CAT LR Arrangement Questions & Solutions

A sample of real CAT LR Arrangement past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 121 LR Arrangement questions in all — sign up free to practise them timed.

Passage
A round table has seven chairs around it. The chairs are numbered 1 through 7 in a clockwise direction. Four friends, Aslam, Bashir, Chhavi, and Davies, sit on four of the chairs. In the starting position, Aslam and Chhavi are sitting next to each other, while for Bashir as well as Davies, there are empty chairs on either side of the chairs that are sitting on.

The friends take turns moving either clockwise or counterclockwise from their chair. The friend who has to move in a turn occupies the first empty chair in whichever direction (s)he chooses to move. Aslam moves first (Turn 1), followed by Bashir, Chhavi, and Davies (Turns 2, 3, and 4, respectively). Then Aslam moves again followed by Bashir, and Chhavi (Turns 5, 6, and 7, respectively).

The following information is known
1. The four friends occupy adjacent chairs only at the end of Turn 2 and Turn 6.
2. Davies occupies Chair 2 after Turn 1 and Chair 4 after Turn 5, and Chhavi occupies Chair 7 after Turn 2.
Q1.CAT 2025
What is the number of the chair initially occupied by Bashir?
Answer: 4
Show solution
Based on the information about the initial positions of Aslam and Chhavi sitting next to each other, while Bashir and Davies have empty chairs on either side of their seats, the possible combinations are,

This has 4 possible combinations.

We are also given that Davies occupies Chair 2 after Turn 1 and Chhavi occupies Chair 7 after Turn 2. We know that the positions of Davies and Chhavi won't be changing after turn 2, as we know that in those turns, only the positions of Aslam and Bashir are being changed in the first 2 turns. Therefore, we can determine that Davies' initial position is chair 2, and Chhavi's initial position is chair 7.

Out of the 4 possible combinations, Davies on chair 2 and Chhavi on chair 7 are only possible in one case, which is,

So, Bashir's initial position is chair 4.

Hence, the correct answer is 4.
Passage
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections.Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.

Q2.CAT 2024
Which of the following two statements is/are DEFINITELY true?
Statement A: Each of R-A, R-B, and R-C has two ATMs.
Statement B: Each of V1, V2, and V3 has two ATMs.
- Only Statement A
- Neither Statement A nor Statement B
- Only Statement B
- Both Statement A and Statement B
Show solution
This is the figure that has been given to us,

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3 of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to 12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).

Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20, and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to either be at (RA, V1) or (RA, V3)

Case 1: 15L ATM is on the intersection (RA, V3)

We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.

Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.

Using the two cases, we can answer the given questions.

We see that in both cases, RA RB and RC have two ATMs
Only in the first case V1 V2 and V3 have two ATM's, in the second case V3 has 3 ATMs and V1 has 1.

Hence only Statement A is correct.
Passage
Twenty five coloured beads are to be arranged in a grid comprising of five rows and five columns. Each cell in the grid must contain exactly one bead. Each bead is coloured either Red, Blue or Green. While arranging the beads along any of the five rows or along any of the five columns, the rules given below are to be followed:

1. Two adjacent beads along the same row or column are always of different colours.
2. There is at least one Green bead between any two Blue beads along the same row or column.
3. There is at least one Blue and at least one Green bead between any two Red beads along the same row or column.

Every unique, complete arrangement of twenty five beads is called a configuration.
Q3.CAT 2020
What is the minimum number of Blue beads in any configuration?
Answer: 6
Show solution
To solve this question we can use the answer of the previous question, since maximum 9 red beads are possible, filling the remaining space with green and blue beads, in such a way that number of blue beads is minimised

Hence number of blue beads is 6
Passage
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.
1. A and B are to be placed in consecutively numbered shelves in increasing order.
2. I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
3. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
4. K is to be placed in shelf number 16.
5. L and J are items of the same type, while H is an item of a different type.
6. C is a candy and is to be placed in a shelf preceded by two empty shelves.
7. L is to be placed in a shelf preceded by exactly one empty shelf.
Q4.CAT 2019
In how many different ways can the items be arranged on the shelves?
- 8
- 4
- 2
- 1
Show solution
The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

The number of arrangements for the first case = 2*2=4

The number of arrangements for the second case = 2*2=4

The total number of arrangements = 4+4=8
Passage
Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:

1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politicians are in the administration committee, and 20% are in the teaching committee.
Q5.CAT 2018
What is the number of educationalists in the research committee?
Answer: 3
Show solution
Let us draw a table according to the information given.

