CAT 2023 Slot 2: Past Year Question Paper (November 3) - CAT MCQ

• The central idea of the passage is the promotion of sustainable shopping practices, particularly second-hand shopping, as a means to combat the detrimental environmental effects of the fashion industry. But, the passage also discusses the need for consumers to be mindful of the environmental impact of their clothing choices, opting for high-quality items that last longer and shed fewer microfibers.
• The passage argues that opting for second clothing might not always be beneficial for the environment by highlighting the microfibre pollution that they can potentially cause. Now, if the second-hand clothes being sold were only of higher quality, it would take care of this problem ([They would be well advised to buy] high-quality items that shed less and last longer [as this] combats both microfibre pollution and excess garments ending up in landfills”)
• So, the correct answer is Option C.
• Option A is more about the purchasing channel than the nature of the clothes so it does not necessarily undermine the central idea of the passage.
• Option B supports the central idea by reducing environmental harm.
• Option D could align with the sustainability goal and support the central idea, so it doesn't necessarily undermine it.

From the context in which the word 'thrifting' is used in the passage, we can conclude that it refers to the purchase of second-hand items at low costs, a practice which is now a trend as consumers get to be 'cool' while also caring for the planet. However, as explained in the passage, the act of thrifting can be considered ironic if, instead of saving the planet, it actually contributes to microfibre pollution of the rivers and oceans. Option A is the correct choice.

Option D is the correct answer because the passage does not mention or suggest that the British don't buy second-hand clothing. Instead, the passage discusses challenges related to luxury brands and their reluctance to circulate their latest season stock globally at a cheaper price. The reasons mentioned include the financial aspect(Option A), concerns about brand image(Option B), and the desire to avoid devaluing their products(Option D). Therefore, the passage does not attribute the slow adoption of companies like ThredUP in the UK to the British not buying second-hand clothing.

All options except B relate to liberalism and the problems caused by its disintegration. Technological advances cannot be considered evidence of the decline of liberalism.

Note the context in which the author talks about the 'Davos elite': 'The biggest enemy of liberalism is not so much atomisation but old-fashioned greed, as members of the Davos elite pile their plates ever higher with perks and share options.' Only option D relates to the greed of the Davos elite. This is the correct answer choice.

• Note the context in which the author talks about archaeology and other the "auxiliary sciences" of history: 'But [to] praise a historian for his accuracy is like praising an architect for using well-seasoned timber or properly mixed concrete in his building. It is a necessary condition of his work, but not his essential function. It is precisely for matters of this kind that the historian is entitled to rely on what have been called the "auxiliary sciences" of history-archaeology, epigraphy, numismatics, chronology, and so forth...'
• The author states auxiliary sciences like archaeology only help historians to ascertain the accuracy of facts. They do not help in the essential function of his work, which is to interpret the facts.
• Option C is the correct choice.

• According to the passage, the "common-sense" view of history is influenced by the positivist view and so it places great weight on facts. In this view, facts are available to the historian in documents, inscriptions, and so on and history can be objective like the sciences if it is derived from historical facts.
• The author's view is in contrast to the common-sense view. The author believes history is a 'selective' system of cognitive orientations to reality. Facts only speak as the historian interprets them. Option A is the correct answer choice.

• Option A is true, based on the lines, 'The economics of European productions are more appealing, too. American audiences are more willing than before to give dubbed or subtitled viewing a chance. This means shows such as "Lupin", a French crime caper on Netflix, can become global hits.'
• Option B is true, too: 'In 2015, about 75% of Netflix's original content was American; now the figure is half, according to Ampere, a media-analysis company.'
• Option C is clearly stated in the passage:'A bigger problem is that national broadcasters still dominate. Streaming services, such as Netflix or Disney+, account for about a third of all viewing hours, even in markets where they are well-established.'
• Only option D is incorrect: 'Now Netflix has offices across Europe. But ultimately the big decisions rest with American executives.'

The author clearly sees Netflix as a unifying force in Europe: 'Now Netflix and friends pump the same content into homes across a continent, making culture a cross-border endeavour, too. If Europeans are to share a currency, bail each other out in times of financial need and share vaccines in a pandemic, then they need to have something in common-even if it is just bingeing on the same series.'

