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CAT 2024 Slot 1 — DILR questions with answers

All 22 questions of the Data Interpretation & Logical Reasoning section (12 MCQs, 10 TITA, 5 sets). Try each one, then open its answer and solution.

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Data Interpretation & Logical Reasoning

CAT 2024 Slot 1 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Data set

Set for questions 25–28

DIRECTIONS for questions 25–28: Read the information given below and answer the question(s) that follow(s).

The chart below shows the price data for seven shares - A, B, C, D, E, F, and G as a candlestick plot for a particular day. The vertical axis shows the price of the share in rupees. A share whose closing price (price at the end of the day) is more than its opening price (price at the start of the day) is called a bullish share; otherwise, it is called a bearish share. All bullish and bearish shares are shown in green and red colour respectively.

Data for questions 25–28, CAT 2024 Slot 1 DILR

Q25MCQBar & Line Charts

Daily Share Price Variability (SPV) is defined as (Day’s high price - Day’s low price) / (Average of the opening and closing prices during the day). Which among the shares A, C, D and F had the highest SPV on that day?
  1. F
  2. A
  3. D
  4. C
Answer and solution

Answer: (C) D

Reading the candlestick chart gives these prices: Solution figure for question 25, CAT 2024 Slot 1 SPV =High−Lowaverage of Opening and Closing= \dfrac{\text{High} - \text{Low}}{\text{average of Opening and Closing}}. F: opening 1800, closing 1600, high 2000, low 1200. SPV =8001700≈0.47= \dfrac{800}{1700} \approx 0.47. A: opening 2200, closing 1800, high 2400, low 1200. SPV =12002000=0.6= \dfrac{1200}{2000} = 0.6. D: opening 500, closing 1000, high 1200, low 300. SPV =900750=1.2= \dfrac{900}{750} = 1.2. C: opening 800, closing 1200, high 1400, low 800. SPV =6001000=0.6= \dfrac{600}{1000} = 0.6. Share D has by far the highest SPV. A has the widest range (1200), but its average price is also high; D's range of 900 is large compared with its low average price of 750. A and C, at 0.6, are only half as large. Hence, option C (D).

Q26TITABar & Line Charts

Daily Share Price Variability (SPV) is defined as (Day’s high price - Day’s low price) / (Average of the opening and closing prices during the day). How many shares had an SPV greater than 0.5 on that day?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

Reading the candlestick chart gives these prices: Solution figure for question 26, CAT 2024 Slot 1 SPV =High−Lowaverage of Opening and Closing= \dfrac{\text{High} - \text{Low}}{\text{average of Opening and Closing}}. For each share: A: 2400−1200(2200+1800)/2=12002000=0.6\dfrac{2400 - 1200}{(2200 + 1800)/2} = \dfrac{1200}{2000} = 0.6 B: 2000−1400(2000+1700)/2=6001850≈0.32\dfrac{2000 - 1400}{(2000 + 1700)/2} = \dfrac{600}{1850} \approx 0.32 C: 1400−800(800+1200)/2=6001000=0.6\dfrac{1400 - 800}{(800 + 1200)/2} = \dfrac{600}{1000} = 0.6 D: 1200−300(500+1000)/2=900750=1.2\dfrac{1200 - 300}{(500 + 1000)/2} = \dfrac{900}{750} = 1.2 E: 1400−1100(1300+1100)/2=3001200=0.25\dfrac{1400 - 1100}{(1300 + 1100)/2} = \dfrac{300}{1200} = 0.25 F: 2000−1200(1800+1600)/2=8001700≈0.47\dfrac{2000 - 1200}{(1800 + 1600)/2} = \dfrac{800}{1700} \approx 0.47 G: 1900−1000(1200+1700)/2=9001450≈0.62\dfrac{1900 - 1000}{(1200 + 1700)/2} = \dfrac{900}{1450} \approx 0.62 The shares with SPV greater than 0.5 are A, C, D and G. F is the closest miss, at about 0.47. The answer is 4.

