LR - Arrangements - Previous Year CAT/MBA Questions
You can practice all previous year CAT questions from the topic LR - Arrangements. This will help you understand the type of questions asked in CAT.
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Answer the next 5 questions based on the information given below:
There are only four neighbourhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1.
2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.
What BEST can be concluded about the total number of new cases in the city on Day 2?
A.
Exactly 7
B.
Either 7 or 8
C.
Either 6 or 7
D.
Exactly 8
Answer: Option
D
Explanation :
We can make the following table:
1. There was at least one new case in every neighborhood on Day 1.
2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
5. Kitmisto is the only place to have 3 new cases on Day 2.
From (6): The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.
Total cases during the 5 days = 12 + 12 + 5 + 14 = 43
On day 5, neighborhoods L, T and K can have maximum 3 cases each, while P can have maximum 2 cases. Hence, maximum possible total cases on day 5 = 3 + 2 + 3 + 3 = 11.
⇒ Maximum possible cases on day 4 = 10.
∴ Maximum possible cases on day 4 + day 5 = 10 + 11 = 21.
Since on day 1 there will be at least 5 cases in the city and every day total cases increase, hence on day 2 total cases must be 6 or more.
Case 1: Total cases on Day 2 = 6
⇒ Total cases on Day 3 = 7
⇒ Total cases on Day 1 = 5
∴ Total cases on day 4 + day 5 = 43 – (5 + 6 + 7) = 25
This is not possible as maximum cases possible on day 4 + day 5 is 21.
Case 2: Total cases on Day 2 = 7
⇒ Total cases on Day 3 = 8
⇒ Total cases on Day 1 = 5 or 6
∴ Total cases on day 4 + day 5 = 43 – (5/6 + 7 + 8) = 23 or 22
This is not possible as maximum cases possible on day 4 + day 5 is 21.
Case 3: Total cases on Day 2 = 8
⇒ Total cases on Day 3 = 9
Since total cases on day 4 cannot be more than 10, hence total cases on day 4 = 10.
Since total cases on day 5 cannot be more than 11, hence total cases on day 5 = 11.
⇒ Total cases on Day 1 = 43 – (8 + 9 + 10 + 11) = 5
Only possibility for total cases of 11 on day 5 is: L – 3, T – 3, P – 2, K – 3.
Only possibility for total cases of 11 on day 5 is: L – 3, T – 3, P – 1, K – 3.
[Note: There is only 1 day when Pesmisto had 2 cases. On remaining days P will have less than 2 cases.]
Total cases in Kitmisto is 14. This is possible when there are 3 cases on 4 days each and 2 cases on the remaining 5th day.
Levmisto and Tyhrmisto have total 12 cases.
Total cases on day 2 + day 3 for these two neighborhoods = 12 – (1 + 3 + 3) = 5
5 cases in 2 days are possible when there are 2 and 3 cases in these 2 days.
On day 2 only Kitmisto had 3 cases, hence we can fill the table accordingly.
Now the remaining 2 slots can be filled.
∴ The total number of new cases in the city on Day 2 = 8
Which of the two statements below is/are necessarily false?
Statement A: There were 2 new cases in Tyhrmisto on Day 3.
Statement B: There were no new cases in Pesmisto on Day 2.
A.
Statement B only
B.
Statement A only
C.
Both Statement A and Statement B
D.
Neither Statement A nor Statement B
Answer: Option
C
Explanation :
Consider the solution to first question of this set.
Answer the following question based on the information given below.
A supermarket has to place 12 items (coded A to L) in shelves numbered 1 to 16. Five of these items are types of biscuits, three are types of candies and the rest are types of savouries. Only one item can be kept in a shelf. Items are to be placed such that all items of same type are clustered together with no empty shelf between items of the same type and at least one empty shelf between two different types of items. At most two empty shelves can have consecutive numbers.
The following additional facts are known.
A and B are to be placed in consecutively numbered shelves in increasing order.
I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
K is to be placed in shelf number 16.
L and J are items of the same type, while H is an item of a different type.
C is a candy and is to be placed in a shelf preceded by two empty shelves.
L is to be placed in a shelf preceded by exactly one empty shelf.
In how many different ways can the items be arranged on the shelves?
A.
1
B.
8
C.
2
D.
4
Answer: Option
B
Explanation :
There are 5 types of biscuits, 3 types of candies and 4 types of savouries. Among 16 shelves, there are 4 empty shelves.
