CAT 2023 Slot 2 — DILR questions with answers

Q25MCQLogical Puzzles

1. A36
2. B30
3. C15
4. D45

Answer and solution

Answer: (D) 45

The nine box averages are distinct integers from 1 to 9, so they are exactly 1, 2, …, 9. A box holds 3 times its average, so all the boxes together hold $3 \times (1 + 2 + \dots + 9) = 3 \times 45 = 135$ coins. The three rows have equal totals, so each row, including the 3rd, holds $135 \div 3 = 45$ coins. The completed grid confirms this: Row 3 has boxes of 7, 8, 9; 1, 8, 9; and 1, 1, 1 coins, which total $24 + 18 + 3 = 45$. Option C (15) is the sum of the three box averages in a row, not the number of coins; each box holds three sacks, so the coins are 3 times that. Hence, option D (45).

Q26MCQLogical Puzzles

1. A3
2. B4
3. C5
4. D8

Answer and solution

Answer: (C) 5

The nine averages are distinct integers from 1 to 9, so exactly 1 to 9; equal row and column totals make each row's and column's averages sum to 15. (r,c) is row r, column c. (3,1): median 8, all sacks above 5, one star, so only 'maximum 9' holds; divisibility by 3 gives 7, 8, 9 (average 8). (2,1): median 2, two stars, so minimum 1 and maximum 9: 1, 2, 9 (average 4). (1,2): median 9 and only two sacks above 5, so 9, 9 and one of 2 to 5 (one star rules out 1); divisibility by 3 gives 3, 9, 9 (average 7). So (1,1) = 3 and (1,3) = 5. Rows 2 and 3 need pairs from 1, 2, 6, 9 summing to 11 and 7; column 2 needs 8: (2,2) = 2, (2,3) = 9, (3,2) = 6, (3,3) = 1. Table 2 gives (1,1) 1, 1, 7 (a 9 cannot fit); (1,3) 1, 6, 8 (a 9 leaves 0); (2,2) 1, 2, 3 (1, 1, 4 earns two stars); (2,3) 9, 9, 9; (3,2) 1, 8, 9; (3,3) 1, 1, 1. Boxes with a 9-coin sack: (1,2), (2,1), (2,3), (3,1), (3,2). The other four top out at 7, 8, 3 and 1; option B (4) misses one. Hence, option C (5).

Q27TITALogical Puzzles

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

Answer and solution

Answer: 4

The nine averages are distinct integers from 1 to 9, so exactly 1 to 9; equal row and column totals make each row's and column's averages sum to 15. (r,c) is row r, column c. (3,1): median 8, all sacks above 5, one star, so only 'maximum 9' holds; divisibility by 3 gives 7, 8, 9 (average 8). (2,1): median 2, two stars, so minimum 1 and maximum 9: 1, 2, 9 (average 4). (1,2): median 9 and only two sacks above 5, so 9, 9 and one of 2 to 5 (one star rules out 1); divisibility by 3 gives 3, 9, 9 (average 7). So (1,1) = 3 and (1,3) = 5. Rows 2 and 3 need pairs from 1, 2, 6, 9 summing to 11 and 7; column 2 needs 8: (2,2) = 2, (2,3) = 9, (3,2) = 6, (3,3) = 1. Table 2 gives (1,1) 1, 1, 7 (a 9 cannot fit); (1,3) 1, 6, 8 (a 9 leaves 0); (2,2) 1, 2, 3 (1, 1, 4 earns two stars); (2,3) 9, 9, 9; (3,2) 1, 8, 9; (3,3) 1, 1, 1. Average equals median in (2,2), (2,3), (3,1) and (3,3): 2 and 2, 9 and 9, 8 and 8, 1 and 1. The other five differ. The answer is 4.