Q27MCQBar & Line Charts

Daily loss for a share is defined as (Opening price - Closing price) / (Opening price). Which among the shares A, B, F and G had the highest daily loss on that day?
  1. G
  2. B
  3. A
  4. F
Answer and solution

Answer: (C) A

Reading the candlestick chart gives these prices: Solution figure for question 27, CAT 2024 Slot 1 Daily loss =Opening−ClosingOpening= \dfrac{\text{Opening} - \text{Closing}}{\text{Opening}}. It is positive only for bearish (red) shares. A: opening 2200, closing 1800. Loss =4002200=211≈0.182= \dfrac{400}{2200} = \dfrac{2}{11} \approx 0.182. B: opening 2000, closing 1700. Loss =3002000=0.15= \dfrac{300}{2000} = 0.15. F: opening 1800, closing 1600. Loss =2001800=19≈0.111= \dfrac{200}{1800} = \dfrac19 \approx 0.111. G is bullish: it opened at 1200 and closed at 1700, so it gained and had no loss. Share A has the highest daily loss; B comes second at 0.15. Hence, option C (A).

Q28MCQBar & Line Charts

What would have been the percentage wealth gain for a trader, who bought equal numbers of all bullish shares at opening price and sold them at their day’s high?
  1. 80%
  2. 50%
  3. 72%
  4. 100%
Answer and solution

Answer: (A) 80%

Reading the candlestick chart gives these prices: Solution figure for question 28, CAT 2024 Slot 1 The bullish (green) shares, which closed above their opening price, are C, D and G. Suppose the trader buys one share of each at the opening price and sells it at the day's high. C: buys at 800, sells at 1400. D: buys at 500, sells at 1200. G: buys at 1200, sells at 1900. Total cost =800+500+1200=2500= 800 + 500 + 1200 = 2500, and total sale value =1400+1200+1900=4500= 1400 + 1200 + 1900 = 4500. Gain =4500−25002500=20002500=0.8= \dfrac{4500 - 2500}{2500} = \dfrac{2000}{2500} = 0.8, that is, 80%. Option D (100%) would need the shares to sell for 5000, double the cost; they sell for 4500. Hence, option A (80%).

Data set

Set for questions 29–32

DIRECTIONS for questions 29–32: Read the information given below and answer the question(s) that follow(s).

Six web surfers M, N, O, P, X, and Y each had 30 stars which they distributed among four bloggers A, B, C, and D. The number of stars received by A and B from the six web surfers is shown in the figure below.

Data for questions 29–32, CAT 2024 Slot 1 DILR

The following additional facts are known regarding the number of stars received by the bloggers from the surfers.

1. The numbers of stars received by the bloggers from the surfers were all multiples of 5 (including 0). 2. The total numbers of stars received by the bloggers were the same. 3. Each blogger received a different number of stars from M. 4. Two surfers gave all their stars to a single blogger. 5. D received more stars than C from Y.

Q29TITABar & Line Charts

What was the total number of stars received by D?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 45

Six surfers each gave 30 stars, so 6×30=1806 \times 30 = 180 stars were given in all. By fact 2, the four bloggers received the same total, so each received 1804=45\dfrac{180}{4} = 45 stars. This matches the chart: A received 10+25+0+5+0+5=4510 + 25 + 0 + 5 + 0 + 5 = 45 and B received 0+0+0+25+0+20=450 + 0 + 0 + 25 + 0 + 20 = 45. So C and D together received the remaining 90. Solution figure for question 29, CAT 2024 Slot 1 Working out the full distribution confirms this: D gets 5 from M, 5 from N, 5 from Y and 30 from one of O and X, which is again 45. The answer is 45.

Q30MCQBar & Line Charts

What was the number of stars received by D from Y?
  1. 5
  2. 10
  3. Can't be determined
  4. 0
Answer and solution

Answer: (A) 5

From the chart, Y gave 5 stars to A and 20 to B, so Y had 30−25=530 - 25 = 5 stars left for C and D. Solution figure for question 30, CAT 2024 Slot 1 Every number of stars given is a multiple of 5 (fact 1), so these 5 stars went entirely to C or entirely to D. By fact 5, D received more stars than C from Y. So D got 5 and C got 0. Option B (10) is impossible, since Y had only 5 stars left. Option D (0) would give C 5 and D 0, breaking fact 5. The value is fixed, so option C (can't be determined) fails too. Hence, option A (5).

