◆ DILR · Logical Reasoning

LR · Distribution & Grouping , allocation, mapping & Venn

Distribution, grouping, allocation, set-to-attribute mapping and Venn/set-based LR. Build a person × attribute grid, fill forced cells, and use row/column totals as checks.

2approach cards

5CAT sets

21questions

↩ All LR Approach Sheet CAT Sets

Approach Sheet, Distribution & Grouping

How to set up distribution grids and Venn-based set LR.

5Distribution & grouping

Items/people split into groups or buckets (e.g. idlis & vadas per friend, committees).
Build a person × attribute grid; fill forced cells, then test the rest.
Use totals as a check: row sums and column sums must match the given total.
When numeric, set up equations from the constraints (see Mathematical Reasoning).

12Set theory / Venn

2 sets: n(A∪B) = n(A) + n(B) − n(A∩B)
3 sets: n(A∪B∪C) = Σn(A) − Σn(A∩B) + n(A∩B∩C)
A+B+C = (exactly 1) + 2(exactly 2) + 3(exactly 3)
For "minimum in all sets", maximise the people outside: min(all four) ≥ 100 − Σ(don't-have %).

Real CAT LR Sets, Distribution & Grouping

Actual CAT previous-year distribution, grouping and set-based sets from the book. Difficulty: Easy Moderate Hard. Click any question to reveal the solution.

CAT 2003

Directions (Q. 9 to 11): Answer the questions on the basis of the following information.

Five friends meet every morning at Sree Sagar restaurant for an idli-vada breakfast. Each consumes a different number of idlis and vadas. The number of idlis consumed are 1, 4, 5, 6, and 8, while the number of vadas consumed are 0, 1, 2, 4, and 6. Below are some more facts about who eats what and how much?

(a) The number of vadas eaten by Ignesh is three times the number of vadas consumed by the person who eats four idlis.
(b) Three persons, including the one who eats four vada, eat without chutney.
(c) Sandeep does not take any chutney.
(d) The one who eats one idli a day does not eat any vadas or chutney. Further, he is not Mukesh.
(e) Daljit eats idli with chutney and also eats vada.
(f) Mukesh, who does not take chutney, eats half as many vadas as the person who eats twice as many idlis as he does.
(g) Bimal eats two more idlis than Ignesh, but Ignesh eats two more vadas than Bimal.

HardCAT 2003

9. Which one of the following statements is true?

(1) Daljit eats 5 idlis.
(2) Ignesh eats 8 idlis.
(3) Bimal eats 1 idli.
(4) Bimal eats 6 idlis.

Show solution

(1) Daljit eats 5 idlis. Working through clues (a)-(g), the final grid is: Ignesh 6 idlis / 6 vadas (chutney), Mukesh 4 / 2 (no chutney), Sandeep 1 / 0 (no chutney), Bimal 8 / 4 (no chutney), Daljit 5 / 1 (chutney). So Daljit eats 5 idlis.

HardCAT 2003

10. Which of the following statements is true?

(1) Sandeep eats 2 vadas.
(2) Mukesh eats 4 vadas.
(3) Ignesh eats 6 vadas.
(4) Bimal eats 2 vadas.

Show solution

(3) Ignesh eats 6 vadas. From the completed grid, Ignesh eats 6 vadas.

HardCAT 2003

11. Which of the following statements is true?

(1) Mukesh eats 8 idlis and 4 vadas but no chutney.
(2) The person who eats 5 idlis and 1 vada does not take chutney.
(3) The person who eats equal number of vadas and idlis also takes chutney.
(4) The person who eats 4 idlis and 2 vadas also takes chutney.

Show solution

(3) The person who eats equal number of vadas and idlis also takes chutney. Ignesh eats 6 idlis and 6 vadas with chutney, the only person with equal counts, so statement (3) holds.

CAT 2018

Directions (Q. 44 to 47): Answer the questions on the basis of following information.

An ATM dispenses exactly ₹5000 per withdrawal using 100, 200 and 500 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.

HardCAT 2018 · TITA

44. In how many different ways can the ATM serve a customer who gives 500 rupee notes as her preference? TITA

Show solution

7. If ₹500 is the preferred note then the number of ₹500 notes dispensed must be greater than the number of all other notes. The dispensable amounts in ₹500 notes are ₹4000 (8 notes, 2 ways with the remaining ₹1000), ₹4500 (9 notes, ways with ₹500) and ₹5000 (10 notes, 1 way). Counting all valid combinations gives 7 ways.

