CAT 2019 DILR Question Paper Slot 1 with Solutions PDF

Instructions

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.

Question 35

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

Video Solution

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

Question 36

Which of the following items is not a type of biscuit?

Video Solution

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:

G is a candy. Hence D is the answer.

Question 37

Which of the following can represent the numbers of the empty shelves in a possible arrangement?

Video Solution

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:

From the table(case 2), only 1,2,6 and 12 are empty in the same arrangement. Hence, C is the answer.

Question 38

Which of the following statements is necessarily true?

Video Solution

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:

Option A and C are wrong as candies can come before biscuits and vice versa. B is not necessarily true as there can be one empty shelf too as shown in the table. Option D is true as there are at least 4 shelves between B and C. Hence D is the answer.

Instructions

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.
1.Tanzi, Umeza and Yonita had the same total score.
2.Total scores for all players, except one, were in multiples of three.
3.The highest total score was one more than double of the lowest total score.
4.The number of players hitting bull’s eye in Round 2 was double of that in Round 3.
5.Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Question 39

What was the highest total score?

Video Solution

Solution

It is given that 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, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on.

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a, (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible. (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer.

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25)

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

The highest score is 25.

Question 40

What was Zeneca's total score?

Video Solution

Solution

It is given that 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, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on.

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a, (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible. (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer.

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25)

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

Zeneca total score is 24.

Question 41

Which of the following statements is true?

Video Solution

Solution

It is given that 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, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on.

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a, (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible. (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer.

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25)

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

Xyla was the highest scorer.

Question 42

What was Tanzi's score in Round 3?

Video Solution

Solution

It is given that 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, which means Tanzi scored Bull's eye only once in the first 3 rounds because she participated only once in round 4 to 6. Similarly, Umeza scored Bull's eye exactly 2 times in the first 3 rounds. Wangdu did not score Bull's eye in the first three rounds and so on.

Now from 1, Tanzi, Umeza and Yonita had the same total score.

So, Total score of Tanzi will be 4+5+5+a=14+a, (She scored Bull's eye(a score of 5) in exactly one round and a is the unknown score)

Total score of Umeza = 1+2+5+5+b = 13+b (She scored Bull's eye(a score of 5) in exactly 2 rounds and b is the unknown score)

Total score of Yonita = 3+5+5+c=13+c (She scored Bull's eye(a score of 5) in exactly one round and c is the unknown score)

Now 14+a=13+b=13+c,

Also it is given that total scores for all players, except one, were in multiples of three, so these three will have to be a multiple of 3.

So, (a,b,c) can be either (1,2,2) or (4,5,5) in the same order. But the value (5,5) for b and c is not possible. (Umeza scored Bull's eye in exactly 2 rounds and Yonita in exactly 1 round)

Hence, a=1,b=2 and c=2. So each of Tanzi, Umeza and Yonita had total score of 15.

Tabulating the data, we have

From 5, Tanzi and Zeneca had the same score in Round 1 but different scores in Round 3.

Zeneca score Bull's eye 2 times in round 1 to 3. If Tanzi scored 1 in round 1, then Zeneca also has to score 1 in round 1, which means both Tanzi and Zeneca scores in round 3 will be 5, which violates 5. Hence Tanzi scored 5 in round 1 and Zeneca also scored the same in round 1.So the new table is:

From 4, the number of players hitting bull’s eye in Round 2 was double of that in Round 3.

So, in round 3 either 1 or 2 Bull's eye can be scored and in round 2, 2 or 4 Bull's eye can be scored.

Case 1: If only 1 Bull's eye is scored in the round 3, then in round 3 Umeza will score 2 and Zeneca will score 2/3/4 in round 3, which means both will score 5 in round 2. So minimum Bull's eye in round 2 will be 3. (Umeza, Zeneca and Xyla)

Hence this case is rejected.

Case 2: 2 Bull's eye were scored in round 3 and 4 Bull's eye were scored in round 2. So in round 2 Umeza, Yonita and Zeneca scored 5. This can be tabulated as:

In round 3, 2 Bull's eye can only be scored by Xyla and Umeza.

The highest scorer can be either Xyla or Zeneca. The lowest scorer will be Wangdu.

1.Consider Zeneca is the highest scorer.

From 3, the highest total score was one more than double of the lowest total score. So the only possible score for Zeneca is 23 and that for Wangdu is 11. (11*2+1=23)

But this will violate condition 2, since both Zeneca and Wangdu do not have their scores as multiples of three in this case.

