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Sample Test: LR & DI- 2 - CAT MCQ


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20 Questions MCQ Test CAT Mock Test Series and 500+ Practice Tests 2024 - Sample Test: LR & DI- 2

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Sample Test: LR & DI- 2 - Question 1

Directions: Study the given information and answer the following question.

Sean's dietitian prepared a diet chat for him for the month of May, which had seven different types of food to be consumed during different days. Types of food included chicken, fish, eggs, green vegetables, wholegrain, milk products, and fruits. Further, on a particular day of the week throughout the month, he ate the same type of food. For example, on all Mondays, he ate the same type of food, and so on for all days of the week. Further, the following information is known about the type of food he ate on some days of May.

(1) On 8th May, Sean did not eat fish, while on 23rd May, Sean ate eggs only.
(2) On Mondays, Sean ate neither fish nor wholegrain and on 17th May, Sean ate wholegrain.
(3) There were at least 2 days between the day he ate fish only and the day he ate milk products only.
(4) There were five Saturdays and five Sundays in that May.
(5) He ate green vegetables only and fruits only on two consecutive days, but he ate neither of these on Thursday.

Q. On which of the following days of the week did Sean eat wholegrain only?

Detailed Solution for Sample Test: LR & DI- 2 - Question 1

It is given that Sean ate the same type of food on a particular day of the week throughout the month. Hence, we need to determine the food items that Sean ate during the first week, i.e. from 1st May to 7th May, and we will know the food items that he ate on different days of the month.
From (1), since Sean ate only eggs on 23rd May, he must have eaten only eggs on 2nd May as well.

Similarly, from (2), he must have eaten wholegrain on 3rd May. From (4), Sunday could only be on 2nd May or 3rd May (since there must be 5 Saturdays and 5 Sundays). If Sunday was on 2nd May, then Monday would be on 3rd May. Since he ate only wholegrain on 3rd May, this violates condition (ii).

Hence, Sunday must be on 3rd May. The remaining days will be 4th - Monday, 5th - Tuesday, 6th - Wednesday, 7th - Thursday. 1st - Friday, 2nd - Saturday.

From (3), he could have eaten fish and milk products on 4th and 7th or on 1st and 4th or on 1st and 5th.

From (5), he must have eaten green vegetables and fruits on 4th and 5th or 5th and 6th or 6th and 7th or 7th and 1st. He could not have eaten these food items on 4th and 5th as it will not leave any possible case for eating fish and milk products (from (3)). He could not have eaten green vegetables and fruits on 1st and 7th for the same reason. He could not have eaten green vegetables and fruits on 6th and 7th because 7th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5th and 6th.

He could have eaten fish and milk products on 4th and 7th. He could not have eaten fish and milk products on 1st and 4th because from (1), he did not eat fish on 1st and he did not eat fish on 4th, which was a Monday. Hence, on 4th, he must have eaten milk products and on 7th, he must have eaten fish. On 1st, he must have eaten chicken. The following table provides this information:

Sample Test: LR & DI- 2 - Question 2

Directions: Study the given information and answer the following question.

Sean's dietitian prepared a diet chat for him for the month of May, which had seven different types of food to be consumed during different days. Types of food included chicken, fish, eggs, green vegetables, wholegrain, milk products, and fruits. Further, on a particular day of the week throughout the month, he ate the same type of food. For example, on all Mondays, he ate the same type of food, and so on for all days of the week. Further, the following information is known about the type of food he ate on some days of May.

(1) On 8th May, Sean did not eat fish, while on 23rd May, Sean ate eggs only.
(2) On Mondays, Sean ate neither fish nor wholegrain and on 17th May, Sean ate wholegrain.
(3) There were at least 2 days between the day he ate fish only and the day he ate milk products only.
(4) There were five Saturdays and five Sundays in that May.
(5) He ate green vegetables only and fruits only on two consecutive days, but he ate neither of these on Thursday.

Q. On which of the following days did Sean eat milk products only?

Detailed Solution for Sample Test: LR & DI- 2 - Question 2

It is given that Sean ate the same type of food on a particular day of the week throughout the month. Hence, we need to determine the food items that Sean ate during the first week, i.e. from 1st May to 7th May, and we will know the food items that he ate on different days of the month.
From (1), since Sean ate only eggs on 23rd May, he must have eaten only eggs on 2nd May as well.

Similarly, from (2), he must have eaten wholegrain on 3rd May. From (4), Sunday could only be on 2nd May or 3rd May (since there must be 5 Saturdays and 5 Sundays). If Sunday was on 2nd May, then Monday would be on 3rd May. Since he ate only wholegrain on 3rd May, this violates condition (ii).

Hence, Sunday must be on 3rd May. The remaining days will be 4th - Monday, 5th - Tuesday, 6th - Wednesday, 7th - Thursday. 1st - Friday, 2nd - Saturday.

From (3), he could have eaten fish and milk products on 4th and 7th or on 1st and 4th or on 1st and 5th.

From (5), he must have eaten green vegetables and fruits on 4th and 5th or 5th and 6th or 6th and 7th or 7th and 1st. He could not have eaten these food items on 4th and 5th as it will not leave any possible case for eating fish and milk products (from (3)). He could not have eaten green vegetables and fruits on 1st and 7th for the same reason. He could not have eaten green vegetables and fruits on 6th and 7th because 7th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5th and 6th.

He could have eaten fish and milk products on 4th and 7th. He could not have eaten fish and milk products on 1st and 4th because from (1), he did not eat fish on 1st and he did not eat fish on 4th, which was a Monday. Hence, on 4th, he must have eaten milk products and on 7th, he must have eaten fish. On 1st, he must have eaten chicken. The following table provides this information:

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Sample Test: LR & DI- 2 - Question 3

Directions: Study the given information and answer the following question.

Sean's dietitian prepared a diet chat for him for the month of May, which had seven different types of food to be consumed during different days. Types of food included chicken, fish, eggs, green vegetables, wholegrain, milk products, and fruits. Further, on a particular day of the week throughout the month, he ate the same type of food. For example, on all Mondays, he ate the same type of food, and so on for all days of the week. Further, the following information is known about the type of food he ate on some days of May.

(1) On 8th May, Sean did not eat fish, while on 23rd May, Sean ate eggs only.
(2) On Mondays, Sean ate neither fish nor wholegrain and on 17th May, Sean ate wholegrain.
(3) There were at least 2 days between the day he ate fish only and the day he ate milk products only.
(4) There were five Saturdays and five Sundays in that May.
(5) He ate green vegetables only and fruits only on two consecutive days, but he ate neither of these on Thursday.

Q. Which of the following food items did Sean eat for the most number of days?

Detailed Solution for Sample Test: LR & DI- 2 - Question 3

It is given that Sean ate the same type of food on a particular day of the week throughout the month. Hence, we need to determine the food items that Sean ate during the first week, i.e. from 1st May to 7th May, and we will know the food items that he ate on different days of the month.
From (1), since Sean ate only eggs on 23rd May, he must have eaten only eggs on 2nd May as well.