It is given that all items of same type are clustered together with no empty shelf between items of the same type.
From (3) and (4), it can be concluded that D, E, F and K are savouries.
From (2) and (5), L, I and J are of one type and H is the other type. Therefore from (6), as C is a candy, L, I J must be types of biscuits and H is a type of candy. Now using (1), we can conclude that A and B are of one type but not candies as there are only 3 types of candies.
Therefore,
Biscuits: A, B, I, J, L Candies: C, H, G Savouries: D, E, F, K
From (3), (4), (6) and (7), there shelf number 12 must be an empty shelf. Also, D, E, F and K are placed in shelves numbered 13, 14, 15 and 16 respectively.
Now from (1), (2) and (7), the sequence (from left to right) in which biscuits are kept is:
(Empty shelf), L, A, B, (I/J), (J/I).
From (6), the candies must be in the following order: (Empty shelf), (Empty shelf), C, (H/G), (G/H)
Thus, we have
In each case, J and I can be arranged in 2 ways and G and H can be arranged among them in 2 ways. Thus, 2 × 2 = 4 ways.
Total number of ways the items can be arranged on the shelves = 4 + 4 = 8
Answer the following question based on the information given below.
In a square layout of size 5m × 5m, 25 equal-sized square platforms of different heights are built.
The heights (in metres) of individual platforms are as shown below:
6 1 2 4 3
9 5 3 2 8
7 8 4 6 5
3 9 5 1 2
1 7 6 3 9
Individuals (all of same height) are seated on these platforms. We say an individual A can reach an individual B if all the three following conditions are met:
(i) A and B are in the same row or column
(ii) A is at a lower height than B
(iii) If there is / are any individual(s) between A and B, such individual(s) must be at a height lower than that of A.
Thus in the table given above, consider the individual seated at height 8 on 3rd row and 2nd column. He can be reached by four individuals. He can be reached by the individual on his left at height 7, by the two individuals on his right at heights of 4 and 6 and by the individual above at height 5.
Rows in the layout are numbered from top to bottom and columns are numbered from left to right.
How many individuals in this layout can be reached by just one individual?
A.
3
B.
5
C.
7
D.
8
Answer: Option
C
Explanation :
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 3 people i.e., 2 and 3 in the same row and 5 in the column below
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
Which of the following is true for any individual at a platform of height 1 m in this layout?
A.
They can be reached by all the individuals in their own row and column.
B.
They can be reached by at least 4 individuals.
C.
They can be reached by at least one individual.
D.
They cannot be reached by anyone.
Answer: Option
D
Explanation :
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 3 people i.e., 2 and 3 in the same row and 5 in the column below
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
A person having a height of 1 m cannot be reached by anyone.
We can find two individuals who cannot be reached anyone in
A.
the last row.
B.
the fourth row.
C.
the fourth column.
D.
the middle column.
Answer: Option
C
Explanation :
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 3 people i.e., 2 and 3 in the same row and 5 in the column below
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
In the last row there is only one individual with height 3 m who cannot be reached.
So option [1] is ruled out. Similarly, in the 4th row there is only one person with height 1 m who cannot be reached by anyone. So option [2] is also incorrect.
In the 4th column, there are 2 people, one with 2 m height and one with 1 m height who cannot be reached by anyone.
So option [3] is correct. In the middle column there is no one who cannot be reached by anyone. So option [4] is also incorrect.
Which of the following statements is true about this layout?
A.
Each row has an individual who can be reached by 5 or more individuals.
B.
Each row has an individual who cannot be reached by anyone.
C.
Each row has at least two individuals who can be reached by an equal number of individuals.
D.
All individuals at the height of 9 m can be reached by at least 5 individuals.
