Solution
Understanding the set :-
This is a minima maxima + quantitative set :
Here you have 100 boxes and a mystery prices in every box, named as a, b, c...
Also, it is given that, there is price of item a is highest , then prices b, then price c, and so on...
Price a - a diamond ring is only in one of the box,
and other prices are atleast double in quantity of previous item i.e.,
we know item a is only 1 , then item b will be atleast two and item c will again be atleast double of item b.
Now, lets try to solve the set with the given information -
Now, we want to maximize the number of items :-
item a - will be only 1.
item b - we know, is atleast double of item a -
therefore, to maximize the number of items we will minimize the number of items b - so b will be 2
item c - minimum of item c = atleast double of item b = 4
item d - minimum of item d = atleast double of item c = 8
item e - minimum of item e = atleast double of item d = 16
item f - minimum of item f = atleast double of item e = 32
item g - minimum of item g = atleast double of item f = 64
Now, let's find the total =
Which, is greater then the number of boxes i.e., 100
Therefore, item g can't there, and item f will be
=> maximum there can be 6 items.
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Question 11