CAT 2019 Slot 1 DILR Question & Solution

Logical ReasoningMedium

Data Set

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.

Question 1

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

Solution:

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the item b should be at least twice. The minimum number of items will be obtained when a=1 and b=99, which means there are only two different types of items.

Question 2

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

Solution:

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the number of items of type b should be at least twice of that of a and the number of items of type c should be at least twice of that of b and so on. So the maximum number of different types of items of a, b and c will be obtained when a=1, b=2, c=4, d=8, e=16, f=69. Hence the maximum number of different types of items will be 6.

If the number of items is 7, then the minimum number of prizes should be 1+2+4+8+16+32+64=127 which is more than 100.

Hence 6 is the answer.

Question 3

Which of the following is not possible?

There are exactly 75 items of type e.

There are exactly 30 items of type b.

There are exactly 45 items of type c.

There are exactly 60 items of type d.

Solution:

Option A: There are exactly 75 items of type e.

a=1,b=2,c=4,d=8, e=85. Here the maximum value of e= 85. Hence it can take the value 75.

An example of such case is a=1,b=2,c=4,d=18, e=75

Option B: There are exactly 30 items of type b.

a=1 b=30 and c=69. Hence this case is also possible.

Option C: There are exactly 45 items of type c.

Since the value of d should be at least 90, it means that d is not present because 45+90 will be more than 100(maximum number of items). Only a,b and c are present.

The maximum value of b = 22 and a =1, but 45+22+1=68, which is less than 100. So this case is not possible.

Option D: There are exactly 60 items of type d.

d=60, c=30, b=9 and a=1. a+b+c+d=100. Hence this case is possible.

C is the answer.

Question 4

You ask for the type of item in box 45. Instead of being given a direct answer, you are told that there are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100.
What is the maximum possible number of different types of items?

Solution:

The total number of items from 1 to 100, which are of same type as in box 45 = 31+1+43=75

Now to maximize the number of items, a=1, b=2, c=4, d=18 and e=75(given)

There can be maximum 5 types of items.

If we consider number of items to be 6, then minimum number of items of 5th type will be 16, 1+2+4+8+16+75=106 which is more than 100.