Q28TITALogical Puzzles

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

Answer and solution

Answer: 9

The nine averages are distinct integers from 1 to 9, so exactly 1 to 9; equal row and column totals make each row's and column's averages sum to 15. (r,c) is row r, column c. (3,1): median 8, all sacks above 5, one star, so only 'maximum 9' holds; divisibility by 3 gives 7, 8, 9 (average 8). (2,1): median 2, two stars, so minimum 1 and maximum 9: 1, 2, 9 (average 4). (1,2): median 9 and only two sacks above 5, so 9, 9 and one of 2 to 5 (one star rules out 1); divisibility by 3 gives 3, 9, 9 (average 7). So (1,1) = 3 and (1,3) = 5. Rows 2 and 3 need pairs from 1, 2, 6, 9 summing to 11 and 7; column 2 needs 8: (2,2) = 2, (2,3) = 9, (3,2) = 6, (3,3) = 1. Table 2 gives (1,1) 1, 1, 7 (a 9 cannot fit); (1,3) 1, 6, 8 (a 9 leaves 0); (2,2) 1, 2, 3 (1, 1, 4 earns two stars); (2,3) 9, 9, 9; (3,2) 1, 8, 9; (3,3) 1, 1, 1. One-coin sacks: 2 in (1,1), one each in (1,3), (2,1), (2,2) and (3,2), and 3 in (3,3): $2 + 4 + 3 = 9$. The answer is 9.

Q29TITALogical Puzzles

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

Answer and solution

Answer: 5

The nine averages are distinct integers from 1 to 9, so exactly 1 to 9; equal row and column totals make each row's and column's averages sum to 15. (r,c) is row r, column c. (3,1): median 8, all sacks above 5, one star, so only 'maximum 9' holds; divisibility by 3 gives 7, 8, 9 (average 8). (2,1): median 2, two stars, so minimum 1 and maximum 9: 1, 2, 9 (average 4). (1,2): median 9 and only two sacks above 5, so 9, 9 and one of 2 to 5 (one star rules out 1); divisibility by 3 gives 3, 9, 9 (average 7). So (1,1) = 3 and (1,3) = 5. Rows 2 and 3 need pairs from 1, 2, 6, 9 summing to 11 and 7; column 2 needs 8: (2,2) = 2, (2,3) = 9, (3,2) = 6, (3,3) = 1. Table 2 gives (1,1) 1, 1, 7 (a 9 cannot fit); (1,3) 1, 6, 8 (a 9 leaves 0); (2,2) 1, 2, 3 (1, 1, 4 earns two stars); (2,3) 9, 9, 9; (3,2) 1, 8, 9; (3,3) 1, 1, 1. All three sacks differ in (1,3), (2,1), (2,2), (3,1) and (3,2); (1,1), (1,2), (2,3) and (3,3) repeat a number. The answer is 5.

Q30MCQMissing Value Tables

1. AOnly Bzygoo and Czechy and Drjbna
2. BOnly Czechy and Drjbna
3. COnly Drjbna
4. DOnly Czechy

Answer and solution

Answer: (B) Only Czechy and Drjbna

Amounts start and end at 1, rise strictly to a peak and then fall strictly, in steps of 1 or 2. Alfloo (2009–2016, 21 crores): the six middle years total 19, with 2 or 3 at each end. A 3 at either end leaves 13 or 14 for the four between, which need at least 16 (such as 4–5–4–3). So both ends are 2 and the four total 15: 3–4–5–3 or 3–5–4–3. Bzygoo (2012–2015): 1–2–3–1 or 1–3–2–1. Czechy (from 2013, total 9): only five years work, 1–2–3–2–1 (four or fewer give at most 7, six or more at least 13). Drjbna (2011–2015, total 10): the middle three total 8 with ends 2 or 3: 2–4–2. Elavalaki (from 2010, total 13): 1–3–5–3–1 (to 2014), or 1–2–3–4–2–1 or 1–2–4–3–2–1 (to 2015); four years or fewer give at most 7, seven at least 16. Only Czechy and Drjbna have a single sequence. Bzygoo can raise 2 then 3 or 3 then 2, so option A fails. Hence, option B (Only Czechy and Drjbna).