Similarly, from (2), he must have eaten wholegrain on 3rd May. From (4), Sunday could only be on 2nd May or 3rd May (since there must be 5 Saturdays and 5 Sundays). If Sunday was on 2nd May, then Monday would be on 3rd May. Since he ate only wholegrain on 3rd May, this violates condition (ii).

Hence, Sunday must be on 3rd May. The remaining days will be 4th - Monday, 5th - Tuesday, 6th - Wednesday, 7th - Thursday. 1st - Friday, 2nd - Saturday.

From (3), he could have eaten fish and milk products on 4th and 7th or on 1st and 4th or on 1st and 5th.

From (5), he must have eaten green vegetables and fruits on 4th and 5th or 5th and 6th or 6th and 7th or 7th and 1st. He could not have eaten these food items on 4th and 5th as it will not leave any possible case for eating fish and milk products (from (3)). He could not have eaten green vegetables and fruits on 1st and 7th for the same reason. He could not have eaten green vegetables and fruits on 6th and 7th because 7th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5th and 6th.

He could have eaten fish and milk products on 4th and 7th. He could not have eaten fish and milk products on 1st and 4th because from (1), he did not eat fish on 1st and he did not eat fish on 4th, which was a Monday. Hence, on 4th, he must have eaten milk products and on 7th, he must have eaten fish. On 1st, he must have eaten chicken. The following table provides this information:

Sample Test: LR & DI- 2 - Question 4

Directions: Study the given information and answer the following question.

Sean's dietitian prepared a diet chat for him for the month of May, which had seven different types of food to be consumed during different days. Types of food included chicken, fish, eggs, green vegetables, wholegrain, milk products, and fruits. Further, on a particular day of the week throughout the month, he ate the same type of food. For example, on all Mondays, he ate the same type of food, and so on for all days of the week. Further, the following information is known about the type of food he ate on some days of May.

(1) On 8th May, Sean did not eat fish, while on 23rd May, Sean ate eggs only.
(2) On Mondays, Sean ate neither fish nor wholegrain and on 17th May, Sean ate wholegrain.
(3) There were at least 2 days between the day he ate fish only and the day he ate milk products only.
(4) There were five Saturdays and five Sundays in that May.
(5) He ate green vegetables only and fruits only on two consecutive days, but he ate neither of these on Thursday.

Q. Which of the following statements is definitely true?

Detailed Solution for Sample Test: LR & DI- 2 - Question 4

It is given that Sean ate the same type of food on a particular day of the week throughout the month. Hence, we need to determine the food items that Sean ate during the first week, i.e. from 1st May to 7th May, and we will know the food items that he ate on different days of the month.
From (1), since Sean ate only eggs on 23rd May, he must have eaten only eggs on 2nd May as well.

Similarly, from (2), he must have eaten wholegrain on 3rd May. From (4), Sunday could only be on 2nd May or 3rd May (since there must be 5 Saturdays and 5 Sundays). If Sunday was on 2nd May, then Monday would be on 3rd May. Since he ate only wholegrain on 3rd May, this violates condition (ii).

Hence, Sunday must be on 3rd May. The remaining days will be 4th - Monday, 5th - Tuesday, 6th - Wednesday, 7th - Thursday. 1st - Friday, 2nd - Saturday.

From (3), he could have eaten fish and milk products on 4th and 7th or on 1st and 4th or on 1st and 5th.

From (5), he must have eaten green vegetables and fruits on 4th and 5th or 5th and 6th or 6th and 7th or 7th and 1st. He could not have eaten these food items on 4th and 5th as it will not leave any possible case for eating fish and milk products (from (3)). He could not have eaten green vegetables and fruits on 1st and 7th for the same reason. He could not have eaten green vegetables and fruits on 6th and 7th because 7th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5th and 6th.

He could have eaten fish and milk products on 4th and 7th. He could not have eaten fish and milk products on 1st and 4th because from (1), he did not eat fish on 1st and he did not eat fish on 4th, which was a Monday. Hence, on 4th, he must have eaten milk products and on 7th, he must have eaten fish. On 1st, he must have eaten chicken. The following table provides this information:

Sample Test: LR & DI- 2 - Question 5

Directions: Study the given information and answer the following question.

Sean's dietitian prepared a diet chat for him for the month of May, which had seven different types of food to be consumed during different days. Types of food included chicken, fish, eggs, green vegetables, wholegrain, milk products, and fruits. Further, on a particular day of the week throughout the month, he ate the same type of food. For example, on all Mondays, he ate the same type of food, and so on for all days of the week. Further, the following information is known about the type of food he ate on some days of May.

(1) On 8th May, Sean did not eat fish, while on 23rd May, Sean ate eggs only.
(2) On Mondays, Sean ate neither fish nor wholegrain and on 17th May, Sean ate wholegrain.
(3) There were at least 2 days between the day he ate fish only and the day he ate milk products only.
(4) There were five Saturdays and five Sundays in that May.
(5) He ate green vegetables only and fruits only on two consecutive days, but he ate neither of these on Thursday.

Q. Which two types of food were consumed by Sean on two consecutive days?

Detailed Solution for Sample Test: LR & DI- 2 - Question 5

It is given that Sean ate the same type of food on a particular day of the week throughout the month. Hence, we need to determine the food items that Sean ate during the first week, i.e. from 1st May to 7th May, and we will know the food items that he ate on different days of the month.
From (1), since Sean ate only eggs on 23rd May, he must have eaten only eggs on 2nd May as well.

Similarly, from (2), he must have eaten wholegrain on 3rd May. From (4), Sunday could only be on 2nd May or 3rd May (since there must be 5 Saturdays and 5 Sundays). If Sunday was on 2nd May, then Monday would be on 3rd May. Since he ate only wholegrain on 3rd May, this violates condition (ii).

Hence, Sunday must be on 3rd May. The remaining days will be 4th - Monday, 5th - Tuesday, 6th - Wednesday, 7th - Thursday. 1st - Friday, 2nd - Saturday.

From (3), he could have eaten fish and milk products on 4th and 7th or on 1st and 4th or on 1st and 5th.

From (5), he must have eaten green vegetables and fruits on 4th and 5th or 5th and 6th or 6th and 7th or 7th and 1st. He could not have eaten these food items on 4th and 5th as it will not leave any possible case for eating fish and milk products (from (3)). He could not have eaten green vegetables and fruits on 1st and 7th for the same reason. He could not have eaten green vegetables and fruits on 6th and 7th because 7th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5th and 6th.

He could have eaten fish and milk products on 4th and 7th. He could not have eaten fish and milk products on 1st and 4th because from (1), he did not eat fish on 1st and he did not eat fish on 4th, which was a Monday. Hence, on 4th, he must have eaten milk products and on 7th, he must have eaten fish. On 1st, he must have eaten chicken. The following table provides this information:

Chicken and eggs were consumed by Sean on two consecutive days, i.e. Friday and Saturday.

