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CAT 2018 Slot 1 DILR · Set 2 of 8Logical ReasoningModerate DifficultyQ5–Q8 · 2 MCQ + 2 TITA

CAT 2018 Slot 1 DILR Set 2: Math Puzzles (4 Questions) with Solutions

The Math Puzzles set from the official CAT 2018 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 2018 Slot 1 DILR · Questions 5–8
An ATM dispenses exactly Rs. 50005000 per withdrawal using 100100, 200200 and 500500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

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

Q5. Math Puzzles

TITA · +3 / 0
In how many different ways can the ATM serve a customer who gives 500 rupee notes as her preference?
Correct Answer
7
Step-by-step solution
📌 Core Concept
Set up the equation for the total amount and count the whole-number solutions that satisfy the preference condition.
🔢 Step-by-Step Solution
1
Let the numbers of Rs. 100, 200 and 500 notes be aa, bb and cc. Then 100a+200b+500c=5000100a+200b+500c=5000, which simplifies to a+2b+5c=50a+2b+5c=50.
2
The preference is for 500 notes, so c>a+bc>a+b.
3
Since a+b≤a+2b=50−5ca+b\le a+2b=50-5c, a large cc satisfies the condition easily, and a small cc cannot satisfy it. Check cc from the top.
4
c=10c=10: a+2b=0a+2b=0, so (a,b)=(0,0)(a,b)=(0,0). This gives 1 way.
5
c=9c=9: a+2b=5a+2b=5, so b=0,1,2b=0,1,2. Here a+b≤5<9a+b\le5<9, so all 3 work. This gives 3 ways.
6
c=8c=8: a+2b=10a+2b=10, and we need a+b<8a+b<8. Checking b=0b=0 to 55 gives a+b=10,9,8,7,6,5a+b=10,9,8,7,6,5. So b=3,4,5b=3,4,5 work. This gives 3 ways.
7
c=7c=7: a+2b=15a+2b=15, so a+b=15−ba+b=15-b. We need 15−b<715-b<7, which means b>8b>8, but b≤7b\le7. No solutions. Smaller cc is worse because the required a+ba+b grows while the bound cc falls.
8
Total: 1+3+3=71+3+3=7.
Quick Check
The valid triples (a,b,c)(a,b,c) are (0,0,10)(0,0,10), (5,0,9)(5,0,9), (3,1,9)(3,1,9), (1,2,9)(1,2,9), (4,3,8)(4,3,8), (2,4,8)(2,4,8) and (0,5,8)(0,5,8). That is 7.
Correct Answer: 7

Q6. Math Puzzles

TITA · +3 / 0
If the ATM could serve only 10 customers with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 10 who could have given 500 rupee notes as their preferences?
Correct Answer
6
Step-by-step solution
📌 Core Concept
Find the fewest 500 rupee notes a customer who prefers 500 must receive. Then see how many such customers fifty notes can cover.
🔢 Step-by-Step Solution
1
With a,b,ca,b,c as the numbers of 100, 200 and 500 notes, a+2b+5c=50a+2b+5c=50. A 500-preferring customer needs c>a+bc>a+b.
2
For c=10c=10, 99 and 88 the valid withdrawals exist, as in the previous question (for example (0,5,8)(0,5,8) is valid).
3
For c=7c=7 we have a+2b=15a+2b=15, so a+b=15−b≥8a+b=15-b\ge 8 because b≤7b\le7. This is not less than 7, so it fails. Smaller cc fails by the same reasoning, since a+ba+b must be at least 50−5c−b50-5c-b while b≤(50−5c)/2b\le(50-5c)/2.
4
So each customer who prefers 500 needs at least 8 notes of Rs.
500
5. With fifty 500 notes, the number of such customers is at most ⌊50/8⌋=6\lfloor 50/8\rfloor=6, since 7×8=56>507\times8=56>50.
6
The other customers prefer 100 or 200 and can be served with no 500 notes. For example, 100 preference with (a,b,c)=(50,0,0)(a,b,c)=(50,0,0) gives 50>050>0, which is valid.
7
Six customers use 6×8=48≤506\times8=48\le50 notes of Rs. 500, and the remaining 4 customers use none. So 6 is achievable.
Correct Answer: 6