Q31MCQMissing Value Tables

1. AIt is either Rs. 7 crores or Rs. 8 crores or Rs. 9 crores.
2. BIt is exactly Rs. 8 crores.
3. CIt is either Rs. 7 crores or Rs. 8 crores.
4. DIt is either Rs. 8 crores or Rs. 9 crores.

Answer and solution

Answer: (C) It is either Rs. 7 crores or Rs. 8 crores.

Amounts start and end at 1, rise strictly to a peak and then fall strictly, in steps of 1 or 2. Alfloo (2009–2016, 21 crores): the six middle years total 19, with 2 or 3 at each end. A 3 at either end leaves 13 or 14 for the four between, which need at least 16 (such as 4–5–4–3). So both ends are 2 and the four total 15: 3–4–5–3 or 3–5–4–3. Bzygoo (2012–2015): 1–2–3–1 or 1–3–2–1. Czechy (from 2013, total 9): only five years work, 1–2–3–2–1 (four or fewer give at most 7, six or more at least 13). Drjbna (2011–2015, total 10): the middle three total 8 with ends 2 or 3: 2–4–2. Elavalaki (from 2010, total 13): 1–3–5–3–1 (to 2014), or 1–2–3–4–2–1 or 1–2–4–3–2–1 (to 2015); four years or fewer give at most 7, seven at least 16. 2015: Alfloo 2, Bzygoo 1, Czechy 3, Drjbna 1, Elavalaki 1 or 0 (closed). Total 8 or 7; 9 is impossible. Hence, option C (It is either Rs. 7 crores or Rs. 8 crores.).

Q32TITAMissing Value Tables

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

Answer and solution

Answer: 17

Amounts start and end at 1, rise strictly to a peak and then fall strictly, in steps of 1 or 2. Alfloo (2009–2016, 21 crores): the six middle years total 19, with 2 or 3 at each end. A 3 at either end leaves 13 or 14 for the four between, which need at least 16 (such as 4–5–4–3). So both ends are 2 and the four total 15: 3–4–5–3 or 3–5–4–3. Bzygoo (2012–2015): 1–2–3–1 or 1–3–2–1. Czechy (from 2013, total 9): only five years work, 1–2–3–2–1 (four or fewer give at most 7, six or more at least 13). Drjbna (2011–2015, total 10): the middle three total 8 with ends 2 or 3: 2–4–2. Elavalaki (from 2010, total 13): 1–3–5–3–1 (to 2014), or 1–2–3–4–2–1 or 1–2–4–3–2–1 (to 2015); four years or fewer give at most 7, seven at least 16. Largest 2013 amounts: Alfloo 5, Bzygoo 3, Czechy 1, Drjbna 4, Elavalaki 4 (the 1–2–3–4–2–1 case). The firms are independent, so the total is $5 + 3 + 1 + 4 + 4 = 17$. The answer is 17.

Q33MCQMissing Value Tables

1. A12
2. B9
3. C11
4. D10

Answer and solution

Answer: (C) 11

Amounts start and end at 1, rise strictly to a peak and then fall strictly, in steps of 1 or 2. Alfloo (2009–2016, 21 crores): the six middle years total 19, with 2 or 3 at each end. A 3 at either end leaves 13 or 14 for the four between, which need at least 16 (such as 4–5–4–3). So both ends are 2 and the four total 15: 3–4–5–3 or 3–5–4–3. Bzygoo (2012–2015): 1–2–3–1 or 1–3–2–1. Czechy (from 2013, total 9): only five years work, 1–2–3–2–1 (four or fewer give at most 7, six or more at least 13). Drjbna (2011–2015, total 10): the middle three total 8 with ends 2 or 3: 2–4–2. Elavalaki (from 2010, total 13): 1–3–5–3–1 (to 2014), or 1–2–3–4–2–1 or 1–2–4–3–2–1 (to 2015); four years or fewer give at most 7, seven at least 16. Elavalaki's 3 in 2013 means 5 or 4 in 2012. Take 4, with Alfloo's smaller 4, Bzygoo 1, Czechy 0 (not founded) and Drjbna 2: $4 + 1 + 0 + 2 + 4 = 11$. Option D (10) needs Elavalaki at 3 in 2012, which comes with 4 in 2013. Hence, option C (11).

