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CAT 2023 Slot 2 DILR Questions & Answer Key
Data Interpretation & Logical Reasoning questions from CAT 2023 Slot 2 (IIM Lucknow, Afternoon Session (12:30 PM – 2:30 PM IST); 20 in the exam) with the answer key and step-by-step solutions
Executive Answer & Section Summary
CAT 2023 Slot 2 Data Interpretation & Logical Reasoning (DILR) was conducted by IIM Lucknow during the Afternoon Session (12:30 PM – 2:30 PM IST). The section comprised 20 questions administered under a strict 40-minute sectional countdown. Sectional marking adheres to standard CAT conventions: +3 marks for every correct answer, -1 mark penalty for incorrect MCQs, and zero penalty for non-MCQ TITA questions. Securing a 99.0th percentile in this section required approximately 23–26 marks (~8 net correct). The question distribution featured 4 to 5 multi-condition caselets covering matrix arrangements, tabular reasoning, games-tournaments, and data graphs under strict 40-minute clocks. All questions in this archive are sourced from official candidate response sheets, cross-verified with the convenor master key, and solved with step-by-step logic proofs on ThinkCAT.
Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.
Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.
The table below provides partial information about the five firms.
Firm
First year of existence
Last year of existence
Total amount raised (Rs. crores)
Alfloo
2009
2016
21
Bzygoo
2012
2015
Czechy
2013
9
Drjbna
2011
2015
10
Elavalaki
2010
13
For which firm(s) can the amounts raised by them be concluded with certainty in each year?
Official Correct Answer: C. ### Core Concept
Each firm's yearly series starts and ends at 1, rises and then falls strictly, and moves by 1 or 2 each year. A firm's amounts are certain only if exactly one series fits its years and total.
### Step-by-Step Solution
1. Drjbna (2011 to 2015, total 10) has the series 1, x, y, z, 1 with x + y + z = 8, where x and z are 2 or 3. If x = 2, then y = 3 forces z = 3, which repeats an amount, so y = 4 and z = 2. If x = 3, then y must be at least 4 and z at most 1, which is impossible. So Drjbna is 1, 2, 4, 2, 1 and is certain.
2. Czechy (starts 2013, total 9) fits 1, 2, 3, 2, 1, which sums to 9. The other short series total 5 (1, 3, 1) or 7 (1, 2, 3, 1 and 1, 3, 2, 1), and 1, 2, 4, 2, 1 totals 10. So Czechy is certain.
3. Bzygoo (2012 to 2015) has the series 1, a, b, 1. Both 1, 2, 3, 1 and 1, 3, 2, 1 obey every rule, so Bzygoo is not certain.
4. Alfloo (eight years, total 21) can be 1, 2, 3, 4, 5, 3, 2, 1 or 1, 2, 3, 5, 4, 3, 2, 1. Both sum to 21, so Alfloo is not certain.
5. Elavalaki (starts 2010, total 13) can be 1, 3, 5, 3, 1 or 1, 2, 4, 3, 2, 1. Both sum to 13, so Elavalaki is not certain.
Only Czechy and Drjbna are fully determined.
### Why Not the Others
- A leaves out Czechy, which is also certain.
- B leaves out Drjbna, which is also certain.
- D wrongly includes Bzygoo, which has two valid series.
**Correct Answer: C**
Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.
Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.
The table below provides partial information about the five firms.
Firm
First year of existence
Last year of existence
Total amount raised (Rs. crores)
Alfloo
2009
2016
21
Bzygoo
2012
2015
Czechy
2013
9
Drjbna
2011
2015
10
Elavalaki
2010
13
What best can be concluded about the total amount of money raised in 2015?
Official Correct Answer: D. ### Core Concept
Fix the 2015 amount of each firm. Czechy and Drjbna are fully determined, and Bzygoo and Alfloo are pinned down for 2015. Only Elavalaki's closing year is open.
### Step-by-Step Solution
1. Drjbna is 1, 2, 4, 2, 1 from 2011, so it raises 1 in 2015. Bzygoo closes in 2015, so it raises 1. Czechy is 1, 2, 3, 2, 1 from 2013, so it raises 3 in 2015.
2. Alfloo's 2015 is its seventh amount, followed by 1 in 2016. It must be 2 or 3. If it were 3, the series would have to peak by the sixth year at 4 or 5. Then the first five years would sum to at least 1 + 2 + 3 + 4 + 5 = 15, and with 4 + 3 + 1 for the last three the total would be at least 23, which is more than 21. So Alfloo raises 2 in 2015.
3. Elavalaki (starts 2010, total 13) can be 1, 3, 5, 3, 1, which closes in 2014 and raises 0 in 2015. It can also be 1, 2, 4, 3, 2, 1 or 1, 2, 3, 4, 2, 1, which close in 2015 and raise 1.
4. The 2015 total is 2 + 1 + 3 + 1 = 7 plus Elavalaki's amount. That gives 7 (it has closed) or 8 (it raises 1).
### Why Not the Others
- A and B require Alfloo to raise 3 or Elavalaki's amount to be fixed, and neither is forced.
- C includes 9, which would need Alfloo at 3 or Elavalaki above 1, and both are impossible.
**Correct Answer: D**
Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.
Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.
The table below provides partial information about the five firms.
Firm
First year of existence
Last year of existence
Total amount raised (Rs. crores)
Alfloo
2009
2016
21
Bzygoo
2012
2015
Czechy
2013
9
Drjbna
2011
2015
10
Elavalaki
2010
13
What is the largest possible total amount of money (in Rs. crores) that could have been raised in 2013?
Official Correct Answer: 17. ### Core Concept
The 2013 total is the sum of each firm's maximum possible 2013 amount. Those maxima can be chosen independently, because the firms' series do not constrain each other.
