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 $100 = 20\text{ pencils}$ 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 : $1-6 ( 5, 6, 6, 9, 9, 9)$ $7 - 16 ( 7, 7, 7, 7, 11, 15, 15, 19, 20, 20 )$ $17 - 20 (20, 29, 31, 33)$