Sample Test: LR & DI- 2 - CAT MCQ

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 1^st May to 7^th 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 23^rd May, he must have eaten only eggs on 2^nd May as well.

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

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

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

From (5), he must have eaten green vegetables and fruits on 4^th and 5^th or 5^th and 6^th or 6^th and 7^th or 7^th and 1^st. He could not have eaten these food items on 4^th and 5^th 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 1^st and 7^th for the same reason. He could not have eaten green vegetables and fruits on 6^th and 7^th because 7^th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5^th and 6^th.

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

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 1^st May to 7^th 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 23^rd May, he must have eaten only eggs on 2^nd May as well.

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

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

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

From (5), he must have eaten green vegetables and fruits on 4^th and 5^th or 5^th and 6^th or 6^th and 7^th or 7^th and 1^st. He could not have eaten these food items on 4^th and 5^th 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 1^st and 7^th for the same reason. He could not have eaten green vegetables and fruits on 6^th and 7^th because 7^th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5^th and 6^th.

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

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 1^st May to 7^th 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 23^rd May, he must have eaten only eggs on 2^nd May as well.

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

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

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

From (5), he must have eaten green vegetables and fruits on 4^th and 5^th or 5^th and 6^th or 6^th and 7^th or 7^th and 1^st. He could not have eaten these food items on 4^th and 5^th 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 1^st and 7^th for the same reason. He could not have eaten green vegetables and fruits on 6^th and 7^th because 7^th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5^th and 6^th.

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

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 1^st May to 7^th 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 23^rd May, he must have eaten only eggs on 2^nd May as well.

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

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

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

From (5), he must have eaten green vegetables and fruits on 4^th and 5^th or 5^th and 6^th or 6^th and 7^th or 7^th and 1^st. He could not have eaten these food items on 4^th and 5^th 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 1^st and 7^th for the same reason. He could not have eaten green vegetables and fruits on 6^th and 7^th because 7^th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5^th and 6^th.

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

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 1^st May to 7^th 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 23^rd May, he must have eaten only eggs on 2^nd May as well.

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

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

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

From (5), he must have eaten green vegetables and fruits on 4^th and 5^th or 5^th and 6^th or 6^th and 7^th or 7^th and 1^st. He could not have eaten these food items on 4^th and 5^th 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 1^st and 7^th for the same reason. He could not have eaten green vegetables and fruits on 6^th and 7^th because 7^th was a Thursday. Hence, he could have eaten green vegetables and fruits on 5^th and 6^th.

He could have eaten fish and milk products on 4^th and 7^th. He could not have eaten fish and milk products on 1^st and 4^th because from (1), he did not eat fish on 1^st and he did not eat fish on 4^th, which was a Monday. Hence, on 4^th, he must have eaten milk products and on 7^th, he must have eaten fish. On 1^st, 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.

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

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

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.

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

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.

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.

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:

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:

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:

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.

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.