*Answer can only contain numeric values
Sample Test: LR & DI- 2 - Question 6

Directions: Study the given information and answer the following question.

The table below gives some information about marks scored by Joey in a written exam for University Professor, which consists of five main categories: Quant, Reasoning, DI, English, and General Knowledge

The following data is also provided:
(1) For every correct answer, the candidate gets one mark, for every un-attempted question, the candidate loses 1/6 of a mark, and for every incorrect answer, the student loses 1/3 of a mark.
(2) Joey scores 67 marks and he attempts 125 questions.
(3) The number of incorrect answers given by Joey in English is 1/6 of total number of incorrect answers and twice the number of incorrect answers in Reasoning.
(4) Joey's net marks in Reasoning are double than those in DI.

Q. Key in the total number of incorrect answers given by Jeoy in Reasoning.


Detailed Solution for Sample Test: LR & DI- 2 - Question 6










*Answer can only contain numeric values
Sample Test: LR & DI- 2 - Question 7

Directions: Study the given information and answer the following question.

The table below gives some information about marks scored by Joey in a written exam for University Professor, which consists of five main categories: Quant, Reasoning, DI, English, and General Knowledge

The following data is also provided:
(1) For every correct answer, the candidate gets one mark, for every un-attempted question, the candidate loses 1/6 of a mark, and for every incorrect answer, the student loses 1/3 of a mark.
(2) Joey scores 67 marks and he attempts 125 questions.
(3) The number of incorrect answers given by Joey in English is 1/6 of total number of incorrect answers and twice the number of incorrect answers in Reasoning.
(4) Joey's net marks in Reasoning are double than those in DI.

Q. The highest number of incorrect answers given by Joey in any category is ____.


Detailed Solution for Sample Test: LR & DI- 2 - Question 7










*Answer can only contain numeric values
Sample Test: LR & DI- 2 - Question 8

Directions: Study the given information and answer the following question.

The table below gives some information about marks scored by Joey in a written exam for University Professor, which consists of five main categories: Quant, Reasoning, DI, English, and General Knowledge

The following data is also provided:
(1) For every correct answer, the candidate gets one mark, for every un-attempted question, the candidate loses 1/6 of a mark, and for every incorrect answer, the student loses 1/3 of a mark.
(2) Joey scores 67 marks and he attempts 125 questions.
(3) The number of incorrect answers given by Joey in English is 1/6 of total number of incorrect answers and twice the number of incorrect answers in Reasoning.
(4) Joey's net marks in Reasoning are double than those in DI.

Q. Key in the number of questions attempted by Joey in Reasoning.


Detailed Solution for Sample Test: LR & DI- 2 - Question 8










Sample Test: LR & DI- 2 - Question 9

Directions: Study the given information and answer the following question.

The table below gives some information about marks scored by Joey in a written exam for University Professor, which consists of five main categories: Quant, Reasoning, DI, English, and General Knowledge

The following data is also provided:
(1) For every correct answer, the candidate gets one mark, for every un-attempted question, the candidate loses 1/6 of a mark, and for every incorrect answer, the student loses 1/3 of a mark.
(2) Joey scores 67 marks and he attempts 125 questions.
(3) The number of incorrect answers given by Joey in English is 1/6 of total number of incorrect answers and twice the number of incorrect answers in Reasoning.
(4) Joey's net marks in Reasoning are double than those in DI.

Q. What is the percentage accuracy of Joey in the examination?

Detailed Solution for Sample Test: LR & DI- 2 - Question 9










Sample Test: LR & DI- 2 - Question 10

Directions: Study the given information and answer the following question.

The table below gives some information about marks scored by Joey in a written exam for University Professor, which consists of five main categories: Quant, Reasoning, DI, English, and General Knowledge

The following data is also provided:
(1) For every correct answer, the candidate gets one mark, for every un-attempted question, the candidate loses 1/6 of a mark, and for every incorrect answer, the student loses 1/3 of a mark.
(2) Joey scores 67 marks and he attempts 125 questions.
(3) The number of incorrect answers given by Joey in English is 1/6 of total number of incorrect answers and twice the number of incorrect answers in Reasoning.
(4) Joey's net marks in Reasoning are double than those in DI.

Q. What is the ratio of the number of questions attempted to the number of incorrect answers?

Detailed Solution for Sample Test: LR & DI- 2 - Question 10











The ratio of the number of questions attempted to the number of incorrect answers is 125 : 36.

Sample Test: LR & DI- 2 - Question 11

Directions: Read the given information carefully and answer the question that follows.

Martin, Taylor, Anderson, Robert and Galileo are playing a card game. Each card bears a letter from A to J on it. In each round of the game, each person independently picks a card and after each of the five persons have picked one card, the five cards are placed in descending alphabetical order of the letters on them such that the card with the letter that appears first in alphabetical order is placed at the end (extreme right) and the card with a letter that appears later in alphabetical order is placed next to the first to the left. In this way, all cards of a round are placed together. The word, so formed, in each round, is labeled as the closing word of that round. After exactly four rounds, the person who picks the leftmost letter of the closing word across the four rounds is declared the winner of the game. No person can pick the same letter in any two rounds.

The closing words of each round are JIFFB, HFDBB, JIHGF and HGFEE, in any order.
It is also known that

(1) Taylor picked cards with consecutive letters across the four rounds, but not necessarily in any order and in none of the rounds did the letter of a card that he picked become the leftmost letter of the closing word
(2) in each round, the letter of the card that Taylor picked was higher in alphabetical order than the one that Robert picked, while in the second round, Anderson and Galileo picked the same lettered cards
(3) the person who was declared the winner picked the card with a letter that appeared earlier in alphabetical order of the closing word in only one round, which was the third round
(4) for only one person is the difference between the highest lettered card that he picked and the lowest lettered card that he picked was 8 according to the numerical value of letter (assigned in ascending order from A to Z, beginning 1 to 26) and this person picked his least lettered card in the first round
(5) no one picked the same lettered card as Robert did in any round and Robert and Galileo did not pick cards with consecutive letters in any round

Q. Who was the winner of the game?

Detailed Solution for Sample Test: LR & DI- 2 - Question 11


Since neither A nor C is a letter picked so no consecutive series with B can be formed, this means, in Round A, Taylor did not pick the letter B
If Taylor picked the letter I in Round A or Round C, the possiblities of 4 consecutive letters are: IJKL, HIJK, GHIJ, FGHINow K is not a letter picked, hence, possibilites are: GHIJ or FGHI.Now H cannot be a Round B or Round D letter picked by Taylor, since these are leftmost letters of the closing words, so H has to be picked in Round C; this means, I was picked by Taylor in Round ASince G is a letter only in Round D out of Rounds B and D, Taylor picked G in Round D and F in Round B

Since no one picked the same letter as Robert in any round, from Condition 5, and in each round, the letter that Taylor picked was higher in alphabetical order than the one that Robert picked

Since F has already been picked by Robert in Round D

Robert and Galileo did not pick consecutive letters in any round.