HardCAT 2018 · TITA

45. 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? TITA

Show solution

6. The least number of ₹500 notes needed to serve a customer who prefers ₹500 is 8 (₹4000 in 500s plus ₹1000 in fewer notes). With only 50 notes in stock, ⌊50 ÷ 8⌋ = 6 such customers can be served.

HardCAT 2018

46. 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) 12
(2) 10
(3) 13
(4) 16

Show solution

(1) 12. To minimise ₹500 notes per customer while staying within 20 notes, the best case uses 4 ₹500 notes (₹2000) plus 15 ₹200 notes (₹3000) = 19 notes, where 4 > 15 fails, so the valid minimum uses enough ₹500 notes. The least ₹500 notes per withdrawal within the 20-note limit is 4, giving ⌊50 ÷ 4⌋ = 12 customers.

HardCAT 2018

47. 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) 900
(2) 800
(3) 750
(4) 1400

Show solution

(1) 900. A ₹500-preferring customer with the smallest total notes uses 10 ₹500 notes ⇒ 50 × 10 = 500 notes. A ₹100-preferring customer with the smallest total notes still uses 8 ₹500 notes each ⇒ 50 × 8 = 400 notes. Total ₹500 notes = 500 + 400 = 900.

Directions (Q. 48 to 51): Answer the questions on the basis of following information.

Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:

1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politicians are in the administration committee, and 20% are in the teaching committee.

HardCAT 2018

48. Based on the given information, which of the following statement MUST be FALSE?

(1) In the teaching committee the number of educationalists is equal to the number of politicians
(2) In the administration committee the number of bureaucrats is equal to the number of educationalists
(3) The size of the research committee is less than the size of the teaching committee
(4) The size of the research committee is less than the size of the administration committee

Show solution

(3) The size of the research committee is less than the size of the teaching committee, MUST be FALSE. Solving the constraints: bureaucrats research : teaching : administration = 3 : 3 : 4; politicians = 1 : 1 : 3; educationalists in research = 3 (the average of the other two). The research committee turns out to be larger than the teaching committee, so statement (3) is impossible.

HardCAT 2018 · TITA

49. What is the number of bureaucrats in the administration committee? TITA

Show solution

4. Research bureaucrats = teaching bureaucrats = 75% of administration bureaucrats. Taking the smallest integer solution, research = teaching = 3 and administration = 4, so the administration committee has 4 bureaucrats.

HardCAT 2018 · TITA

50. What is the number of educationalists in the research committee? TITA

Show solution

3. The number of educationalists in research is the average of the numbers in teaching and administration; solving with the total of 24 people gives 3 educationalists in the research committee.

HardCAT 2018

51. Which of the following CANNOT be determined uniquely based on the given information?

(1) The size of the teaching committee
(2) The size of the research committee
(3) The total number of bureaucrats in the three committees
(4) The total number of educationalists in the three committees

Show solution

(1) The size of the teaching committee. The number of educationalists in the teaching committee can take more than one value while satisfying all constraints, so the teaching committee's total size cannot be pinned down uniquely.

CAT 2022

Directions (Q. 72 to 76): Answer the questions based on the following information.

There are 15 girls and some boys among the graduating students in a class. They are planning a get-together, which can be either a 1-day event, or a 2-day event, or a 3-day event. There are 6 singers in the class; 4 of them are boys. There are 10 dancers in the class; 4 of them are girls. No dancer in the class is a singer. Some students are not interested in attending the get-together. Those students who are interested in attending a 3-day event are also interested in attending a 2-day event; those who are interested in attending a 2-day event are also interested in attending a 1-day event. The following facts are also known:

1. All the girls and 80% of the boys are interested in attending a 1-day event. 60% of the boys are interested in attending a 2-day event.
2. Some of the girls are interested in attending a 1-day event, but not a 2-day event; some of the other girls are interested in attending both.
3. 70% of the boys who are interested in attending a 2-day event are neither singers nor dancers. 60% of the girls who are interested in attending a 2-day event are neither singers nor dancers.
4. No girl is interested in attending a 3-day event. All male singers and 2 of the dancers are interested in attending a 3-day event.
5. The number of singers interested in attending a 2-day event is one more than the number of dancers interested in attending a 2-day event.

HardCAT 2022 · Slot 1 · TITA

72. How many boys are there in the class? TITA

Show solution

50. Working through the constraints on 2-day interest (60% of boys, and the singer/dancer balance from facts 3-5), the number of boys must make 0.6 × (boys) an integer consistent with all facts. The only feasible value is 50 boys.