• Talking about which shows have better appeal, the passage states, 'Not everything works across borders. Comedy sometimes struggles. Whodunits and bloodthirsty maelstroms between arch Romans and uppity tribesmen have a more universal appeal...'. So, a murder mystery drama set in North Africa and France is likely, according to the passage, to be successful with audiences across the EU.
• Based on the lines above, option D is easily eliminated. The passage declares 'German television is not always built for export', so option B is also ruled out. The passage focuses on translations of European productions and their success. Option A does not relate to this.

• Sentence 3 is the odd one out because it introduces a different perspective. While the other sentences discuss the negative outcomes associated with self-care in children, Sentence 3 suggests a need to consider that the bulk of research has yet to track long-term consequences. This sentence introduces a more neutral or balanced viewpoint that doesn't align with the general theme of the other sentences, which focus on the negative aspects of self-care for children.
• The remaining sentences (1, 2, 4, and 5) can be put together to form a coherent paragraph discussing the concept of self-care in relation to children, particularly those referred to as "Latchkey children."

• Sentence 4 introduces the conflict between Western Christian concepts and indigenous African tradition.This gives context for the discussion to be followed.
• Now we can see that Sentence 3 starts with “However….” which implies that this sentence is challenging the views presented in Sentence 4 by asserting the existence of a rich indigenous African cultural ethos. Therefore Sentence 3 must be following Sentence 4.
• Sentence 2 introduces Chuwkuemeka Ike and provides additional information about Chukwuemeka Ike's exploration of the conflict, adding depth to the discussion. Sentence 1 introduces an example of contemporary African writing that explores the conflict. It makes sense that first we give an example and then elaborate on it. Sentence 1 introduces the book; it is a logical inferences that Ike must be the author of this book.
• Therefore Sentence 2 must be following Sentence 1.

Therefore the correct order is 4-3-1-2.

• The passage discusses the increasing frequency and intensity of heatwaves due to climate change, with vulnerable groups experiencing uneven impacts. It emphasizes that adaptation to heatwaves is a significant public policy concern. The research findings suggest that even vulnerable individuals may not perceive themselves as at risk of extreme heat, highlighting the importance of understanding how extreme heat is narrated. The passage specifically mentions the central role of the news media in warning people about the potential danger of heatwaves and their impacts on infrastructure and society. Option C effectively conveys the primary focus on heatwaves posing a substantial risk and the critical role of the media in alerting the public to this danger.
• Option A implies a general importance of protection without specifically highlighting the role of the media in alerting people to the risks of heatwaves.
• Option B acknowledges the vulnerability to heatwaves but it does not emphasize the role of the media in alerting people and suggests a broader critique of measures taken.
• Option D mentions the need for news stories to become more effective but does not emphasize the central role of the media in alerting people to the risk of heatwaves, as the passage does.

• The passage discusses the phenomenon of counterfactual thinking, highlighting that people spontaneously create counterfactual alternatives to reality for various reasons. These reasons include explaining the past, preparing for the future, implicating various relations (including causal ones), affecting emotions, and supporting moral judgments. Additionally, the passage mentions that the ability to create counterfactuals develops throughout childhood and contributes to reasoning about other people's beliefs. Option C effectively encompasses these key points, making it the most accurate summary of the passage.
• Option A focuses primarily on the preparation for the future aspect, neglecting the broader reasons for creating counterfactual alternatives.
• Option B does not emphasize the developmental aspect and various reasons for creating
• Option D inaccurately suggests that counterfactual thinking helps reverse past and future actions, which is not the main point of the passage, and it oversimplifies the role of counterfactuals.

We are given that 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.

=> The total number of coins in a box range from 3 (1 + 1 + 1) to 27 (9 + 9 + 9)

Since, it is given that the average number of coins per sack in the boxes are all distinct integers => The total number of coins in a box would be 3, 6, 9, 12, 15, 18, 21, 24, 27 => averages of 1, 2, 3, 4,....,9 => Sum = 45.

=> Sum of averages coins in a box in a row or column = 45/3 = 15 [The total number of coins in each row is the same. The total number of coins in each column is also the same.] ==> (1)

Let us represent the final configuration of the sacks in boxes as follows:

Also a bag (x,y) => bag in xth row and yth column.

