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FN
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Should be 'E'.

No clue about the exact number of books.
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OA is C..

here is why...

suppose 100 students total..

25 bought 8 books or less 35 bought 10 books or more..

list em out suppose the 25% all bought 8 books each..8888888888888 suppose the 35% all bought 10 books..1010101010101010

25+35=60

so we know 60% have 8 books and 10 books..the median has to be those who bought 9..
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key is 10 or more books..so think of 35% on the upper end of the list..

25% bought 8 or less books..so they are at the lower end of the list..

only those who bought 9 books are in the median..
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[quote="fresinha12"]key is 10 or more books..so think of 35% on the upper end of the list..

25% bought 8 or less books..so they are at the lower end of the list..

only those who bought 9 books are in the median..[/quote]

To satisfy both of the statements, the distribution will be like this:

--------25%--------|--------40%-------|--------65%-------|
Not more than 8.....|....should be 9.......|........Atleast 10......|
(8 is the highest)

hence median is 9.

OA is justified.
Very good question. +1
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fresinha12
key is 10 or more books..so think of 35% on the upper end of the list..

25% bought 8 or less books..so they are at the lower end of the list..

only those who bought 9 books are in the median..

To satisfy both of the statements, the distribution will be like this:

--------25%--------|--------40%-------|--------65%-------|
Not more than 8.....|....should be 9.......|........Atleast 10......|
(8 is the highest)

hence median is 9.

OA is justified.
Very good question. +1

Good justification!



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