Q7. Math Puzzles

MCQ · +3 / −1
What is the maximum number of customers that the ATM can serve with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?
  1. A) 12
  2. B) 10
  3. C) 13
  4. D) 16
Correct Answer
A) 12
Step-by-step solution
📌 Core Concept
With at most 20 notes per withdrawal, find the fewest 500 notes any valid withdrawal can contain. The stock of fifty 500 notes then caps the number of customers.
🔢 Step-by-Step Solution
1
Use a+2b+5c=50a+2b+5c=50. The total number of notes is n=a+b+c=50−b−4cn=a+b+c=50-b-4c.
2
The limit n≤20n\le20 gives b+4c≥30b+4c\ge30.
3
Every customer must use at least one 500 note if this is possible only with c≥1c\ge1. Check small cc against the upper bound b≤⌊(50−5c)/2⌋b\le\lfloor(50-5c)/2\rfloor:
c=0c=0: need b≥30b\ge30, but b≤25b\le25. Impossible.
c=1c=1: need b≥26b\ge26, but b≤22b\le22. Impossible.
c=2c=2: need b≥22b\ge22, but b≤20b\le20. Impossible.
c=3c=3: need b≥18b\ge18, but b≤17b\le17. Impossible.
c=4c=4: need b≥14b\ge14, and b≤15b\le15. Possible.
4
For c=4c=4, b=15b=15, a=0a=0 is a withdrawal of 19 notes with 200 as the preference (15>415>4). It is valid. Likewise (a,b,c)=(2,14,4)(a,b,c)=(2,14,4) has 20 notes and 14>614>6, which is also valid.
5
So every customer needs at least four 500 notes, and the maximum number of customers is ⌊50/4⌋=12\lfloor 50/4\rfloor=12.
Why Not the Others
10 is below what the stock allows.
13 and 16 would need fewer than four 500 notes per customer, which the 20-note limit rules out.
Correct Answer: A

Q8. Math Puzzles

MCQ · +3 / −1
What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?
  1. A) 900
  2. B) 800
  3. C) 750
  4. D) 1400
Correct Answer
A) 900
Step-by-step solution
📌 Core Concept
For each preference, find the withdrawal with the fewest notes. The 500-note count follows from those two withdrawals.
🔢 Step-by-Step Solution
1
With a,b,ca,b,c as the numbers of 100, 200 and 500 notes, a+2b+5c=50a+2b+5c=50 and the total is n=50−b−4cn=50-b-4c. To minimise nn, make b+4cb+4c as large as possible.
2
500 preference (c>a+bc>a+b): c=10c=10 gives (0,0,10)(0,0,10) with 10 notes. c=9c=9 gives at best (1,2,9)(1,2,9) with 12 notes, and c=8c=8 gives at best (0,5,8)(0,5,8) with 13 notes. The minimum is 10 notes, using ten 500 notes.
3
100 preference (a>b+ca>b+c): substituting a=50−2b−5ca=50-2b-5c gives 50−2b−5c>b+c50-2b-5c>b+c, so 3(b+2c)<503(b+2c)<50, so b+2c≤16b+2c\le16.
4
Maximise b+4cb+4c under b+2c≤16b+2c\le16. Each unit of cc costs 2 of the budget and gains 4, while each unit of bb costs 1 and gains 1. So cc is the better use, and c=8c=8, b=0b=0 gives b+4c=32b+4c=32.
5
Then a=50−40=10a=50-40=10, and 10>810>8 is valid. The withdrawal is (10,0,8)(10,0,8) with 18 notes, and 18 is the minimum, since c=7,b=2c=7,b=2 gives only 30 for b+4cb+4c.
6
Rs. 500 notes needed: 50×10+50×8=500+400=90050\times10+50\times8=500+400=900.
Why Not the Others
800 and 750 are below the 900 that these minimum-note withdrawals need.
1400 would come from using more 500 notes per customer than the minimum-note choice requires.
Correct Answer: A
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CAT 2018 Slot 1 DILR at a Glance

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

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