It is given that the numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. Let '4x' be the number of bureaucrats in Administration committee.

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees. Let us assume that 'y' is the number of educationalists in the research committee and 'd' be the difference in the number of educationalists in Research and teaching committees.

60% of the politicians are in the administration committee, and 20% are in the teaching committee. Let '5z' be the number of total number of politicians.

We can say that

$\Rightarrow$ 10x+3y+5z = 24

We can see that each of x, y and z has to a natural number integer. If x > 1, then both y and z can't take any natural number.

Hence, we can say that x = 1.

At x = 1, 3y+5z = 14. If y = 1 or 2, Z is not an integer.

At x = 1 and y = 3, z = 1 which is the only possible solution.

We can see that 'd' can assume two possible values. d = 1 or 2.

From the table, we can see that the number of educationalists in the research committee = 3.
Passage
Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata 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:
1. The eight friends were seated in six different rows.
2. They occupied 3 Window seats, 4 Aisle seats and 1 Middle seat.
3. 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 for his/her choice of seat.
4. 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.
5. Gargi was sitting next to Kikira, and Manik was sitting next to Jayanta.
6. 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.
Q6.CAT 2017
In which row was Manik sitting?
- 10
- 11
- 12
- 13
Show solution
We are given that 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.
Let us see how the friends are supposed to pay for the seats they choose:-
In row 1-1000
In row 2-10 - 300 for window and 500 for aisle
In row 11 - 200 for window and 400 for aisle
In row 12,13 - 1000
In row 14-20 - 200 for window and 400 for aisle
In row 21-30 - 0
Thus, As we can see 10, 11 and 12 are the only consecutive seats in which the amounts is different.
Thus, Jayanth, Ajit and Byomkesh sat in row 10, row 11 and row 12.
Manik sat beside Jayantha and thus Manik is also sitting in row 10.
Now we are given that 7 of the 8 friends paid a total of 4600 Rs.
Let's start with the cases:-
It is obvious that 5 friends cannot pay 1000 Rs for their seat because the amount will exceed 4600
Case 1:- 4 friends pay 1000 Rs each. Thus, the remaining friends will pay 600 Rs.
This is possible only when each of them pay 200 Rs.
So the case is- 1000*4 , 200*3
Case 2 :- 3 friends pay 1000 Rs each. Thus, the remaining friends will pay 1600 Rs.
There are 2 cases where this is possible:-
1000*3, 500*2, 400, 200
1000*3, 400*4
Case 3:- 2 friends pay 1000 Rs each. Thus, the remaining 5 friends will pay 2600 Rs.
This is not possible as each friend can pay a maximum of 500 Rs.
Thus, the possible cases are
1000*4 , 200*3
1000*3, 500*2, 400, 200
1000*3, 400*4
As there is no case in which a friend has to pay 300 Rs thus, Jayantha must be sitting in row 10 aisle seat.
Thus, Jayantha paid 500 Rs.
Thus, the case is:-
1000*3, 500*2, 400, 200
Thus, Manik must have also paid 500 sitting in row 10 aisle seat
Ajit must be sitting in row 11 aisle seat paying 400 Rs.
Byomyesh must be sitting row 12 aisle seat paying 1000 Rs.
Thus, among Gargi, Kikira, Pradosh and Tapesh 2 must have paid 1000, 1 must have paid 200 and the remaining person must have paid nothing.
Now we know Gargi and Kikira are sitting adjacent to each other and thus, either both or none of them must have paid 1000 Rs.
Among Pradosh and Tapesh a maximum of 1 person could have paid 1000 Rs.
Thus, the only possible case here is :-
Gargi and Kikira paid 1000 each.
Pradosh is sitting ahead of Tapesh and one of them paid 200 Rs.
Since, both of them were sitting in seats marked by the same letter, in consecutive rows thus, the only possibility is Pradosh sitting in row 20 window seat and paying 200 and Tapesh sitting in row 21 paying nothing.
Thus, the amount paid by each friend is as shown below:

Manik is sitting in row 10.
Passage
A tea taster was assigned to rate teas from six different locations — Munnar, Wayanad, 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:
a: Cup 6 contained tea from Himachal.
2. Tea from Ooty got the highest rating, but it was not in Cup 3.
3. The rating of tea in Cup 3 was double the rating of the tea in Cup 5.
4. Only two cups got ratings in even numbers.
5. Cup 2 got the minimum rating and this rating was an even number.
6. Tea in Cup 3 got a higher rating than that in Cup 1.
7. The rating of tea from Wayanad was more than the rating of tea from Munnar, but less than that from Assam.
Q7.CAT 2017
If the tea from Munnar did not get the minimum rating, what was the rating of the tea from Wayanad?
- 3
- 5
- 1
- 6
Show solution
Now we are given that the lowest rating is an even number and only 2 cups got an even number rating.
Let's take cases:-
1. The lowest rating is 4
If the lowest rating in 4 then the other ratings will be in the range 5-10.
From this we need 4 odd and 1 even numbers.
This is not possible as there are only 3 odd numbers from 5-10.
Thus, the lowest rating is not 4.
2. The lowest rating is 2.
If the lowest rating in 4 then the other ratings will be in the range 5-10.
From this we need 4 odd and 1 even numbers.
This is possible when the odd ratings are 3,5,7 and 9.
We are given that the highest rating is not even. Thus, 10 rating is not possible.
We are also given that the rating of tea in Cup 3 was double the rating of the tea in Cup 5.
Thus, the rating of the tea in cup 3 is an even number.
Thus, the rating of the tea in cup 5 must be an odd number.
Only 1 such pair is possible of 3 and 6.
Thus, the tea in cup 2 got the rating of 2.
The tea in cup 3 got a rating of 6 and the tea in cup 5 got a rating of 3.
We are given that:-
Tea in Cup 3 got a higher rating than that in Cup 1.
Thus, the tea in cup 1 got a rating of 5.
Cup 6 contained tea from Himachal and the Tea from Ooty got the highest rating.
Thus, cup 6 got a rating of 7 and cup 4 got a rating of 9.
The table is as shown below:-

If the tea from Munnar did not get the minimum rating then it must have got the 2^nd lowest rating as we know, Assam>Wyanand>Munnar.
Thus, Wyanand must have got a rating of 5.
Passage
DIRECTIONS for the following two questions: Answer the questions on the basis of the information given below.

The Head of a newly formed government desires to appoint five of the six elected members A, B, C, D, E and F to portfolios of Home, Power, Defence, Telecom and Finance. F does not want any portfolio if D gets one of the five. C wants either Home or Finance or no portfolio. B says that if D gets either Power or Telecom then she must get the other one. E insists on a portfolio if A gets one.
Q8.CAT 2003
Which is a valid assignment?
- A-Home, B-Power, C-Defence, D-Telecom, E-Finance.
- C-Home, D-Power, A-Defence, B-Telecom, E-Finance.
- A-Home, B-Power, E-Defence, D-Telecom, F-Finance.
- B-Home, F-Power, E-Defence, C-Telecom, A-Finance.
Show solution
Since C wants either home or finance or none so options A and D are eliminated.
Since F does not want any portfolio if D gets one, Option C is eliminated.
Passage
DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.

Five friends meet every morning at Sree Sagar restaurant for an idli-vada breakfast. Each consumes a different number of idlis and vadas. The number of idlis consumed are 1, 4, 5, 6, and 8, while the number of vadas consumed are 0, 1, 2, 4, and 6. Below are some more facts about who eats what and how much.

i. The number of vadas eaten by Ignesh is three times the number of vadas consumed by the person who eats four idlis.

ii. Three persons, including the one who eats four vadas eat without chutney.

iii. Sandeep does not take any chutney.

iv. The one who eats one idli a day does not eat any vadas or chutney. Further, he is not Mukesh.

v. Daljit eats idli with chutney and also eats vada.

vi. Mukesh, who does not take chutney, eats half as many vadas as the person who eats twice as many idlis as he does.

vii. Bimal eats two more idlis than Ignesh, but Ignesh eats two more vadas than Bimal.
Q9.CAT 2003
Which of the following statements is true?
- Mukesh eats 8 idlis and 4 vadas but no chutney.
- The person who eats 5 idlis and 1 vada does not take chutney.
- The person who eats equal number of vadas and idlis also takes chutney.
- The person who eats 4 idlis and 2 vadas also takes chutney.
Show solution
Considering (i), Ignesh has to eat 6 vadas, since 6 is the only multiple of 3.

Also, using the same information, we can say that a person consumes 2 vadas and 4 idlis.

Using (vii), Bimal eats 2 more idlis than Ignesh.