We are given 2 clues => Table-1 & Table-2

Consider bag (3,1)

=> From Table-1 => Median = 8 & From Table-2 all 3 sacks have more than 5 coins. Also * => There is a 9 in one of the sacks.

=> c, 8, 9 are the coins in bag (3,1), now c > 5 & c + 8 + 9 should be a multiple of 3 => c = 7 is the only possiblility.

=> bag (3,1) has 7, 8, 9 coins with average = 8.

Consider bag (2,1)

Median = 2 and 1 sack has more than 5 coins. Also ** => conditions i & iii should be satisfied.

=> 1, 2, 9 are the coins in bag (2,1) with average = 4

Consider bag (1,2)

Median = 9 and 2 elements are more than 5. Also * => (9 is present & 1 is not present)

=> c, 9, 9 are the coins in bag (1,2) and c is not equal to 1 and less than 5 => c = 3 for c + 18 to be a multiple of 3.

=> 3, 9, 9 are the coins in bag (1,2) with average = 7.

Capturing this info. in the table:

From (1), The average in bag (1,1) is 15 - 4 - 8 = 3.

From (1), The average in bag (1,3) is 15 - 3 - 7 = 5.

Consider bag (1,1)

Avg = 3, 1 sack has more than 5 and ** => 2 conditions are being satisfied. => (can't be condition-3 => 9 coins as the total sum of coins is it self 3*3 = 9)

=> bag (1,1) has 1, 1, 7 coins with average = 3.

Consider bag (1,3)

Avg. = 5 => Sum = 15.

Median = 6 and 2 sacks have more than 5 and * => (1 condition is satisfied)

Not condition ii as the median is 6 & Not condition iii as the sum of 2 sacks itself will become 6 + 9 = 15

=> 1, 6, c are the coins => For sum = 15 => c = 15 - 1 - 6 = 8

=> bag (1,3) has 1, 6, 8 coins with average = 5.

Consider bag (3,3)

0 sacks have more than 5 coins and ** => conditions i & ii are being satisfied.

=> 1,1,c are the coins. Now c = 1 or 2 or 3 or 4 => c = 1 or 4 for number of coins to be a multiple of 3.

But c = 1 as no other bag has the possibility to get avg. = 1 as bag (2,2) should have 1, b, c coins and b and c should be more than 1 as only 1*

=> bag (3,3) has 1, 1, 1 coins with average = 1.

Now, we can fill the averages in all the bags.

In bag (2,3) Avg. = 9 => 9, 9, 9 are the coins.

In bag (2,2) => Avg. = 2 => Sum = 6 and only 1* => smallest elemens=t should be 1.

=> 1, b, c are the coins where b + c = 5 and b,c can't be equal to 1 and less than 5 => 2 + 3 = 5 is the only possibility.

=> 1, 2, 3 are the coins with average = 2.

Considering bag (3,2)

Avg. = 6 => Sum = 18.

2 sacks more than 5 coins and ** => 2 sacks have 1 and 9 coins.

=> bag (3,2) has 1, c, 9 coins and c = 18 - 1 - 9 = 8

=> bag (3,2) has 1, 8, 9 coins with average = 6 coins.

==> Final required table, bracket number => average coins per sack in the bag.

Average = Median in boxes (3,1), (2,2), (2,3) and (3,3) => 4 boxes.

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively. 

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4. 

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that 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.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4. 

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2. 

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the score of Akhil is 7 on day 1.

The correct option is B

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that 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.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil's on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the maximum score is obtained by Chatur. 

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that 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.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the minimum score obtained by Bimal is 25.

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1± 1 or ± 2± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

We see that only for C and D, we can conclude the amounts raised with certainty.

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Money raised in 2015 is 2 + 1 + 3 + 1 + 0/1 = 7 or 8.

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Maximum money raised in 2013 is 5 + 3 + 1 + 4 + 4 = 17.

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.

The increase or decrease can be ± 1 or ± 2. => (1)

We are also told that each firm raised Rs. 1 crore in its first and last year of existence

Consider A:

It raised money for 8 years

=> The raising pattern looks like follows:

1, a, b, c, d, e, f, 1 => where a, b, c,..,, f are the unknown amounts raised.