Q31TITABar & Line Charts

How many surfers distributed their stars among exactly 2 bloggers?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 2

Each surfer gave 30 stars in multiples of 5. From the chart, the stars given to A and B are: Solution figure for question 31, CAT 2024 Slot 1 M gave A 10 and B 0, leaving 20 for C and D. By fact 3, M gave each blogger a different number; 0 and 10 are taken, so C and D got two different multiples of 5 adding to 20, other than 0 and 10: 5 and 15. M gave to 3 bloggers. N gave A 25 and B 0, so its last 5 stars went to one of C or D: 2 bloggers. P gave A 5 and B 25, all 30 stars: 2 bloggers. Y gave A 5 and B 20, and by fact 5 its last 5 went to D: 3 bloggers. O and X gave nothing to A or B. Every other surfer has already given to at least two bloggers, so by fact 4 O and X are the two who gave all 30 stars to one blogger. They cannot pick the same one, as each blogger's total is 180÷4=45180 \div 4 = 45. The two possible tables: Solution figure for question 31, CAT 2024 Slot 1 Solution figure for question 31, CAT 2024 Slot 1 In both, exactly N and P gave to 2 bloggers. The answer is 2.

Q32MCQBar & Line Charts

Which of the following can be determined with certainty? I. The number of stars received by C from M II. The number of stars received by D from O
  1. Neither I or II
  2. Only I
  3. Only II
  4. Both I and II
Answer and solution

Answer: (B) Only I

All 180 stars are shared equally, so each blogger received 45. From the chart, A and B already have 45 each: Solution figure for question 32, CAT 2024 Slot 1 Y has 5 stars left after A and B; by fact 5 they go to D. O and X gave nothing to A or B, and every other surfer has already split its stars, so by fact 4 O and X each gave all 30 to one blogger. They cannot both choose the same blogger (60>4560 > 45), so one gave C 30 and the other gave D 30. M gave A 10 and B 0, so by fact 3 C and D got 5 and 15 in some order. If D got 15 from M, D would have at least 15+5+30=50>4515 + 5 + 30 = 50 > 45. So M gave C 15 and D 5, and N's last 5 goes to D, making both totals 45. Solution figure for question 32, CAT 2024 Slot 1 Solution figure for question 32, CAT 2024 Slot 1 Statement I is determined: C got 15 from M in both tables. Statement II is not: D got 0 from O in one table and 30 in the other, since nothing fixes whether O or X chose D. Hence, option B (Only I).

Data set

Set for questions 33–37

DIRECTIONS for questions 33–37: Read the information given below and answer the question(s) that follow(s).

The game of QUIET is played between two teams. Six teams, numbered 1, 2, 3, 4, 5, and 6, play in a QUIET tournament. These teams are divided equally into two groups. In the tournament, each team plays every other team in the same group only once, and each team in the other group exactly twice. The tournament has several rounds, each of which consists of a few games. Every team plays exactly one game in each round.

The following additional facts are known about the schedule of games in the tournament.

1. Each team played against a team from the other group in Round 8. 2. In Round 4 and Round 7, the match-ups, that is the pair of teams playing against each other, were identical. In Round 5 and Round 8, the match-ups were identical. 3. Team 4 played Team 6 in both Round 1 and Round 2. 4. Team 1 played Team 5 ONLY once and that was in Round 2. 5. Team 3 played Team 4 in Round 3. Team 1 played Team 6 in Round 6. 6. In Round 8, Team 3 played Team 6, while Team 2 played Team 5.

Q33TITAGames & Tournaments

How many rounds were there in the tournament?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 8

The six teams form two groups of three. Games within a group: each pair plays once. A group of three has (32)=3\binom{3}{2} = 3 pairs, so the two groups give 2×3=62 \times 3 = 6 games. Games across the groups: each of the 3 teams in one group plays each of the 3 teams in the other twice, giving 3×3×2=183 \times 3 \times 2 = 18 games. Total games =6+18=24= 6 + 18 = 24. In each round every team plays exactly one game, so each round has 6÷2=36 \div 2 = 3 games. The number of rounds is 243=8\dfrac{24}{3} = 8. The full schedule, worked out from the given facts, fills exactly these 8 rounds: Solution figure for question 33, CAT 2024 Slot 1 The answer is 8.

Q34TITAGames & Tournaments

What is the number of the team that played Team 1 in Round 5?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

By fact 6, Round 8 has Team 3 vs Team 6 and Team 2 vs Team 5. Every team plays exactly one game in each round, so the third game of Round 8 is between the two teams left over: Team 1 vs Team 4. By fact 2, Round 5 has the same match-ups as Round 8: 3 vs 6, 2 vs 5 and 1 vs 4. So Team 1 played Team 4 in Round 5. This agrees with the complete schedule built from all the facts: Solution figure for question 34, CAT 2024 Slot 1 The answer is 4.

