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A new game show on TV has 100 boxes numbered 1, 2,..., 100 in a row, each containing a mystery prize. The prizes are items of different types, a, b, c, ..., in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes.

What is the maximum possible number of different types of prizes?

Entered answer:

Solution

✅ Correct Answer: 6
Slide 1/2

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 = 2+4+8+16+32+64=1272 + 4 +8+16+32+ 64 = 127

Which, is greater then the number of boxes i.e., 100

Therefore, item g can't there, and item f will be 100−1−2−4−8−16=69100 - 1- 2-4-8-16=69

=> maximum there can be 6 items.

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