Xavier Aptitude Test 2022 Paper

Boxes 7 to 16 contain a total of 100 pencils each. The minimum number of broken pencils the box can hold is 5% of the total pencils and a maximum of 20% of the total pencils.
5 percent of $100=5$ and $20\%$ of $100=20\mathrm{pencils}$
Hence the number of broken pencils must be in the range of 5 to 20.
The frequency of the different number of broken pencils is :
$5-1$
$6-2$
$7-4$
$9-3$
$11-1$
$15-2$
$19-1$
$20-3$
$29-1$
$31-1$
$33-1$
The boxes cannot contain $29,31,33$ to be the number of broken pencils because they are beyond 20 percent.
Since boxes $1-6$ can contain only between $2.5$ to 10 pencils. The remaining boxes which include broken pencils of numbers less than 10 must be a part of $7-16$. Because boxes $17-20$ cannot contain broken pencils of numbers less than 10.
Hence $7-16$ must have 4 boxes that contain less than 10 broken pencils.
Going through the options :
Option A fails because this includes only 3 boxes with less than 10 pencils.
Option B fails because we only have 2 boxes with 6 broken pencils but this includes 3.
Option D fails because it does not include a box of 15 and a box of 20 pencils which can only be a part of boxes with 100 or boxes with 200 pencils. Since boxes $17-20$ can include only one among $15$ or $20$ because $29,31,33$ are a part of this group. Hence this case fails.
Option E fails because this includes 5 boxes with broken pencils less than 10 but this is not possible because this must exactly contain 4 boxes with less than 10 pencils.
Option C is a feasible case containing :
$1-6\left(5,6,6,9,9,9\right)$
$7-16\left(7,7,7,7,11,15,15,19,20,20\right)$
$17-20\left(20,29,31,33\right)$