CAT 2019 Slot 2 — DILR questions with answers

Q35MCQGames & Tournaments

1. A3, 6, 3
2. B3, 3, 3
3. C3, 3, 0
4. D3, 0, 3

Answer and solution

Answer: (B) 3, 3, 3

In Rounds 1 to 6, Joshin plays only 5 and 6, so his 14 is 7 + 7 there. Gordon plays Rounds 2 to 6 but never scores in consecutive rounds, so he scores in at most three (2, 4, 6); 17 then needs 7 + 7 + 3, with the 3 in Round 6, whose 7 is Joshin's. Amita's 8 is 7 in Round 1 and 1 in Round 6. Bala's 2 is 1 in Rounds 1 and 2, Ikea's 2 is 1 in Rounds 4 and 5 (Round 6's 1 is Amita's), and Hansa's 1 is then Round 3's. Left: a 3 in each of Rounds 1, 2, 4 and 5, and 7 and 3 in Round 3, for Chen (3), David (6), Eric (3) and Fatima (10). Fatima needs 7 + 3, so she wins Round 3. Eric scored in a round with Fatima, so his 3 is in Round 3. Round 5's 3 is Fatima's, since Chen and David do not play it. Chen does not play Round 4, so David takes its 3, and Chen and David share Rounds 1 and 2. After Round 3: Chen 3, David 3, Eric 3. Option A gives David 6, but his second 3 comes in Round 4. Hence, option B (3, 3, 3).

Q36MCQGames & Tournaments

1. ABala, Ikea, Joshin
2. BBala, Hansa, Ikea
3. CBala, Chen, Gordon
4. DHansa, Ikea, Joshin

Answer and solution

Answer: (D) Hansa, Ikea, Joshin

Each round awards 7, 3 and 1 points. Round $r$ has players $r$ to $r + 5$, counting past 10 back to 1. Joshin plays only Rounds 5 and 6 of the first six, so his 14 is 7 + 7 there and he has 0 after Round 4. Gordon plays Rounds 2 to 6 but never scores in consecutive rounds, so he scores in at most three (2, 4, 6); 17 then needs 7 + 7 + 3, with the 3 in Round 6, whose 7 is Joshin's. Amita's 8 is 7 in Round 1 and 1 in Round 6. Bala's 2 is 1 in Rounds 1 and 2. Ikea's 2 is 1 in Rounds 4 and 5, as Round 6's 1 is Amita's. Hansa's single point is then Round 3's 1. With every 1 used, Fatima's 10 needs Round 3's 7, the only 7 still free. Eric scored in a round with Fatima, so his 3 is also in Round 3. After Round 4: Joshin 0, Hansa 1, Ikea 1, Bala 2, Chen 3, Eric 3, David 6, Amita 7, Fatima at least 7, Gordon 14. The bottom three are Joshin, Hansa and Ikea. Option A includes Bala, but his 2 points are more than Hansa's 1. Hence, option D (Hansa, Ikea, Joshin).

Q37MCQGames & Tournaments

1. AJoshin
2. BChen
3. CAmita
4. DIkea

Answer and solution

Answer: (D) Ikea

Round $r$ has players $r$ to $r + 5$, counting past 10 back to 1. Points are 7, 3 and 1. Rounds 1 to 6: Joshin's 14 is 7 + 7, Amita's 8 is 7 + 1 and Ikea's 2 is 1 + 1, two scoring rounds each. Chen's 3 cannot be 1 + 1 + 1 in Rounds 1 to 3, because Bala, who plays only Rounds 1 and 2, needs both their 1s for his 2; so Chen scored once. Rounds 7 to 10, using each player's gain after Round 6: Ikea gains 15 in Rounds 7 to 9, which needs 7 + 7 + 1. Chen gains 3 in Rounds 8 to 10 and is one of the two players who scored in three consecutive rounds, so he scores 1 in each; Ikea's 1 is therefore in Round 7. With all the 1s taken, Amita's gain of 10 is 7 + 3, two rounds, and Joshin's gain of 3 is a single 3, in Round 7. Scoring rounds: Ikea 2 + 3 = 5; Amita 2 + 2 = 4; Chen 1 + 3 = 4; Joshin 2 + 1 = 3. The nearest rivals, Amita and Chen, reach only 4. Hence, option D (Ikea).

