Answer the following question based on the information given below.
A supermarket has to place 12 items (coded A to L) in shelves numbered 1 to 16. Five of these items are types of biscuits, three are types of candies and the rest are types of savouries. Only one item can be kept in a shelf. Items are to be placed such that all items of same type are clustered together with no empty shelf between items of the same type and at least one empty shelf between two different types of items. At most two empty shelves can have consecutive numbers.
The following additional facts are known.
A and B are to be placed in consecutively numbered shelves in increasing order.
I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.
D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.
K is to be placed in shelf number 16.
L and J are items of the same type, while H is an item of a different type.
C is a candy and is to be placed in a shelf preceded by two empty shelves.
L is to be placed in a shelf preceded by exactly one empty shelf.
1.
CAT 2019 LRDI Slot 1
| LR - Arrangements
In how many different ways can the items be arranged on the shelves?
There are 5 types of biscuits, 3 types of candies and 4 types of savouries. Among 16 shelves, there are 4 empty shelves.
It is given that all items of same type are clustered together with no empty shelf between items of the same type.
From (3) and (4), it can be concluded that D, E, F and K are savouries.
From (2) and (5), L, I and J are of one type and H is the other type. Therefore from (6), as C is a candy, L, I J must be types of biscuits and H is a type of candy. Now using (1), we can conclude that A and B are of one type but not candies as there are only 3 types of candies.
Therefore,
Biscuits: A, B, I, J, L Candies: C, H, G Savouries: D, E, F, K
From (3), (4), (6) and (7), there shelf number 12 must be an empty shelf. Also, D, E, F and K are placed in shelves numbered 13, 14, 15 and 16 respectively.
Now from (1), (2) and (7), the sequence (from left to right) in which biscuits are kept is:
(Empty shelf), L, A, B, (I/J), (J/I).
From (6), the candies must be in the following order: (Empty shelf), (Empty shelf), C, (H/G), (G/H)
Thus, we have
In each case, J and I can be arranged in 2 ways and G and H can be arranged among them in 2 ways. Thus, 2 × 2 = 4 ways.
Total number of ways the items can be arranged on the shelves = 4 + 4 = 8
Answer the following question based on the information given below.
The Ministry of Home Affairs is analysing crimes committed by foreigners in different states and union territories (UT) of India. All cases refer to the ones registered against foreigners in 2016.
The number of cases – classified into three categories: IPC crimes, SLL crimes and other crimes – for nine states/UTs are shown in the figure below. These nine belong to the top ten states/UTs in terms of the total number of cases registered. The remaining state (among top ten) is West Bengal, where all the 520 cases registered were SLL crimes.
The table below shows the ranks of the ten states/UTs mentioned above among ALL states/UTs of India in terms of the number of cases registered in each of the three category of crimes. A state/UT is given rank r for a category of crimes if there are (r−1) states/UTs having a larger number of cases registered in that category of crimes. For example, if two states have the same number of cases in a category, and exactly three other states/UTs have larger numbers of cases registered in the same category, then both the states are given rank 4 in that category. Missing ranks in the table are denoted by *.
5.
CAT 2019 LRDI Slot 1
| DI - Tables & Graphs
What is the rank of Kerala in the ‘IPC crimes’ category?
The approximate values in the bar chart can be tablated as follows.
All the above states/UTs are arranged in descending order according to the number of cases. The 'Rank' column in the table lists the rank given to each state/UT among all states/UTs of India.
Note that Maharashtra and Karnataka are ranked 3. This means the two states with higher number of cases than Maharashtra and Karnataka have to be ranked as 1 and 2. So Delhi and Goa are ranked 1 and 2 respectively.
Since two states have been given the same rank (3), so the next state will be ranked 5.
In the two states where the highest total number of cases are registered, the ratio of the total number of cases in IPC crimes to the total number in SLL crimes is closest to ?
The approximate values in the bar chart can be tablated as follows.
All the above states/UTs are arranged in descending order according to the number of cases. The 'Rank' column in the table lists the rank given to each state/UT among all states/UTs of India.
Since Telangana and Maharashtra are both ranked 8, so all the states having more than 6 cases (Other crimes) are ranked 1 to 7.
So the final ranking is as follows.
So, Tamil Nadu is ranked 2 and Puducherry is ranked 3. Hence both statements (i) and (ii) are true.
The approximate values in the bar chart can be tablated as follows.
Following table is sorted list for all three crime categories where in each category, states are arrranged in descending order (from top to bottom) according to the number of cases. The 'Rank' column determines the rank of each state among all states/UTs of India.
Observe that in IPC crimes category, Maharashtra and Karnataka are ranked 3. So Delhi and Goa would be ranked 1 and 2 respectively.
In Other crimes category, Delhi is clearly ranked 1 as it has the highest number of cases.
In SLL crimes category, West Bengal and Karnataka are ranked 1 and 2 respectively. Since Goa and Maharashtra have been ranked 4. Also Delhi's number of cases (36) is lesser than that of Karnataka (49), so Delhi cannot be ranked 2. So it has to be ranked 3. Please note that Delhi having 36 cases was an approximate value.
