CAT LR Arrangements Questions
Master CAT LR Arrangements Questions with practice questions and detailed solutions.
Instructions
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.
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.
Question 1.
Which of the following statements is correct?
Which of the following statements is correct?
Text Explanation:
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 cash requirement 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.
The only statement that is correct considering both the cases is Option B
"The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs."
Question 2.
How many ATMs have cash requirements of Rs. 10 Lakhs or more?
How many ATMs have cash requirements of Rs. 10 Lakhs or more?
Text Explanation:
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 cash requirement 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.
Three ATM's with 10L or more: 15L, 12L and 11L. Answer is 3.
Question 3.
What best can be said about the road distance (in km) between the ATMs having the second highest and the second lowest cash requirements?
What best can be said about the road distance (in km) between the ATMs having the second highest and the second lowest cash requirements?
Text Explanation:
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 cash requirement 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.
Second highest is 12L and second lowest is 8L
In Case 1 the distance between 12L and 8L ATM is 4km and in Case 2 the distance between 12L and 8L ATM is 7km.
Hence the answer is either 4km or 7km.
Question 4.
What is the number of ATMs whose locations and cash requirements can both be uniquely determined?
What is the number of ATMs whose locations and cash requirements can both be uniquely determined?
Text Explanation:
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 cash requirement 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.
ATMs that can be uniquely determines are the ATMs with cash 9L, 11L and 12L. Hence the answer is 3.
Question 5.
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.
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.
Text Explanation:
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 cash requirement 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.
Instructions
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.
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.
Question 6.
What is the maximum possible number of Red beads that can appear in any conguration?
What is the maximum possible number of Red beads that can appear in any conguration?
Text Explanation:
Maximum 9 red beads are possible as shown here
Question 7.
The total number of possible congurations using beads of only two colours is:
The total number of possible congurations using beads of only two colours is:
Text Explanation:
Since we are required to use only two colours, these can be either
1. Green + Blue
2. Blue + Red
3. Green + Red
But we know that Between any two Red coloured beads in a row or a column, there have to be both a Green and a Blue coloured bead. Hence without both Green and Blue, Red beads cannot be used to fill the grid.Thus if we use only Green and Blue beads, the two configurations that are possible are:
There are only 2 configurations possible
Instructions
The schematic diagram below shows 12 rectangular houses in a housing complex. House numbers are mentioned in the rectangles representing the houses. The houses are located in six columns - Column-A through Column-F, and two rows - Row-1 and Row-2. The houses are divided into two blocks - Block XX and Block YY. The diagram also shows two roads, one passing in front of the houses in Row-2 and another between the two blocks.
Some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The road adjacency value of a house is the number of its sides adjacent to a road. For example, the road adjacency values of C2, F2, and B1 are 2, 1, and 0, respectively. The neighbour count of a house is the number of sides of that house adjacent to occupied houses in the same block. For example, E1 and C1 can have the maximum possible neighbour counts of 3 and 2, respectively.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbour count). The following information is also known.
1. The maximum quoted price of a house in Block XX is Rs. 24 lakhs. The minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column-E.
2. Row-1 has two occupied houses, one in each block.
3. Both houses in Column-E are vacant. Each of Column-D and Column-F has at least one occupied house.
4. There is only one house with parking space in Block YY.
The schematic diagram below shows 12 rectangular houses in a housing complex. House numbers are mentioned in the rectangles representing the houses. The houses are located in six columns - Column-A through Column-F, and two rows - Row-1 and Row-2. The houses are divided into two blocks - Block XX and Block YY. The diagram also shows two roads, one passing in front of the houses in Row-2 and another between the two blocks.
Some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The road adjacency value of a house is the number of its sides adjacent to a road. For example, the road adjacency values of C2, F2, and B1 are 2, 1, and 0, respectively. The neighbour count of a house is the number of sides of that house adjacent to occupied houses in the same block. For example, E1 and C1 can have the maximum possible neighbour counts of 3 and 2, respectively.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbour count). The following information is also known.
1. The maximum quoted price of a house in Block XX is Rs. 24 lakhs. The minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column-E.
2. Row-1 has two occupied houses, one in each block.
3. Both houses in Column-E are vacant. Each of Column-D and Column-F has at least one occupied house.
4. There is only one house with parking space in Block YY.
Question 8.
What is the maximum possible quoted price (in lakhs of Rs.) for a vacant house in Column-E?