Q35MCQGames & Tournaments

Which team among the teams numbered 2, 3, 4, and 5 was not part of the same group?
  1. 5
  2. 3
  3. 4
  4. 2
Answer and solution

Answer: (A) 5

Teams in the same group meet once; teams in different groups meet twice. Fact 4: Team 1 played Team 5 only once, so 1 and 5 are in the same group. Fact 3: Team 4 played Team 6 twice (Rounds 1 and 2), so 4 and 6 are in different groups. Fact 1: every Round 8 game is between the groups. By fact 6, those games include 3 vs 6 and 2 vs 5, so 3 and 6 are in different groups, and so are 2 and 5. As 3 and 6 are split, exactly one of them joins 1 and 5. If it were 3, the other group would be 2, 4 and 6, putting 4 and 6 together, which is impossible. So 6 joins 1 and 5. Solution figure for question 35, CAT 2024 Slot 1 The groups are 1, 5, 6 and 2, 3, 4. Among Teams 2, 3, 4 and 5, Teams 2, 3 and 4 share a group; Team 5 does not. Hence, option A (5).

Q36TITAGames & Tournaments

What is the number of the team that played Team 1 in Round 7?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 3

Groups first. Team 1 met Team 5 only once, so they share a group; Team 4 met Team 6 twice, so they are in different groups. All Round 8 games are between groups and include 3 vs 6, so exactly one of 3 and 6 joins 1 and 5; if it were 3, 4 and 6 would be left together. So the groups are 1, 5, 6 and 2, 3, 4. Team 1 therefore plays 5 and 6 once each, and 2, 3 and 4 twice each: 8 games, one per round. Known games of Team 1: Round 2 vs 5 and Round 6 vs 6. Round 8 has 3 vs 6 and 2 vs 5, so 1 vs 4; Round 5 repeats Round 8, so 1 vs 4 again. In Round 3, Team 3 plays Team 4, and Team 1 has used up its games against 4, 5 and 6, so Team 1 plays Team 2. Rounds 1, 4 and 7 remain, for one game against 2 and two against 3. Rounds 4 and 7 have identical match-ups, so Team 1 meets the same team in both; that must be Team 3, as only one game against 2 is left. Solution figure for question 36, CAT 2024 Slot 1 The answer is 3.

Q37TITAGames & Tournaments

What is the number of the team that played Team 6 in Round 3?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 5

Groups first. Team 1 met Team 5 only once, so they share a group; Team 4 met Team 6 twice, so they are in different groups. All Round 8 games are between groups and include 3 vs 6, so exactly one of 3 and 6 joins 1 and 5; if it were 3, 4 and 6 would be left together. So the groups are 1, 5, 6 and 2, 3, 4. Team 6 therefore plays 1 and 5 once each, and 2, 3 and 4 twice each: 8 games, one per round. Known games of Team 6: Rounds 1 and 2 vs 4, Round 6 vs 1, Round 8 vs 3, and Round 5 (same as Round 8) vs 3. Rounds 3, 4 and 7 remain, for two games against 2 and one against 5. Rounds 4 and 7 have identical match-ups, so Team 6 meets the same team in both; that must be Team 2, as Team 5 is played only once. So Team 6 plays Team 5 in Round 3, alongside 3 vs 4 from fact 5. Solution figure for question 37, CAT 2024 Slot 1 The answer is 5.