Also a + b + c + d + e + f = 21 - 2 = 19.

We can observe that 19/6 is slightly greater than 3 => The average amount raised should be around 3.

If a = 3 and f = 3 => b + c + d + e = 13 (not possible) as the minimum case would be (4, 5, 6, 4) => Not possible.

If a = 3 and f = 2 => b + c + d + e = 14 (not possible) as the minimum case would be (4, 5, 4, 3) => Not possible.

=> a = 2 and f = 2 => b + c + d + e = 15 the minimum case is (3, 4, 5, 3) or (3, 5, 4, 3) which gives a sum of 15.

So, the possible cases for A are:

Consider B:

The patterns looks as follows:

1, a, b, 1

If a = 2, b has to be equal to 3 to satisfy (1)

if a = 3, b has to be equal to 2 to satisfy (1)

=> The possible cases for B are:

Consider C:

The pattern looks as follows:

1, ..., 1

Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1

Let us assume there are 3 gaps between => a + b + c = 7, the minimum case possible is 1, 2, 3, 2, 1 => Satisfies.

Now, if there are 4 gaps => a + b + c + d = 7 => The average value is 7/4 which is less than 2 => Not possible.

=> The possible cases for C are:

Consider D:

The pattern looks as follows:

1, a, b, c, 1

=> a + b + c = 8

When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies.

When a = 2 and c = 3, b should be 3 (Not satisfying (1))

When a = 3 and c = 3, b should be 2 (Not satisfying (1))

=> The possible cases for D are:

Consider E:

The pattern looks as follows:

1,.....,1

For 1 or 2 gaps, we can't get a sum of 11.

Assume 3 gaps => a + b + c = 11, the maximum case is 3, 5, 3 => Satisfies.

Now, assume 4 gaps

=> a + b + c + d = 11, the minimum case is 2, 3, 4, 2 or 2, 4, 3, 2 which satisfies (1) and 2 + 3 + 4 + 2 = 11.

=> The possible cases for E are:

In summary, the possible cases for all 5 companies is:

Given that total amount raised in 2014 is 12

=> 3 + 3/2 + 2 + 2 + 1/2 = 12 => 

=> possible case is 3 + 3 + 2 + 2 + 2 = 12.

A) In 2013, B raised 2 crores and E also raised 3/4 crores => Not Possible.

B) In 2013, A could have raised 5/4 and D raised 4 => Possible.

C) In 2014, A raised 3 and B raised 3 => Possible.

D) In 2014, B raised 3 where as E raised 2 => 3 > 2 => Possible.

Consider Statement 2: Anjali took Ride-1 at 11 am after waiting for 30 minutes for Chitra to complete it. It was the only ride where Anjali waited.

This implies that Chitra took Ride 1 at 10 am. Now we also know that she spent Rs 50 and that she left at 11 am. Now, since she did one ride costing Rs 20 at 10, she must have taken Ride-3 at 9 am.

So we get the following table for Chitra.

Now we know that Chitra and Anjali spent Rs 50 before 12:15 pm. It is not possible for Anjali to go on Ride-3 at 10 am as we know that she was waiting for 30 minutes before taking Ride-1 (She was waiting from 10:30 am).

Now, since we know that Ride-1 was the only ride for which she waited, we can say that she took Ride-1 at 11 am and started Ride-3 at 12 am

So we get the following table for Anjali.

Now, we know that Bipasha started her first ride at 11:30 am. We also know that they all spent Rs 50 before 12:15 pm.

Therefore, the first ride Bipasha takes will be Ride-2, costing Rs 50.

So we get the following table for Bipasha.

We know that Ride 3 stops at 1 pm. So the last ride taken by Anjali will either be Ride-2 or Ride-4. Now, considering Statement 4,we know that the last ride taken by Anjali and Bipasha was same and that Bipasha rode it after Anjali. So their last ride can’t be 2.

So the last ride of both Bipasha and Anjali will be 4.

Now if we assume that immediately after ending Ride-3, Anjali goes to Ride-4, then the last ride of Bipasha will be Ride-4 from 2 pm - 3 pm. But we know that Bipasha rode 3 rides. So this case is not possible.