Now since B-J combination is to make difference of 8; this is possible only for Round A or Round B, as only these two rounds have the letter B.
But B in Round A has already been used. So the letter B should belong to Round B to form a B-J combination. Since this should be Round 1 (with least number letter B in it), hence, Round B is Round 1


Anderson and Galileo picked the same letter in Round 2.
2 same letters exist for Round A and Round D
Round B is already Round 1
So either Round A is Round 2 or Round D is Round 2
Now E cannot be picked by Galileo in Round D, so Round D is NOT Round 2
This means Round A is Round 2




Since for only 1 person was the difference 8

From Condition 3, the person who was declared the winner pickedd the alphabet that appears earlier in alphabetical order of the Closing word in only one round, which was the third round.
The winner either picked letter H or letter J (the leftmost letter of Round C or D)
Now J never appears earlier in any round, while H appears earlier only once in Round C, hence, Round C is Round 3; and Round D is Round 4

Thus the winner is the person in Round 4 who picked the letter H, i.e. Galileo

Sample Test: LR & DI- 2 - Question 12

Directions: Read the given information carefully and answer the question that follows.

Martin, Taylor, Anderson, Robert and Galileo are playing a card game. Each card bears a letter from A to J on it. In each round of the game, each person independently picks a card and after each of the five persons have picked one card, the five cards are placed in descending alphabetical order of the letters on them such that the card with the letter that appears first in alphabetical order is placed at the end (extreme right) and the card with a letter that appears later in alphabetical order is placed next to the first to the left. In this way, all cards of a round are placed together. The word, so formed, in each round, is labeled as the closing word of that round. After exactly four rounds, the person who picks the leftmost letter of the closing word across the four rounds is declared the winner of the game. No person can pick the same letter in any two rounds.

The closing words of each round are JIFFB, HFDBB, JIHGF and HGFEE, in any order.
It is also known that

(1) Taylor picked cards with consecutive letters across the four rounds, but not necessarily in any order and in none of the rounds did the letter of a card that he picked become the leftmost letter of the closing word
(2) in each round, the letter of the card that Taylor picked was higher in alphabetical order than the one that Robert picked, while in the second round, Anderson and Galileo picked the same lettered cards
(3) the person who was declared the winner picked the card with a letter that appeared earlier in alphabetical order of the closing word in only one round, which was the third round
(4) for only one person is the difference between the highest lettered card that he picked and the lowest lettered card that he picked was 8 according to the numerical value of letter (assigned in ascending order from A to Z, beginning 1 to 26) and this person picked his least lettered card in the first round
(5) no one picked the same lettered card as Robert did in any round and Robert and Galileo did not pick cards with consecutive letters in any round

Q. Which of the following letters did Anderson pick in the first round?

Detailed Solution for Sample Test: LR & DI- 2 - Question 12


Since neither A nor C is a letter picked so no consecutive series with B can be formed, this means, in Round A, Taylor did not pick the letter B
If Taylor picked the letter I in Round A or Round C, the possiblities of 4 consecutive letters are: IJKL, HIJK, GHIJ, FGHINow K is not a letter picked, hence, possibilites are: GHIJ or FGHI.Now H cannot be a Round B or Round D letter picked by Taylor, since these are leftmost letters of the closing words, so H has to be picked in Round C; this means, I was picked by Taylor in Round ASince G is a letter only in Round D out of Rounds B and D, Taylor picked G in Round D and F in Round B

Since no one picked the same letter as Robert in any round, from Condition 5, and in each round, the letter that Taylor picked was higher in alphabetical order than the one that Robert picked

Since F has already been picked by Robert in Round D

Robert and Galileo did not pick consecutive letters in any round.

Now since B-J combination is to make difference of 8; this is possible only for Round A or Round B, as only these two rounds have the letter B.
But B in Round A has already been used. So the letter B should belong to Round B to form a B-J combination. Since this should be Round 1 (with least number letter B in it), hence, Round B is Round 1


Anderson and Galileo picked the same letter in Round 2.
2 same letters exist for Round A and Round D
Round B is already Round 1
So either Round A is Round 2 or Round D is Round 2
Now E cannot be picked by Galileo in Round D, so Round D is NOT Round 2
This means Round A is Round 2




Since for only 1 person was the difference 8

From Condition 3, the person who was declared the winner pickedd the alphabet that appears earlier in alphabetical order of the Closing word in only one round, which was the third round.
The winner either picked letter H or letter J (the leftmost letter of Round C or D)
Now J never appears earlier in any round, while H appears earlier only once in Round C, hence, Round C is Round 3; and Round D is Round 4

Thus Anderson pick letter H in first round.

Sample Test: LR & DI- 2 - Question 13

Directions: Read the given information carefully and answer the question that follows.

Martin, Taylor, Anderson, Robert and Galileo are playing a card game. Each card bears a letter from A to J on it. In each round of the game, each person independently picks a card and after each of the five persons have picked one card, the five cards are placed in descending alphabetical order of the letters on them such that the card with the letter that appears first in alphabetical order is placed at the end (extreme right) and the card with a letter that appears later in alphabetical order is placed next to the first to the left. In this way, all cards of a round are placed together. The word, so formed, in each round, is labeled as the closing word of that round. After exactly four rounds, the person who picks the leftmost letter of the closing word across the four rounds is declared the winner of the game. No person can pick the same letter in any two rounds.

The closing words of each round are JIFFB, HFDBB, JIHGF and HGFEE, in any order.
It is also known that

(1) Taylor picked cards with consecutive letters across the four rounds, but not necessarily in any order and in none of the rounds did the letter of a card that he picked become the leftmost letter of the closing word
(2) in each round, the letter of the card that Taylor picked was higher in alphabetical order than the one that Robert picked, while in the second round, Anderson and Galileo picked the same lettered cards
(3) the person who was declared the winner picked the card with a letter that appeared earlier in alphabetical order of the closing word in only one round, which was the third round
(4) for only one person is the difference between the highest lettered card that he picked and the lowest lettered card that he picked was 8 according to the numerical value of letter (assigned in ascending order from A to Z, beginning 1 to 26) and this person picked his least lettered card in the first round
(5) no one picked the same lettered card as Robert did in any round and Robert and Galileo did not pick cards with consecutive letters in any round

Q. Who picked the least ranked alphabet according to alphabetical order in the fourth round?

Detailed Solution for Sample Test: LR & DI- 2 - Question 13


Since neither A nor C is a letter picked so no consecutive series with B can be formed, this means, in Round A, Taylor did not pick the letter B
If Taylor picked the letter I in Round A or Round C, the possiblities of 4 consecutive letters are: IJKL, HIJK, GHIJ, FGHINow K is not a letter picked, hence, possibilites are: GHIJ or FGHI.Now H cannot be a Round B or Round D letter picked by Taylor, since these are leftmost letters of the closing words, so H has to be picked in Round C; this means, I was picked by Taylor in Round ASince G is a letter only in Round D out of Rounds B and D, Taylor picked G in Round D and F in Round B

Since no one picked the same letter as Robert in any round, from Condition 5, and in each round, the letter that Taylor picked was higher in alphabetical order than the one that Robert picked

Since F has already been picked by Robert in Round D

Robert and Galileo did not pick consecutive letters in any round.