Q38MCQGames & Tournaments

1. AAmita, Eric, Joshin
2. BAmita, Chen, David
3. CAmita, Bala, Chen
4. DAmita, Chen, Eric

Answer and solution

Answer: (D) Amita, Chen, Eric

Compare each player's total after Round 6 with the total after Round 10. Round 10 has players 10, 1, 2, 3, 4 and 5: Joshin, Amita, Bala, Chen, David and Eric. Eric goes from 3 to 10, gaining 7. Among Rounds 7 to 10 he plays only Round 10, so he won it. Chen goes from 3 to 6, gaining 3, and among Rounds 7 to 10 he plays only Rounds 8, 9 and 10. He is one of the two players who scored in three consecutive rounds, so he scored in all three, 1 point each. So he took 1 point in Round 10. Amita is given to have scored in Round 10, and only the 3-point place is left, so she took it. Round 10's three scoring places are therefore Eric (7), Amita (3) and Chen (1). Option A has Joshin in place of Chen, but no place is left for Joshin; David, in option B, gained nothing after Round 6. Hence, option D (Amita, Chen, Eric).

Q39MCQPie & Scatter Charts

1. A42.86
2. B75.00
3. C57.14
4. D14.29

Q40MCQPie & Scatter Charts

1. A-10%
2. B10%
3. C-20%
4. D-30%

Q41MCQPie & Scatter Charts

1. A367 mm
2. B500 mm
3. C450 mm
4. D460 mm

Q42MCQPie & Scatter Charts

1. A475 mm
2. B505 mm
3. C490 mm
4. D525 mm

Q43MCQLogical Puzzles

1. AChetan
2. BBeti
3. CEsha
4. DDave

Answer and solution

Answer: (D) Dave

A paper needs at least four 5-mark, three 10-mark and two 15-mark questions: 9 questions worth 80 marks. The other 20 marks need 2, 3 or 4 more questions (for example $\{10,10\}$, $\{5,5,10\}$ or $\{5,5,5,5\}$), so a paper has 11, 12 or 13 questions. Everyone set the same number of questions, so MT and ET together have a multiple of 6 questions. ET is longer than MT, so only $11 + 13 = 24$ works, and each person set 4. In MT, Annie set one of the 11 questions and the other five set the remaining 10, at least 2 each, so exactly 2 each. In ET, Annie therefore set 3 and everyone else 2. ET in blocks: Dave set the first question, so Q1–Q2. Chetan has Q3, so Q3–Q4. Annie's three include Q5, so Q5–Q7. Esha has Q8, so Q8–Q9. Fakir set the last question, so Q12–Q13, leaving Q10–Q11 for Beti. The completed ET table shows these blocks: So the second ET question is Dave's. Chetan (option A) set Q3–Q4, Beti (option B) Q10–Q11 and Esha (option C) Q8–Q9. Hence, option D (Dave).

Q44MCQLogical Puzzles

1. A13
2. B12
3. C10
4. DCannot be determined

Answer and solution

Answer: (A) 13

A paper needs at least four 5-mark, three 10-mark and two 15-mark questions: 9 questions worth 80 marks. The other 20 marks come from $\{5,5,5,5\}$, $\{5,5,10\}$, $\{10,10\}$ or $\{5,15\}$, giving 13, 12, 11 or 11 questions. Everyone set the same number of questions, so MT and ET together have a multiple of 6. ET is longer, so only $11 + 13 = 24$ works: MT has 11 questions and ET has 13. ET's 13 questions mean its extra 20 marks are $\{5,5,5,5\}$, so ET has $4 + 4 = 8$ five-mark questions. MT's 11 questions mean its extra is $\{10,10\}$ (four 5-mark questions) or $\{5,15\}$ (five). With only four, MT's fifth question would carry 10 marks, but Annie's fifth MT question carries 5. So MT has five. The completed MT and ET tables show these marks: Total $= 5 + 8 = 13$. Option B (12) is the count if MT had only four 5-mark questions, which statement (i) rules out. Both papers are fully fixed, so option D fails too. Hence, option A (13).