Q34MCQMissing Value Tables

1. ABzygoo raised the same amount of money as Elavalaki in 2013.
2. BAlfloo raised the same amount of money as Drjbna in 2013.
3. CAlfloo raised the same amount of money as Bzygoo in 2014.
4. DBzygoo raised more money than Elavalaki in 2014.

Answer and solution

Answer: (A) Bzygoo raised the same amount of money as Elavalaki in 2013.

Amounts start and end at 1, rise strictly to a peak and then fall strictly, in steps of 1 or 2. Alfloo (2009–2016, 21 crores): the six middle years total 19, with 2 or 3 at each end. A 3 at either end leaves 13 or 14 for the four between, which need at least 16 (such as 4–5–4–3). So both ends are 2 and the four total 15: 3–4–5–3 or 3–5–4–3. Bzygoo (2012–2015): 1–2–3–1 or 1–3–2–1. Czechy (from 2013, total 9): only five years work, 1–2–3–2–1 (four or fewer give at most 7, six or more at least 13). Drjbna (2011–2015, total 10): the middle three total 8 with ends 2 or 3: 2–4–2. Elavalaki (from 2010, total 13): 1–3–5–3–1 (to 2014), or 1–2–3–4–2–1 or 1–2–4–3–2–1 (to 2015); four years or fewer give at most 7, seven at least 16. In 2014 Alfloo, Czechy and Drjbna raised $3 + 2 + 2 = 7$, so Bzygoo and Elavalaki raised 5: 3 and 2. Then in 2013 Bzygoo raised 2 and Elavalaki 4 or 3, so A is impossible. B can hold (both 4); C and D hold. Hence, option A (Bzygoo raised the same amount of money as Elavalaki in 2013.).

Q35MCQLogical Puzzles

1. A5
2. B7
3. C6
4. D8

Answer and solution

Answer: (B) 7

From Table 1, consecutive day totals sum to 30, 31, 32 and 34. Days 3 and 4 are equal, so each is 16; then Days 1, 2 and 5 are 15, 15 and 18. Chatur scores multiples of 3. His Day 2 score, the unique highest, is 9 (at 6, his other days would all be 3, yet his minimum is only on Day 1); no one else scores 9. His Day 1 score, his unique minimum, is 3 (at 6, his other days would need 9), so Days 3 to 5 are 6, and Akhil's Day 4 is 3. Day 4: Bimal $= 16 - 3 - 6 = 7$. Day 3: Akhil and Bimal tie at rank 2, so each has $(16 - 6) \div 2 = 5$. Day 1: Bimal repeats his 5, so Akhil $= 15 - 3 - 5 = 7$. Day 2: Akhil and Bimal share 6, Akhil ahead: 5 and 1, or 4 and 2. Day 5: they share 12, Bimal above 6 but below 9, Akhil below 6: 5 and 7, or 4 and 8. So Akhil scored 7 on Day 1; option A (5) is Bimal's score. Hence, option B (7).