### Step-by-Step Solution
1. Czechy starts in 2013, so it raises 1. Drjbna is 1, 2, 4, 2, 1 from 2011, so it raises 4 in 2013.
2. Bzygoo (2012 to 2015) is 1, 2, 3, 1 or 1, 3, 2, 1, so its 2013 amount is at most 3.
3. Elavalaki (starts 2010, total 13) can be 1, 3, 5, 3, 1, 1, 2, 4, 3, 2, 1 or 1, 2, 3, 4, 2, 1. Its 2013 amount is 3, 3 or 4, so the maximum is 4.
4. Alfloo's 2013 is its fifth amount. Reaching 6 in the fifth year needs at least 1 + 2 + 3 + 4 + 6 = 16, and the last three years need at least 4 + 2 + 1 = 7. That totals 23, which is more than 21, so 6 is impossible. A fifth amount of 5 works with 1, 2, 3, 4, 5, 3, 2, 1, which sums to 21. So the maximum is 5.
5. Largest 2013 total = 5 + 3 + 1 + 4 + 4 = 17.
### Quick Check
The maxima come from Alfloo 1, 2, 3, 4, 5, 3, 2, 1, Bzygoo 1, 3, 2, 1 and Elavalaki 1, 2, 3, 4, 2, 1. These are all valid, and 5 + 3 + 1 + 4 + 4 = 17.
**Correct Answer: 17**
Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.
Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.
The table below provides partial information about the five firms.
Firm
First year of existence
Last year of existence
Total amount raised (Rs. crores)
Alfloo
2009
2016
21
Bzygoo
2012
2015
Czechy
2013
9
Drjbna
2011
2015
10
Elavalaki
2010
13
If Elavalaki raised Rs. 3 crores in 2013, then what is the smallest possible total amount of money (in Rs. crores) that could have been raised by all the companies in 2012?
Official Correct Answer: A. ### Core Concept
The condition on Elavalaki's 2013 amount narrows its possible series. After that, the 2012 total is found by taking each firm's smallest possible 2012 amount.
### Step-by-Step Solution
1. Elavalaki's possible series are 1, 3, 5, 3, 1, 1, 2, 4, 3, 2, 1 and 1, 2, 3, 4, 2, 1, starting in 2010. Their 2013 amounts are 3, 3 and 4. Raising 3 in 2013 leaves the first two series, so its 2012 amount is 5 or 4. The minimum is 4.
2. Czechy starts in 2013, so it raises 0 in 2012. Bzygoo starts in 2012, so it raises 1. Drjbna is 1, 2, 4, 2, 1 from 2011, so it raises 2 in 2012.
3. Alfloo's 2012 is its fourth amount. On the way up it is at least 4, because 1, then at least 2, then at least 3, then at least 4. After the peak it must exceed three further decreasing amounts ending at 1, so it is at least 5. The minimum is 4, achieved by 1, 2, 3, 4, 5, 3, 2, 1, which sums to 21.
4. Smallest 2012 total = 4 + 1 + 0 + 2 + 4 = 11.
### Why Not the Others
- 9 and 10 would need Alfloo or Elavalaki below the minimum amounts worked out above.
- 12 is attainable, but it is not the smallest.
**Correct Answer: A**
Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.
Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.
The table below provides partial information about the five firms.
Firm
First year of existence
Last year of existence
Total amount raised (Rs. crores)
Alfloo
2009
2016
21
Bzygoo
2012
2015
Czechy
2013
9
Drjbna
2011
2015
10
Elavalaki
2010
13
If the total amount of money raised in 2014 is Rs. 12 crores, then which of the following is not possible?
Official Correct Answer: D. ### Core Concept
A 2014 total of 12 fixes Bzygoo's and Elavalaki's series. Each option can then be tested against the series that remain.
### Step-by-Step Solution
1. In 2014, Czechy raises 2 (1, 2, 3, 2, 1 from 2013) and Drjbna raises 2 (1, 2, 4, 2, 1 from 2011). That leaves 8 for Alfloo, Bzygoo and Elavalaki.
2. Alfloo's 2014 is its sixth amount, followed by 2 in 2015 and 1 in 2016. It is 3 or 4. If it were 4, the first five years would sum to 21 - 4 - 2 - 1 = 14, but a fifth amount above 4 forces at least 15, so it is impossible. Alfloo's 2014 amount is therefore 3, and its 2013 amount is 5 (1, 2, 3, 4, 5, 3, 2, 1) or 4 (1, 2, 3, 5, 4, 3, 2, 1).
3. Bzygoo plus Elavalaki must be 5 in 2014. Bzygoo is 3 or 2 and Elavalaki is 1 or 2, so Bzygoo is 3 and Elavalaki is 2. That means Bzygoo is 1, 2, 3, 1, with 2 in 2013. Elavalaki is 1, 2, 4, 3, 2, 1 or 1, 2, 3, 4, 2, 1, with 3 or 4 in 2013.
4. Test the options. A: Alfloo 3 equals Bzygoo 3, so it is possible. B: Bzygoo 3 is greater than Elavalaki 2, so it is possible. C: Alfloo's 2013 amount can be 4, which equals Drjbna's 4, so it is possible. D: Bzygoo raised 2 in 2013, but Elavalaki raised 3 or 4, so they cannot be equal.
**Correct Answer: D**
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9, both inclusive. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5
Day 2
Day 3
Day 4
Day 5
15
15.5
16
17
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day
Day 1
Day 2
Day 3
Day 4
Day 5
Akhil
1
2
2
3
3
Bimal
2
3
2
1
1
Chatur
3
1
1
2
2
The following information is also known.
1
Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
2
The total score on Day 3 is the same as the total score on Day 4.
Official Correct Answer: A. ### Core Concept
The 2-day averages fix every daily total. The ranks and clues then fix individual scores.
### Step-by-Step Solution
1. Let the daily totals be $T_1$ to $T_5$. The averages give $T_1+T_2=30$, $T_2+T_3=31$, $T_3+T_4=32$ and $T_4+T_5=34$.
2. Clue 2 says $T_3=T_4$. From the first three equations, $T_3=T_1+1$ and $T_4=31-T_1$. Setting these equal gives $T_1=15$, so $T_2=15$, $T_3=16$, $T_4=16$ and $T_5=18$.
