CAT Mock Test- 3 (October 27) - CAT MCQ

{Finding your true self means self-reflection, engaging in candid self-appraisal and seeking ‘genuine self-knowledge’, in the words of the American philosopher Charles Guignon."}. So, self-reflection, self-appraisal and seeking self-knowledge are the three activities associated with finding one's true self, according to Charles Guignon.

Options A, B and C convey these activities. Option D has not been implied explicitly in the passage. Hence, option D is the answer. 

Option A talks about bonafide knowledge of one's characteristics, which is the same as, genuine self-knowledge.

Option B discusses introspection, part of self-realisation.

Self-appraisal refers to one's worth and significance. Option C conveys this

In the second paragraph, the author makes the following observation about inner mode - {Authenticity in this ethical sense also had a critical edge, standing against and challenging the utilitarian practices and conformist tendencies of the conventional social and economic order.} Furthermore, in the fourth paragraph, the author states that "Performative authenticity is tied to economic success and social prestige" Hence, option B can be inferred.

{Performative authenticity is tied to economic success and social prestige, which means... that your specialness and self-realisation have to be performed...The inner ideal aims at a way of being that is unfeigned and without illusions.} Option C can be inferred from these lines.

In the penultimate paragraph, the author remarks the following about performative mode-{Self-elaboration still requires self-examination, but not necessarily of any inner or even aesthetic kind.} Option D can be inferred as well.

This leaves us with option A. The former part of option A is true- {...they must ...positively affect others with their self-representations, personal characteristics and quality of life.}

However, the latter part has not been implied in the passage. The inner mode resists the cultivation of an affirming audience, true, but this cannot be equated to a lack of desire to affect others positively.

Option A is the answer.

In the first paragraph, the author states that the meteorite strike wiped out non-avian dinosaurs. However, the author does not discuss the fate of avian dinosaurs in the passage. Option A cannot be inferred.

The author states the following: "But researchers have had no idea how the event changed the tropical rain forests of Africa or, until now, those of South America." In the penultimate paragraph, the author notes that "Within moments of the Chicxulub meteorite’s impact, however, this ecosystem was irrevocably altered." This description, however, is only limited to South America's forest ecosystems. The author then extends on how these ecosystems changed drastically. However, no information is provided concerning forests in Africa; the author focuses solely on the alterations in the tropical rainforests of South America. Hence, Option B can be eliminated.

"Yet Western European and North American palaeontologists have paid little attention to tropical forests". This statement, however, does not imply that Asian and African palaeontologists majorly carried out the research. Option C can be eliminated as well.

{" Prior to the meteorite, the authors determined, South America’s forests featured many conifers and a brightly lit open canopy...As flowering species competed for light, they formed dense canopies of leaves and created the layered Amazon rain forest we know today.}

So, before the meteorite strike, the forests had open canopies, while currently, they have dense canopies. Option D can be inferred.

Hence, Option D is the correct choice.

In the penultimate paragraph, the author states that dinosaurs played a crucial role in maintaining the pre-meteoritic ecosystem. However, it has not been implied that their extinction was necessary for the new ecosystem to develop. It was merely a chain of events. Statment I cannot be inferred.

{Scientists already knew that the effects of the meteorite collision and its aftermath—at least in temperate zones—varied with local conditions and distance from the Chicxulub impact crater in Mexico’s Yucatán Peninsula.} Hence, the effects varied with distance from the crater. So the location of the crater is important. Statement II can be inferred.

{But researchers have had no idea how the event changed the tropical rain forests of Africa or, until now, those of South America.} The study is centred on the effects of the collision on South American forests. The effects on African tropical forests remain a mystery. Statement III is wrong.

Though the soil was deficient in nutrients for a long-time after the collision, the exact cause for this depletion has not been discussed in the passage. Statement iv cannot be inferred.

The passage does not discuss the impact of rainfall anywhere. Hence, option C is the answer.