Q36MCQLogical Puzzles

1. ACannot be determined
2. BAkhil
3. CBimal
4. DChatur

Answer and solution

Answer: (D) Chatur

From Table 1, consecutive day totals sum to 30, 31, 32 and 34. Days 3 and 4 are equal, so each is 16; then Days 1, 2 and 5 are 15, 15 and 18. Chatur scores multiples of 3. His Day 2 score, the unique highest, is 9 (at 6, his other days would all be 3, yet his minimum is only on Day 1); no one else scores 9. His Day 1 score, his unique minimum, is 3 (at 6, his other days would need 9), so Days 3 to 5 are 6, and Akhil's Day 4 is 3. Day 4: Bimal $= 16 - 3 - 6 = 7$. Day 3: Akhil and Bimal tie at rank 2, so each has $(16 - 6) \div 2 = 5$. Day 1: Bimal repeats his 5, so Akhil $= 15 - 3 - 5 = 7$. Day 2: Akhil and Bimal share 6, Akhil ahead: 5 and 1, or 4 and 2. Day 5: they share 12, Bimal above 6 but below 9, Akhil below 6: 5 and 7, or 4 and 8. Chatur totals $3 + 9 + 6 + 6 + 6 = 30$, Akhil 23 to 25 and Bimal 25 to 27, so Chatur always leads; option A fails. Hence, option D (Chatur).

Q37TITALogical Puzzles

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

Answer and solution

Answer: 25

From Table 1, consecutive day totals sum to 30, 31, 32 and 34. Days 3 and 4 are equal, so each is 16; then Days 1, 2 and 5 are 15, 15 and 18. Chatur scores multiples of 3. His Day 2 score, the unique highest, is 9 (at 6, his other days would all be 3, yet his minimum is only on Day 1); no one else scores 9. His Day 1 score, his unique minimum, is 3 (at 6, his other days would need 9), so Days 3 to 5 are 6, and Akhil's Day 4 is 3. Day 4: Bimal $= 16 - 3 - 6 = 7$. Day 3: Akhil and Bimal tie at rank 2, so each has $(16 - 6) \div 2 = 5$. Day 1: Bimal repeats his 5, so Akhil $= 15 - 3 - 5 = 7$. Day 2: Akhil and Bimal share 6, Akhil ahead: 5 and 1, or 4 and 2. Day 5: they share 12, Bimal above 6 but below 9, Akhil below 6: 5 and 7, or 4 and 8. Bimal's lowest choices, 1 and 7, are both allowed: $5 + 1 + 5 + 7 + 7 = 25$. The answer is 25.

Q38MCQLogical Puzzles

1. ACannot be determined
2. B5
3. C6
4. D4

Answer and solution

Answer: (D) 4

From Table 1, consecutive day totals add to $30, 31, 32$ and $34$. Days 3 and 4 are equal, so each is $16$, and Day 2 $= 15$, Day 1 $= 15$, Day 5 $= 18$. Chatur scores only 3, 6 or 9. His Day-2 score, the unique highest, is 9; nobody else scores 9. His minimum comes only on Day 1, so he scores 3 on Day 1 and 6 on Days 3, 4 and 5. So Akhil's Day-4 score is 3. Day 3: Akhil and Bimal tie in Table 2, so each scores $(16 - 6)/2 = 5$. Bimal's Day 1 is also 5 (fact 3). Day 4: Bimal $= 16 - 6 - 3 = 7$. Day 2: Akhil ranks above Bimal and they total $15 - 9 = 6$, so Bimal scores 1 or 2. Day 5: Bimal $> 6 >$ Akhil and they total 12, so Bimal scores 7 or 8. Bimal's total is $5 + b_2 + 5 + 7 + b_5$, with $b_2$ and $b_5$ his Day-2 and Day-5 scores: 25, 26 or 27. Only 27 is a multiple of 3, so $b_2 = 2$, and Akhil's Day-2 score is $6 - 2 = 4$. Option B (5) needs $b_2 = 1$, giving Bimal 25 or 26. Hence, option D (4).