3. Table 2 ranks Akhil and Bimal equal on Day 3. Clue 3 says Bimal scores the same $b$ on Days 1 and 3, so Akhil also scores $b$ on Day 3. Chatur is rank 1 on Day 3, so his score $c_3>b$. Then $2b+c_3=16$.
4. Chatur scores only multiples of 3, and his minimum occurs only on Day 1. Hence $c_3\ge 6$. If $c_3=6$, then $b=5$. If $c_3=9$, then $2b=7$, which is impossible. So $c_3=6$ and $b=5$.
5. Chatur's Day 1 score is his unique minimum and is a multiple of 3 below Bimal's 5, so it is 3.
6. On Day 1 the total is 15 and the ranks are Akhil 1, Bimal 2, Chatur 3. So $A_1=15-5-3=7$, which is above Bimal's 5 as the ranking requires.
### Why Not the Others
- 5, 6 and 8 give a Day 1 total different from 15 once Bimal has 5 and Chatur has 3.
### Quick Check
$7+5+3=15=T_1$.
**Correct Answer: A**
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9, both inclusive. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5
Day 2
Day 3
Day 4
Day 5
15
15.5
16
17
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day
Day 1
Day 2
Day 3
Day 4
Day 5
Akhil
1
2
2
3
3
Bimal
2
3
2
1
1
Chatur
3
1
1
2
2
The following information is also known.
1
Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
2
The total score on Day 3 is the same as the total score on Day 4.
Official Correct Answer: D. ### Core Concept
Fix every score that is forced. Then compare the best possible totals of Akhil and Bimal with Chatur's total.
### Step-by-Step Solution
1. The daily totals are 15, 15, 16, 16 and 18. Bimal scores 5 on Days 1 and 3, Akhil scores 7 on Day 1 and 5 on Day 3, and Chatur scores 3 on Day 1 and 6 on Day 3.
2. Day 4 ranks are Bimal 1, Chatur 2, Akhil 3. Akhil's Day 4 score equals Chatur's minimum, 3. Chatur's Day 4 score is 6 or 9. If it were 9, Bimal would need more than 9, which is impossible. So Chatur scores 6 and Bimal scores $16-3-6=7$.
3. Chatur's Day 2 score is the unique highest in the competition, so it exceeds Bimal's 7. It is a multiple of 3, so it is 9. Day 5 works the same way: Chatur has 6, Bimal exceeds 6 but stays below 9, and Akhil is below 6.
4. Chatur's total is $3+9+6+6+6=30$.
5. On Day 2, Chatur has 9 and the other two sum to 6, with Akhil above Bimal. So Bimal scores at most 2 and Akhil at most 5. Day 5 gives $A_5+B_5=12$ with $A_5\le5$ and $B_5\le8$.
6. Bimal's maximum is $5+2+5+7+8=27$. Akhil's maximum is $7+5+5+3+5=25$. Both are below 30.
### Why Not the Others
- Akhil and Bimal cannot reach 30 under any allowed split.
- The total is determinable, so "Cannot be determined" is wrong.
**Correct Answer: D**
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9, both inclusive. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5
Day 2
Day 3
Day 4
Day 5
15
15.5
16
17
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day
Day 1
Day 2
Day 3
Day 4
Day 5
Akhil
1
2
2
3
3
Bimal
2
3
2
1
1
Chatur
3
1
1
2
2
The following information is also known.
1
Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
2
The total score on Day 3 is the same as the total score on Day 4.
3
Bimal's scores are the same on Day 1 and Day 3.
What is the minimum possible total score of Bimal?
Official Correct Answer: 25. ### Core Concept
The forced scores leave only Bimal's Day 2 and Day 5 scores open. Minimise each within the rank and highest-score constraints.
### Step-by-Step Solution
1. The daily totals are 15, 15, 16, 16 and 18. Bimal scores 5 on Days 1 and 3 and 7 on Day 4. Chatur scores 3, 9, 6, 6 and 6 on Days 1 to 5.
2. Day 2: Chatur has 9, so Akhil and Bimal sum to 6. Ranks put Akhil above Bimal, so Bimal's options are 1 or 2, giving $(A_2,B_2)=(5,1)$ or $(4,2)$. The minimum for Bimal is 1.
3. Day 5: the total is 18 and Chatur has 6, so $A_5+B_5=12$. Ranks put Akhil below Chatur, so $A_5\le5$. Hence $B_5\ge7$, and the minimum is 7 with $A_5=5$.
4. Minimum total for Bimal is $5+1+5+7+7=25$.
### Quick Check
These scores keep every rank in Table 2, and Day 2 sums to $5+1+9=15$.
**Correct Answer: 25**
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9, both inclusive. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5
Day 2
Day 3
Day 4
Day 5
15
15.5
16
17
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day
Day 1
Day 2
Day 3
Day 4
Day 5
Akhil
1
2
2
3
3
Bimal
2
3
2
1
1
Chatur
3
1
1
2
2
The following information is also known.
1
Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
2
The total score on Day 3 is the same as the total score on Day 4.
3
Bimal's scores are the same on Day 1 and Day 3.
If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?
Official Correct Answer: B. ### Core Concept
Bimal's open scores are $B_2\in\{1,2\}$ and $B_5\in\{7,8\}$. A divisibility condition on his total selects one pair.
### Step-by-Step Solution
1. Bimal scores 5, 5 and 7 on Days 1, 3 and 4, a sum of 17. His total is $17+B_2+B_5$.
2. Day 2: Chatur scores 9, and Akhil and Bimal sum to 6 with Akhil above Bimal. So $B_2$ is 1 or 2, with $A_2=5$ or 4 respectively.
3. Day 5: $A_5+B_5=12$ with $A_5\le5$, so $B_5\ge7$. Chatur's 9 on Day 2 is the unique highest score, so $B_5<9$. Hence $B_5$ is 7 or 8.