All the other options can be inferred from the information presented in the last paragraph. {This influx of nitrogen, along with phosphorus from the meteorite’s ash, enabled other flowering plants to thrive alongside the legumes and to displace conifers. As flowering species competed for light, they formed dense canopies of leaves and created the layered Amazon rain forest we know today.}

After reading the given collection of statements, we identify that the core of the discussion appears to be liberal democracies and the inherent limitations associated with them. Statement 2 (independent sentence) first introduces us to this subject and sets the context, comparing liberal democracy to capitalism and subsequently to a competition/game. Sentence 1 follows 2 because the phrase 'such systems' (in 1) refers to liberal democracy and capitalism, discussed in the opening sentence. The author also describes the limitation originating from such competition: "imprisons its players to stay within their roles". Sentence 4 follows 1- 'that competition' (in 4) refers to the competition described in 1. Additionally, it adds to the preceding point that discusses the limitations. Finally, 3 ends the paragraph as it discusses a consequence of the drawback mentioned in 4. Hence, 2143 is the answer.

Sentence 1 is the opening line, as it introduces the subject of discussion-parasites. 3 follows 1. Because of the behaviour mentioned in 1, parasites are seen as pariahs or outcasts. 42 form a bloc. 4 discusses the other side associated with parasites and why they may not be all bad. And 2 builds on 4, urging us to consider this other side.  Hence, the correct order is 1342

While arranging the parts of the paragraph, we should find some grammatical or contextual connections between them -

• 4 is the first sentenceas it is an independent sentence which is introducing the subject 'The First World War'. It means that the First World War is widely regarded as a war in which the majority of significant operations took place on mainland Europe, and only a few naval battles were fought across the world's oceans. 
• 1 is the second sentence as it says that ' this is only partially true '. From the phrase 'thought of as' in sentence 4, we can understand that it was just a perception or an opinion and it may or may not be true. 
• 3 is the third sentence as the phrase 'this operation' is referring to the major operation that was conducted at sea. It means that the operation at sea was the British blockade that sought to obstruct Germany's ability to import goods. 
• 2 is the last sentence as it says that the war that took place on land contributed to the victory of the allies (Britain, France, Italy, and the United States) in 1918, and it resulted in Germany's destruction.

Thus, the correct order is 4132:

• 4: The First World War is largely thought of as a conflict where the majority of the significant operations took place almost exclusively on mainland Europe with the exception of a handful of naval clashes fought throughout the world's oceans. 
• 1: This is only partially true because while most of the fighting did take place on the continent, one of the largest and most sophisticated undertakings of the war was conducted mainly at sea. 
• 3: This operation was the British blockade from 1914-1919 which sought to obstruct Germany's ability to import goods, and thus in the most literal sense starve the German people and military into submission.
• 2: While the land war certainly contributed to the Entente's (Britain, France, Italy, U.S.) victory in 1918, it was the blockade that truly broke Germany's back.

Option A succinctly summarizes the passage, highlighting the key theme: the unintended rise of protectionism and nationalism as a reaction to the negative repercussions of globalization.

The total number of times balls were picked is 4*4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

Therefore, it cannot be determined who picked a Purple ball in round 3.
Hence, option 4 is the correct answer.

The total number of times balls were picked is 4 × 4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

Points scored by P = 1 + 5 + 3 + 7 = 16
Points scored by Q = 3 × 3 + 7 = 16
Points scored by R = 5 + 9 + 7 + 3 = 24
Points scored by S = 7 + 5 + 1 + 3 = 16
Therefore, R is the winner.
Hence, option C is the correct answer.

The total number of times balls were picked is 4 × 4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

It is clear that even if we know which coloured ball P, R or S picked in round 1 or 3, we'll know who picked what in each round.
Therefore, 1 more condition is required.