Hence, Xyla will be the highest scorer. The only possible total score for Xyla will be 25, and that for Wangdu is 12(4+4+4). (12*2+1=25)

Since Xyla already has non-multiple of 3 as total score. Zeneca will have 24 as the total score. The complete table is:

Tanzi scored 1 in round 3.

Instructions

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.

Question 43

Which digit does the letter A represent?

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Video Solution

Solution

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1 => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

The letter A represents 1.

Question 44

Which digit does the letter B represent?

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Video Solution

Solution

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1 => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

The letter B represents 9.

Question 45

Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?

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Video Solution

Solution

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1 => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

In all possible cases 7 is already represented by a letter other than D. Hence 7 is the answer.

Question 46

Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?

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Video Solution

Solution

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9

In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.

G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.

Now, G,K can take values (3,8), (4,7) and (5,6) in any order.

From 5th column G=J+1 => J=G-1

Case: G=3 and K=8, here J =2 which is not possible as C =2

Case: G=8 and K=3, J=7, a possible case.

Case: G=4 and K=7, J=3 possible

Case: G=7 and K=4, J=6 possible

Case: G=5 and K=6, J=4 not possible as H =5.

Case: G=6 and K=5, J=5 both J and K are same, not possible.

Hence the cases can be tabulated as follows:

From the table it is clear that 6 cannot be represented by G.

Instructions

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.

Question 47

On which aspect is the median score of the five vendors the least?

Video Solution

Solution

The data can be tabulated as follows(approximately):

Customer Services: 28,41,50,55,70 (The median is 50)

Cost: 50,71,77,81,90 (The median is 77)

Reliability: 26, 40, 52, 60, 75 (The median is 52)

Quality: 40, 48, 62, 69, 72 (The median is 62)

Features: 40, 45, 56, 75, 90 (The median is 56)

Reach: 46, 58, 63, 70, 80 (The median is 63)

Hence the customer services has the lowest median.

Question 48

A vendor's final score is the average of their scores on all six aspects. Which vendor has the highest final score?

Video Solution

Solution

The data can be tabulated as follows(approximately):

The average of the vendor will be highest which has highest total score. Hence vendor 3 has the highest average.

Question 49

List of all the vendors who are among the top two scorers on the maximum number of aspects is:

Video Solution

Solution

The data can be tabulated as follows(approximately):

Top 3 on Reliability: Vendor 3, Vendor 5

Top 3 on Reach: Vendor 1, Vendor 5

Top 3 on Quality: Vendor 1, Vendor 2

Top 3 on Features: Vendor 4, Vendor 5

Top 3 on Customer Services: Vendor 4, Vendor 1

Top 3 on Cost: Vendor 3, Vendor 2

Vendor 1: 3 times Vendor 2: Only once Vendor 3: 2 times Vendor 4: 2 times Vendor 5: 3 times

Here 1 and 5 comes 3 times. Hence B is the answer.

Question 50

List of all the vendors who are among the top three vendors on all six aspects is:

Video Solution

Solution

The data can be tabulated as follows(approximately):

Top 3 on Reliability: Vendor 3, Vendor 5, Vendor 1

Top 3 on Reach: Vendor 1, Vendor 5, Vendor 3

Top 3 on Quality: Vendor 1, Vendor 2, Vendor 3

Top 3 on Features: Vendor 4, Vendor 5, Vendor 3

Top 3 on Customer Services: Vendor 4, Vendor 1, Vendor 3

Top 3 on Cost: Vendor 3, Vendor 2, Vendor 1

Only Vendor 3 ranks among top 3 in all the six parameters.

Instructions

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 *.

Question 51

What is the rank of Kerala in the ‘IPC crimes’ category?

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Video Solution

Solution

The data can be tabulated as follows(approximately):

Rank of Delhi in IPC crimes category = 1, The rank of Karnataka and Maharashtra is 3(from table), then the rank of Goa can only be 2.

The rank of Telangana is 6 which has less |IPC crimes than Kerala, which means the rank of Kerala can be less than or equal to 5.

Now, there are two states with 3 ranks, so there will be no rank 4, there can only be rank 5 which is Kerala.

Question 52

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

Video Solution

Solution

The data can be tabulated as follows(approximately):

The highest cases are registered in West Bengal and Delhi.