4. The four possible totals are 25 $(1,7)$, 26 $(1,8)$, 26 $(2,7)$ and 27 $(2,8)$. Only 27 is a multiple of 3, so $B_2=2$.
5. Then $A_2=6-2=4$.
### Why Not the Others
- 5 would need $B_2=1$, which gives a total of 25 or 26, not a multiple of 3.
- 6 would equal Chatur's score minus 3 and leave $B_2=0$, which is not a valid score.
- The value is fixed, so "Cannot be determined" is wrong.
**Correct Answer: B**
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9, both inclusive. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5
Day 2
Day 3
Day 4
Day 5
15
15.5
16
17
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day
Day 1
Day 2
Day 3
Day 4
Day 5
Akhil
1
2
2
3
3
Bimal
2
3
2
1
1
Chatur
3
1
1
2
2
The following information is also known.
1
Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
2
The total score on Day 3 is the same as the total score on Day 4.
3
Bimal's scores are the same on Day 1 and Day 3.
If Akhil attains a total score of 24, then what is the total score of Bimal?
Official Correct Answer: 26. ### Core Concept
Akhil's total of 24 restricts his two open scores. Both consistent cases give Bimal the same total.
### Step-by-Step Solution
1. Akhil's fixed scores are 7, 5 and 3 on Days 1, 3 and 4, a sum of 15. So $A_2+A_5=24-15=9$.
2. Day 2 gives $A_2\in\{4,5\}$, with $B_2=6-A_2$. Day 5 gives $A_5\le5$ and $B_5=12-A_5$, so $A_5\in\{4,5\}$ since $B_5<9$.
3. A sum of 9 needs one score to be 4 and the other 5.
4. Case 1: $A_2=4$, $A_5=5$. Then $B_2=2$ and $B_5=7$, so Bimal totals $5+2+5+7+7=26$.
5. Case 2: $A_2=5$, $A_5=4$. Then $B_2=1$ and $B_5=8$, so Bimal totals $5+1+5+7+8=26$.
6. Both cases give 26.
### Quick Check
The three totals are $24+26+30=80$. The daily totals are $15+15+16+16+18=80$.
**Correct Answer: 26**
There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a, if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by. if two or more of these conditions are satisfied.
i
The minimum among the numbers of coins in the three sacks in the box is 1.
ii
The median of the numbers of coins in the three sacks is 1.
iii
The maximum among the numbers of coins in the three sacks in the box is 9.
Official Reference Chart - CAT 2023 Slot 2
What is the total number of coins in all the boxes in the 3rd row?
Official Correct Answer: B. **Clue Breakdown**:
- Each box has three sacks with 1-9 coins.
- Each box's average coins per sack is a distinct integer.
- Total coins in each row and column are equal.
- Table 1 provides medians for some boxes.
- Table 2 indicates the number of sacks with >5 coins and conditions met.
**Step-by-Step Deduction**:
1. **Total Coins Calculation**:
- Each box's total coins must be a multiple of 3 (since average is integer).
- There are 9 boxes, each with a unique total, so totals are 3,6,9,...,27.
- Sum of all box totals = 3*(1+2+...+9) = 3*45 = 135.
- Since each row and column sums to T, and there are 3 rows, 3T = 135 ⇒ T = 45.
**Question Resolution**:
- The total coins in each row, including the third row, is 45.
**Final Answer**:
Correct Answer: B) 45
There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a, if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by. if two or more of these conditions are satisfied.
i
The minimum among the numbers of coins in the three sacks in the box is 1.
ii
The median of the numbers of coins in the three sacks is 1.
iii
The maximum among the numbers of coins in the three sacks in the box is 9.
Official Reference Chart - CAT 2023 Slot 2
How many boxes have at least one sack containing 9 coins?
Official Correct Answer: D. ### Core Concept
Each box is pinned down by combining its median (Table 1), its count of sacks above 5 and condition marker (Table 2), the integer-average requirement and equal row and column totals.
### Step-by-Step Solution
1. Box (3,3) has 0 sacks above 5 and two or more conditions. Condition (iii) is impossible, so the min and median are both 1. The box is $\{1,1,x\}$ with $x\le5$.
2. Box (1,2) has median 9 and two sacks above 5, so it is $\{a,9,9\}$ with $a\le5$. It satisfies exactly one condition, so $a\ne1$. An integer average needs $a+18$ divisible by 3, so $a=3$. The box is $\{3,9,9\}$ with total 21.
3. Box (3,1) has median 8 and three sacks above 5, so it is $\{a,8,c\}$ with $a\ge6$ and $c\ge8$. Exactly one condition forces $c=9$. An integer average needs $a+17$ divisible by 3, so $a=7$. The box is $\{7,8,9\}$ with average 8 and total 24.
4. Box (2,1) has median 2, one sack above 5 and two conditions. Min 1 and max 9 give $\{1,2,9\}$ with average 4 and total 12.
5. Box (3,2) has two sacks above 5 and two conditions, so it is $\{1,x,9\}$ with $x\ge6$. An integer average needs $10+x$ divisible by 3, so $x=8$. The box is $\{1,8,9\}$ with average 6 and total 18.
6. Box (1,3) has median 6, so it is $\{3,6,9\}$ (average 6) or $\{1,6,8\}$ (average 5). Averages must be distinct and 6 is already used, so it is $\{1,6,8\}$ with total 15.
7. Box (1,1) has one sack above 5 and two conditions, and 4 and 5 are taken. That leaves $\{1,1,7\}$ with total 9. Row 1 totals $9+21+15=45$.
8. Column 1 gives $9+12+24=45$, so every row and column totals 45. Row 3 gives box (3,3) $=45-24-18=3$, which is $\{1,1,1\}$. Column 2 gives box (2,2) $=45-21-18=6$, which is $\{1,2,3\}$. Row 2 gives box (2,3) $=45-12-6=27$, which is $\{9,9,9\}$.