Q45MCQLogical Puzzles

1. AOnly Beti, Dave, Esha and Fakir
2. BOnly Dave and Fakir
3. COnly Esha and Fakir
4. DOnly Dave, Esha and Fakir

Answer and solution

Answer: (D) Only Dave, Esha and Fakir

A paper needs at least four 5s, three 10s and two 15s: 9 questions worth 80 marks. The other 20 marks add 2, 3 or 4 questions, so a paper has 11, 12 or 13. Equal shares make MT plus ET a multiple of 6, and ET is longer, so only $11 + 13 = 24$ works: each person set 4. An 11-question paper has four 5s (extra $\{10,10\}$) or five (extra $\{5,15\}$). MT's fifth question carries 5 marks, so MT has five 5s, three 10s and three 15s (Q9–Q11). ET's 13 questions need extra $\{5,5,5,5\}$: eight 5s, three 10s and two 15s (Q12–Q13). MT: Annie set only Q5, so the others set 2 each: Fakir Q1–Q2, Chetan Q3–Q4, Dave Q10–Q11. Esha has Q8; taking Q7–Q8 would split Beti's pair, so Esha has Q8–Q9 and Beti Q6–Q7. ET: Annie set 3 and the others 2: Dave Q1–Q2, Chetan Q3–Q4, Annie Q5–Q7, Esha Q8–Q9, Beti Q10–Q11, Fakir Q12–Q13. The 15-mark questions came from Esha (MT Q9), Dave (MT Q10–Q11) and Fakir (ET Q12–Q13). Option A adds Beti, whose questions all carry 10 marks; options B and C each leave someone out. Hence, option D (Only Dave, Esha and Fakir).

Q46MCQLogical Puzzles

1. ASeventh question
2. BFourth question
3. CNinth question
4. DTenth question

Answer and solution

Answer: (D) Tenth question

A paper needs at least four 5-mark, three 10-mark and two 15-mark questions: 9 questions worth 80 marks. The other 20 marks need 2, 3 or 4 more questions, so a paper has 11, 12 or 13 questions. Everyone set the same number, so MT and ET together have a multiple of 6 questions. ET is longer, so only $11 + 13 = 24$ works, and each person set 4. In MT, Annie set one of the 11 questions and the other five set the remaining 10, at least 2 each, so exactly 2 each. In ET, Annie therefore set 3 and everyone else 2. ET in blocks: Dave set the first question, so Q1–Q2. Chetan has Q3, so Q3–Q4. Annie's three include Q5, so Q5–Q7. Esha has Q8, so Q8–Q9. Fakir set the last question, so Q12–Q13. That leaves Q10–Q11 for Beti. The completed ET table shows these blocks: So Beti set the tenth question. The seventh (option A) is Annie's, the fourth (option B) Chetan's and the ninth (option C) Esha's. Hence, option D (Tenth question).

Q47TITAMissing Value Tables

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

Answer and solution

Answer: 13

Each row and each column has nine pouches with an integer average, so every row total and column total is a multiple of 9. Three pouches with minimum $a$ and maximum $b$ hold between $2a + b$ and $a + 2b$ coins, depending on the middle pouch. Column 1: the row 1 slot (minimum 2, maximum 4) holds 8 to 10 coins, the row 2 slot (3, 5) holds 11 to 13, and the row 3 slot holds exactly 4 (given). So the column total lies between $8 + 11 + 4 = 23$ and $10 + 13 + 4 = 27$. The only multiple of 9 in that range is 27, and reaching it needs both slots at their maximum: row 1 holds $2 + 4 + 4 = 10$ and row 2 holds $3 + 5 + 5 = 13$. The completed grid, with each slot's total and its three pouches, shows these in column 1: The answer is 13.

Q48TITAMissing Value Tables

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

Answer and solution

Answer: 8

Each row and column has nine pouches with an integer average, so its total is a multiple of 9. Three pouches with minimum $a$ and maximum $b$ hold between $2a + b$ and $a + 2b$ coins. Column 1: 8 to 10, 11 to 13 and the given 4 total 23 to 27, so 27. The pouches are $2, 4, 4$ and $3, 5, 5$, and the 4 coins (minimum 1, maximum 2) are $1, 1, 2$. Column 2: 20 to 22, the centre's 3 (every pouch holds 1) and 4 to 5 total 27 to 30, so 27. The pouches are $6, 6, 8$, then $1, 1, 1$, then $1, 1, 2$. Column 3 follows from the rows. Row 1: $10 + 20$ plus 5 to 7 must be 36, so $1, 2, 3$. Row 2: $13 + 3$ plus 32 to 46 must be 54, so $6, 12, 20$. Row 3: $4 + 4$ plus 9 to 12 must be 18, so $2, 3, 5$. The completed grid shows every slot's pouches: Pouches with exactly one coin: 2 in row 3 column 1, 3 in the centre, 2 in row 3 column 2 and 1 in row 1 column 3, so $2 + 3 + 2 + 1 = 8$. The answer is 8.

