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