9. Boxes containing a 9 are (1,2), (2,1), (2,3), (3,1) and (3,2), which is 5 boxes.
### Why Not the Others
- A (3), B (8) and C (4) all miss at least one of the five boxes above. Box (3,3) $\{1,1,1\}$, box (1,1) $\{1,1,7\}$, box (1,3) $\{1,6,8\}$ and box (2,2) $\{1,2,3\}$ have no 9, so 8 is far too many.
**Correct Answer: D**
There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a, if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by. if two or more of these conditions are satisfied.
i
The minimum among the numbers of coins in the three sacks in the box is 1.
ii
The median of the numbers of coins in the three sacks is 1.
iii
The maximum among the numbers of coins in the three sacks in the box is 9.
Official Reference Chart - CAT 2023 Slot 2
For how many boxes are the average and median of the numbers of coins contained in the three sacks in that box the same?
Official Correct Answer: 4. ### Core Concept
Reconstruct every box from Table 1 and Table 2, using integer averages and equal row and column totals, then compare each box's average with its median.
### Step-by-Step Solution
1. Box (3,3) has 0 sacks above 5 and two conditions. Condition (iii) fails, so min and median are both 1, giving $\{1,1,x\}$.
2. Box (1,2) has median 9 and two sacks above 5, so it is $\{a,9,9\}$. An integer average and exactly one condition give $a=3$. The box is $\{3,9,9\}$.
3. Box (3,1) has median 8 and three sacks above 5, with exactly one condition (max 9). An integer average gives $\{7,8,9\}$ with average 8.
4. Box (2,1) has median 2, one sack above 5 and two conditions, so it is $\{1,2,9\}$ with average 4.
5. Box (3,2) has two sacks above 5 and two conditions, so it is $\{1,x,9\}$. An integer average gives $x=8$, so the box is $\{1,8,9\}$ with average 6.
6. Average 6 is taken, so box (1,3) with median 6 is $\{1,6,8\}$ with average 5. Box (1,1) must be $\{1,1,7\}$ with average 3, since 4 and 5 are taken.
7. Row 1 totals $9+21+15=45$, so all rows and columns total 45. The remaining boxes follow: (3,3) $=\{1,1,1\}$, (2,2) $=\{1,2,3\}$ and (2,3) $=\{9,9,9\}$.
8. Compare average with median box by box:
- (1,1): average 3, median 1, no
- (1,2): average 7, median 9, no
- (1,3): average 5, median 6, no
- (2,1): average 4, median 2, no
- (2,2): average 2, median 2, yes
- (2,3): average 9, median 9, yes
- (3,1): average 8, median 8, yes
- (3,2): average 6, median 8, no
- (3,3): average 1, median 1, yes
### Quick Check
The averages $3,7,5,4,2,9,8,6,1$ are all distinct, and four boxes match.
**Correct Answer: 4**
There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a, if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by. if two or more of these conditions are satisfied.
i
The minimum among the numbers of coins in the three sacks in the box is 1.
ii
The median of the numbers of coins in the three sacks is 1.
iii
The maximum among the numbers of coins in the three sacks in the box is 9.
Official Correct Answer: 9. ### Core Concept
Rebuild all nine boxes from the medians, the Table 2 counts and markers, the integer-average condition and the equal row and column totals. Then count the 1s.
### Step-by-Step Solution
1. Box (3,3) has 0 sacks above 5 and two conditions. Condition (iii) fails, so min and median are both 1, giving $\{1,1,x\}$.
2. Box (1,2) has median 9 and two sacks above 5, so it is $\{a,9,9\}$. An integer average and exactly one condition give $\{3,9,9\}$, with total 21.
3. Box (3,1) has median 8 and three sacks above 5, with exactly one condition (max 9). An integer average gives $\{7,8,9\}$ with total 24.
4. Box (2,1) has median 2, one sack above 5 and two conditions, so it is $\{1,2,9\}$ with total 12.
5. Box (3,2) has two sacks above 5 and two conditions, so it is $\{1,x,9\}$. An integer average gives $x=8$, so the box is $\{1,8,9\}$ with average 6 and total 18.
6. Average 6 is taken, so box (1,3) with median 6 is $\{1,6,8\}$ with total 15. Box (1,1) must be $\{1,1,7\}$ with total 9, since averages 4 and 5 are taken.
7. Row 1 totals $9+21+15=45$, so every row and column totals 45. Row 3 gives box (3,3) $=3$, which is $\{1,1,1\}$. Column 2 gives box (2,2) $=6$, which is $\{1,2,3\}$. Row 2 gives box (2,3) $=27$, which is $\{9,9,9\}$.
8. Count the 1s box by box:
- (1,1): 2
- (1,2): 0
- (1,3): 1
- (2,1): 1
- (2,2): 1
- (2,3): 0
- (3,1): 0
- (3,2): 1
- (3,3): 3
The total is $2+0+1+1+1+0+0+1+3=9$.
### Quick Check
The averages $3,7,5,4,2,9,8,6,1$ are distinct, and every row and column totals 45.
**Correct Answer: 9**
There are nine boxes arranged in a 3×3 array as shown in Tables 1 and 2. Each box contains three sacks. Each sack has a certain number of coins, between 1 and 9, both inclusive.
The average number of coins per sack in the boxes are all distinct integers. The total number of coins in each row is the same. The total number of coins in each column is also the same.
Table 1 gives information regarding the median of the numbers of coins in the three sacks in a box for some of the boxes. In Table 2 each box has a number which represents the number of sacks in that box having more than 5 coins. That number is followed by a, if the sacks in that box satisfy exactly one among the following three conditions, and it is followed by. if two or more of these conditions are satisfied.
i
The minimum among the numbers of coins in the three sacks in the box is 1.
ii
The median of the numbers of coins in the three sacks is 1.
iii
The maximum among the numbers of coins in the three sacks in the box is 9.
Official Reference Chart - CAT 2023 Slot 2
In how many boxes do all three sacks contain different numbers of coins?
