convention has already been applied , $u$ and $v$ here are magnitude of object distance and image distance)

hence ,
${{dvover v^2}}+ {{duover u^2}}={{dfover f^2}}$

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?