Answer: Option
C
Explanation :
Let us check how many people in the grid can reach individuals. Let us do this by moving row wise
Row 1
6 – Can be reached by 1, 2 and 4 in the same row i.e., a total of 3 people
1 – Cannot be reached by anyone
2 – Can be reached by only 1 in the same row on the left
4 – Can be reached by 3 people i.e., 2 and 3 in the same row and 2 in the column below
3 – Cannot be reached by anyone
In a similar way we proceed for the other rows
Row 2
9 – Can be reached by 6, 5, 8 and 7 i.e., of total of 4 people
5 – Can be reached by 1 and 3 i.e., a total of 2 people
3 – Can be reach by 2 and 2 i.e., a total of 2 people
2 – Cannot be reached by anyone
8 – Can be reached by 3 people i.e., 2 and 3 in the same row and 5 in the column below
Row 3
7 – Can be reached by only 3 in the row below
8 – Can be reached 5, 7, 4, and 6 i.e., total of 4 people
4 – Can be reached by only 3 in the above row
6 – Can be reached only by 4 and 5 in the same row and 4, 2, 1 and 3 in the same column i.e., a total of 6 people
5 – Can only be reached by 2 in the same row below
Row 4
3 – Can only be reached by 1 in the row below
9 – Can be reached by 3 and 5 in the same row and 7 & 8 in the row below and above i.e., a total of 4 people
5 – Can be reached by 1 & 2 in the same row and 4 in the row above
1 – Cannot be reached by anyone
2 – Can be reached only by 1 in the same row
Row 5
1 – Cannot be reached by anyone
7 – Can be reached by 1 and 6 in the same row i.e., a total of 2 people
6 – Can be reached by 5 in the row above and 3 in the same row. i.e., a total of 2 people
3 – Can be reached only by 1 in the row above
9 – Can be reached by 3, 6 and 7 in the same row and 2, 5 and 8 in the rows above i.e., a total of 6 people
Looking at the table we can see the person having height 2 m in the 1st row, people having height 7 m, 4 m and 5 m in the 3rd row, people having height 3 m and 2 m in the 4th row and the person having height 3 m in the 5th row i.e., a total of 7 people can be reached by only one person.
In row 1, the maximum number of people who can reach a particular individual is 3, which is for the person with height 6 m and for the person with height 4 m. So statement (1) is false.
Row 3 does not have any individual who cannot be reached by anyone. Hence statement [2] is false.
In row 1, person with height 6 m and person with height 4 m can each be reached by 3 people. In row 2 people with height 5 m and 3 m can each be reached by 2 people.
In Row 3, people with height 7 m and 3 m can be reached by only 1 person. In Row 4, people with height 3 m and 2 m can be reached by only 1 person. In Row 5, people with height 7 m and 6 m can be reached by 2 people each. So in each row we have 2 individuals who can be reached by an equal number of individuals.
So statement 3 is true. Individual with height 9 m in row 2 can be reached by 4 people.
Answer the following question based on the information given below.
A tea taster was assigned to rate teas from six different locations – Munnar, Wayanand, Ooty, Darjeeling, Assam and Himachal. These teas were placed in six cups, numbered 1 to 6, not necessarily in the same order. The tea taster was asked to rate these teas on the strength of their flavour on a scale of 1 to 10. He gave a unique integer rating to each tea. Some other information is given below:
Cup 6 contained tea from Himachal.
Tea from Ooty got the highest rating, but it was not in Cup 3.
The rating of tea in Cup 3 was double the rating of the tea in Cup 5.
Only two cups got ratings in even numbers.
Cup 2 got the minimum rating and this rating was an even number.
Tea in Cup 3 got a higher rating than that in Cup 1.
The rating of tea from Wayanad was more than the rating of tea from Munnar, but less than that from Assam.
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
As can be seen from the table, 2nd highest rating given was 7.
What was the number of the cup that contained tea from Ooty?
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Answer:4
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
The number of the cup that contained the tea from Ooty is 4.
If the tea from Munnar did not get the minimum rating, what was the rating of the tea from Wayanad?
A.
3
B.
5
C.
1
D.
6
Answer: Option
B
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
If tea from Munnar did not get the least rating of 2, it would imply that the tea from Munnar got the 2nd least rating of 3, since the tea from Assam and Wayanad have a better rating than the tea from Munnar. As we already know that tea from Ooty and Himachal are the ones with the top 2 ratings, it would mean that tea from Assam is rated 3rd highest with a rating of 6 and tea from Wayanad is rated 4th highest with a rating of 5.
If cups containing teas from Wayanad and Ooty had consecutive numbers, which of the following statements may be true?
A.
Cup 5 contains tea from Assam
B.
Cup 1 contains tea from Darjeeling
C.
Tea from Wayanad has got a rating of 6
D.