What is the maximum possible quoted price (in lakhs of Rs.) for a vacant house in Column-E?
Text Explanation:
It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs
Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space:
=> 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count)
The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.
Case 2: House without parking space:
=> 10+5a+3b = 24 => 5a+3b = 14
=> (a, b) = (1, 3)
Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2. Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)
Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2:
We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b)
If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible)
Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0
b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.
Case 2: The minimum quoted house is E1:
We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b).
i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value)
Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case of house D1 being occupied and F1 being empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours)
Here, even if we take the number of neighbors to be 1, which is the maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest-value house in block YY. Therefore, F1 cannot be empty.
Since F1 is occupied and we know that there is only one house occupied in row 1 of each block, D1 becomes unoccupied. D2 becomes occupied because it is given in the question that each of Column-D and Column-F has at least one occupied house.
Here, the value of D1 is 18 as D2 is occupied.
We do not know the status of house F2.
Therefore, the final diagram is given below:
From the diagram, the vacant house with the maximum possible quoted price in column E is E2 when F2 is occupied.
The maximum possible quoted price of E2 is 10+5*1+3*2 = 21 Lacs. ( E2 has no parking space because E1 has the parking space, and it is given that there is only one house with parking space in Block YY.)
Question 9.
Which of the following houses is definitely occupied?
Which of the following houses is definitely occupied?
Text Explanation:
It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs
Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space:
=> 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count)
The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.
Case 2: House without parking space:
=> 10+5a+3b = 24 => 5a+3b = 14
=> (a, b) = (1, 3)
Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2. Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)
Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2:
We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b)
If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible)
Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0
b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.
Case 2: The minimum quoted house is E1:
We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b).
i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value)
Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case of house D1 being occupied and F1 being empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours)
Here, even if we take the number of neighbors to be 1, which is the maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest-value house in block YY. Therefore, F1 cannot be empty.
Since F1 is occupied and we know that there is only one house occupied in row 1 of each block, D1 becomes unoccupied. D2 becomes occupied because it is given in the question that each of Column-D and Column-F has at least one occupied house.
Here, the value of D1 is 18 as D2 is occupied.
We do not know the status of house F2.
Therefore, the final diagram is given below:
From the diagram, we can see that B1 and D2 are definitely occupied. The rest of the options are not definitely correct.
The correct options are both B and C.
Question 10.
Which house in Block YY has parking space?
Which house in Block YY has parking space?
Text Explanation:
It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs
Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space:
=> 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count)
The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.
Case 2: House without parking space:
=> 10+5a+3b = 24 => 5a+3b = 14
=> (a, b) = (1, 3)
Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2. Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)
Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2:
We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b)
If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible)
Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0
b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.
Case 2: The minimum quoted house is E1:
We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b).
i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value)
Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case of house D1 being occupied and F1 being empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours)
Here, even if we take the number of neighbors to be 1, which is the maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest-value house in block YY. Therefore, F1 cannot be empty.
Since F1 is occupied and we know that there is only one house occupied in row 1 of each block, D1 becomes unoccupied. D2 becomes occupied because it is given in the question that each of Column-D and Column-F has at least one occupied house.
Here, the value of D1 is 18 as D2 is occupied.
We do not know the status of house F2.
Therefore, the final diagram is given below:
From the diagram, we can see that E1 has the parking space (case 2).
The correct option is A
Question 11.
Which of the following options best describes the number of vacant houses in Row-2?
Which of the following options best describes the number of vacant houses in Row-2?
Text Explanation:
It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs
Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space:
=> 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count)
The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.
Case 2: House without parking space:
=> 10+5a+3b = 24 => 5a+3b = 14
=> (a, b) = (1, 3)
Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2. Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)
Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2:
We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b)
If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible)
Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0
b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.
Case 2: The minimum quoted house is E1:
We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b).
i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value)
Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case of house D1 being occupied and F1 being empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours)
Here, even if we take the number of neighbors to be 1, which is the maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest-value house in block YY. Therefore, F1 cannot be empty.
Since F1 is occupied and we know that there is only one house occupied in row 1 of each block, D1 becomes unoccupied. D2 becomes occupied because it is given in the question that each of Column-D and Column-F has at least one occupied house.
Here, the value of D1 is 18 as D2 is occupied.
We do not know the status of house F2.
Therefore, the final diagram is given below:
From the diagram, we can say that the number of vacant houses in Row 2 in Block XX is 1, and the number of vacant houses in Row 2 in Block YY is either 1 or 2.