Answer the following question based on the information given below.
Six players – Tanzi, Umeza, Wangdu, Xyla, Yonita and Zeneca competed in an archery tournament. The tournament had three compulsory rounds, Rounds 1 to 3. In each round every player shot an arrow at a target. Hitting the centre of the target (called bull’s eye) fetched the highest score of 5. The only other possible scores that a player could achieve were 4, 3, 2 and 1. Every bull’s eye score in the first three rounds gave a player one additional chance to shoot in the bonus rounds, Rounds 4 to 6. The possible scores in Rounds 4 to 6 were identical to the first three. A player’s total score in the tournament was the sum of his/her scores in all rounds played by him/her. The table below presents partial information on points scored by the players after completion of the tournament. In the table, NP means that the player did not participate in that round, while a hyphen means that the player participated in that round and the score information is missing.
The following facts are also known.
Tanzi, Umeza and Yonita had the same total score.
Total scores for all players, except one, were in multiples of three.
The highest total score was one more than double of the lowest total score.
The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.
9.
CAT 2019 LRDI Slot 1
| DI - Games & Tournaments
Tanzi, Umeza, Xyla, Yonita and Zeneca got 1,2,3,1,2 chances to shoot in the bonus rounds respectively. Therefore, in the compulsory round, 9 bull’s eye were hit. From (4), it can be concluded that number of bull’s eye hit in Rounds 1, 2 and 3 were 3, 4 and 2 respectively.
Note that Xyla got three chances in the bonus round. So she must have hit three bull’s eye in the compulsory rounds. Therefore, Xyla’s minimum score is 5 × 4 + 1 + 1 = 22.
Zeneca’s maximum score could be 5 × 4 + 4 = 24. Her minimum score could be 5 × 4 + 1 = 21. If her score was the maximum score, it must be an odd number. So, 23 is the maximum score. Now using (3), the lowest score = 11. But both 11 and 23 are not divisible by 3. Using (2), we can conclude that 23 is not the maximum score. So, Xyla’s must have scored maximum. And Zeneca’s score was either 21(5 × 4 + 1) or 24(5 × 4 + 4).
Thus, Xyla scored 5 in each of the compulsory rounds and 4 in round 6.
Tanzi hit one bull’s eye either in round 1 or in round 3. So her minimum score = 5 + 4 + 1 + 5 = 15 and her maximum score = 5 + 4 + 4 + 5 = 18.
Yonita’s maximum score = 5 + 4 + 3 + 5 = 17.
So from (1) and (2), Tanzi, Umeza and Yonita each had total score 15. And hence, Wangdu scored least points i.e., 12 points. The only possible combination is 4 points in each of the compulsory rounds.
So, Tanzi scored 1 and 5 in rounds 1 and 3 in some order. Assume that she scored 1 in round 1 and 5 in round 3. From (5), Zeneca scored 1 in round 1. But then she must also have scored 5 in round 3 as she hit bull’s eye twice in the compulsory rounds. But this is contradiction to (5). So, Tanzi and Zeneca scored 5 in round 1. Tanzi scored 1 in round 3.
Thus, Tanzi, Zeneca and Xyla hit bull’s eye in round 1. Therefore Yonita (total score =15) must have hit bull’s eye in the second round and scored 2 points in the first round.
Umeza must have hit two bull’s eye in rounds 2 and 3. Also, she must have scored 2 points in the first round.
Thus, Umeza and Xyla hit the bull’s eye in the third round. Therefore, Zeneca hit bull’s eye in the second round and scored 4 points in the third round.
Answer the following question based on the information given below.
A new game show on TV has 100 boxes numbered 1, 2, . . . , 100 in a row, each containing a mystery prize. The prizes are items of different types, a, b, c, . . . , in decreasing order of value. The most expensive item is of type a, a diamond ring, and there is exactly one of these. You are told that the number of items at least doubles as you move to the next type. For example, there would be at least twice as many items of type b as of type a, at least twice as many items of type c as of type b and so on. There is no particular order in which the prizes are placed in the boxes.
As we need to find maximum possible different types of prizes, number of prizes of type b has to be minimum possible and hence must be 2, number of items of type c = 4 …and so on.
1(type a) + 2(type b) + 4(type c) + 8(type d) + 16(type e) + 32(type e) = 63
Suppose there is prize of type f then number of items has to be at least 64. But then there are more than 100 items, which is not true. So there cannot be prize of type f.
[1]. If there are 30 items of type b, then items of type c = 100 – 30 −1 = 69. So this case is possible.
[2]. There are 75 items of type e, then items of type b, c and d = 100 – 75 −1 = 24.
Some of the values of (b, c, d) are (2, 4, 18) or (2, 5, 17) or (3, 6, 15). So this case is possible.
[3]. If there are 60 items of type d, then items of type b and c = 100 – 60 −1 = 39. So this case is possible as we can find many combinations for (b, c).