Now since B-J combination is to make difference of 8; this is possible only for Round A or Round B, as only these two rounds have the letter B.
But B in Round A has already been used. So the letter B should belong to Round B to form a B-J combination. Since this should be Round 1 (with least number letter B in it), hence, Round B is Round 1


Anderson and Galileo picked the same letter in Round 2.
2 same letters exist for Round A and Round D
Round B is already Round 1
So either Round A is Round 2 or Round D is Round 2
Now E cannot be picked by Galileo in Round D, so Round D is NOT Round 2
This means Round A is Round 2




Since for only 1 person was the difference 8

From Condition 3, the person who was declared the winner pickedd the alphabet that appears earlier in alphabetical order of the Closing word in only one round, which was the third round.
The winner either picked letter H or letter J (the leftmost letter of Round C or D)
Now J never appears earlier in any round, while H appears earlier only once in Round C, hence, Round C is Round 3; and Round D is Round 4

Sample Test: LR & DI- 2 - Question 14

Directions: Read the given information carefully and answer the question that follows.

Martin, Taylor, Anderson, Robert and Galileo are playing a card game. Each card bears a letter from A to J on it. In each round of the game, each person independently picks a card and after each of the five persons have picked one card, the five cards are placed in descending alphabetical order of the letters on them such that the card with the letter that appears first in alphabetical order is placed at the end (extreme right) and the card with a letter that appears later in alphabetical order is placed next to the first to the left. In this way, all cards of a round are placed together. The word, so formed, in each round, is labeled as the closing word of that round. After exactly four rounds, the person who picks the leftmost letter of the closing word across the four rounds is declared the winner of the game. No person can pick the same letter in any two rounds.

The closing words of each round are JIFFB, HFDBB, JIHGF and HGFEE, in any order.
It is also known that

(1) Taylor picked cards with consecutive letters across the four rounds, but not necessarily in any order and in none of the rounds did the letter of a card that he picked become the leftmost letter of the closing word
(2) in each round, the letter of the card that Taylor picked was higher in alphabetical order than the one that Robert picked, while in the second round, Anderson and Galileo picked the same lettered cards
(3) the person who was declared the winner picked the card with a letter that appeared earlier in alphabetical order of the closing word in only one round, which was the third round
(4) for only one person is the difference between the highest lettered card that he picked and the lowest lettered card that he picked was 8 according to the numerical value of letter (assigned in ascending order from A to Z, beginning 1 to 26) and this person picked his least lettered card in the first round
(5) no one picked the same lettered card as Robert did in any round and Robert and Galileo did not pick cards with consecutive letters in any round

Q. Who among the following picked the highest letter according to alphabetical order in more than one rounds?

Detailed Solution for Sample Test: LR & DI- 2 - Question 14


Since neither A nor C is a letter picked so no consecutive series with B can be formed, this means, in Round A, Taylor did not pick the letter B
If Taylor picked the letter I in Round A or Round C, the possiblities of 4 consecutive letters are: IJKL, HIJK, GHIJ, FGHINow K is not a letter picked, hence, possibilites are: GHIJ or FGHI.Now H cannot be a Round B or Round D letter picked by Taylor, since these are leftmost letters of the closing words, so H has to be picked in Round C; this means, I was picked by Taylor in Round ASince G is a letter only in Round D out of Rounds B and D, Taylor picked G in Round D and F in Round B

Since no one picked the same letter as Robert in any round, from Condition 5, and in each round, the letter that Taylor picked was higher in alphabetical order than the one that Robert picked

Since F has already been picked by Robert in Round D

Robert and Galileo did not pick consecutive letters in any round.

Now since B-J combination is to make difference of 8; this is possible only for Round A or Round B, as only these two rounds have the letter B.
But B in Round A has already been used. So the letter B should belong to Round B to form a B-J combination. Since this should be Round 1 (with least number letter B in it), hence, Round B is Round 1


Anderson and Galileo picked the same letter in Round 2.
2 same letters exist for Round A and Round D
Round B is already Round 1
So either Round A is Round 2 or Round D is Round 2
Now E cannot be picked by Galileo in Round D, so Round D is NOT Round 2
This means Round A is Round 2




Since for only 1 person was the difference 8

From Condition 3, the person who was declared the winner pickedd the alphabet that appears earlier in alphabetical order of the Closing word in only one round, which was the third round.
The winner either picked letter H or letter J (the leftmost letter of Round C or D)
Now J never appears earlier in any round, while H appears earlier only once in Round C, hence, Round C is Round 3; and Round D is Round 4

Anderson picked H and J in rounds 1 and 3, respectively.

Sample Test: LR & DI- 2 - Question 15

Directions: Read the given information carefully and answer the question that follows.

Martin, Taylor, Anderson, Robert and Galileo are playing a card game. Each card bears a letter from A to J on it. In each round of the game, each person independently picks a card and after each of the five persons has picked one card, the five cards are placed in the descending alphabetical order of the letters on them such that the card with the letter that appears first in the alphabetical order is placed at the end (extreme right) and the card with a letter that appears later in the alphabetical order is placed next to the first to the left. In this way, all cards of a round are placed together. The word, so formed, in each round, is labelled as the closing word of that round. After exactly four rounds, the person who picks the leftmost letter of the closing word across the four rounds is declared the winner of the game. No person can pick the same letter in any two rounds.

The closing words of each round are JIFFB, HFDBB, JIHGF and HGFEE, in any order.
It is also known that:

(1) Taylor picked cards with consecutive letters across the four rounds, but not necessarily in any order and in none of the rounds did the letter of a card that he picked become the leftmost letter of the closing word.
(2) In each round, the letter of the card that Taylor picked was higher in the alphabetical order than the one that Robert picked, while in the second round, Anderson and Galileo picked the same lettered cards.
(3) The person who was declared the winner picked the card with a letter that appeared earlier in the alphabetical order of the closing word in only one round, which was the third round.
(4) For only one person is the difference between the highest lettered card that he picked and the lowest lettered card that he picked was 8 according to the numerical value of letter (assigned in ascending order from A to Z, beginning 1 to 26) and this person picked his least lettered card in the first round.
(5) No one picked the same lettered card as Robert did in any round and Robert and Galileo did not pick cards with consecutive letters in any round.