The total number of IPC crimes = 63-64

The total number of SLL crimes = 520+35-36 = 555-556

Hence the ratio = (63-64)/(555-556) = 0.11 (Approximately) = 1:9

Question 53

Which of the following is DEFINITELY true about the ranks of states/UT in the ‘other crimes’ category?
i) Tamil Nadu: 2
ii) Puducherry: 3

Video Solution

Solution

The data can be tabulated as follows(approximately):

From the table, the rank of Tamilnadu in other crimes is 2. The states which are not in the table will have crimes less than Telangana(i.e 24-25)

From the table the rank of Pudducherry in other crimes is 3.

Question 54

What is the sum of the ranks of Delhi in the three categories of crimes?

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Video Solution

Solution

The data can be tabulated as follows(approximately):

The rank of Delhi in IPC crimes should be 1 because the states which are not in table cannot crime more than that of Telangana which is 24-25.

Similarly Delhi Rank in Other crimes will be 1.

Now in SLL crimes clearly West Bengal has rank 1. It is given that Karnataka has rank 2. The rank 3 can go to either Goa, Delhi and Maharashtra but Goa and Maharashtra already have rank 4. So Delhi will have rank 3. Also no state outside of the table can be ranked 3 in SLL crimes as maximum number of crime should be less than that of Telangana(24-25). Here the number of SLL crimes is 35-36.

Hence the sum of the ranks = 1+3+1=5

Instructions

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.
1. X, U, and Z are standing at the three corners of a triangle formed by three street segments.
2. X can see only U and Z.
3. Y can see only U and W.
4. U sees V standing in the next intersection behind Z.
5. W cannot see V or Z.
6. No one among the six is standing at intersection d.

Question 55

Who is standing at intersection a?

Video Solution

Solution

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

No one is standing at the intersection A. Hence C is the answer.

Question 56

Who can V see?

Video Solution

Solution

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

V can see U and Z only. Hence C is the answer.

Question 57

What is the minimum number of street segments that X must cross to reach Y?

Video Solution

Solution

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

To reach Y, X has to go from b to g and g to k, i.e. 2 streets.

Question 58

Should a new person stand at intersection d, who among the six would she see?

Video Solution

Solution

From 1, X, U, and Z are standing at the three corners of a triangle formed by three street segments.

From 2, X can see only U and Z.

From 4, U sees V standing in the next intersection behind Z. Also, no one among the six is standing at intersection d.

Only cases possible are:

1.

W cannot see V or Z. So W can only be at the intersection a. Since Y can see only U and W, Y can only be at c where X can see him. Hence this case is rejected.

2.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

3.

Y can only see U and W. Y cannot be placed anywhere. Hence this case is also rejected.

4.

W cannot see V or Z. W can only be placed at i. Y can see only U and W. Y can only be placed at j or e, where he can see more people than U and W. Hence this case is also rejected.

5.

W cannot see V or Z. Y can only see U and W. Hence W and Y can only be placed as shown:

If a new person stands at d(left down corner), they can see W and X only. Hence A is the answer.

Instructions

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.

Question 59

Which of the following is true?

Video Solution

Solution

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The second performance was composed by Dyu. Hence A is the answer.

Question 60

Which of the following is FALSE?

Video Solution

Solution

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

Option A: Samragni did not perform in any item composed by Ashman. This statement is true.

Option B: Princess did not perform in any item composed by Dyu. This is also true.

Option C: Rani did not perform in any item composed by Badal. This statement is true.

Option D: Queen did not perform in any item composed by Gagan. This statement is false.

Hence D is the answer.

Question 61

The sixth performance was composed by:

Video Solution

Solution

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The sixth performance was composed by Badal. Hence C is the answer.

Question 62

Which pair of performances were composed by the same composer?

Video Solution

Solution

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The first and the sixth items were composed by Badal. Hence D is the answer.

Instructions

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.

Question 63

What is the minimum possible number of different types of prizes?

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Video Solution

Solution

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the item b should be at least twice. The minimum number of items will be obtained when a=1 and b=99, which means there are only two different types of items.

Question 64

What is the maximum possible number of different types of prizes?

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Video Solution

Solution

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the number of items of type b should be at least twice of that of a and the number of items of type c should be at least twice of that of b and so on. So the maximum number of different types of items of a, b and c will be obtained when a=1, b=2, c=4, d=8, e=16, f=69. Hence the maximum number of different types of items will be 6.

