Test: CAT Logical Reasoning & Data Interpretation- 3 - CAT MCQ

To win the world championship, a player must win all the five matches in that tournament.
In the four preceding tournaments there are 128 players.
So in an individual tournament.
64 will win 0 match.
32 will win 1 match.
16 will win 2 matches.
8 will win 3 matches.
4 will win 4 matches
2 will win 5 matches.
1 (finalist) will win 6 matches.
1 (champion) will win 7 matches.
There are 31 places which are already confirmed and 97 players are for the 32nd position.
We must maximize the number of matches won by the other 31 players
⇒ remaining 97 players must win least number of matches.
In the first tounament
Thus for 97 players, 64 will win 0 matches 32 will win 1 match and the remaining 1 player will win 2 matches. 
n the first tournament total number of matches won by these 97 players= 64x0+32x1+1x2= 34 wins
In all the 4 tournaments total number of wins will be 4x34=136
There are 97 players, so they will distribute victories among themselves. As 136/97=1.402 some players can have 2 wins and others will have 1 win. 
Suppose X players win 2 matches and Y win only 1 match.
So X+Y=97
2X+Y=136
X= 39
Y=58
So 39 players have 2 wins, and 58 players will have 1 win.
Among these players one will be selected for the world championship based on certain criteria.
Hence minimum number of wins required = 5 + 2 = 7
 

To get the maximum number, we need to take the case where 33 players won maximum number of matches, of which exactly 32 were selected for the World Championship based on certain criteria.
Consider these 33 players. Say each of them won at least n matches each.
In every tournament, 64 players win at least one match. Suppose the same set of 64 win the first round of each tournament. Our set of 33 goes on to win more than one match on average in the 4 tournaments.
Hence, 64 - 33 = 31 players win exactly one match.
Hence, players 1-33 win >= n matches, 34 - 64 win exactly one match and 65-128 win no matches.
Total number of wins in 4 tournaments = 4 * 127 = 508
Wins accounted for by players 34-64 = 31 * 4 = 124
Wins remaining = 508 - 124 = 384. These 384 wins need to be distributed over the remaining 33 players in the most equal way possible i.e difference in wins of player 1 and player 33 is the minimum possible.
The largest multiple of 33 <=384 is 33*11 i.e. 363. Suppose the first 33 players have 11 wins each. This accounts for 33*11 = 363 wins.
Hence, number of wins left = 384 - 363 = 21. Let these 21 wins go to the first 21 players.
Hence, players 1-21 win 12 matches, 22 - 33 win 11 matches, 34-64 win 1 match and 65 -128 win 0 matches.
Thus, the maximum number of wins a player can have and still not be selected is 11 wins.

To minimize the number of matches won by that player, we must maximize the number of matches won by the other 31 players => remaining 97 players must win least number of matches
In every round, 64 players win at least one match and 64 players win 0 matches.
Let the person who entered the world championship with least number of wins be X.
In the first tournament, of the 64 members who win at least one match, 32 players win exactly 1 match, 31 players win more than won match and X wins 2 matches.
From the second tournament to the fourth tournament, different 32 players win exactly 1 match and X won 0 matches.
From this we can say that after all the four tournaments, 31 players won maximum number of matches, and a few others, along with X, won exactly 2 matches.
Of these people who won exactly 2 matches, X was selected for the world championship based on certain criteria.

Maximum number of wins is possible if these four players are the semifinalists in all the four tournaments and each of them won exactly one tournament.
Wins in each tournament = 5 + 5 + 6 + 7 = 23
Wins in all four tournaments = 23 * 4 = 92
So, none of these four players can win the World Championship.
⇒ All four are quarterfinalists, 3 are semifinalists and 1 is a finalist
⇒ Matches won in World
Championship = 2 + 3 + 3 + 4 = 12
Total wins = 92 + 12 = 104

Let’s solve this set by converting the times into 24-hour format.
If they start at 8, 9, 10 hours, they will take 18 hours. If they start at 19, 20, 21 hours, they will take 20 hours. Else, they take 15 hours.
 They start at 20:00 hours from Paris, they’ll reach Luxembourg at 16:00 hours.
Let’s continuously represent their journeys:
20:00 (P) - 16:00 (L) = 20 hours
17:00 (L) - 08:00 (P) = 15 hours
9:00 (P) - 3:00 (L) = 18 hours
4:00 (L) - 19:00 (P) = 15 hours
20:00 (P) -
Now, this cycle will continue.
In 80 trips, there will be 20 such cycles. 40 trips from Paris to Luxembourg.
2 trips each of 20 and 18 hours from Paris to Luxembourg.
Thus, average time from Paris to Luxembourg = 20*(20 + 18)/40​ = 19 hours.

Let’s solve this set by converting the times into 24-hour format.
If they start at 8, 9, 10 hours, they will take 18 hours. If they start at 19, 20, 21 hours, they will take 20 hours. Else, they take 15 hours.
They start at 6:00 hours from Luxembourg. They’ll reach Paris at 21:00 hours.
Let’s represent their trips -
6:00 (P) - 21:00 (L) = 15 hours
22:00 (L) - 13:00 (P) = 15 hours
14:00 (P) - 5:00 (L) = 15 hours
6:00 (L) - 21:00(P) = 15 hours
22:00 (P) - 13:00 (L) = 15 hours
14:00 (L) - 5:00 (P) = 15 hours
6:00(P) -
Now, this cycle will repeat. Since all the trips take 15 hours, the average trip time will also be 15 hours.