Hence, the total number of vacant houses is either 2 or 3
The correct option is D
Question 12.
How many houses are vacant in Block XX?
How many houses are vacant in Block XX?
Text Explanation:
It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) + 5 × (road adjacency value) + 3 × (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs
Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space:
=> 12+5a+3b = 24 => 5a+3b = 12 ( a = road adjacency value, b= neighbor count)
The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3.
Hence, case 1 is invalid.
Case 2: House without parking space:
=> 10+5a+3b = 24 => 5a+3b = 14
=> (a, b) = (1, 3)
Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2. Therefore, the neighbor houses of B2, which are (B1, A2, and C2) are occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1, and C1 are vacant.
Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)
Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.
It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2:
We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b)
If the house has parking space, then: 12+5*1+3*b = 15 => 3b = -2 (which is not possible)
Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0
b = 0 implies all the neighbor house of E2 is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.
Case 2: The minimum quoted house is E1:
We know that the road adjacency of E1 is 0, hence we can calculate whether the house has parking space or not, and the neighbor count (b).
i) If the house has no working space, then: 10+5*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value)
Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1.
Let's take the case of house D1 being occupied and F1 being empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours)
Here, even if we take the number of neighbors to be 1, which is the maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest-value house in block YY. Therefore, F1 cannot be empty.
Since F1 is occupied and we know that there is only one house occupied in row 1 of each block, D1 becomes unoccupied. D2 becomes occupied because it is given in the question that each of Column-D and Column-F has at least one occupied house.
Here, the value of D1 is 18 as D2 is occupied.
We do not know the status of house F2.
Therefore, the final diagram is given below:
From the diagram, we can see that 3 houses are vacant in block XX.
Instructions
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.
- The four friends occupy adjacent chairs only at the end of Turn 2 and Turn 6.
- Davies occupies Chair 2 after Turn 1 and Chair 4 after Turn 5, and Chhavi occupies Chair 7 after Turn 2.
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.
- The four friends occupy adjacent chairs only at the end of Turn 2 and Turn 6.
- Davies occupies Chair 2 after Turn 1 and Chair 4 after Turn 5, and Chhavi occupies Chair 7 after Turn 2.
Question 13.
What is the number of the chair initially occupied by Bashir?
What is the number of the chair initially occupied by Bashir?
Text Explanation:
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.
Question 14.
Who sits on the chair numbered 4 at the end of Turn 3?
Who sits on the chair numbered 4 at the end of Turn 3?
Text Explanation:
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,
We now have the initial position, and we are told that all the members are adjacent only after turn 2 and turn 6.We know that A changed his position in turn 1.
Turn 1: Aslam moves
If A changed his position anticlockwise, he would go to chair 5, and if he changed his position clockwise, he would go to chair 1.
If A ends in chair 5 after turn 1, then it is not possible for all of them to be adjacent to each other after B changes his position in turn 2.
So, A has to end in chair 1 after turn 1, so that B will have a chance in turn 2 for all of them to be adjacent.
Turn 2: Bashir movesFor all of them to be adjacent after turn 2, B must change his position anticlockwise, and after turn 2, he would end up on chair 3.
Turn 3: Chhavi movesWe are given that Davies occupies chair 4 after turn 5, and for this to happen, chair 4 must be empty and cannot be occupied by anyone who does not move before turn 5. So, in turn 3, C must go anticlockwise and occupy chair 6, as if he goes clockwise, he would occupy chair 4, which must be empty for Davies.
Turn 4: Davies movesDavies will not move in turn 5, and for him to occupy chair 4 after turn 5, he must occupy it on turn 4. So, D has to go clockwise in turn 4 and must occupy chair 4.
Turn 5: Aslam movesWe know that all of them are again adjacent after turn 6. If A goes clockwise in turn 5, then he will occupy the chair 2, and in that case, it is not possible for them to be adjacent after turn 6 in either of the cases of B going clockwise or anticlockwise.
So, A must go anticlockwise and occupy chair 7 for them to be adjacent after turn 6.
Turn 6: Bashir movesNow, for them to be adjacent after turn 6, B must go clockwise and must occupy chair 5.
Turn 7: Chavvi movesWe do not have any information to determine whether C went clockwise or anticlockwise. So, both cases are possible after turn 7.
After turn 3, chair 4 is occupied by no one.Hence, the correct answer is option D.
Question 15.
Which of the chairs are occupied at the end of Turn 6?