Official Correct Answer: 5. ### Core Concept
Rebuild each box from Table 1 and Table 2, using integer averages and equal row and column totals. Then count the boxes whose three values are all different.
### Step-by-Step Solution
1. Box (3,3) has 0 sacks above 5 and two conditions. Condition (iii) fails, so min and median are both 1, giving $\{1,1,x\}$.
2. Box (1,2) has median 9 and two sacks above 5, so it is $\{a,9,9\}$. An integer average and exactly one condition give $\{3,9,9\}$, with total 21.
3. Box (3,1) has median 8 and three sacks above 5, with exactly one condition (max 9). An integer average gives $\{7,8,9\}$ with total 24.
4. Box (2,1) has median 2, one sack above 5 and two conditions, so it is $\{1,2,9\}$ with total 12.
5. Box (3,2) has two sacks above 5 and two conditions, so it is $\{1,x,9\}$. An integer average gives $\{1,8,9\}$ with total 18.
6. Average 6 is taken, so box (1,3) with median 6 is $\{1,6,8\}$ with total 15. Box (1,1) must be $\{1,1,7\}$ with total 9, since averages 4 and 5 are taken.
7. Row 1 totals $9+21+15=45$, so every row and column totals 45. Row 3 gives (3,3) $=\{1,1,1\}$. Column 2 gives (2,2) $=\{1,2,3\}$. Row 2 gives (2,3) $=\{9,9,9\}$.
8. The final boxes are:
- (1,1) $\{1,1,7\}$: repeats
- (1,2) $\{3,9,9\}$: repeats
- (1,3) $\{1,6,8\}$: all different
- (2,1) $\{1,2,9\}$: all different
- (2,2) $\{1,2,3\}$: all different
- (2,3) $\{9,9,9\}$: repeats
- (3,1) $\{7,8,9\}$: all different
- (3,2) $\{1,8,9\}$: all different
- (3,3) $\{1,1,1\}$: repeats
Five boxes have three different values.
**Correct Answer: 5**
Anjali, Bipasha, and Chitra visited an entertainment park that has four rides. Each ride lasts one hour and can accommodate one visitor at one point. All rides begin at 9 am and must be completed by 5 pm except for Ride-3, for which the last ride has to be completed by 1 pm.
Ride gates open every 30 minutes, e.g. 10 am, 10:30 am, and so on. Whenever a ride gate opens, and there is no visitor inside, the first visitor waiting in the queue buys the ticket just before taking the ride. The ticket prices are Rs. 20, Rs. 50, Rs. 30 and Rs. 40 for Rides 1 to 4, respectively. Each of the three visitors took at least one ride and did not necessarily take all rides. None of them took the same ride more than once. The movement time from one ride to another is negligible, and a visitor leaves the ride immediately after the completion of the ride. No one takes a break inside the park unless mentioned explicitly.
The following information is also known.
1
Chitra never waited in the queue and completed her visit by 11 am after spending Rs. 50 to pay for the ticket(s).
2
Anjali took Ride-1 at 11 am after waiting for 30 mins for Chitra to complete it. It was the only ride where Anjali waited.
3
Bipasha began her first of three rides at 11:30 am. All three visitors incurred the same amount of ticket expense by 12:15 pm.
4
The last ride taken by Anjali and Bipasha was the same, where Bipasha waited 30 mins for Anjali to complete her ride. Before standing in the queue for that ride, Bipasha took a 1-hour coffee break after completing her previous ride.
What was the total amount spent on tickets (in Rs.) by Bipasha?
Official Correct Answer: C. ### Core Concept
Ticket spending is fixed by the clues, so Bipasha's three rides can be identified from her timings, the equal-expense clue and the ride-closing rules.
### Step-by-Step Solution
1. Chitra spent Rs. 50 with no waiting and finished by 11 am. Anjali waited for Chitra to finish Ride-1 at 11 am, so Chitra was on Ride-1 from 10 to 11. Rs. 50 as a total with Ride-1 (Rs. 20) means Ride-3 (Rs. 30). Chitra's spend is therefore Rs. 50.
2. By 12:15 pm all three had spent the same amount, so Bipasha had spent Rs. 50 by 12:15.
3. Bipasha began her first ride at 11:30 am. That ride runs until 12:30, so her second ticket cannot be bought before 12:30. Her only ticket by 12:15 is the 11:30 ride, so her first ride costs Rs. 50 and is Ride-2 (11:30 to 12:30).
4. Clue 4 says Bipasha took a 1-hour break after her second ride, then queued behind Anjali for the last ride. Her second ride ends no earlier than 1:30, so she joins that queue no earlier than 2:30. Her last ride cannot be Ride-3, which must finish by 1 pm, and cannot be Ride-2, which she has already taken.
5. It also cannot be Ride-1. Anjali took Ride-1 at 11 am, which is long before Bipasha could join a queue, and Bipasha would have to wait for Anjali's Ride-1 ride. So the ride they share last is Ride-4 (Rs. 40).
6. Bipasha's middle ride cannot be Ride-3, because the last Ride-3 gate is 12:00 and she is still on Ride-2 until 12:30. It cannot be Ride-2 or Ride-4, which she takes first and last. So it is Ride-1 (Rs. 20).
7. Bipasha's spend is 50 + 20 + 40 = Rs. 110.
### Why Not the Others
- 100, 120 and 90 do not match the ride set Ride-2, Ride-1 and Ride-4 that the clues force.
### Quick Check
Bipasha rides Ride-2 from 11:30 to 12:30, Ride-1 from 12:30 to 1:30, takes a break until 2:30, then queues for Ride-4. This is consistent with all four clues.
**Correct Answer: C**
Anjali, Bipasha, and Chitra visited an entertainment park that has four rides. Each ride lasts one hour and can accommodate one visitor at one point. All rides begin at 9 am and must be completed by 5 pm except for Ride-3, for which the last ride has to be completed by 1 pm.