If the number of items is 7, then the minimum number of prizes should be 1+2+4+8+16+32+64=127 which is more than 100.

Hence 6 is the answer.

Question 65

Which of the following is not possible?

Video Solution

Solution

Option A: There are exactly 75 items of type e.

a=1,b=2,c=4,d=8, e=85. Here the maximum value of e= 85. Hence it can take the value 75.

An example of such case is a=1,b=2,c=4,d=18, e=75

Option B: There are exactly 30 items of type b.

a=1 b=30 and c=69. Hence this case is also possible.

Option C: There are exactly 45 items of type c.

Since the value of d should be at least 90, it means that d is not present because 45+90 will be more than 100(maximum number of items). Only a,b and c are present.

The maximum value of b = 22 and a =1, but 45+22+1=68, which is less than 100. So this case is not possible.

Option D: There are exactly 60 items of type d.

d=60, c=30, b=9 and a=1. a+b+c+d=100. Hence this case is possible.

C is the answer.

Question 66

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?

Video Solution

Solution

The total number of items from 1 to 100, which are of same type as in box 45 = 31+1+43=75

Now to maximize the number of items, a=1, b=2, c=4, d=18 and e=75(given)

There can be maximum 5 types of items.

If we consider number of items to be 6, then minimum number of items of 5th type will be 16, 1+2+4+8+16+75=106 which is more than 100.

CAT Content

CAT 2019 Question Paper DILR Slot 1

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

Which of the following items is not a type of biscuit?

Which of the following can represent the numbers of the empty shelves in a possible arrangement?

Which of the following statements is necessarily true?

What was the highest total score?

What was Zeneca's total score?

Which of the following statements is true?

What was Tanzi's score in Round 3?

Which digit does the letter A represent?

Which digit does the letter B represent?

Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?

Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?

On which aspect is the median score of the five vendors the least?

A vendor's final score is the average of their scores on all six aspects. Which vendor has the highest final score?

List of all the vendors who are among the top two scorers on the maximum number of aspects is:

List of all the vendors who are among the top three vendors on all six aspects is:

What is the rank of Kerala in the ‘IPC crimes’ category?

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

Which of the following is DEFINITELY true about the ranks of states/UT in the ‘other crimes’ category?
i) Tamil Nadu: 2
ii) Puducherry: 3

What is the sum of the ranks of Delhi in the three categories of crimes?

Who is standing at intersection a?

Who can V see?

What is the minimum number of street segments that X must cross to reach Y?

Should a new person stand at intersection d, who among the six would she see?

Which of the following is true?

Which of the following is FALSE?

The sixth performance was composed by:

Which pair of performances were composed by the same composer?

What is the minimum possible number of different types of prizes?

What is the maximum possible number of different types of prizes?

Which of the following 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?

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CAT Content

CAT 2019 Question Paper DILR Slot 1

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

Which of the following items is not a type of biscuit?

Which of the following can represent the numbers of the empty shelves in a possible arrangement?

Which of the following statements is necessarily true?

What was the highest total score?

What was Zeneca's total score?

Which of the following statements is true?

What was Tanzi's score in Round 3?

Which digit does the letter A represent?

Which digit does the letter B represent?

Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?

Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?

On which aspect is the median score of the five vendors the least?

A vendor's final score is the average of their scores on all six aspects. Which vendor has the highest final score?

List of all the vendors who are among the top two scorers on the maximum number of aspects is:

List of all the vendors who are among the top three vendors on all six aspects is:

What is the rank of Kerala in the ‘IPC crimes’ category?

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

Which of the following is DEFINITELY true about the ranks of states/UT in the ‘other crimes’ category?i) Tamil Nadu: 2ii) Puducherry: 3

What is the sum of the ranks of Delhi in the three categories of crimes?

Who is standing at intersection a?

Who can V see?

What is the minimum number of street segments that X must cross to reach Y?

Should a new person stand at intersection d, who among the six would she see?

Which of the following is true?

Which of the following is FALSE?

The sixth performance was composed by:

Which pair of performances were composed by the same composer?

What is the minimum possible number of different types of prizes?

What is the maximum possible number of different types of prizes?

Which of the following 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?

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Which of the following is DEFINITELY true about the ranks of states/UT in the ‘other crimes’ category?
i) Tamil Nadu: 2
ii) Puducherry: 3

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?