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CAT 2019 Slot 1 DILR · Set 3 of 8Logical ReasoningModerate DifficultyQ9–Q12 · 2 MCQ + 2 TITA

CAT 2019 Slot 1 DILR Set 3: Math Puzzles (4 Questions) with Solutions

The Math Puzzles set from the official CAT 2019 Slot 1 DILR paper: the caselet, all 4 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2019 Slot 1 DILR · Questions 9–12
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.

In the exam this set competed with 7 others for 60 minutes. Budget about 8 minutes if you attempt it.

Q9. Math Puzzles

TITA · +3 / 0
What is the minimum possible number of different types of prizes?
Correct Answer
2
Step-by-step solution
📌 Core Concept
The counts must add up to 100, and each type needs at least twice as many items as the type before it. Fewer types means a larger count for each, so check whether 2 types can work.
🔢 Step-by-Step Solution
1
Type a has exactly one item, so one type alone gives only 1 box. This cannot fill 100 boxes.
2
Try two types. Type a has 1 item, so type b has the remaining 100−1=99100 - 1 = 99 items.
3
The doubling rule needs b≥2×1=2b \ge 2 \times 1 = 2, and 99≥299 \ge 2 holds. So two types are enough.
4
One type is impossible and two types are possible, so the minimum is 2.
Quick Check
1+99=1001 + 99 = 100 and 99≥2×199 \ge 2 \times 1, so the arrangement is valid.
Correct Answer: 2

Q10. Math Puzzles

TITA · +3 / 0
What is the maximum possible number of different types of prizes?
Correct Answer
6
Step-by-step solution
Clue Breakdown:
100 boxes with prizes of types a, b, c, etc.
Type 'a' has exactly 1 item, the most valuable.
Each subsequent type has at least twice as many items as the previous type.
Step-by-Step Deduction Grid:
TypeMinimum Items RequiredCumulative Items
a11
b23
c47
d815
e1631
f3263
g64127 (Exceeds 100)
Question Resolution:
To maximize the number of types, each subsequent type must have the minimum required items (exactly double). Adding type 'g' exceeds 100 boxes, so it's not possible.
Final Answer Confirmation:
The maximum number of different types is 6.
Correct Answer: 6

Q11. Math Puzzles

MCQ · +3 / −1
Which of the following is not possible?
  1. A) There are exactly 60 items of type d.
  2. B) There are exactly 30 items of type b
  3. C) There are exactly 45 items of typе с
  4. D) There are exactly 75 items of type e
Correct Answer
C) There are exactly 45 items of typе с
Step-by-step solution
📌 Core Concept
Write the counts as a=1a = 1, b≥2ab \ge 2a, c≥2bc \ge 2b, d≥2cd \ge 2c, e≥2de \ge 2d, with total 100. For each option, check whether the remaining boxes can be filled without breaking the doubling rule.
🔢 Step-by-Step Solution
1
Option C: take c=45c = 45. Then b≤22b \le 22 and a=1a = 1, so a+b+c≤68a + b + c \le 68, which leaves at least 32 boxes for later types.
2
Those boxes must hold a type d, and the rule needs d≥2c=90d \ge 2c = 90. That makes the total more than 100.
3
The other way is to make c the last type. Then b=100−1−45=54b = 100 - 1 - 45 = 54, but c≥2b=108c \ge 2b = 108 fails.
4
So exactly 45 items of type c is impossible.
5
The other options can all be built:
A: 1+13+26+60=1001 + 13 + 26 + 60 = 100.
B: 1+30+69=1001 + 30 + 69 = 100, and 69≥6069 \ge 60.
D: 1+2+4+18+75=1001 + 2 + 4 + 18 + 75 = 100, with 18≥818 \ge 8 and 75≥3675 \ge 36.
Why Not the Others
A: d=60d = 60 works with a,b,c=1,13,26a, b, c = 1, 13, 26.
B: b=30b = 30 works with c=69c = 69.
D: e=75e = 75 works with 1,2,4,181, 2, 4, 18 before it.
Correct Answer: C

Q12. Math Puzzles

MCQ · +3 / −1
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?
  1. A) 6
  2. B) 4
  3. C) 3
  4. D) 5
Correct Answer
D) 5
Step-by-step solution
📌 Core Concept
The clue gives the exact size of the type in box 45. That fixes the last type, and the remaining boxes are shared by the earlier types, which must satisfy the doubling rule.
🔢 Step-by-Step Solution
1
Box 45's type has 31+1+43=7531 + 1 + 43 = 75 items.
2
Any type with more items than this would need at least 2×75=1502 \times 75 = 150 items, which is impossible. So box 45's type is the last (rarest-valued) type, with 75 items.
3
The other types share 100−75=25100 - 75 = 25 items, and type a has exactly 1.
4
To get the most types, make the earlier counts as small as the rule allows: 1,2,4,8,16,…1, 2, 4, 8, 16, \dots
5
Five types before the last one need at least 1+2+4+8+16=31>251 + 2 + 4 + 8 + 16 = 31 > 25, which is too many.
6
Four types before the last one need at least 1+2+4+8=15≤251 + 2 + 4 + 8 = 15 \le 25, so this fits. Raise the largest count to use all 25 boxes: 1,2,4,181, 2, 4, 18. Then 18≥818 \ge 8 and 75≥3675 \ge 36 both hold.
7
The total is 4 earlier types plus the last type, so 5 types.
Why Not the Others
6: needs at least 31 items in the earlier types, but only 25 are available.
4 and 3: possible, but not the maximum, since 5 types work.
Correct Answer: D
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CAT 2019 Slot 1 DILR at a Glance

Sets in this section8
Questions in this section32
Section time60 minutes
Convenor IIMIIM Kozhikode

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