Ride gates open every 30 minutes, e.g. 10 am, 10:30 am, and so on. Whenever a ride gate opens, and there is no visitor inside, the first visitor waiting in the queue buys the ticket just before taking the ride. The ticket prices are Rs. 20, Rs. 50, Rs. 30 and Rs. 40 for Rides 1 to 4, respectively. Each of the three visitors took at least one ride and did not necessarily take all rides. None of them took the same ride more than once. The movement time from one ride to another is negligible, and a visitor leaves the ride immediately after the completion of the ride. No one takes a break inside the park unless mentioned explicitly.
The following information is also known.
1
Chitra never waited in the queue and completed her visit by 11 am after spending Rs. 50 to pay for the ticket(s).
2
Anjali took Ride-1 at 11 am after waiting for 30 mins for Chitra to complete it. It was the only ride where Anjali waited.
3
Bipasha began her first of three rides at 11:30 am. All three visitors incurred the same amount of ticket expense by 12:15 pm.
4
The last ride taken by Anjali and Bipasha was the same, where Bipasha waited 30 mins for Anjali to complete her ride. Before standing in the queue for that ride, Bipasha took a 1-hour coffee break after completing her previous ride.
Which were all the rides that Anjali completed by 2:00 pm?
Official Correct Answer: C. ### Core Concept
Anjali's timeline comes from the ticket-spend clue and the rule that she waited only once.
### Step-by-Step Solution
1. Clue 2: Anjali arrived at Ride-1 at 10:30, waited until 11:00, and rode from 11 to 12. This was the only ride where she waited, so she took no ride before it (see step 2).
2. Clue 3: Her ticket spend by 12:15 equals Chitra's Rs. 50. Ride-1 cost Rs. 20, so exactly Rs. 30 more was spent by 12:15. A ride before Ride-1 would also have been paid for by then, so the Rs. 30 is not a prior ride. It is Ride-3 starting at 12:00. The 12:00 start also respects Ride-3's cutoff of a 1 pm finish.
3. Anjali then goes straight on without waiting. Ride-2 is open at 1:00 (Bipasha left it at 12:30), so she rides Ride-2 from 1 to 2. Buying Ride-2 at 12:00 instead would push her spend past Rs. 50 by 12:15.
4. Clue 4: Her last ride, which Bipasha takes after her, is Ride-4 from 2 to 3. It starts at 2:00, so it is not completed by 2 pm.
5. By 2:00 pm Anjali has completed Ride-1 (12:00), Ride-3 (1:00) and Ride-2 (2:00).
### Why Not the Others
- Options A and B include Ride-4, which Anjali only finishes at 3:00.
- Option D omits Ride-2, which she finishes at 2:00.
**Correct Answer: C**
Anjali, Bipasha, and Chitra visited an entertainment park that has four rides. Each ride lasts one hour and can accommodate one visitor at one point. All rides begin at 9 am and must be completed by 5 pm except for Ride-3, for which the last ride has to be completed by 1 pm.
Ride gates open every 30 minutes, e.g. 10 am, 10:30 am, and so on. Whenever a ride gate opens, and there is no visitor inside, the first visitor waiting in the queue buys the ticket just before taking the ride. The ticket prices are Rs. 20, Rs. 50, Rs. 30 and Rs. 40 for Rides 1 to 4, respectively. Each of the three visitors took at least one ride and did not necessarily take all rides. None of them took the same ride more than once. The movement time from one ride to another is negligible, and a visitor leaves the ride immediately after the completion of the ride. No one takes a break inside the park unless mentioned explicitly.
The following information is also known.
1
Chitra never waited in the queue and completed her visit by 11 am after spending Rs. 50 to pay for the ticket(s).
2
Anjali took Ride-1 at 11 am after waiting for 30 mins for Chitra to complete it. It was the only ride where Anjali waited.
3
Bipasha began her first of three rides at 11:30 am. All three visitors incurred the same amount of ticket expense by 12:15 pm.
4
The last ride taken by Anjali and Bipasha was the same, where Bipasha waited 30 mins for Anjali to complete her ride. Before standing in the queue for that ride, Bipasha took a 1-hour coffee break after completing her previous ride.
Official Correct Answer: D. ### Core Concept
Once each visitor's ride list is built from the clues, the common ride can be read off directly.
### Step-by-Step Solution
1. Chitra: Clue 1 gives Rs. 50 with no waiting, and Clue 2 shows she was on Ride-1 from 10 to 11. So Chitra took Ride-3 (9 to 10) and Ride-1 (10 to 11), which cost 30 + 20 = 50.
2. Anjali: Ride-1 (11 to 12), Ride-3 (12 to 1), Ride-2 (1 to 2) and Ride-4 (2 to 3). The Rs. 50 spent by 12:15 (Clue 3) forces Ride-3 at 12:00, and Clue 4 makes Ride-4 her last ride.
3. Bipasha: Ride-2 (11:30 to 12:30), which is her only ticket by 12:15 and costs Rs. 50. Then Ride-1 (12:30 to 1:30), a 1-hour coffee break, and Ride-4 (3 to 4) after waiting for Anjali, as Clue 4 says.
4. Compare the lists:
- Ride-1: Chitra, Anjali, Bipasha.
- Ride-2: Anjali, Bipasha.
- Ride-3: Chitra, Anjali.
- Ride-4: Anjali, Bipasha.
5. Only Ride-1 appears in all three lists.
### Why Not the Others
- Ride-4 and Ride-2 were not taken by Chitra.
- Ride-3 was not taken by Bipasha, since its last ride must finish by 1 pm and her free slot started at 12:30.
**Correct Answer: D**
Anjali, Bipasha, and Chitra visited an entertainment park that has four rides. Each ride lasts one hour and can accommodate one visitor at one point. All rides begin at 9 am and must be completed by 5 pm except for Ride-3, for which the last ride has to be completed by 1 pm.