Tea from Darjeeling got the minimum rating
Answer: Option
B
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
Given that cups containing tea from Wayanad and Ooty have consecutive numbers, tea from Wayanad can either be in cup 3 or cup 5. However, as tea in cup 3 has the 3rd highest rating it cannot be from Wayanad. This is because the tea from Assam is better than the tea from Wayanad. So then tea from Wayanad will be in cup no 5 and will have a rating of 3 and tea in cup 2 which has the lowest rating of 2 is from Munnar. This means that in cup 1 and cup 3 we can have tea from Assam and Darjeeling (not essentially in that order). If we now examine the statements in each of the 4 options we will see that statements in options (1), (3) and (4) are not true. However statement in option (2) may be true.
Answer the following question based on the information given below.
Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are goin to Delhi from Kolkatta by a flight operated by Cheap Air. In the flight, sitting is arranged in 30 rows, numbered 1 to 30, each consisting of 6 seats, marked by letters A to F from left to right, respectively. Seats A to C are to the left of the aisle (the passage running from the front of the aircraft to the back), and seats D to F are to the right of the aisle. Seats A and F are by the windows and referred to as Window seats, C and D are by the aisle and are referred to as Aisle seats while B and E are referred to as Middle seats. Seats marked by consecutive letters are called consecutive seats (or seats next to each other).A seat number is a combination of the row number, followed by the letter indicating the position in the row, e.g, 1A is the left window seat in the first row, while 12E is the right middle seat in the 12th row.
Cheap Air charges Rs. 1000 extra for any seats in Rows 1, 12 and 13 as those have extra legroom. For Rows 2-10, it charges Rs. 300 extra for Window seats and Rs. 500 extra for Aisle seats. For Rows 11 and 14 to 20, it charges Rs. 200 extra for Window seats and Rs. 400 extra for Aisle seats. All other seats are available at no extra charge.
The following are known:
The eight friends were seated in six different ways.
They occupied 3 Window seats, 4 Aisle seats and 1 Middle seat.
Seven of them had to pay extra amounts, totaling to Rs. 4600, for their choices of seat. One of them did not pay any additional amount of his/her choice of seat.
Jayanta, Ajit and Byomkesh were sitting in seats marked by the same letter, in consecutive rows in increasing order of row numbers; but all of them paid different amounts for their choices of seat. One of these amounts may be zero.
Gargi was sitting next to Kikira, and Manik was sitting next to Jayanta.
Prodosh and Tapesh were sitting in seats marked by the same letter, in consecutive rows in increasing order of row numbers; but they paid different amounts for their choices of seat. One of these amounts may be zero.
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Using this data let us answer the question.
As can be seen from the table, Manik was sitting in row 10.
How much extra did Jayanta pay for his choice of seat?
A.
Rs. 300
B.
Rs. 400
C.
Rs. 500
D.
Rs. 1000
Answer: Option
C
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
How much extra did Gargi pay for her choice of seat?
A.
0
B.
Rs. 300
C.
Rs. 400
D.
Rs. 1000
Answer: Option
D
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Who among the following did not pay any extra amount for his/her choice of seat?
A.
Kikira
B.
Manik
C.
Gargi
D.
Tapesh
Answer: Option
D
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Using this data let us answer the question.
As Tapesh was sitting in row 21, he did not pay any extra amount for his seat.
Study the following information carefully and answer the questions.
A word arrangement machine, when given a particular input, rearranges it using a particular rule. The following is the illustration and the steps of the arrangement
Answer the following question based on the information given below.
i. There are three houses on each side of the road.
ii. These six houses are labeled as P, Q, R, S, T and U.
iii. The houses are of different colours, namely, Red, Blue, Green, Orange, Yellow and White.
iv. The houses are of different heights.
v. T, the tallest house, is exactly opposite to the Red coloured house.
vi. The shortest house is exactly opposite to the Green coloured house.
vii. U, the Orange coloured house, is located between P and S.
viii. R, the Yellow coloured house, is exactly opposite to P.
ix. Q, the Green coloured house, is exactly opposite to U.
x. P, the White coloured house, is taller than R, but shorter than S and Q.
We have to arrange six houses on opposite sides of a road.
From condition (vii), we can say that P, U and S lie on one side of the road as follows:
From condition (viii) and (ix) we can further complete the arrangement as follows. We have also used the color of the house P from statement (x).
The only left house is definitely T. From conditions (v) and we can complete the arrangement as follows.
From condition (vi) it can be deduced that U is the shortest house. Also from the last condition it can be deduced that P is the fourth tallest, R is the fifth tallest and S and Q are second and third tallest not in that order.
Filling all this data we can see the arrangement as follows:
The color of the tallest house (T) is Blue.
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