Q49TITAMissing Value Tables

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

Answer and solution

Answer: 2

A slot's average is an integer exactly when its three-pouch total is a multiple of 3, so find all nine totals. Each row and column has nine pouches with an integer average, so its total is a multiple of 9. Three pouches with minimum $a$ and maximum $b$ hold between $2a + b$ and $a + 2b$ coins. Column 1: 8 to 10, 11 to 13 and the given 4 total 23 to 27, so 27, giving 10, 13 and 4. Column 2: 20 to 22, then 3 in the centre (every pouch holds 1), then 4 to 5 total 27 to 30, so 27, giving 20, 3 and 4. Column 3 follows from the rows. Row 1: $10 + 20$ plus 5 to 7 must be 36, so 6. Row 2: $13 + 3$ plus 32 to 46 must be 54, so 38. Row 3: $4 + 4$ plus 9 to 12 must be 18, so 10. The completed grid gives each slot's total with its three pouches below it: The nine totals are 10, 20, 6, 13, 3, 38, 4, 4 and 10. Only 3 (the centre) and 6 (row 1, column 3) are multiples of 3. The answer is 2.

Q50TITAMissing Value Tables

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

Answer and solution

Answer: 3

Each row and column has nine pouches with an integer average, so its total is a multiple of 9. Three pouches with minimum $a$ and maximum $b$ hold between $2a + b$ and $a + 2b$ coins. Column 1: 8 to 10, 11 to 13 and the given 4 total 23 to 27, so 27: the slots hold 10, 13 and 4. Column 2: 20 to 22, then 3 in the centre (every pouch holds 1), then 4 to 5 total 27 to 30, so 27: the slots hold 20, 3 and 4. Column 3 follows from the rows. Row 1: $10 + 20$ plus 5 to 7 must be 36, so 6. Row 2: $13 + 3$ plus 32 to 46 must be 54, so 38. Row 3: $4 + 4$ plus 9 to 12 must be 18, so 10. The completed grid gives each slot's total with its three pouches below it: Totals above Rs. 10: 13, 20 and 38. The two slots with exactly Rs. 10 (row 1 column 1 and row 3 column 3) do not count, because the total must strictly exceed Rs. 10. The answer is 3.

Q51MCQTeam Selection & Scheduling

1. A12
2. B30
3. C31
4. D15

Q52MCQTeam Selection & Scheduling

1. ADr. Wayne
2. BDr. Kane
3. CDr. Ben
4. DBoth Dr. Wayne and Dr. Kane

Q53MCQTeam Selection & Scheduling

1. ASame duration on all three days
2. BFriday
3. CMonday
4. DWednesday

Answer and solution

Answer: (C) Monday

Mr. Singh arrived at 8:50 with token 13, so patients 1 to 13 were all waiting when the clinic opened at 9 a.m. The timings below therefore hold on all three days; only the rooms differ. Dr. Ben (10 minutes) becomes free at 9:10, 9:20, 9:30, 9:40 and 9:50; Dr. Kane (15 minutes) at 9:15, 9:30 and 9:45; Dr. Wayne (25 minutes) at 9:25 and 9:50. Patients 1 to 3 go in at 9:00, then patient 4 at 9:10, 5 at 9:15, 6 at 9:20, 7 at 9:25, 8 and 9 at 9:30, 10 at 9:40 and 11 at 9:45. At 9:50 Ben and Wayne are both free: patient 12 takes the lower-numbered room and patient 13 the other. Monday: Ben is in room 1 and Wayne in room 3, so Mr. Singh sees Wayne from 9:50 to 10:15. Wednesday (Wayne in room 1, Ben in room 2) and Friday (Wayne in room 2, Ben in room 3): he sees Ben from 9:50 to 10:00. Timelines for Monday, Wednesday and Friday, top to bottom (doctors in room order): He stays 85 minutes on Monday but 70 minutes on the other two days, so the durations are not equal (option A), and Friday (option B) and Wednesday (option D) are shorter. Hence, option C (Monday).