Which of the chairs are occupied at the end of Turn 6?
Text Explanation:
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,
We now have the initial position, and we are told that all the members are adjacent only after turn 2 and turn 6.We know that A changed his position in turn 1.
Turn 1: Aslam moves
If A changed his position anticlockwise, he would go to chair 5, and if he changed his position clockwise, he would go to chair 1.
If A ends in chair 5 after turn 1, then it is not possible for all of them to be adjacent to each other after B changes his position in turn 2.
So, A has to end in chair 1 after turn 1, so that B will have a chance in turn 2 for all of them to be adjacent.
Turn 2: Bashir movesFor all of them to be adjacent after turn 2, B must change his position anticlockwise, and after turn 2, he would end up on chair 3.
Turn 3: Chhavi movesWe are given that Davies occupies chair 4 after turn 5, and for this to happen, chair 4 must be empty and cannot be occupied by anyone who does not move before turn 5. So, in turn 3, C must go anticlockwise and occupy chair 6, as if he goes clockwise, he would occupy chair 4, which must be empty for Davies.
Turn 4: Davies movesDavies will not move in turn 5, and for him to occupy chair 4 after turn 5, he must occupy it on turn 4. So, D has to go clockwise in turn 4 and must occupy chair 4.
Turn 5: Aslam movesWe know that all of them are again adjacent after turn 6. If A goes clockwise in turn 5, then he will occupy the chair 2, and in that case, it is not possible for them to be adjacent after turn 6 in either of the cases of B going clockwise or anticlockwise.
So, A must go anticlockwise and occupy chair 7 for them to be adjacent after turn 6.
Turn 6: Bashir movesNow, for them to be adjacent after turn 6, B must go clockwise and must occupy chair 5.
Turn 7: Chavvi movesWe do not have any information to determine whether C went clockwise or anticlockwise. So, both cases are possible after turn 7.
The chairs occupied after turn 6 are 4, 5, 6 and 7.Hence, the correct answer is option A.
Question 16.
Which of the following BEST describes the friends sitting on chairs adjacent to the one occupied by Bashir at the end of Turn 7?
Which of the following BEST describes the friends sitting on chairs adjacent to the one occupied by Bashir at the end of Turn 7?
Text Explanation:
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,
We now have the initial position, and we are told that all the members are adjacent only after turn 2 and turn 6.We know that A changed his position in turn 1.
Turn 1: Aslam moves
If A changed his position anticlockwise, he would go to chair 5, and if he changed his position clockwise, he would go to chair 1.
If A ends in chair 5 after turn 1, then it is not possible for all of them to be adjacent to each other after B changes his position in turn 2.
So, A has to end in chair 1 after turn 1, so that B will have a chance in turn 2 for all of them to be adjacent.
Turn 2: Bashir movesFor all of them to be adjacent after turn 2, B must change his position anticlockwise, and after turn 2, he would end up on chair 3.
Turn 3: Chhavi movesWe are given that Davies occupies chair 4 after turn 5, and for this to happen, chair 4 must be empty and cannot be occupied by anyone who does not move before turn 5. So, in turn 3, C must go anticlockwise and occupy chair 6, as if he goes clockwise, he would occupy chair 4, which must be empty for Davies.
Turn 4: Davies movesDavies will not move in turn 5, and for him to occupy chair 4 after turn 5, he must occupy it on turn 4. So, D has to go clockwise in turn 4 and must occupy chair 4.
Turn 5: Aslam movesWe know that all of them are again adjacent after turn 6. If A goes clockwise in turn 5, then he will occupy the chair 2, and in that case, it is not possible for them to be adjacent after turn 6 in either of the cases of B going clockwise or anticlockwise.
So, A must go anticlockwise and occupy chair 7 for them to be adjacent after turn 6.
Turn 6: Bashir movesNow, for them to be adjacent after turn 6, B must go clockwise and must occupy chair 5.
Turn 7: Chavvi movesWe do not have any information to determine whether C went clockwise or anticlockwise. So, both cases are possible after turn 7.
In either of the cases after turn 7, the seat adjacent to Bashir is occupied only by Davies, and the other one is empty.Hence, the correct answer is option B.
Instructions
Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
• Immediately to the left;
• Immediate to the right;
• Second to the left; or
• Second to the right.
The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as ‘?’.
Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
• Immediately to the left;
• Immediate to the right;
• Second to the left; or
• Second to the right.
The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as ‘?’.
Question 17.
