Xavier Aptitude Test 2022 Paper

Going by the options :
Option A: Boxes $7-16$ contain a total of 124 pencils. Boxes $\left(1-6\right)$ has 6 boxes with broken pencils which can be included from :$\left(5,6,6,7,7,7,7,9,9,9\right)$. The minimum possible sum of the 6 pencils is : $\left(5+6+6+7\right)=24$ and the maximum possible sum is $\left(7+9+9+9\right)=34$.
Boxes $7-16$ contains all the boxes with broken pencils except one among the boxes with broken pencils among $11-20$ and hence ;This can contain : $\left(11+15+15+19+20+20\right)$ or $\left(15+15+19+20+20+20\right)$ or $\left(11+15+15+20+20+20\right)$ or $\left(11+15+19+20+20+20\right)=\frac{\frac{\frac{100}{109}}{101}}{105}$
The only possible case to contain 124 pencils is by considering the case : $\left(24+100\right)=\left(5,6,6,7,11+15+15+19+20+20\right)=124$ .
Hence box $17-20$ must contain $\left(20,29,31,33\right)$.
Option B: Boxes $17-20$ contain a total of 108 pencils. Since $29,31,33$ pencils must be a part of $17-20$ boxes. The remaining box must contain $108-\left(29+31+33\right)=15\mathrm{pencils}$. Hence the order is $\left(15,29,31,33\right)$
Option C: Boxes $11-16$ contain a total of $101$ defective pencils. This is only possible if the boxes here contain : $\left(20,20,20,15,15,11\right)$pencils. Hence the box containing 19 pencils must be a part of boxes $17-20$ and the remaining three contain $29,31,33.\left(19,29,31,33\right)$
Option D : Box number 17 containing more pencils than any box from box number $1-14$. Hence this only possible if Box $15,16,17$contains 20 pencils each and $18,19,20$ contain $\left(29,31,33\right)$.
Option E : Box $7-16$ containing 133 broken pencils :
Boxes $\left(1-6\right)$ has 6 boxes with broken pencils which can be included from :
$\left(5,6,6,7,7,7,7,9,9,9\right)$. The minimum possible sum of the 6 pencils is : $\left(5+6+6+7\right)=24$ and the maximum possible sum is $\left(7+9+9+9\right)=34.$
Boxes $7-16$ contains all the boxes with broken pencils except one among the boxes with broken pencils among $11-20$ and hence ;
This can contain : $\left(11+15+15+19+20+20\right)$ or $\left(15+15+19+20+20+20\right)$ or $\left(11+15+15+20+20+20\right)$ or $\left(11+15+19+20+20+20\right)=\frac{\frac{\frac{100}{109}}{101}}{105}$
This has multiple possibilities which include $\left(109+24\right)$ or $\left(101+32\right)$ or $\left(105+28\right)$. Hence cannot be uniquely determined