The total number of times balls were picked is 4 × 4 = 16, i.e. all the balls Red, Blue, Yellow, Green and Purple were together picked a total of 16 times. Total points earned because of Yellow ball was 18 and available points are 1, 3, 5, 7 and 11 from these points. There is only possibility of making 18, which is when 3 points ball was picked 6 times. Therefore, Yellow ball was picked 6 times and carries 3 points. All balls were picked a total of 16 times out of which Yellow ball was picked 6 times and hence rest of the 4 balls were picked a total of 10 times. If each ball was picked at least once, then the only one way in which balls could have been picked is 1, 2, 3 and 4 times in any order. Red ball was picked in each round; therefore, Red ball was picked 4 times. Purple ball was picked more times than Green ball but less times than Blue ball; therefore, Purple ball was picked 2 times whereas Green and Blue balls were picked 1 and 3 times, respectively. The following table shows the points and the number of times a ball was picked:

Since Yellow ball was picked 6 times, this is the ball which was picked more than once in the last round. If in every other round, no 2 persons had picked the same ball. Then, there are two cases: 1. Yellow ball was picked by everyone in the last round and 2 other times in 2 other rounds 2. Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. But Red ball was picked in every round; hence, case 1 is not possible. Therefore, Yellow ball was picked by 3 friends in the last round and 3 other times in 3 other rounds. P picked a Yellow ball in round 2; hence, P cannot pick a Yellow ball in round 4 because P, R and S picked different balls in each round. Therefore, Q, R and S picked a Yellow ball in the last round; and since Red ball was picked in every round, P picked a Red ball in the last round. P picked a Yellow ball in round 2. R and S picked different balls in each round; therefore, they could not have picked a Yellow ball in any other round but still Yellow ball is to be picked 2 times. Therefore, Yellow ball was picked 2 more times by Q and since Q picked Red ball in round 2, he would have picked a Yellow ball in round 1 and round 3. This information is given in the table shown below:

R picked the ball which was picked only in round 2, i.e. R picked a Green ball in round 2. Blue ball was picked 3 times; hence, it was picked in round 1, 2 and 3. In round 2, only S is the one whose ball colour is not known, so it should be Blue. P and R picked a Blue ball in round 1 and round 3 in any order. R and S picked a Red ball in round 1 and round 3 in any order. Purple ball was picked by P and S in round 1 and round 3 in any order. The final table is as given below:

If R scored 5 points in round 3, then he must have picked Blue in round 3.
Therefore, S must have picked a Purple coloured ball in round 1.
Hence, option 1 is the correct answer.

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person.

Four persons (Finch, Harper, Nick and Cody) have more red cards than Joana.

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person.

Cody has 3 blue cards.

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person and the cards they have.

Harper has 3 more red cards than Bill.

Given that the sum of the number of cards with any person is an odd number, then, for each person, the number of one of the two colored cards with him must be an odd one and the other an even one.

From (4), Finch has 3 red cards. Since he has odd number of red cards, then he must be having even number of blue cards. Hence, his blue cards can be 2/4/6. However, he has less blue cards than at least 3 persons. Hence, he must have 2 blue cards.
From (6), Nick has 4 blue cards. Hence, his red cards can be 1/3/5. Since Finch has 3 red cards, Nick can have 1/5 red cards. From (6), he has more red cards than Harper. Hence, he cannot have 1 red card. Hence, Nick must have 5 red cards.
From (2), Cody cannot have 1 or 2 blue cards (since Finch has 2 blue cards).
From (3), if Bill has 1 blue card, Harper must have 3 red cards. This is not possible as Finch has 3 red cards. Hence, Bill cannot have 1 blue card.
From (5), Harper can have 4/5/6 blue cards. Since Nick has 4 blue cards, Harper can have 5/6 blue cards.
Joana is the only person who can have 1 blue card.
From (5), Harper cannot have 1 red card and from (6), Harper cannot have 6 red cards. He also cannot have 3 or 5 red cards because Finch and Nick have 3 and 5 red cards, respectively. Hence, Harper can have 2/4 red cards. He must have odd number of blue cards (since he has even number of red cards). Therefore, Harper must have 5 blue cards.
If Harper has 2 red cards, Bill must have 4 blue cards, which is not possible as Nick has 4 blue cards. If Harper has 4 red cards, Bill must be having 6 blue cards (as Bill cannot have 4). This is the only possibility. Hence, Harper has 4 red cards and Bill has 6 blue cards.
Since Joana has 1 blue card, she must be having 2/4/6 red cards. From (5), Joana must have 2 red cards (since Harper has more red cards than Joana).
Since Bill has 6 blue cards, he must have 1 red card.
Cody has 3 blue cards and must have 6 red cards.