Who is sitting immediately to the right of Bina?
Who is sitting immediately to the right of Bina?
Text Explanation:
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8: The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7 and the only person left to be assigned is Eshan making the person at position 7 to be Eshan.
This is the final arrangement. The person immediately to the right of Bina is Eshan.
Hence, the correct answer is option B.
Question 18.
Who is sitting third to the left of Eshan?
Who is sitting third to the left of Eshan?
Text Explanation:
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8: The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
This is the final arrangement. The person third to the left of Eshan is Chirag.
Hence, the correct answer is option C.
Question 19.
For which of the following pass types can the total number of occurrences be uniquely determined?
For which of the following pass types can the total number of occurrences be uniquely determined?
Text Explanation:
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8: The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
In rounds 4 and 5, the possibilities were1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
Here, Buck could have been passed second to the right, second to the left or immediately to the left in rounds 4 and 5, which is unknown to us. The only pass type that we are certain to have happened is immediately to the right.
Hence, the correct answer is option C.
Question 20.
For which of the following children is it possible to determine how many times they received the Buck?
For which of the following children is it possible to determine how many times they received the Buck?
Text Explanation:
Round 1:
The holder of the buck is Bina. The buck is passed immediately to the left, which means it is given to position 2 from 1, and it is received by Aarav. So, Aarav's position is 2.
Round 2:
The holder of the buck is Aarav. The buck is passed second to the right, which means it is given to the position 7 from 2, and the person receiving it is unknown.
Round 3:
The holder of the buck is unknown, but it is at position 7. The buck is passed immediately to the right, which means it is given to the position 6 from 7, and the person receiving it is Diya. Therefore, Diya's position is 6.
Round 4 and 5:
The buck is at position 6 at the start of round 4 with Diya, and the pass types of rounds 4 and 5 are not provided to us. The only information known to us is that after round 5, the buck is with Aarav, who is at position 2.
The possible ways for Aarav to receive the buck after rounds 4 and 5 are:
1) Passing the buck second to the right in round 4 and second to the right in round 5.
2) Passing the buck second to the left in round 4 and to the left in round 5.
3) Passing the buck to the left in round 4 and second to the left in round 5.
These are the only possible ways for the buck to reach Aarav after round 6.
Round 6:
The holder of the buck is Aarav. The buck is passed second to the left, which means it is given to the position 4 from 2, and the person receiving it is unknown.
Round 7:
The holder of the buck is unknown, but it is at position 4. The buck is passed immediately to the left, which means it is given to position 5 from 4, and the person receiving it is Gaurav. Therefore, Gaurav's position is 5.
Round 8:The holder of the buck is Gaurav. The buck is passed immediately to the left, which means it is given to position 6 from 5, and the person receiving it is unknown. But we know that the person in position 6 is Diya. So, the buck is currently at position 6 with Diya.
Round 9 and 10:
The pass types of rounds 9 and 10 are unknown to us, but we are told that the buck is passed to Farhan in round 9 and to Chirag in round 10. The positions of Farhan and Chirag are currently not assigned, and they are out of 3, 4, and 7. From position 6, the only positions that Buck can reach out of the three are 4 or 7.
If the buck is passed to position 7 in round 9, then it has to reach either 3 or 4 in round 10 to get to Chirag, but it is not possible to pass the buck from 7 to 3 or 4 in a single round. So, we can eliminate the case of Buck going to 7 in round 9.
The only case left is for the buck to go to position 4 in round 9 and to position 3 in round 10, as from 4, the buck can only be passed to 3 out of 3 and 7 in one round. Hence, we can conclude that Farhan is placed at position 4 and Chirag is placed at position 3. This leaves out only position 7, and the only person left to be assigned is Eshan, making the person at position 7 Eshan.
In rounds 4 and 5, the possibilities were1) Passing the buck second to the right in round 4 and second to the right in round 5. Here, the buck would have gone to Farhan in round 4.
2) Passing the buck second to the left in round 4 and to the left in round 5. Here, the buck would have gone to Bina in round 4.
3) Passing the buck to the left in round 4 and second to the left in round 5. Here, the buck would have gone to Eshan in round 4.
In round 4, the buck might have gone to either Farhan, Bina or Eshan, which is not known to us. Therefore, the count of the number of times the buck was received by them cannot be determined uniquely, whereas it can be uniquely determined in the case of Gaurav, which is 1 in round 7.
Hence, the correct answer is option D.
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