Data set

Set for questions 38–42

DIRECTIONS for questions 38-42: Read the information given below and answer the question(s) that follow(s). Two students, Amiya and Ramya are the only candidates in an election for the position of class representative. Students will vote based on the intensity level of Amiya's and Ramya's campaigns and the type of campaigns they run. Each campaign is said to have a level of 1 if it is a staid campaign and a level of 2 if it is a vigorous campaign. Campaigns can be of two types, they can either focus on issues, or on attacking the other candidate. If Amiya and Ramya both run campaigns focusing on issues, then • The percentage of students voting in the election will be 20 times the sum of the levels of campaigning of the two students. For example, if Amiya and Ramya both run vigorous campaigns, then 20×(2+2)%20 \times (2+2)\%, that is, 80%80\% of the students will vote in the election. • Among voting students, the percentage of votes for each candidate will be proportional to the levels of their campaigns. For example, if Amiya runs a staid (i.e., level 1) campaign while Ramya runs a vigorous (i.e., level 2) campaign, then Amiya will receive 13\frac{1}{3} of the votes cast, and Ramya will receive the other 23\frac{2}{3}. The above-mentioned percentages change as follows if at least one of them runs a campaign attacking their opponent. • If Amiya runs a campaign attacking Ramya and Ramya runs a campaign focusing on issues, then 10%10\% of the students who would have otherwise voted for Amiya will vote for Ramya, and another 10%10\% who would have otherwise voted for Amiya, will not vote at all. • If Ramya runs a campaign attacking Amiya and Amiya runs a campaign focusing on issues, then 20%20\% of the students who would have otherwise voted for Ramya will vote for Amiya, and another 5%5\% who would have otherwise voted for Ramya, will not vote at all. • If both run campaigns attacking each other, then 10%10\% of the students who would have otherwise voted for them had they run campaigns focusing on issues, will not vote at all.

Q38MCQTables & Caselets

If both of them run staid campaigns attacking the other, then what percentage of students will vote in the election?
  1. 40%
  2. 64%
  3. 60%
  4. 36%
Answer and solution

Answer: (D) 36%

Both run staid campaigns, so each has level 1. Had both focused on issues, turnout would be 20×(1+1)%=40%20 \times (1+1)\% = 40\% of the students, split in the ratio 1:11 : 1, so each would get 20%20\%. Because both attack, 10%10\% of each candidate's would-be voters do not vote. Each loses 10%10\% of 20%20\%, which is 2%2\%, and keeps 18%18\%. Turnout =18%+18%=36%= 18\% + 18\% = 36\%. Option A, 40%40\%, is the turnout for staid campaigns on issues; it ignores the 2%+2%=4%2\% + 2\% = 4\% of students who stop voting when both attack. Hence, option D (36%).

Q39MCQTables & Caselets

What is the minimum percentage of students who will vote in the election?
  1. 32%
  2. 40%
  3. 38%
  4. 36%
Answer and solution

Answer: (D) 36%

With issue campaigns at levels aa (Amiya) and rr (Ramya), turnout is 20(a+r)%20(a+r)\%, split in the ratio a:ra : r. So Amiya would get 20a%20a\% and Ramya 20r%20r\% of all students. Every case below grows with the levels, so the minimum needs both staid (a=r=1a = r = 1): 20%20\% each, 40%40\% in all. Votes that switch candidates are still cast; only voters who drop out lower turnout. Compare the four campaign types: Both on issues: nobody drops out, turnout 40%40\%. Only Amiya attacks: 10%10\% of her 20%20\% drop out, which is 2%2\%, so turnout is 38%38\%. Only Ramya attacks: 5%5\% of her 20%20\% drop out, which is 1%1\%, so turnout is 39%39\%. Both attack: 10%10\% of each candidate's 20%20\% drop out, 2%+2%=4%2\% + 2\% = 4\%, so turnout is 36%36\%. The lowest turnout is 36%36\%. Option C, 38%38\%, is the case where only Amiya attacks; it loses 2%2\% fewer voters than a mutual attack. Hence, option D (36%).

Q40MCQTables & Caselets

If Amiya runs a campaign focusing on issues, then what is the maximum percentage of votes that she can get?
  1. 48%
  2. 44%
  3. 40%
  4. 36%
Answer and solution

Answer: (A) 48%

With issue campaigns at levels aa (Amiya) and rr (Ramya), turnout is 20(a+r)%20(a+r)\%, split in the ratio a:ra : r, so Amiya would get 20a%20a\% and Ramya 20r%20r\% of all students. Amiya focuses on issues. If Ramya also focuses on issues, Amiya gets 20a%20a\%, at most 40%40\%. If Ramya attacks, 20%20\% of Ramya's would-be voters switch to Amiya. Amiya then gets 20a%+0.2×20r%=(20a+4r)%20a\% + 0.2 \times 20r\% = (20a + 4r)\%. This is largest when both are vigorous (a=r=2a = r = 2): Amiya's own 40%40\% plus 20%20\% of Ramya's 40%40\%, which is 8%8\%. Amiya gets 40%+8%=48%40\% + 8\% = 48\%. Option B, 44%44\%, is Amiya's share when Ramya's attacking campaign is staid: only 20%20\% of 20%20\%, which is 4%4\%, switches over. Hence, option A (48%).