The following table provides the cards of each person and the cards they have.

The difference between the number of blue cards that Joana has and the number of red cards that Cody has is 5.

We represent IIM-A, IIM-B, IIM-C and IIM-I as A, B, C and I, respectively.
Now, from the given data, the following diagram can be formed:

If the third ranker voted for IIM-I, then the fifth ranker should have voted for either IIM-B or IIM-C.

Total Players in Camp: 500 players

Percentage of Batsmen: 60% of 500 = 300 players

Percentage of Bowlers: 70% of 500 = 350 players

Step 1: Calculating All-rounders
All-rounders: 300 batsmen + 350 bowlers - 500 total players = 150 all-rounders

Step 2: Calculating Pure Batsmen and Pure Bowlers
Pure Batsmen: 300 - 150 = 150 players

Pure Bowlers: 350 - 150 = 200 players

Step 3: Calculating Left-handed Players
Left-handed Pure Batsmen: 50% of 150 = 75 players

Left-handed Pure Bowlers: 60% of 200 = 120 players

Left-handed All-rounders: 20% of 150 = 30 players

Step 4: Total Left-handed Players
Total Left-handed Players: 75 (batsmen) + 120 (bowlers) + 30 (all-rounders) = 225 players

Step 5: Calculating Right-handed Players
Total Right-handed Players: 500 total players - 225 left-handed players = 275 right-handed players

Answer:
The total number of right-handed players in the cricket camp is 275.

The given equation can be written as (x² + 2x − A) = 0

=> x = 0 is one such root which lie between 3 and -1.

=> x² + 2x − A = 0 must not have a single root between 3 and -1 => f(3)*f(-1) > 0

=> (15-A) (-1-A) > 0

=> (A-15)(A+1) > 0

=> A ∈ (−∞ ,−1) U (15, ∞ )

But the given equation x(x² + 2x−A) = 0 has unequal roots => x² + 2x−A = 0 must have real and unequal roots => 4 + 4A > 0 => A > -1.  

=> A ∈ [15,∞ )

Let the thickness of the small, medium and large pizzas be t₁​, t₂​, t₃​ respectively.

The diameter of small, medium and large pizzas = 8, 11 and 14 inches.

Volume of any pizza = Area × thickness = π r² × t 

Volume of small pizza = π(8​)² × t₁​ = 16πt₁​  

Volume of the medium pizza =

Volume of the large pizza = 

P₁​:P₂​:P₃​ = V₁​:V₂​:V_3​ 

=> 2:3:4 = 16πt₁ ​: 121/4​πt_2​ : 49πt₃ ​= 2:3:4   

64t1 = 2x
121t2 = 3x
196t3 = 4x
t1 : t2 : t3 = 1/32​ : 3/121 ​: 1​/49

Total volume of the dumbbell = 2 x Volume of hemispheres + Volume of cylinder

Volume of one small sphere =  

Hence, the number of spheres = 

First, we will deal with the inner modulus. The critical point of the inner modulus is -2.

Case 1: x ≥ -2

|(x-2)-|x+2||-5 becomes |(x-2)-x-2|-5=|-4|-5=-1 which is always <0

Therefore for x ≥ -2, the inequality does not holds.

Case 2: x<-2

|(x-2)-|x+2||-5 becomes |(x-2)+x+2|-5=|2x|-5.