Q54MCQTeam Selection & Scheduling

1. A30
2. B10
3. C15
4. D0

Answer and solution

Answer: (D) 0

On Thursday, room 1 has Dr. Wayne (25 minutes), room 2 Dr. Ben (10 minutes) and room 3 Dr. Kane (15 minutes). 9:00: the two waiting patients take rooms 1 and 2: Wayne (till 9:25) and Ben (till 9:10). Kane stays free. 9:15: two arrive. Wayne is busy, so they go to Ben (till 9:25) and Kane (till 9:30). 9:30: two arrive just as Kane's patient leaves, and take rooms 1 and 2: Wayne (till 9:55) and Ben (till 9:40). 9:45: two arrive. Wayne is busy, so Ben (till 9:55) and Kane (till 10:00) take them. At 10:00 the situation is the same as at 9:30, so this 30-minute cycle repeats until closing. The timeline shows the pattern: Wayne is free only from 9:25 to 9:30, 9:55 to 10:00, and so on, and Kane is busy during each of these gaps (till 9:30, till 10:00, and so on). Kane's own free spells, such as 9:00 to 9:15 and 9:30 to 9:45, all fall while Wayne is busy, so option C (15) fails. The three doctors are never free together. Hence, option D (0).

Q55MCQTeam Selection & Scheduling

1. APaula
2. BTerence
3. CQuentin
4. DSally

Q56MCQTeam Selection & Scheduling

1. APaula
2. BQuentin
3. CSally
4. DRobert

Q57MCQTeam Selection & Scheduling

1. ATerence
2. BSally
3. CPaula
4. DNo one

Q58MCQTeam Selection & Scheduling

1. AArabic and French
2. BBasque and French
3. CArabic and Basque
4. DBasque and Dutch

Q59TITARadar & Special Graphs

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

Answer and solution

Answer: 25

Take the store's 2016 cost as 100. A percentage profit of 100% means profit equals cost, so 2016 revenue was 200. Cost doubled from 2016 to 2018, so the 2018 cost is 200. The chart shows Produce incurred 50% of the 2018 cost, which is $0.5 \times 200 = 100$. In 2018, Clothing's revenue equalled Produce's cost, so Clothing earned 100. Clothing brought in 40% of the 2018 revenue, so the total revenue was $100 \div 0.4 = 250$. The completed cost and revenue tables (2016 cost taken as 100) show these values: Percentage profit in 2018 $= \frac{250 - 200}{200} \times 100 = 25\%$. The answer is 25.

Q60MCQRadar & Special Graphs

1. A16 : 9
2. B4 : 3
3. C9 : 16
4. D8 : 5

Answer and solution

Answer: (D) 8 : 5

Take the 2016 cost as 100. A 100% profit means 2016 revenue was 200. 2017: revenue doubled to 400, and the chart shows Produce earned 40% of it, which is 160. 2018: cost doubled from 2016 to 200. Produce incurred 50% of it, which is 100, and this equals Clothing's 2018 revenue. Clothing earned 40% of the 2018 revenue, so the total was $100 \div 0.4 = 250$. Produce also earned 40% of it, which is 100. The completed cost and revenue tables (2016 cost taken as 100) show these values: Ratio $= 160 : 100 = 8 : 5$. Option A (16 : 9) would need 2018 Produce revenue of 90, and option B (4 : 3) would need 120, but it is 100. Hence, option D (8 : 5).

Q61TITARadar & Special Graphs

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

Answer and solution

Answer: 70

Take the store's 2016 cost as 100. A percentage profit of 100% means profit equals cost, so revenue was 200 and profit was 100. The chart shows Electronics earned 50% of the 2016 revenue, which is $0.5 \times 200 = 100$, and incurred 30% of the 2016 cost, which is $0.3 \times 100 = 30$. The completed cost and revenue tables (2016 cost taken as 100) show these values: Electronics profit $= 100 - 30 = 70$. Share of the store's profit $= \frac{70}{100} \times 100 = 70\%$. The answer is 70.