Possibilities:

Bimal - 8, Ignesh - 6

Bimal - 6, Ignesh - 4

But Ignesh cannot have 4 idlis because the person who eats 4 idlis eats 2 vadas.

Hence we take Bimal - 8 and Ignesh - 6.

Also, we get that Bimal eats 6 - 2 = 4 vadas.

So far, we get the following information.

Using (vi), there is a person who eats twice as many idlis as Mukesh. The only pair satisfying is 8, 4.

So, Mukesh eats 4 idlis. Plus the person who eats 4 idlis eats 2 vadas. Hence, we get

Daljit also eats Vada as per info (v), so we get the following

(iv) gives us the information that the one who eats 1idli does not have vada.

Considering the persons who had chutney and those who didn't, 3 persons do not have chutney. Bimal is one of them(the one eating 4 vadas).

Mukesh is the second one to not take chutney(last hint). Also, Sandip does not take chutney. Hence, we get this information as well.

Hence, Ignesh eating 6 vadas and 6 idlis eat Chutney.
Passage
Directions for the following four questions: Answer the questions based on the following information.

A and B are two sets (e.g. A = Mothers, B = Women).

The elements that could belong to both the sets (e.g. women who are mothers) is given by the set $C = A \cap B$ .

The elements which could belong to either A or B, or both, is indicated by the set $D = A \cup B$ .

A set that does not contain any elements is known as a null set represented by $\phi$ (e.g. if none of the women in the set B is a mother, then $C = A \cap B$ is a null set, or C = $\phi$ ).

Let ‘V’ signify the set of all vertebrates, ‘M’ the set of all mammals, ‘D’ dogs, ‘F’ fish, ‘A’ alsatian and ‘P’, a dog named Pluto.
Q10.CAT 2001
Given that $X = M \cap D$ is such that $X = D$ . Which of the following is true?
- All dogs are mammals
- Some dogs are mammals
- $X = \phi$
- All mammals are dogs
Show solution
It is given that $X = M \cap D$ is such that $X = D$ , which means D is a subset of M . Which means all dogs are mammals. Hence , option A.
Passage
Directions for the next 2 questions: There are three bottles of water, A, B, C, whose capacities are 5 litres, 3 litres, and 2 litres respectively. For transferring water from one bottle to another and to drain out the bottles, there exists a piping system. The flow through these pipes is computer controlled. The computer that controls the flow through these pipes can be fed with three types of instructions, as explained below:

Initially, A is full with water, and B and C are empty.
Q11.CAT 2000
Consider the same sequence of three instructions ‘and the same initial state mentioned above. Three more instructions are added at the end of the above sequence to have A contain 4 litres of water. In this total sequence of six instructions, the fourth one is DRAIN (A). This is the only DRAIN instruction in the entire sequence. At the end of the execution of the above sequence, how much water (in litres) is contained in C?
- One
- Two
- Zero
- None of these
Show solution
We know that after 3 operations A contain 1 ltr and 4 th operation is Drain(a) , so 1 ltr is drained and only 4 ltrs remain in all there cylinders. Also after 6 operations A should have all the 4 ltrs in it. hence the other 2 vessels must contain 0 ltrs because there is only one drain operation in the entire sequence.
Passage
A, B, C and D are to be seated in a row. But C and D cannot be together. Also B cannot be at the third place.
Q12.CAT 1998
Which of the following must be false?
- A is at the first place
- A is at the second place
- A is at the third place
- A is at the fourth place
Show solution
Among all positions for A, if it has first position then either B will be at the 3rd position or C and D will be together which is not possible Hence answer will be A) .

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CAT LR Arrangement Questions & Solutions

Passage

A round table has seven chairs around it. The chairs are numbered 1 through 7 in a clockwise direction. Four friends, Aslam, Bashir, Chhavi, and Davies, sit on four of the chairs. In the starting position, Aslam and Chhavi are sitting next to each other, while for Bashir as well as Davies, there are empty chairs on either side of the chairs that are sitting on.

The friends take turns moving either clockwise or counterclockwise from their chair. The friend who has to move in a turn occupies the first empty chair in whichever direction (s)he chooses to move. Aslam moves first (Turn 1), followed by Bashir, Chhavi, and Davies (Turns 2, 3, and 4, respectively). Then Aslam moves again followed by Bashir, and Chhavi (Turns 5, 6, and 7, respectively).

The following information is known
1. The four friends occupy adjacent chairs only at the end of Turn 2 and Turn 6.
2. Davies occupies Chair 2 after Turn 1 and Chair 4 after Turn 5, and Chhavi occupies Chair 7 after Turn 2.