Q. The round in which no vowel got picked was

Detailed Solution for Sample Test: LR & DI- 2 - Question 15


Since neither A nor C is a letter picked, no consecutive series with B can be formed; this means, in Round A, Taylor did not pick the letter B. If Taylor picked the letter I in Round A or Round C, the possibilities of 4 consecutive letters are: IJKL, HIJK, GHIJ, FGHI. Now, K is not a letter picked, so possibilities are: GHIJ or FGHI. Now, H cannot be a Round B or Round D letter picked by Taylor, since these are leftmost letters of the closing words. So, H has to be picked in Round C; this means, I was picked by Taylor in Round A. Since G is a letter only in Round D out of Rounds B and D, Taylor picked G in Round D and F in Round B.

Since no one picked the same letter as Robert in any round, from Condition 5, and in each round, the letter that Taylor picked was higher in the alphabetical order than the one that Robert picked:

Since F has already been picked by Robert in Round D:

Robert and Galileo did not pick consecutive letters in any round.

Now, since the B-J combination is to make difference of 8, this is possible only for Round A or Round B, as only these two rounds have the letter B.
But B in Round A has already been used. So, the letter B should belong to Round B to form the B-J combination. Since this should be Round 1 (with least number letter B in it), Round B is Round 1.


Anderson and Galileo picked the same letter in Round 2.
2 same letters exist for Round A and Round D.
Round B is already Round 1.
So, either Round A is Round 2 or Round D is Round 2.
Now, E cannot be picked by Galileo in Round D, so Round D is NOT Round 2.
This means Round A is Round 2.




Since for only 1 person was the difference 8:

From Condition 3, the person who was declared the winner picked the letter that appears earlier in the alphabetical order of the closing word in only one round, which was the third round.
The winner picked either letter H or letter J (the leftmost letter of Round C or D).
Now, J never appears earlier in any round, while H appears earlier only once in Round C, so Round C is Round 3, and Round D is Round 4.

Sample Test: LR & DI- 2 - Question 16

Directions: Study the given information and answer the following question.

On a particular day, five buses - Bus 1, Bus 2, Bus 3, Bus 4 and Bus 5 - destined for the same bus terminus started from different starting points and travelled for different times: 1 hour and 40 minutes, 2 hours and 5 minutes, 2 hours and 40 minutes, 3 hours, and 3 hours and 25 minutes to reach the bus terminus. The journey of buses started at different times, i.e. 11:00, 11:25, 11:45, 11:50 and 12:15.

It is also known that
(1) no two buses reached the bus terminus at the same time
(2) Bus 1, whose travel time was 2 hours and 5 minutes, did not reach the bus terminus before 13:20
(3) among the five buses, the last bus to reach the bus terminus was Bus 3
(4) Bus 4 reached the bus terminus after 14:45, but the travel time was not 3 hours and 25 minutes
(5) among the five buses, at least three buses reached the bus terminus before Bus 5 did

Q. What was the travel time for the bus that reached the bus terminus before 1:00?

Detailed Solution for Sample Test: LR & DI- 2 - Question 16

1 hour and 40 minutes

From (1), Bus 1's travel time was 2 hours and 5 minutes and it reached the bus terminus after 13:20. It could not have the starting time of 11:00.
From (2), Bus 3 was the last bus to reach the bus terminus and from (4), Bus 5 must be the fourth bus to reach the bus terminus.
Bus 4 reached after 14:45. Since it did not have a travel time of 3 hours and 25 minutes, the bus 4 could have started at 12:15 and have a travel time of 2 hours and 40 minutes or 3 hours, or, it could have started at 11:50 and have a travel time of 3 hours. The earliest that Bus 4 could reach the bus terminus is 14:50.

Bus 3 and Bus 5 could not have started at 11:00, as the bus that would have started at 11:00, must have reached the bus terminus by at most 14:25. They also could not have started at 11:25 as then they would have reached the bus terminus by 14:50. In either case, they could not be the fourth and fifth buses to reach the bus terminus.
Since, except for Bus 2, none of the others could have started at 11:00, Bus 2 must have started at 11:00. Also, apart from Bus 1 (and Bus 2), none of the others could have started at 11:25. Hence, Bus 1 must have started at 11:25.

If Bus 4 had started at 12:15 and had a travel time of 3 hours, it would have reached the bus terminus by 15:15. Bus 3 and Bus 5 must have starting times of 11:45 and 11:50 for 3 hours and 25 minutes and 2 hours and 40 minutes, in any order. The bus which had a travel time of 2 hours and 40 minutes would have reached by at most 14:30. Either of Bus 3 and Bus 5 can be this bus because they reached after Bus 4. Hence, this case is not possible.

If Bus 4 started at 12:15 and had a travel time of 160 minutes, it would have reached by 14:55. Bus 3 and Bus 5 must have started at 11:45 and 11:50 with travel times of 180 minutes and 205 minutes, in any order. The bus whose travel time was 180 minutes would have reached by at most 14:50. This case is also not possible because one of Bus 5 and Bus 3 would have reached the bus terminus before Bus 4.

If Bus 4 started at 11:50 with travel time of 3 hours, it would reach bus terminus by 14:50. Bus 3 and Bus 5 must have starting times of 11:45 and 12:15 with travel times of 2 hours and 40 minutes and 3 hours and 25 minutes, in any order. The bus which has a travel time of 2 hours and 40 minutes must have started at 12:15 for it to reach after Bus 4. This bus would have reached by 14:55. The bus whose travel time is 3 hours and 25 minutes must have started by 11:45 and would have reached by 15:10. Hence, the former must be Bus 5 and the latter must be Bus 3.

The following table presents, for each bus, the travel time, the start time and the time at which it reaches the bus terminus:

Sample Test: LR & DI- 2 - Question 17

Directions: Study the given information and answer the following question.

On a particular day, five buses - Bus 1, Bus 2, Bus 3, Bus 4 and Bus 5 - destined for the same bus terminus started from different starting points and travelled for different times: 1 hour and 40 minutes, 2 hours and 5 minutes, 2 hours and 40 minutes, 3 hours, and 3 hours and 25 minutes to reach the bus terminus. The journey of buses started at different times, i.e. 11:00, 11:25, 11:45, 11:50 and 12:15.

It is also known that
(1) no two buses reached the bus terminus at the same time
(2) Bus 1, whose travel time was 2 hours and 5 minutes, did not reach the bus terminus before 13:20
(3) among the five buses, the last bus to reach the bus terminus was Bus 3
(4) Bus 4 reached the bus terminus after 14:45, but the travel time was not 3 hours and 25 minutes
(5) among the five buses, at least three buses reached the bus terminus before Bus 5 did

Q. At what time did Bus 4 reach the terminus?

Detailed Solution for Sample Test: LR & DI- 2 - Question 17

Bus 4 reached the terminus at 14:50.