Q62MCQRadar & Special Graphs

1. A15.5
2. B25.0
3. C8.3
4. D33.3

Answer and solution

Answer: (C) 8.3

Take the 2016 cost as 100. A 100% profit means 2016 revenue was 200. 2017: revenue doubled to 400. Electronics earned 30% of it, which is 120, and made no profit, so its cost was also 120. The chart puts Electronics at 40% of the 2017 cost, so total cost $= 120 \div 0.4 = 300$. Profit percentage $= \frac{400 - 300}{300} \times 100 \approx 33.33\%$. 2018: cost doubled from 2016 to 200. Produce's 50% share of that cost is 100, which equals Clothing's revenue; Clothing earned 40% of revenue, so revenue $= 100 \div 0.4 = 250$. Profit percentage $= \frac{250 - 200}{200} \times 100 = 25\%$. The completed cost and revenue tables (2016 cost taken as 100) show these totals: Difference $= 33.33 - 25 \approx 8.3$ percentage points. Options B (25.0) and D (33.3) are the two years' profit percentages themselves, not their difference. Hence, option C (8.3).

Q63TITASet Theory

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

Answer and solution

Answer: 64

Every student prefers exactly one of the two candidates. 200 prefer Sunita, so $500 - 200 = 300$ prefer Ragini. By statement 2, 80% of Sunita's 200, which is 160, support proposal A. By statement 3, 30% of Ragini's 300, which is 90, support proposal A. Together $160 + 90 = 250$, matching the 250 supporters of proposal A in statement 1. The solved diagrams for Sunita's and Ragini's supporters show these as A(160) and A(90); their split of proposal B is not needed here. Share of proposal A's supporters who prefer Sunita $= \frac{160}{250} \times 100 = 64\%$. The answer is 64.

Q64TITASet Theory

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

Answer and solution

Answer: 84

Every student prefers exactly one candidate: 200 prefer Sunita, so $500 - 200 = 300$ prefer Ragini. By statement 1, 250 students supported proposal A, so $500 - 250 = 250$ did not. By statement 3, $30\%$ of Ragini's 300 students, that is 90, supported A. So $300 - 90 = 210$ of Ragini's students did not support A. Check with statement 2: $80\%$ of Sunita's 200, that is 160, supported A, and $160 + 90 = 250$, as statement 1 says. So $200 - 160 = 40$ of Sunita's students did not support A, and $40 + 210 = 250$. The completed diagrams, built from all seven statements, show the same split: 40 of Sunita's students and $150 + 60 = 210$ of Ragini's lie outside A. Statements 4 to 7 fill in the proposal B regions but are not needed here. Required percentage $= \frac{210}{250} \times 100 = 84\%$. The answer is 84.

Q65MCQSet Theory

1. A40
2. B25
3. C20
4. D50

Answer and solution

Answer: (D) 50

Every student prefers one candidate: 200 prefer Sunita, so $500 - 200 = 300$ prefer Ragini. Sunita: by statement 4, $20\%$ of the 250 supporters of proposal B prefer Sunita, which is 50. By statement 6, all 50 also support proposal A, so 50 of Sunita's students support both proposals. Ragini: by statement 3, $30\%$ of her 300 students, that is 90, support proposal A. By statement 5, $40\%$ of the 250 students who do not support proposal B, that is 100, prefer Ragini. By statement 7, $20\%$ of her 300 students, that is 60, support neither proposal. So $100 - 60 = 40$ of them support proposal A only, and $90 - 40 = 50$ support both. So $50 + 50 = 100$ students support both proposals, and 50 of them prefer Sunita: $\frac{50}{100} \times 100 = 50\%$. Option C (20) is the share of all proposal B supporters who prefer Sunita (statement 4), not the share among those who support both. Hence, option D (50).

Q66MCQSet Theory

1. A150
2. B210
3. C200
4. D40

Answer and solution

Answer: (A) 150

Every student prefers one candidate: 200 prefer Sunita, so $500 - 200 = 300$ prefer Ragini. By statement 4, $20\%$ of the 250 supporters of proposal B prefer Sunita, which is 50, so $250 - 50 = 200$ of Ragini's students support proposal B. By statement 3, $30\%$ of Ragini's 300 students, that is 90, support proposal A. By statement 5, $40\%$ of the 250 students who do not support proposal B, that is 100, prefer Ragini. By statement 7, $20\%$ of her 300 students, that is 60, support neither proposal. So $100 - 60 = 40$ support proposal A only, and $90 - 40 = 50$ support both proposals. Ragini's students who support proposal B but not proposal A $= 200 - 50 = 150$. Option C (200) counts every Ragini student who supports proposal B, including the 50 who also support proposal A. Hence, option A (150).