Q1.CAT 2025

What is the number of the chair initially occupied by Bashir?

Answer: 4

Show solution

Based on the information about the initial positions of Aslam and Chhavi sitting next to each other, while Bashir and Davies have empty chairs on either side of their seats, the possible combinations are,

This has 4 possible combinations.

We are also given that Davies occupies Chair 2 after Turn 1 and Chhavi occupies Chair 7 after Turn 2. We know that the positions of Davies and Chhavi won't be changing after turn 2, as we know that in those turns, only the positions of Aslam and Bashir are being changed in the first 2 turns. Therefore, we can determine that Davies' initial position is chair 2, and Chhavi's initial position is chair 7.

Out of the 4 possible combinations, Davies on chair 2 and Chhavi on chair 7 are only possible in one case, which is,

So, Bashir's initial position is chair 4.

Hence, the correct answer is 4.

Passage

The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections.Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.

Q2.CAT 2024

Which of the following two statements is/are DEFINITELY true?
Statement A: Each of R-A, R-B, and R-C has two ATMs.
Statement B: Each of V1, V2, and V3 has two ATMs.

Only Statement A
Neither Statement A nor Statement B
Only Statement B
Both Statement A and Statement B

Show solution

This is the figure that has been given to us,

We are given the information that, out of the 9 intersections in the figure, 6 of them have ATMs. That means, 3 of these intersections are empty.
We are also told that, the ATMs with the highest and lowest capacity are on the same road, highest capacity being 15L and lowest being 7L.
This information not only gives us clues about the location of these two ATMs but also, now we know the upper and lower bounds for cash in the six ATMs with distinct cash.
Next piece of information that is given is that, road distance between the ATM with the second highest cashrequirement and the ATM located at the intersection of R-C and V3 is12 km. Since we can only traverse on the roads, from (RC, V3) we have to either traverse the 5km road or the 7km road. The only way it can add up to 12 is 5+7. That means, ATM with the second highest capacity is at (RB, V2).

Now, let us start arranging the ATM's.
We are told that 15 and 7 are on the same road. Since we are given the total capacities on the roads, we need to identify the roads with capacity higher or equal to 22.
There are only two possible choice, either RA or V3.
Looking at V3, we see that 15L ATM cannot come at (RB, V3) or (RC, V3) since the RB and RC capacity is 20, and the minimum ATM limit is 7L, if a 15L ATM is on a road with total capacity 20L, this is a situation that is not possible since there cannot be an ATM with 5L capacity.
The same is the case with the intersection (RA, V2). So, we can narrow down the fact that the 15L ATM has to either be at (RA, V1) or (RA, V3)

Case 1: 15L ATM is on the intersection (RA, V3)

We see that, for V3 to add upto 26, there has to be an ATM with cash of 11L, there cannot be two ATM's since the minimum capacity is 7L.
We can place the 11L ATM at (RB, V3) or (RC, V3), if we place them at either of these intersections, the remaining ATM has to have a capacity of 9L for the same reason. 9L cannot be at (RC, V1) or (RB, V1) since the total capacity of V1 is 15L and there cannot be an ATM with 6L. And it also cannot be at (RB, V2) since there is already an ATM with 11L that means the ATM with the second highest capacity cannot be 9. So that means 9L has to be at (RC, V2). And then filling in the rest of the numbers we get the final arrangement for Case-1.

Case 2: When 15L is at (RA, V1)
There can be no other ATM on V1 in this scenario, 7L ATM which is on RA cannot be on (RA, V2) considering the sum of the numbers on V2 is 21, and there cannot be 7+7 or a 14L ATM since the capacity of both RB and RC is 20. So, 7L has to be on V3, and since there cannot be a single ATM of 19L on V3, there has to be two other ATMs on V3 adding up to 19. Rearranging the numbers, we get the scenario for the second case.

Using the two cases, we can answer the given questions.

We see that in both cases, RA RB and RC have two ATMs
Only in the first case V1 V2 and V3 have two ATM's, in the second case V3 has 3 ATMs and V1 has 1.

Hence only Statement A is correct.

Passage

Twenty five coloured beads are to be arranged in a grid comprising of five rows and five columns. Each cell in the grid must contain exactly one bead. Each bead is coloured either Red, Blue or Green. While arranging the beads along any of the five rows or along any of the five columns, the rules given below are to be followed:

1. Two adjacent beads along the same row or column are always of different colours.
2. There is at least one Green bead between any two Blue beads along the same row or column.
3. There is at least one Blue and at least one Green bead between any two Red beads along the same row or column.