From (1), Bus 1's travel time was 2 hours and 5 minutes and it reached the bus terminus after 13:20. It could not have the starting time of 11:00.
From (2), Bus 3 was the last bus to reach the bus terminus and from (4), Bus 5 must be the fourth bus to reach the bus terminus.
Bus 4 reached after 14:45. Since it did not have a travel time of 3 hours and 25 minutes, the bus 4 could have started at 12:15 and have a travel time of 2 hours and 40 minutes or 3 hours, or, it could have started at 11:50 and have a travel time of 3 hours. The earliest that Bus 4 could reach the bus terminus is 14:50.

Bus 3 and Bus 5 could not have started at 11:00, as the bus that would have started at 11:00, must have reached the bus terminus by at most 14:25. They also could not have started at 11:25 as then they would have reached the bus terminus by 14:50. In either case, they could not be the fourth and fifth buses to reach the bus terminus.
Since, except for Bus 2, none of the others could have started at 11:00, Bus 2 must have started at 11:00. Also, apart from Bus 1 (and Bus 2), none of the others could have started at 11:25. Hence, Bus 1 must have started at 11:25.

If Bus 4 had started at 12:15 and had a travel time of 3 hours, it would have reached the bus terminus by 15:15. Bus 3 and Bus 5 must have starting times of 11:45 and 11:50 for 3 hours and 25 minutes and 2 hours and 40 minutes, in any order. The bus which had a travel time of 2 hours and 40 minutes would have reached by at most 14:30. Either of Bus 3 and Bus 5 can be this bus because they reached after Bus 4. Hence, this case is not possible.

If Bus 4 started at 12:15 and had a travel time of 160 minutes, it would have reached by 14:55. Bus 3 and Bus 5 must have started at 11:45 and 11:50 with travel times of 180 minutes and 205 minutes, in any order. The bus whose travel time was 180 minutes would have reached by at most 14:50. This case is also not possible because one of Bus 5 and Bus 3 would have reached the bus terminus before Bus 4.

If Bus 4 started at 11:50 with travel time of 3 hours, it would reach bus terminus by 14:50. Bus 3 and Bus 5 must have starting times of 11:45 and 12:15 with travel times of 2 hours and 40 minutes and 3 hours and 25 minutes, in any order. The bus which has a travel time of 2 hours and 40 minutes must have started at 12:15 for it to reach after Bus 4. This bus would have reached by 14:55. The bus whose travel time is 3 hours and 25 minutes must have started by 11:45 and would have reached by 15:10. Hence, the former must be Bus 5 and the latter must be Bus 3.

The following table presents, for each bus, the travel time, the start time and the time at which it reaches the bus terminus:

*Answer can only contain numeric values
Sample Test: LR & DI- 2 - Question 18

Directions: Study the given information and answer the following question.

On a particular day, five buses - Bus 1, Bus 2, Bus 3, Bus 4 and Bus 5 - destined for the same bus terminus started from different starting points and travelled for different times: 1 hour and 40 minutes, 2 hours and 5 minutes, 2 hours and 40 minutes, 3 hours, and 3 hours and 25 minutes to reach the bus terminus. The journey of buses started at different times, i.e. 11:00, 11:25, 11:45, 11:50 and 12:15.

It is also known that
(1) no two buses reached the bus terminus at the same time
(2) Bus 1, whose travel time was 2 hours and 5 minutes, did not reach the bus terminus before 13:20
(3) among the five buses, the last bus to reach the bus terminus was Bus 3
(4) Bus 4 reached the bus terminus after 14:45, but the travel time was not 3 hours and 25 minutes
(5) among the five buses, at least three buses reached the bus terminus before Bus 5 did

Q. What was the travel time (in minutes) for the bus that started at 11:45?


Detailed Solution for Sample Test: LR & DI- 2 - Question 18

The travel time for the bus that started at 11:45 was 205 minutes.

From (1), Bus 1's travel time was 2 hours and 5 minutes and it reached the bus terminus after 13:20. It could not have the starting time of 11:00.
From (2), Bus 3 was the last bus to reach the bus terminus and from (4), Bus 5 must be the fourth bus to reach the bus terminus.
Bus 4 reached after 14:45. Since it did not have a travel time of 3 hours and 25 minutes, the bus 4 could have started at 12:15 and have a travel time of 2 hours and 40 minutes or 3 hours, or, it could have started at 11:50 and have a travel time of 3 hours. The earliest that Bus 4 could reach the bus terminus is 14:50.

Bus 3 and Bus 5 could not have started at 11:00, as the bus that would have started at 11:00, must have reached the bus terminus by at most 14:25. They also could not have started at 11:25 as then they would have reached the bus terminus by 14:50. In either case, they could not be the fourth and fifth buses to reach the bus terminus.
Since, except for Bus 2, none of the others could have started at 11:00, Bus 2 must have started at 11:00. Also, apart from Bus 1 (and Bus 2), none of the others could have started at 11:25. Hence, Bus 1 must have started at 11:25.

If Bus 4 had started at 12:15 and had a travel time of 3 hours, it would have reached the bus terminus by 15:15. Bus 3 and Bus 5 must have starting times of 11:45 and 11:50 for 3 hours and 25 minutes and 2 hours and 40 minutes, in any order. The bus which had a travel time of 2 hours and 40 minutes would have reached by at most 14:30. Either of Bus 3 and Bus 5 can be this bus because they reached after Bus 4. Hence, this case is not possible.

If Bus 4 started at 12:15 and had a travel time of 160 minutes, it would have reached by 14:55. Bus 3 and Bus 5 must have started at 11:45 and 11:50 with travel times of 180 minutes and 205 minutes, in any order. The bus whose travel time was 180 minutes would have reached by at most 14:50. This case is also not possible because one of Bus 5 and Bus 3 would have reached the bus terminus before Bus 4.

If Bus 4 started at 11:50 with travel time of 3 hours, it would reach bus terminus by 14:50. Bus 3 and Bus 5 must have starting times of 11:45 and 12:15 with travel times of 2 hours and 40 minutes and 3 hours and 25 minutes, in any order. The bus which has a travel time of 2 hours and 40 minutes must have started at 12:15 for it to reach after Bus 4. This bus would have reached by 14:55. The bus whose travel time is 3 hours and 25 minutes must have started by 11:45 and would have reached by 15:10. Hence, the former must be Bus 5 and the latter must be Bus 3.

The following table presents, for each bus, the travel time, the start time and the time at which it reaches the bus terminus:

Sample Test: LR & DI- 2 - Question 19

Directions: Study the given information and answer the following question.

On a particular day, five buses - Bus 1, Bus 2, Bus 3, Bus 4 and Bus 5 - destined for the same bus terminus started from different starting points and travelled for different times: 1 hour and 40 minutes, 2 hours and 5 minutes, 2 hours and 40 minutes, 3 hours, and 3 hours and 25 minutes to reach the bus terminus. The journey of buses started at different times, i.e. 11:00, 11:25, 11:45, 11:50 and 12:15.