So, |2x ≥ 5 is satisfied under 2 conditions:

(i) 2x ≤-5 => x ≤ −5/2​. We have already asumed that x<-2 and upon solving, we get x ≤ −5/2​. 

.'. x ≤ −5/2​ holds true.

(ii) 2x>5 => x>5/2​. We have already assumed that x<-2 and upon solving, we get x>5/2​ which contradicts our main assumption. So, we reject this.

So, the range of x is (−∞ , −5/2​]

Given, 2^(1+3+5+...x) = (0.0625)⁻²⁵

Taking log on both sides, we will get,

(1+3+5+...x)log2 = −25log0.0625 

=>(1+3+5+...x)log2 = −25log(1/16​) 

=> (1+3+5+...x)log2 = −25log(2⁻⁴) 

=>(1 + 3 + 5 +...x)log2 = 100log2 

=> Sum of odd numbers upto x= 100.

Let the total odd numbers be n, then sum of n odd numbers starting from 1= n²

So, n² = 100 or n, Number of odd terms= 10.

.'. x is the 10th odd number from the beginning = 19.

Selecting squares of side 1 unit each:

We can do so by selecting 1 unit from the 8 squares along the length and 1 unit from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= ⁸C₁​×⁸C₁​= 8×8= 64  

Selecting squares of side 2 units each:

We can do so by selecting 2 continuous units from the 8 squares along the length and 2 continuous units from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= 7×7=49

Selecting squares of side 3 units each:

We can do so by selecting 3 continuous units from the 8 squares along the length and 3 continuous units from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= 6×6=36

Selecting squares of side 4 units each:

We can do so by selecting 4 continuous units from the 8 squares along the length and 4 continuous units from 8 squares along the breadth. So, total ways of selecting a square of length 1 unit= 5×5=25

.'. Total ways of selecting a square of side length at most 4 units from a chess board= 64+49+36+25= 174

S = 1³ + 2³ + 3³ +... 1000³ is

We know that the sum of cubes of natural number starting from 1 upto n is given by: 

So, for even factors, we need at least one 2 among the four 2s available with S. We can select either one, two, three or four 2s. For the other prime factors, we also have the option to choose 0 out of the available number of 5s, 11s, 7s and 13s.

So, total even factors= 4×7×3×3×3 = 756.

Let the first term of the A.P. be 'a', and common difference be 'd'.

Then, the first, second and fourth terms of the A.P. are 'a', 'a+d' and 'a+3d' respectively.

Since, they are in G.P., (a+d)² = a(a+3d)

=> a²+d²+2ad=a²+3ad

=> d²=ad

=> d²−ad = 0

=> d(d-a) = 0

.'. d=0 or d=a.

 Since, it is mentioned that the A.P is increasing, we will take the value of d= a.

Hence, a+4d= 13 reduces to 5a= 13 or a= 13/5=d.

Sum of first 10 terms of the A.P.= 10​/2 [2a+9d] = 5[11a] = 55a = 55 × 513​ = 143 

let ABCDEF be the regular hexagon.
Diagonals AD, BE, CF will intersect in a regular hexagon
Number of diagonals = n(n-3)/2
Hence, there will be 9 diagonsl.

From the figure it is easily seen that the number of points of intersection is 13.

The big pile of apples contains the same amount of large apples of half a rupee each (from Mrs. Jones), as smaller apples of one-third rupee each (from Mrs. Smith). The average price is therefore (1/2 + 1/3)/2 = 5/12 rupee.
But the apples were sold for 2/5 rupees each (5 apples for 2 rupees). Or: 25/60 and 24/60 rupee respectively. This means that per sold apple, there is a shortage of 1/60 rupee. The total shortage is Rs.7. so the ladies together started out with 420 apples. These are worth 2/5x 420 = 168 rupees, or with equal division, 84 rupees for each. If Mrs. Jones would have sold her apples herself, she would have received 105 rupees. Conclusion: Mrs. Jones loses 21 rupees in this deal.