Since Anjali didn’t have any break or waiting time, the only ride she can ride at 1 pm will be Ride 2 and then she will go on Ride-4 from 2 pm to 3 pm.

So we get the following table for Anjali:

Now we know that the last ride that Bipasha took was Ride-4 and that she had a gap of 1.5 hrs before it. This is only possible when she takes one ride between Ride-2 and Ride-4. Since Ride-3 is closed at 1 pm, she can only take Ride 1. So we get the following table for her.

Anjali completed a total of 4 rides, 3 of which were completed at 2. Therefore the answer is Option B: Ride-1, Ride 3, and Ride -2

A positive integer less than 50, having exactly two distinct factors other than 1 and itself, is either a perfect cube below 50 or an integer that is a product of exactly two distinct primes.
Case i)
Perfect cubes below 50 are 2³ and 3³. So, two numbers here
Case ii)
For the product of two primes to be below 50, the individual primes should be below 25.
(Because, the smallest prime is 2 and multiplying 2 with anything greater than or equal to 25 yields a number greater than or equal to 50.)
2, 3, 5, 7, 11, 13, 17, 19, 23 are prime numbers less than 25.

2, 3, 5, 7 are the primes less than √50, any product of two numbers among them yields a product less than 50.
So, there are 4C₂ = 6 pairs here.
11, 13, 17, 19, 23 are the primes greater than √50, any product of two numbers among them yields a product greater than 50.
So, there are 0 pairs here.
Between the two lists 11 and 13 can pair with 2 and 3, while 17, 19, and 23 can only pair with 2.
So, there are 7 pairs here.
So, totally, there are 2 + 6 + 0 + 7 = 15 such numbers.

” In a company, 20% of the employees work in the manufacturing department.”
Ratio of number of manufacturing to non-manufacturing employees = 1 : 4
So, let the number of manufacturing to non-manufacturing employees be x and 4x respectively.
“the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company,”
Ratio of total salaries of manufacturing to non-manufacturing employees = 1 : 5
So, let the total salaries of manufacturing to non-manufacturing employees be y and 5y respectively.
So, the ratio of average salaries of manufacturing to non-manufacturing employees will be 

The price of a precious stone is directly proportional to the square of its weight.”
If W is the weight of the stone and P is the price of that stone, then P = k × W²
For the entire, unbroken stone, the price will be 18² × k = 324 k.
“If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000.”
The minimum profit is achieved when the weights of the broken stones are close to each other, that is, the weights are 3, 4, 5, and 6 units.
In this case the combines worth of the four stones =(3² + 4² + 5² + 6²)k = 86k
The maximum profit is achieved when the weights of the broken stones are far from each other, that is, the weights are 1, 2, 3, and 12 units.
In this case the combines worth of the four stones =(1² + 2² + 3² + 12²)k = 158k
The difference in the total value = 2,88,000.
158k – 86k = 72k = 2,88,000
k = 4,000
So, the price of the original stone = 324 k = 12,96,000

Let the number of white and black shirts bought by Jayant be w and b respectively.
Then the total Cost Price (CP) =1000 × w + 1125 × b =1000(w + b) + 125 × b
Since the goods are marked up by 25% and then offered at a discount of 10%, the total Selling Price (SP)
= CP × 1.25 × 0.9 = 1.125 CP
This implies that there was a 12.5% of Profit, which is given to be 51, 000
12.5%(CP) = 51,000
CP = 4,08,000
1000(w + b) + 125 × b = 4,08,000
w and b are positive integers (since at least one shirt of each color needs to be purchased.)
To purchase maximum number of shirts, you need to purchase minimum number of the costlier shirts, which are the blue ones…
Observe that the total CP is a multiple of 1000. And for that to happen b should be a multiple of 8 in
1000(w + b) + 125 × b = 4,08,000.
So the minimum value of b = 8, in which case, w = 399.
Hence, the maximum number of shirts that can be purchased = 399 + 8 = 407

Let the total amount be equal to T.
T = n × 352
“However, if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receive less than or equal to Rs 330”
 
So, the maximum value that n can take is 16.

Let the common differences of the series a₁, a₂, a₃,… and b₁, b₂, b₃ be p and q respectively.
We are told that p and q are prime numbers.