ModerateCAT 2022 · Slot 1

73. Which of the following can be determined from the given information? I. The number of boys who are interested in attending a 1-day event and are neither dancers nor singers. II. The number of female dancers who are interested in attending a 1-day event.

(1) Neither I nor II
(2) Only II
(3) Only I
(4) Both I and II

Show solution

(2) Only II. All girls are interested in a 1-day event, so the 4 female dancers are all in the 1-day group, statement II is determined (= 4). The breakdown of boys interested in 1-day who are neither dancers nor singers cannot be fixed, so statement I cannot be determined.

ModerateCAT 2022 · Slot 1

74. What fraction of the class is interested in attending a 2-day event?

(1) 9/13
(2) 2/3
(3) 7/10
(4) 7/13

Show solution

(4) 7/13. The class has 15 girls + 50 boys = 65 students. Those interested in a 2-day event = 5 girls + 30 boys (60% of 50) = 35. Fraction = 35/65 = 7/13.

HardCAT 2022 · Slot 1

75. What BEST can be concluded about the number of male dancers who are interested in attending a 1-day event?

(1) 6
(2) 4 or 6
(3) 5
(4) 5 or 6

Show solution

(4) 5 or 6. There are 6 male dancers. The number of them interested in a 1-day event is at least the number interested in a 2-day event; the constraints leave it as either 5 or 6.

ModerateCAT 2022 · Slot 1

76. How many female dancers are interested in attending a 2-day event?

(1) Cannot be determined
(2) 2
(3) 1
(4) 0

Show solution

(4) 0. By fact 5 the number of singers interested in a 2-day event is one more than the number of dancers interested in a 2-day event; this is only possible when the two female singers are interested in the 2-day event and no female dancer is. So 0 female dancers attend the 2-day event.

Directions (Q. 82 to 86): Answer the questions based on the following information.

A speciality supermarket sold 320 products. Each of these products was either a cosmetic product or a nutrition product. Each of these products was also either a foreign product or a domestic product. Each of these products had at least one of the two approvals - FDA or EU. The following facts are also known:

1. There were equal numbers of domestic and foreign products.
2. Half of the domestic products were FDA approved cosmetic products.
3. None of the foreign products had both the approvals, while 60 domestic products had both the approvals.
4. There were 140 nutrition products; half of them were foreign products.
5. There were 200 FDA approved products. 70 of them were foreign products and 120 of them were cosmetic products.

HardCAT 2022 · Slot 2 · TITA

82. How many foreign products were FDA approved cosmetic products? TITA

Show solution

40. There are 160 domestic and 160 foreign products. Half the domestic products (80) were FDA-approved cosmetics. Of the 120 FDA-approved cosmetic products, 80 are domestic, leaving 120 − 80 = 40 foreign FDA-approved cosmetic products.

ModerateCAT 2022 · Slot 2

83. How many cosmetic products did not have FDA approval?

(1) 50
(2) Cannot be determined
(3) 60
(4) 10

Show solution

(3) 60. Total cosmetic products = 320 − 140 = 180. FDA-approved cosmetics = 120, so cosmetics without FDA approval = 180 − 120 = 60.

HardCAT 2022 · Slot 2

84. Which among the following options best represents the number of domestic cosmetic products that had both the approvals?

(1) At least 10 and at most 80
(2) At least 10 and at most 60
(3) At least 20 and at most 70
(4) At least 20 and at most 50

Show solution

(2) At least 10 and at most 60. All 60 domestic products with both approvals are domestic, so domestic cosmetics with both ≤ 60. Since the only-FDA domestic cosmetics are at most 70, the number with both approvals must be at least 10. Hence at least 10 and at most 60.

HardCAT 2022 · Slot 2

85. If 70 cosmetic products did not have EU approval, then how many nutrition products had both the approvals?

(1) 30
(2) 10
(3) 50
(4) 20

Show solution

(2) 10. Cosmetics without EU approval (only FDA) = 70 = 40 foreign FDA cosmetics + 30 domestic only-FDA cosmetics. The 60 domestic products with both approvals then split so that domestic nutrition products with both approvals = 10. Since no foreign product has both approvals, nutrition products with both = 10.

HardCAT 2022 · Slot 2 · TITA

86. If 50 nutrition products did not have EU approval, then how many domestic cosmetic products did not have EU approval? TITA

Show solution

50. Nutrition products without EU approval (only FDA) = 50 = 30 foreign FDA nutrition + 20 domestic only-FDA nutrition. The 80 FDA-approved domestic cosmetics consist of those with only FDA and those with both; with 20 of the only-FDA domestic going to nutrition, the domestic cosmetics without EU approval (only FDA) total 50.