Every unique, complete arrangement of twenty five beads is called a configuration.

Q3.CAT 2020

What is the minimum number of Blue beads in any configuration?

Answer: 6

Show solution

To solve this question we can use the answer of the previous question, since maximum 9 red beads are possible, filling the remaining space with green and blue beads, in such a way that number of blue beads is minimised

Hence number of blue beads is 6

Passage

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.
1. A and B are to be placed in consecutively numbered shelves in increasing order.
2. I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
3. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
4. K is to be placed in shelf number 16.
5. L and J are items of the same type, while H is an item of a different type.
6. C is a candy and is to be placed in a shelf preceded by two empty shelves.
7. L is to be placed in a shelf preceded by exactly one empty shelf.

Q4.CAT 2019

In how many different ways can the items be arranged on the shelves?

Show solution

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

The number of arrangements for the first case = 2*2=4

The number of arrangements for the second case = 2*2=4

The total number of arrangements = 4+4=8

Passage

Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:

1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politicians are in the administration committee, and 20% are in the teaching committee.

Q5.CAT 2018

What is the number of educationalists in the research committee?

Answer: 3

Show solution

Let us draw a table according to the information given.

It is given that the numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee. Let '4x' be the number of bureaucrats in Administration committee.

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees. Let us assume that 'y' is the number of educationalists in the research committee and 'd' be the difference in the number of educationalists in Research and teaching committees.

60% of the politicians are in the administration committee, and 20% are in the teaching committee. Let '5z' be the number of total number of politicians.

We can say that

$\Rightarrow$ 10x+3y+5z = 24

We can see that each of x, y and z has to a natural number integer. If x > 1, then both y and z can't take any natural number.

Hence, we can say that x = 1.

At x = 1, 3y+5z = 14. If y = 1 or 2, Z is not an integer.

At x = 1 and y = 3, z = 1 which is the only possible solution.

We can see that 'd' can assume two possible values. d = 1 or 2.

From the table, we can see that the number of educationalists in the research committee = 3.

Passage

Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata 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:
1. The eight friends were seated in six different rows.
2. They occupied 3 Window seats, 4 Aisle seats and 1 Middle seat.
3. 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 for his/her choice of seat.
4. 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.
5. Gargi was sitting next to Kikira, and Manik was sitting next to Jayanta.
6. 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.

Q6.CAT 2017

In which row was Manik sitting?

Show solution

We are given that 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.
Let us see how the friends are supposed to pay for the seats they choose:-
In row 1-1000
In row 2-10 - 300 for window and 500 for aisle
In row 11 - 200 for window and 400 for aisle
In row 12,13 - 1000
In row 14-20 - 200 for window and 400 for aisle
In row 21-30 - 0
Thus, As we can see 10, 11 and 12 are the only consecutive seats in which the amounts is different.
Thus, Jayanth, Ajit and Byomkesh sat in row 10, row 11 and row 12.
Manik sat beside Jayantha and thus Manik is also sitting in row 10.
Now we are given that 7 of the 8 friends paid a total of 4600 Rs.
Let's start with the cases:-
It is obvious that 5 friends cannot pay 1000 Rs for their seat because the amount will exceed 4600
Case 1:- 4 friends pay 1000 Rs each. Thus, the remaining friends will pay 600 Rs.
This is possible only when each of them pay 200 Rs.
So the case is- 1000*4 , 200*3
Case 2 :- 3 friends pay 1000 Rs each. Thus, the remaining friends will pay 1600 Rs.
There are 2 cases where this is possible:-
1000*3, 500*2, 400, 200
1000*3, 400*4
Case 3:- 2 friends pay 1000 Rs each. Thus, the remaining 5 friends will pay 2600 Rs.
This is not possible as each friend can pay a maximum of 500 Rs.
Thus, the possible cases are
1000*4 , 200*3
1000*3, 500*2, 400, 200
1000*3, 400*4
As there is no case in which a friend has to pay 300 Rs thus, Jayantha must be sitting in row 10 aisle seat.
Thus, Jayantha paid 500 Rs.
Thus, the case is:-
1000*3, 500*2, 400, 200
Thus, Manik must have also paid 500 sitting in row 10 aisle seat
Ajit must be sitting in row 11 aisle seat paying 400 Rs.
Byomyesh must be sitting row 12 aisle seat paying 1000 Rs.
Thus, among Gargi, Kikira, Pradosh and Tapesh 2 must have paid 1000, 1 must have paid 200 and the remaining person must have paid nothing.
Now we know Gargi and Kikira are sitting adjacent to each other and thus, either both or none of them must have paid 1000 Rs.
Among Pradosh and Tapesh a maximum of 1 person could have paid 1000 Rs.
Thus, the only possible case here is :-
Gargi and Kikira paid 1000 each.
Pradosh is sitting ahead of Tapesh and one of them paid 200 Rs.
Since, both of them were sitting in seats marked by the same letter, in consecutive rows thus, the only possibility is Pradosh sitting in row 20 window seat and paying 200 and Tapesh sitting in row 21 paying nothing.
Thus, the amount paid by each friend is as shown below:

Manik is sitting in row 10.

Passage

A tea taster was assigned to rate teas from six different locations — Munnar, Wayanad, 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:
a: Cup 6 contained tea from Himachal.
2. Tea from Ooty got the highest rating, but it was not in Cup 3.
3. The rating of tea in Cup 3 was double the rating of the tea in Cup 5.
4. Only two cups got ratings in even numbers.
5. Cup 2 got the minimum rating and this rating was an even number.
6. Tea in Cup 3 got a higher rating than that in Cup 1.
7. The rating of tea from Wayanad was more than the rating of tea from Munnar, but less than that from Assam.

Q7.CAT 2017

If the tea from Munnar did not get the minimum rating, what was the rating of the tea from Wayanad?

Show solution

Now we are given that the lowest rating is an even number and only 2 cups got an even number rating.
Let's take cases:-
1. The lowest rating is 4
If the lowest rating in 4 then the other ratings will be in the range 5-10.
From this we need 4 odd and 1 even numbers.
This is not possible as there are only 3 odd numbers from 5-10.
Thus, the lowest rating is not 4.
2. The lowest rating is 2.
If the lowest rating in 4 then the other ratings will be in the range 5-10.
From this we need 4 odd and 1 even numbers.
This is possible when the odd ratings are 3,5,7 and 9.
We are given that the highest rating is not even. Thus, 10 rating is not possible.
We are also given that the rating of tea in Cup 3 was double the rating of the tea in Cup 5.
Thus, the rating of the tea in cup 3 is an even number.
Thus, the rating of the tea in cup 5 must be an odd number.
Only 1 such pair is possible of 3 and 6.
Thus, the tea in cup 2 got the rating of 2.
The tea in cup 3 got a rating of 6 and the tea in cup 5 got a rating of 3.
We are given that:-
Tea in Cup 3 got a higher rating than that in Cup 1.
Thus, the tea in cup 1 got a rating of 5.
Cup 6 contained tea from Himachal and the Tea from Ooty got the highest rating.
Thus, cup 6 got a rating of 7 and cup 4 got a rating of 9.
The table is as shown below:-

Passage

DIRECTIONS for the following two questions: Answer the questions on the basis of the information given below.

The Head of a newly formed government desires to appoint five of the six elected members A, B, C, D, E and F to portfolios of Home, Power, Defence, Telecom and Finance. F does not want any portfolio if D gets one of the five. C wants either Home or Finance or no portfolio. B says that if D gets either Power or Telecom then she must get the other one. E insists on a portfolio if A gets one.

Q8.CAT 2003

Which is a valid assignment?

A-Home, B-Power, C-Defence, D-Telecom, E-Finance.
C-Home, D-Power, A-Defence, B-Telecom, E-Finance.
A-Home, B-Power, E-Defence, D-Telecom, F-Finance.
B-Home, F-Power, E-Defence, C-Telecom, A-Finance.

Show solution

Since C wants either home or finance or none so options A and D are eliminated.
Since F does not want any portfolio if D gets one, Option C is eliminated.

Passage

DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.

Five friends meet every morning at Sree Sagar restaurant for an idli-vada breakfast. Each consumes a different number of idlis and vadas. The number of idlis consumed are 1, 4, 5, 6, and 8, while the number of vadas consumed are 0, 1, 2, 4, and 6. Below are some more facts about who eats what and how much.

i. The number of vadas eaten by Ignesh is three times the number of vadas consumed by the person who eats four idlis.

ii. Three persons, including the one who eats four vadas eat without chutney.

iii. Sandeep does not take any chutney.

iv. The one who eats one idli a day does not eat any vadas or chutney. Further, he is not Mukesh.

v. Daljit eats idli with chutney and also eats vada.

vi. Mukesh, who does not take chutney, eats half as many vadas as the person who eats twice as many idlis as he does.

vii. Bimal eats two more idlis than Ignesh, but Ignesh eats two more vadas than Bimal.