Ride gates open every 30 minutes, e.g. 10 am, 10:30 am, and so on. Whenever a ride gate opens, and there is no visitor inside, the first visitor waiting in the queue buys the ticket just before taking the ride. The ticket prices are Rs. 20, Rs. 50, Rs. 30 and Rs. 40 for Rides 1 to 4, respectively. Each of the three visitors took at least one ride and did not necessarily take all rides. None of them took the same ride more than once. The movement time from one ride to another is negligible, and a visitor leaves the ride immediately after the completion of the ride. No one takes a break inside the park unless mentioned explicitly.
The following information is also known.
1
Chitra never waited in the queue and completed her visit by 11 am after spending Rs. 50 to pay for the ticket(s).
2
Anjali took Ride-1 at 11 am after waiting for 30 mins for Chitra to complete it. It was the only ride where Anjali waited.
3
Bipasha began her first of three rides at 11:30 am. All three visitors incurred the same amount of ticket expense by 12:15 pm.
4
The last ride taken by Anjali and Bipasha was the same, where Bipasha waited 30 mins for Anjali to complete her ride. Before standing in the queue for that ride, Bipasha took a 1-hour coffee break after completing her previous ride.
How many rides did Anjali and Chitra take in total?
Official Correct Answer: 6. ### Core Concept
The number of rides for each visitor follows from their timelines. Chitra's is fixed by Clues 1 and 2, and Anjali's by Clues 2 to 4.
### Step-by-Step Solution
1. Chitra spent Rs. 50 without waiting. Anjali waited from 10:30 to 11:00 for Chitra to finish Ride-1, so Chitra was on Ride-1 from 10 to 11. Rs. 50 therefore means Ride-1 (20) plus Ride-3 (30, from 9 to 10). Chitra took 2 rides.
2. Anjali rode Ride-1 from 11 to 12 and had spent Rs. 50 by 12:15, the same as Chitra. The extra Rs. 30 is Ride-3 at 12:00, since a ride before Ride-1 would conflict with her only waiting at Ride-1. This gives Ride-3 from 12 to 1.
3. She continues without waiting: Ride-2 from 1 to 2.
4. Clues 3 and 4 make Ride-4 her last ride, shared with Bipasha. Bipasha waits 30 minutes for Anjali to finish it, so Anjali rides Ride-4 from 2 to 3. Anjali took 4 rides.
5. Total for Anjali and Chitra: 4 + 2 = 6.
### Quick Check
Anjali's four rides are Ride-1, Ride-3, Ride-2 and Ride-4. Each starts as soon as the previous one ends, so she waits only at Ride-1, as Clue 2 requires.
**Correct Answer: 6**
Anjali, Bipasha, and Chitra visited an entertainment park that has four rides. Each ride lasts one hour and can accommodate one visitor at one point. All rides begin at 9 am and must be completed by 5 pm except for Ride-3, for which the last ride has to be completed by 1 pm.
Ride gates open every 30 minutes, e.g. 10 am, 10:30 am, and so on. Whenever a ride gate opens, and there is no visitor inside, the first visitor waiting in the queue buys the ticket just before taking the ride. The ticket prices are Rs. 20, Rs. 50, Rs. 30 and Rs. 40 for Rides 1 to 4, respectively. Each of the three visitors took at least one ride and did not necessarily take all rides. None of them took the same ride more than once. The movement time from one ride to another is negligible, and a visitor leaves the ride immediately after the completion of the ride. No one takes a break inside the park unless mentioned explicitly.
The following information is also known.
1
Chitra never waited in the queue and completed her visit by 11 am after spending Rs. 50 to pay for the ticket(s).
2
Anjali took Ride-1 at 11 am after waiting for 30 mins for Chitra to complete it. It was the only ride where Anjali waited.
3
Bipasha began her first of three rides at 11:30 am. All three visitors incurred the same amount of ticket expense by 12:15 pm.
4
The last ride taken by Anjali and Bipasha was the same, where Bipasha waited 30 mins for Anjali to complete her ride. Before standing in the queue for that ride, Bipasha took a 1-hour coffee break after completing her previous ride.
What was the total amount spent on tickets (in Rs.) by Anjali?
Official Correct Answer: 140. ### Core Concept
The spend-matching clue and the waiting rule fix Anjali's rides, and her ticket total is the sum of their prices.
### Step-by-Step Solution
1. Clue 2: Anjali rode Ride-1 from 11 to 12 (Rs. 20) after waiting from 10:30. This was the only ride where she waited.
2. Clue 3: By 12:15 all three had spent the same, and Chitra's spend was Rs. 50. So Anjali had spent Rs. 30 beyond Ride-1 by 12:15. A ride before 10:30 would also have been paid for by then, so the Rs. 30 is Ride-3 at 12:00 (12 to 1). This is also a valid time for Ride-3, which must finish by 1 pm.
3. Ride-2 is the next ride: she goes in at 1:00 with no waiting, after Bipasha left at 12:30, and rides until 2:00 (Rs. 50).
4. Clue 4: Her last ride is the one Bipasha later takes after waiting 30 minutes. Ride-1 and Ride-2 were already taken by Anjali, and Ride-3 is closed after 1 pm. So it is Ride-4 from 2 to 3 (Rs. 40).
5. Total: 20 + 30 + 50 + 40 = 140.
### Quick Check
Anjali's Rs. 50 by 12:15 matches Chitra's, and 140 is the sum of all four ticket prices (20 + 50 + 30 + 40).
**Correct Answer: 140**
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How many questions were asked in CAT 2023 Slot 2 DILR?
CAT 2023 Slot 2 DILR contained 20 questions convened by IIM Lucknow. Marking was +3 marks for correct answers, -1 for incorrect MCQs, and zero penalty for TITA questions.
What score was required for a 99th percentile in CAT 2023 Slot 2 DILR?
Achieving a 99.0th percentile in DILR required approximately 23–26 marks (~8 net correct), reflecting the high difficulty (heavy number systems, grid logic dilr).
Can I practice other sections of CAT 2023 Slot 2?
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