Q39TITALogical Puzzles

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

Answer and solution

Answer: 26

From Table 1, consecutive day totals sum to 30, 31, 32 and 34. Days 3 and 4 are equal, so each is 16; then Days 1, 2 and 5 are 15, 15 and 18. Chatur scores multiples of 3. His Day 2 score, the unique highest, is 9 (at 6, his other days would all be 3, yet his minimum is only on Day 1); no one else scores 9. His Day 1 score, his unique minimum, is 3 (at 6, his other days would need 9), so Days 3 to 5 are 6, and Akhil's Day 4 is 3. Day 4: Bimal $= 16 - 3 - 6 = 7$. Day 3: Akhil and Bimal tie at rank 2, so each has $(16 - 6) \div 2 = 5$. Day 1: Bimal repeats his 5, so Akhil $= 15 - 3 - 5 = 7$. Day 2: Akhil and Bimal share 6, Akhil ahead: 5 and 1, or 4 and 2. Day 5: they share 12, Bimal above 6 but below 9, Akhil below 6: 5 and 7, or 4 and 8. All days total 80 and Chatur has 30, so Akhil and Bimal always total 50; Bimal has $50 - 24 = 26$. The answer is 26.

Q40MCQTeam Selection & Scheduling

1. A90
2. B120
3. C110
4. D100

Answer and solution

Answer: (C) 110

Anjali waited from 10:30 for Chitra to finish Ride-1, so Chitra rode it 10 to 11 am (Rs 20). Her other Rs 30 was Ride-3, 9 to 10 am. Anjali rode Ride-1 from 11 am to 12 pm and, taking no break, started another ride at 12 pm. She had spent Rs 50 by 12:15, so that ride cost Rs 30: Ride-3, 12 to 1 pm. Bipasha's first ride, at 11:30, must cost Rs 50 by 12:15: Ride-2, until 12:30. Their shared last ride is not Ride-1 or Ride-3 (Anjali's) or Ride-2 (Bipasha's first), so it is Ride-4. If Anjali rode it at 1 pm, Bipasha would queue at 1:30 after a break from 12:30, leaving her only two rides. So Anjali rode Ride-2 from 1 to 2 pm and Ride-4 from 2 to 3 pm. Bipasha queued at 2:30 after a break from 1:30, so her middle ride ran 12:30 to 1:30. Ride-3 closes at 1 pm, so it was Ride-1. She rode Ride-4 from 3 to 4 pm. Bipasha paid $50 + 20 + 40 = 110$. Option B (120) takes Ride-3 as her middle ride, which Ride-3's 1 pm closing rules out. Hence, option C (110).

Q41MCQTeam Selection & Scheduling

1. ARide-1 and Ride-3
2. BRide-1, Ride-2, and Ride-3
3. CRide-1, Ride-2, and Ride-4
4. DRide-1 and Ride-4

Answer and solution

Answer: (B) Ride-1, Ride-2, and Ride-3

Anjali waited from 10:30 for Chitra to finish Ride-1, so Chitra rode it 10 to 11 am (Rs 20). Her other Rs 30 was Ride-3, 9 to 10 am. Anjali rode Ride-1 from 11 am to 12 pm and, taking no break, started another ride at 12 pm. She had spent Rs 50 by 12:15, so that ride cost Rs 30: Ride-3, 12 to 1 pm. Bipasha's first ride, at 11:30, must cost Rs 50 by 12:15: Ride-2, until 12:30. Their shared last ride is not Ride-1 or Ride-3 (Anjali's) or Ride-2 (Bipasha's first), so it is Ride-4. If Anjali rode it at 1 pm, Bipasha would queue at 1:30 after a break from 12:30, leaving her only two rides. So Anjali rode Ride-2 from 1 to 2 pm and Ride-4 from 2 to 3 pm. Bipasha queued at 2:30 after a break from 1:30, so her middle ride ran 12:30 to 1:30. Ride-3 closes at 1 pm, so it was Ride-1. She rode Ride-4 from 3 to 4 pm. By 2 pm Anjali had finished Ride-1 (12 pm), Ride-3 (1 pm) and Ride-2 (2 pm). Option A misses Ride-2, which ends exactly at 2 pm. Hence, option B (Ride-1, Ride-2, and Ride-3).