Q41MCQTables & Caselets

If Ramya runs a campaign attacking Amiya, then what is the minimum percentage of votes that she is guaranteed to get?
  1. 12%
  2. 15%
  3. 30%
  4. 18%
Answer and solution

Answer: (B) 15%

With issue campaigns at levels aa (Amiya) and rr (Ramya), turnout is 20(a+r)%20(a+r)\%, split in the ratio a:ra : r, so Ramya would get 20r%20r\% of all students, whatever Amiya's level. Ramya attacks, so there are two cases for Amiya's campaign. Amiya on issues: 20%20\% of Ramya's would-be voters switch to Amiya and 5%5\% do not vote, so Ramya keeps 75%75\% of 20r%20r\%, which is 15r%15r\%. Amiya also attacks: only 10%10\% of Ramya's would-be voters stop voting, so she keeps 90%90\% of 20r%20r\%, which is 18r%18r\%. The worst case is a staid campaign (r=1r = 1) with Amiya on issues: 15%15\%. Whatever Amiya does, Ramya gets at least 15%15\%, so that is her guaranteed minimum. Option D, 18%18\%, is her share when both attack; she can fall below it when Amiya sticks to issues. Hence, option B (15%).

Q42MCQTables & Caselets

What is the maximum possible voting margin with which one of the candidates can win?
  1. 20%
  2. 29%
  3. 28%
  4. 26%
Answer and solution

Answer: (B) 29%

With issue campaigns at levels aa (Amiya) and rr (Ramya), turnout is 20(a+r)%20(a+r)\% split in the ratio a:ra : r, so Amiya would get 20a%20a\% and Ramya 20r%20r\% of all students. For a big margin the winner is vigorous and the loser staid (equal levels give at most 18%18\%). Amiya wins (a=2a = 2, r=1r = 1): Both on issues: 40%−20%=20%40\% - 20\% = 20\%. Ramya attacks: she keeps 75%75\% of 20%20\%, which is 15%15\%, and 20%20\% of 20%20\%, which is 4%4\%, switches to Amiya, who gets 44%44\%. Margin 44%−15%=29%44\% - 15\% = 29\%. Amiya attacks: she keeps 80%80\% of 40%40\%, which is 32%32\%, and Ramya gets 20%+4%=24%20\% + 4\% = 24\%. Margin 8%8\%. Both attack: 36%−18%=18%36\% - 18\% = 18\%. Ramya wins (r=2r = 2, a=1a = 1): her best case is Amiya attacking while Ramya stays on issues. Amiya keeps 80%80\% of 20%20\%, which is 16%16\%, and Ramya gets 40%+2%=42%40\% + 2\% = 42\%, a margin of 26%26\% (option D). Her other cases give 20%20\%, 18%18\% and 2%2\%. The largest margin is 29%29\%. Hence, option B (29%).

Data set

Set for questions 43–46

DIRECTIONS for questions 43–46: Read the information given below and answer the question(s) that follow(s).

The chart below provides complete information about the number of countries visited by Dheeraj, Samantha and Nitesh, in Asia, Europe and the rest of the world (ROW).

Data for questions 43–46, CAT 2024 Slot 1 DILR

The following additional facts are known about the countries visited by them.

1. 32 countries were visited by at least one of them. 2. USA (in ROW) is the only country that was visited by all three of them. 3. China (in Asia) is the only country that was visited by both Dheeraj and Nitesh, but not by Samantha. 4. France (in Europe) is the only country outside Asia, which was visited by both Dheeraj and Samantha, but not by Nitesh. 5. Half of the countries visited by both Samantha and Nitesh are in Europe.