[4]. If there are 45 items of type c, then items of type a, b and c in all cannot be more than 1 + 22 + 45 = 68. Now items of type d has to be more than 90. But then total number of items exceed 100. So this case is not possible.
You ask for the type of item in box 45. Instead of being given a direct answer, you are told that there are 31 items of the same type as box 45 in boxes 1 to 44 and 43 items of the same type as box 45 in boxes 46 to 100.
What is the maximum possible number of different types of items?
Answer the following question based on the information given below.
The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.
17.
CAT 2019 LRDI Slot 1
| LR - Coding Decoding
| LR - Mathematical Reasoning
Answer the following question based on the information given below.
Princess, Queen, Rani and Samragni were the four finalists in a dance competition. Ashman, Badal, Gagan and Dyu were the four music composers who individually assigned items to the dancers. Each dancer had to individually perform in two dance items assigned by the different composers. The first items performed by the four dancers were all assigned by different music composers. No dancer performed her second item before the performance of the first item by any other dancers. The dancers performed their second items in the same sequence of their performance of their first items.
The following additional facts are known.i) No composer who assigned item to Princess, assigned any item to Queen.
ii) No composer who assigned item to Rani, assigned any item to Samragni.
iii) The first performance was by Princess; this item was assigned by Badal.
iv) The last performance was by Rani; this item was assigned by Gagan.
v) The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.
21.
CAT 2019 LRDI Slot 1
| LR - Selection & Distribution
Four music composers assigned one item; among items 1 to 4; to each of the four finalist. These four composers assigned one item; among items 5 to 8; to different finalists.
From (v), as items assigned by Ashman were performed consecutively, these must be the fourth and the fifth performances.
From (iii), (iv) and (v), the first, the fourth, the fifth and the eighth performances were composed by Badal, Ashman, Ashman, Gagan respectively. Also as the number of performances between items assigned by each of the remaining composers was the same, Badal and Gagan must have composed 6th and 3rd items. Therefore, the second and the seventh items were composed by Dyu. As Ashman and Gagan composed items for Rani, using (ii), we can concluded that Gagan’s other composition was performed by Queen. Also, using (i), Badal’s other composition was performed by Samragni.
Thus, we have
It can be seen that the second performance was composed by Dyu.
Answer the following question based on the information given below.
Five vendors are being considered for a service. The evaluation committee evaluated each vendor on six aspects – Cost, Customer Service, Features, Quality, Reach, and Reliability. Each of these evaluations are on a scale of 0 (worst) to 100 (perfect). The evaluation scores on these aspects are shown in the radar chart. For example, Vendor 1 obtains a score of 52 on Reliability, Vendor 2 obtains a score of 45 on Features and Vendor 3 obtains a score of 90 on Cost.
25.
CAT 2019 LRDI Slot 1
| DI - Tables & Graphs
On which aspect is the median score of the five vendors the least?
Following table lists down the top two vendors (highlighted in blue) in each aspect.
As observed, Vendor 1 is among the top two in 3 aspects (Reach, Quality and Customer Service). Vendor 5 is also among top two in 3 aspects (Reliability, Reach and Features).
All other vendors are among top two in 2 aspects only.
Answer the following question based on the information given below.
The figure below shows the street map for a certain region with the street intersections marked from a through l. A person standing at an intersection can see along straight lines to other intersections that are in her line of sight and all other people standing at these intersections. For example, a person standing at intersection g can see all people standing at intersections b, c, e, f, h, and k. In particular, the person standing at intersection g can see the person standing at intersection e irrespective of whether there is a person standing at intersection f.
Six people U, V, W, X, Y, and Z, are standing at different intersections. No two people are standing at the same intersection.
The following additional facts are known.
X, U, and Z are standing at the three corners of a triangle formed by three street segments.
X can see only U and Z.
Y can see only U and W.
U sees V standing in the next intersection behind Z.
W cannot see V or Z. 6. No one among the six is standing at intersection d.
From (1), X, U, Z are at b,c, g or at b, f, g in some order. Thus, X, U or Z is definitely at g.
Let X is at g.
Case (i): U and Z at b and f.
From (4), U has to be at b, Z at f and V at j. From (2), no one is at c, e, k and h. As Y sees both U and W, Y must be at a and W at i. But then W sees V, which contradicts (5).
Thus, this case is not valid.
Case (ii): U and Z at b and c.
From (4), U has to be at c, Z at b and V at a. From (2), no one is at c, e,f, k and h. But then Y must be at I, j or l. But in that case Y cannot see U. Thus, this case is not valid.
Therefore, X cannot be at g.
Let Z is at g.
From (4), U is at c or f.
Case (i) U is at c and hence V is at k and X is at a. Again there is no place for V. This case is invalid.
Case (ii) U is at f, X is at b and V at h. V will be at e as he sees U. But then he will be able to see Z also. So this case is also invalid.
Thus, U is at g. Therefore, Z is at f, V at e and X at b. So, from (2) no one will be at a, c and j. From (3) and (5), it can be concluded that Y is at k and W at l. Thus, we have
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