Q42MCQTeam Selection & Scheduling

1. ARide-1
2. BRide-4
3. CRide-3
4. DRide-2

Answer and solution

Answer: (A) Ride-1

Anjali waited from 10:30 for Chitra to finish Ride-1, so Chitra rode it 10 to 11 am (Rs 20). Her other Rs 30 was Ride-3, 9 to 10 am. Anjali rode Ride-1 from 11 am to 12 pm and, taking no break, started another ride at 12 pm. She had spent Rs 50 by 12:15, so that ride cost Rs 30: Ride-3, 12 to 1 pm. Bipasha's first ride, at 11:30, must cost Rs 50 by 12:15: Ride-2, until 12:30. Their shared last ride is not Ride-1 or Ride-3 (Anjali's) or Ride-2 (Bipasha's first), so it is Ride-4. If Anjali rode it at 1 pm, Bipasha would queue at 1:30 after a break from 12:30, leaving her only two rides. So Anjali rode Ride-2 from 1 to 2 pm and Ride-4 from 2 to 3 pm. Bipasha queued at 2:30 after a break from 1:30, so her middle ride ran 12:30 to 1:30. Ride-3 closes at 1 pm, so it was Ride-1. She rode Ride-4 from 3 to 4 pm. Ride-1 was taken by Chitra, Anjali and Bipasha. Chitra rode only Rides 1 and 3, and Bipasha never rode Ride-3, so Rides 2, 3 and 4 each miss someone. Hence, option A (Ride-1).

Q43TITATeam Selection & Scheduling

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

Answer and solution

Answer: 6

Anjali waited from 10:30 for Chitra to finish Ride-1, so Chitra rode it 10 to 11 am (Rs 20). Her other Rs 30 was Ride-3, 9 to 10 am. Anjali rode Ride-1 from 11 am to 12 pm and, taking no break, started another ride at 12 pm. She had spent Rs 50 by 12:15, so that ride cost Rs 30: Ride-3, 12 to 1 pm. Bipasha's first ride, at 11:30, must cost Rs 50 by 12:15: Ride-2, until 12:30. Their shared last ride is not Ride-1 or Ride-3 (Anjali's) or Ride-2 (Bipasha's first), so it is Ride-4. If Anjali rode it at 1 pm, Bipasha would queue at 1:30 after a break from 12:30, leaving her only two rides. So Anjali rode Ride-2 from 1 to 2 pm and Ride-4 from 2 to 3 pm. Bipasha queued at 2:30 after a break from 1:30, so her middle ride ran 12:30 to 1:30. Ride-3 closes at 1 pm, so it was Ride-1. She rode Ride-4 from 3 to 4 pm. Anjali took 4 rides and Chitra 2, so together $4 + 2 = 6$. The answer is 6.

Q44TITATeam Selection & Scheduling

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

Answer and solution

Answer: 140

Anjali waited from 10:30 for Chitra to finish Ride-1, so Chitra rode it 10 to 11 am (Rs 20). Her other Rs 30 was Ride-3, 9 to 10 am. Anjali rode Ride-1 from 11 am to 12 pm and, taking no break, started another ride at 12 pm. She had spent Rs 50 by 12:15, so that ride cost Rs 30: Ride-3, 12 to 1 pm. Bipasha's first ride, at 11:30, must cost Rs 50 by 12:15: Ride-2, until 12:30. Their shared last ride is not Ride-1 or Ride-3 (Anjali's) or Ride-2 (Bipasha's first), so it is Ride-4. If Anjali rode it at 1 pm, Bipasha would queue at 1:30 after a break from 12:30, leaving her only two rides. So Anjali rode Ride-2 from 1 to 2 pm and Ride-4 from 2 to 3 pm. Bipasha queued at 2:30 after a break from 1:30, so her middle ride ran 12:30 to 1:30. Ride-3 closes at 1 pm, so it was Ride-1. She rode Ride-4 from 3 to 4 pm. Anjali paid $20 + 30 + 50 + 40 = 140$. The answer is 140.