Q9.CAT 2003

Which of the following statements is true?

Mukesh eats 8 idlis and 4 vadas but no chutney.
The person who eats 5 idlis and 1 vada does not take chutney.
The person who eats equal number of vadas and idlis also takes chutney.
The person who eats 4 idlis and 2 vadas also takes chutney.

Show solution

Considering (i), Ignesh has to eat 6 vadas, since 6 is the only multiple of 3.

Also, using the same information, we can say that a person consumes 2 vadas and 4 idlis.

Using (vii), Bimal eats 2 more idlis than Ignesh.

Possibilities:

Bimal - 8, Ignesh - 6

Bimal - 6, Ignesh - 4

But Ignesh cannot have 4 idlis because the person who eats 4 idlis eats 2 vadas.

Hence we take Bimal - 8 and Ignesh - 6.

Also, we get that Bimal eats 6 - 2 = 4 vadas.

So far, we get the following information.

Using (vi), there is a person who eats twice as many idlis as Mukesh. The only pair satisfying is 8, 4.

So, Mukesh eats 4 idlis. Plus the person who eats 4 idlis eats 2 vadas. Hence, we get

Daljit also eats Vada as per info (v), so we get the following

(iv) gives us the information that the one who eats 1idli does not have vada.

Considering the persons who had chutney and those who didn't, 3 persons do not have chutney. Bimal is one of them(the one eating 4 vadas).

Mukesh is the second one to not take chutney(last hint). Also, Sandip does not take chutney. Hence, we get this information as well.

Hence, Ignesh eating 6 vadas and 6 idlis eat Chutney.

Passage

Directions for the following four questions: Answer the questions based on the following information.

A and B are two sets (e.g. A = Mothers, B = Women).

The elements that could belong to both the sets (e.g. women who are mothers) is given by the set $C = A \cap B$ .

The elements which could belong to either A or B, or both, is indicated by the set $D = A \cup B$ .

A set that does not contain any elements is known as a null set represented by $\phi$ (e.g. if none of the women in the set B is a mother, then $C = A \cap B$ is a null set, or C = $\phi$ ).

Let ‘V’ signify the set of all vertebrates, ‘M’ the set of all mammals, ‘D’ dogs, ‘F’ fish, ‘A’ alsatian and ‘P’, a dog named Pluto.

Q10.CAT 2001

Given that $X = M \cap D$ is such that $X = D$ . Which of the following is true?

All dogs are mammals
Some dogs are mammals
$X = \phi$
All mammals are dogs

Show solution

It is given that $X = M \cap D$ is such that $X = D$ , which means D is a subset of M . Which means all dogs are mammals. Hence , option A.

Passage

Directions for the next 2 questions: There are three bottles of water, A, B, C, whose capacities are 5 litres, 3 litres, and 2 litres respectively. For transferring water from one bottle to another and to drain out the bottles, there exists a piping system. The flow through these pipes is computer controlled. The computer that controls the flow through these pipes can be fed with three types of instructions, as explained below:

Initially, A is full with water, and B and C are empty.

Q11.CAT 2000

Consider the same sequence of three instructions ‘and the same initial state mentioned above. Three more instructions are added at the end of the above sequence to have A contain 4 litres of water. In this total sequence of six instructions, the fourth one is DRAIN (A). This is the only DRAIN instruction in the entire sequence. At the end of the execution of the above sequence, how much water (in litres) is contained in C?

One
Two
Zero
None of these

Show solution

We know that after 3 operations A contain 1 ltr and 4 th operation is Drain(a) , so 1 ltr is drained and only 4 ltrs remain in all there cylinders. Also after 6 operations A should have all the 4 ltrs in it. hence the other 2 vessels must contain 0 ltrs because there is only one drain operation in the entire sequence.

Passage

A, B, C and D are to be seated in a row. But C and D cannot be together. Also B cannot be at the third place.

Q12.CAT 1998

Which of the following must be false?

A is at the first place
A is at the second place
A is at the third place
A is at the fourth place

Show solution

Among all positions for A, if it has first position then either B will be at the 3rd position or C and D will be together which is not possible Hence answer will be A) .

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