Q43TITASet Theory

How many countries in Asia were visited by at least one of Dheeraj, Samantha and Nitesh?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 3

From the chart, Dheeraj visited 3, 7 and 1 countries in Asia, Europe and ROW; Samantha 0, 9 and 4; Nitesh 1, 6 and 12. That is 43 visits to 32 distinct countries. If pp, qq, ss countries were visited by exactly one, two and three people, then p+q+s=32p + q + s = 32 and p+2q+3s=43p + 2q + 3s = 43, so q+2s=11q + 2s = 11. Only USA was visited by all three, so s=1s = 1 and q=9q = 9. Of these 9, China is the only Dheeraj–Nitesh country and France the only Dheeraj–Samantha one (Samantha visited no Asian country). So 7 were visited by Samantha and Nitesh only. With USA they share 8, half of them (4) in Europe; none is in Asia, so the other 3 are in ROW. Solution figure for question 43, CAT 2024 Slot 1 In Asia, Samantha visited none. Dheeraj's 3 are China and 3−1=23 - 1 = 2 of his own, and Nitesh's only one is China. So 2+1=32 + 1 = 3 Asian countries were visited. The answer is 3.

Q44TITASet Theory

How many countries in Europe were visited only by Nitesh?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 2

From the chart, Dheeraj visited 3, 7 and 1 countries in Asia, Europe and ROW; Samantha 0, 9 and 4; Nitesh 1, 6 and 12. That is 43 visits to 32 distinct countries. If pp, qq, ss countries were visited by exactly one, two and three people, then p+q+s=32p + q + s = 32 and p+2q+3s=43p + 2q + 3s = 43, so q+2s=11q + 2s = 11. Only USA was visited by all three, so s=1s = 1 and q=9q = 9. Of these 9, China is the only Dheeraj–Nitesh country and France the only Dheeraj–Samantha one (Samantha visited no Asian country). So 7 were visited by Samantha and Nitesh only. With USA they share 8, half of them (4) in Europe; none is in Asia, so the other 3 are in ROW. Solution figure for question 44, CAT 2024 Slot 1 Nitesh's 6 European countries include the 4 he shares with Samantha; he shares none in Europe with Dheeraj, and USA is in ROW. So 6−4=26 - 4 = 2 were visited by Nitesh alone. The answer is 2.

Q45TITASet Theory

How many countries in ROW were visited by both Nitesh and Samantha?

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

From the chart, Dheeraj visited 3, 7 and 1 countries in Asia, Europe and ROW; Samantha 0, 9 and 4; Nitesh 1, 6 and 12. That is 43 visits to 32 distinct countries. If pp, qq, ss countries were visited by exactly one, two and three people, then p+q+s=32p + q + s = 32 and p+2q+3s=43p + 2q + 3s = 43, so q+2s=11q + 2s = 11. Only USA was visited by all three, so s=1s = 1 and q=9q = 9. Of these 9, China is the only Dheeraj–Nitesh country and France the only Dheeraj–Samantha one (Samantha visited no Asian country). So 7 were visited by Samantha and Nitesh only. With USA they share 8, half of them (4) in Europe; none is in Asia, so the other 3 are in ROW. Solution figure for question 45, CAT 2024 Slot 1 In ROW, Nitesh and Samantha share the 3 countries only the two of them visited, plus USA. That gives 3+1=43 + 1 = 4. The answer is 4.

Q46MCQSet Theory

How many countries in Europe were visited by exactly one of Dheeraj, Samantha and Nitesh?
  1. 10
  2. 5
  3. 14
  4. 12
Answer and solution

Answer: (D) 12

From the chart, Dheeraj visited 3, 7 and 1 countries in Asia, Europe and ROW; Samantha 0, 9 and 4; Nitesh 1, 6 and 12. That is 43 visits to 32 distinct countries. If pp, qq, ss countries were visited by exactly one, two and three people, then p+q+s=32p + q + s = 32 and p+2q+3s=43p + 2q + 3s = 43, so q+2s=11q + 2s = 11. Only USA was visited by all three, so s=1s = 1 and q=9q = 9. Of these 9, China is the only Dheeraj–Nitesh country and France the only Dheeraj–Samantha one (Samantha visited no Asian country). So 7 were visited by Samantha and Nitesh only. With USA they share 8, half of them (4) in Europe; none is in Asia, so the other 3 are in ROW. Solution figure for question 46, CAT 2024 Slot 1 In Europe, the shared countries are France (Dheeraj–Samantha) and the 4 Samantha–Nitesh ones. Countries with one visitor: Dheeraj 7−1=67 - 1 = 6, Samantha 9−1−4=49 - 1 - 4 = 4, Nitesh 6−4=26 - 4 = 2, in all 6+4+2=126 + 4 + 2 = 12. Option A, 10, leaves out Nitesh's 2. Hence, option D (12).