It is also known that
(1) no two buses reached the bus terminus at the same time
(2) Bus 1, whose travel time was 2 hours and 5 minutes, did not reach the bus terminus before 13:20
(3) among the five buses, the last bus to reach the bus terminus was Bus 3
(4) Bus 4 reached the bus terminus after 14:45, but the travel time was not 3 hours and 25 minutes
(5) among the five buses, at least three buses reached the bus terminus before Bus 5 did

Q. Which bus should a person take after 11:30 so that his travel time is minimum?

Detailed Solution for Sample Test: LR & DI- 2 - Question 19

From (1), Bus 1's travel time was 2 hours and 5 minutes and it reached the bus terminus after 13:20. It could not have the starting time of 11:00.
From (2), Bus 3 was the last bus to reach the bus terminus and from (4), Bus 5 must be the fourth bus to reach the bus terminus.
Bus 4 reached after 14:45. Since it did not have a travel time of 3 hours and 25 minutes, the bus 4 could have started at 12:15 and have a travel time of 2 hours and 40 minutes or 3 hours, or, it could have started at 11:50 and have a travel time of 3 hours. The earliest that Bus 4 could reach the bus terminus is 14:50.

Bus 3 and Bus 5 could not have started at 11:00, as the bus that would have started at 11:00, must have reached the bus terminus by at most 14:25. They also could not have started at 11:25 as then they would have reached the bus terminus by 14:50. In either case, they could not be the fourth and fifth buses to reach the bus terminus.
Since, except for Bus 2, none of the others could have started at 11:00, Bus 2 must have started at 11:00. Also, apart from Bus 1 (and Bus 2), none of the others could have started at 11:25. Hence, Bus 1 must have started at 11:25.

If Bus 4 had started at 12:15 and had a travel time of 3 hours, it would have reached the bus terminus by 15:15. Bus 3 and Bus 5 must have starting times of 11:45 and 11:50 for 3 hours and 25 minutes and 2 hours and 40 minutes, in any order. The bus which had a travel time of 2 hours and 40 minutes would have reached by at most 14:30. Either of Bus 3 and Bus 5 can be this bus because they reached after Bus 4. Hence, this case is not possible.

If Bus 4 started at 12:15 and had a travel time of 160 minutes, it would have reached by 14:55. Bus 3 and Bus 5 must have started at 11:45 and 11:50 with travel times of 180 minutes and 205 minutes, in any order. The bus whose travel time was 180 minutes would have reached by at most 14:50. This case is also not possible because one of Bus 5 and Bus 3 would have reached the bus terminus before Bus 4.

If Bus 4 started at 11:50 with travel time of 3 hours, it would reach bus terminus by 14:50. Bus 3 and Bus 5 must have starting times of 11:45 and 12:15 with travel times of 2 hours and 40 minutes and 3 hours and 25 minutes, in any order. The bus which has a travel time of 2 hours and 40 minutes must have started at 12:15 for it to reach after Bus 4. This bus would have reached by 14:55. The bus whose travel time is 3 hours and 25 minutes must have started by 11:45 and would have reached by 15:10. Hence, the former must be Bus 5 and the latter must be Bus 3.

The following table presents, for each bus, the travel time, the start time and the time at which it reaches the bus terminus:

The starting time of Bus 5 is 12:15 and its travel time is 2 hours and 40 minutes.

Sample Test: LR & DI- 2 - Question 20

Directions: Study the given information and answer the following question.

On a particular day, five buses - Bus 1, Bus 2, Bus 3, Bus 4 and Bus 5 - destined for the same bus terminus started from different starting points and travelled for different times: 1 hour and 40 minutes, 2 hours and 5 minutes, 2 hours and 40 minutes, 3 hours, and 3 hours and 25 minutes to reach the bus terminus. The journey of buses started at different times, i.e. 11:00, 11:25, 11:45, 11:50 and 12:15.

It is also known that
(1) no two buses reached the bus terminus at the same time
(2) Bus 1, whose travel time was 2 hours and 5 minutes, did not reach the bus terminus before 13:20
(3) among the five buses, the last bus to reach the bus terminus was Bus 3
(4) Bus 4 reached the bus terminus after 14:45, but the travel time was not 3 hours and 25 minutes
(5) among the five buses, at least three buses reached the bus terminus before Bus 5 did

Q. Which of the following is true?

Detailed Solution for Sample Test: LR & DI- 2 - Question 20

From (1), Bus 1's travel time was 2 hours and 5 minutes and it reached the bus terminus after 13:20. It could not have the starting time of 11:00.
From (2), Bus 3 was the last bus to reach the bus terminus and from (4), Bus 5 must be the fourth bus to reach the bus terminus.
Bus 4 reached after 14:45. Since it did not have a travel time of 3 hours and 25 minutes, the bus 4 could have started at 12:15 and have a travel time of 2 hours and 40 minutes or 3 hours, or, it could have started at 11:50 and have a travel time of 3 hours. The earliest that Bus 4 could reach the bus terminus is 14:50.

Bus 3 and Bus 5 could not have started at 11:00, as the bus that would have started at 11:00, must have reached the bus terminus by at most 14:25. They also could not have started at 11:25 as then they would have reached the bus terminus by 14:50. In either case, they could not be the fourth and fifth buses to reach the bus terminus.
Since, except for Bus 2, none of the others could have started at 11:00, Bus 2 must have started at 11:00. Also, apart from Bus 1 (and Bus 2), none of the others could have started at 11:25. Hence, Bus 1 must have started at 11:25.

If Bus 4 had started at 12:15 and had a travel time of 3 hours, it would have reached the bus terminus by 15:15. Bus 3 and Bus 5 must have starting times of 11:45 and 11:50 for 3 hours and 25 minutes and 2 hours and 40 minutes, in any order. The bus which had a travel time of 2 hours and 40 minutes would have reached by at most 14:30. Either of Bus 3 and Bus 5 can be this bus because they reached after Bus 4. Hence, this case is not possible.

If Bus 4 started at 12:15 and had a travel time of 160 minutes, it would have reached by 14:55. Bus 3 and Bus 5 must have started at 11:45 and 11:50 with travel times of 180 minutes and 205 minutes, in any order. The bus whose travel time was 180 minutes would have reached by at most 14:50. This case is also not possible because one of Bus 5 and Bus 3 would have reached the bus terminus before Bus 4.

If Bus 4 started at 11:50 with travel time of 3 hours, it would reach bus terminus by 14:50. Bus 3 and Bus 5 must have starting times of 11:45 and 12:15 with travel times of 2 hours and 40 minutes and 3 hours and 25 minutes, in any order. The bus which has a travel time of 2 hours and 40 minutes must have started at 12:15 for it to reach after Bus 4. This bus would have reached by 14:55. The bus whose travel time is 3 hours and 25 minutes must have started by 11:45 and would have reached by 15:10. Hence, the former must be Bus 5 and the latter must be Bus 3.

The following table presents, for each bus, the travel time, the start time and the time at which it reaches the bus terminus:

Bus 2 has a travel time of 1 hour and 40 minutes.

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