Let’s solve this set by converting the times into 24-hour format.
If they start at 8, 9, 10 hours, they will take 18 hours. If they start at 19, 20, 21 hours, they will take 20 hours. Else, they take 15 hours.
They start at 4:00 hours from Paris. They’ll reach at 19:00 hours in Luxembourg.
Let’s represent the bus journey:
4:00 (P) - 19:00 (L)
20:00 (L) - 16:00 (P)
17:00(P) - 8:00 (L)
9:00(L) - 3:00 (P)
4:00(P) - 19:00 (L)
This cycle will repeat.
They’ll never start at 8:00 AM from Luxembourg.

Let’s solve this set by converting the times into 24-hour format.
If they start at 8, 9, 10 hours, they will take 18 hours. If they start at 19, 20, 21 hours, they will take 20 hours. Else, they take 15 hours.
Let’s compute their trip times.
14:00 (L) - 5: 00 (P)
6: 00 (P) - 21: 00 (L)
22:00 (L) - 13:00 (P)
14:00 (P) - 5:00 (L)
6:00 (L) - 21:00 (P)
22:00(P) - 13:00 (L)
This cycle will repeat itself.
In each cycle they will start twice between 1:00 PM and 11:00 PM from Paris.
The cycle will repeat 16 times. Thus, 32 times they will start between the given times.
Further, in the next 4 trips, they’ll start once. Thus, they’ll start 33 times.

Let us start by tabulating the data available.
We have no information about the number of persons in each club. 
The sports-sports cell in the table represents the number of persons from sports club who stayed in sports club. Since we do not have this information (we have information only regarding the movement from one club to another), let us mark all such cells with X. 
The cell sports-dramatics (row-column) represents the number of students from sports club who left for dramatics club. Therefore, the cell sports-total will provide the number of students who left the sports club and the cell total-sports will provide the number of students who left other clubs for the sports club. 
The number of persons who moved from dramatics club to the sports club is the same as the number of persons who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well. Let us use ‘a’ to denote the number of persons who moved from the dramatics club to the sports club and ‘b’ to denote the number of persons who moved from sports club to literary club. The number of persons who moved out of sports and quiz clubs is the same. Let us denote it by ‘c’. 
 A total of 21 students left one club for another. No person moved from the quiz club to sports club (quiz-sports = 0).

The number of persons who moved to the literary club is one more than the number of persons who moved to the sports club. Let the number of persons who moved to the literary club be ‘d+1’ and the number of persons who moved to the sports club be ‘d’.
The number of persons who left the quiz club for the dramatics club and the literary club is the same. Let us denote the number of persons who left the quiz club for dramatics club by ‘e’.  4 students moved out of dramatics club and 5 students moved out of literary club.

As we can see from the table, c+c+4+5 =21
⇒ c = 6 
The number of students who left other clubs for dramatics club is 4 more than the number of students who left other clubs for the quiz club.
Let the number of students who left other clubs for quiz club be ‘f’.
⇒ Number of students who left other clubs for dramatics club = f + 4.
The number of students who joined sports club is exactly half the number of students who left it. We know that 6 students left the sports club. Therefore, 3 students should have joined the sports club.  ⇒ d = 3
We can see from the table that e + e = 6 
⇒ e = 3 
Let us fill the vacant cells with variables from g to k. We get the following table. 

f + f + 4 + 3 + 4 = 21
⇒ 2f = 10
f = 5
a+b = 3
a+b+i = 6
⇒ i = 3
a+h+j = 4 --------------(1)
b+h = 1 ----------------(2)
a+b = 3 ----------------(3)
b+g+k = 5 -------------(4)
j + k = 2 --------------(5)
a+g = 6 ---------------(6)
Let us rewrite every variable in terms of 'a'.
b = 3-a
g =6-a
Substituting these values in (4), we get,
3-a+6-a+k = 5
9-2a+k = 5
k = 2a-4
Substituting the value of 'k' in (5), we get,
j+2a-4 = 2
j = 6-2a
(1)⇒ a+h+j = 4
a + h + 6 - 2a = 4
⇒ h = a - 2
It has been given that at least one student moved from Sports to Literary club. Therefore, the value of 'a' cannot be 3.
We know that k=2a-4. Therefore, the value of 'a' should be at least 2. 
2 is the only value that falls within the range.
Solving the equations using a=2, we get the following table:

Maximum number of people left literary club for dramatics club. Therefore, option A is the right answer.

Let us start by tabulating the data available.
We have no information about the number of persons in each club. 
The sports-sports cell in the table represents the number of persons from sports club who stayed in sports club. Since we do not have this information (we have information only regarding the movement from one club to another), let us mark all such cells with X. 
The cell sports-dramatics (row-column) represents the number of students from sports club who left for dramatics club. Therefore, the cell sports-total will provide the number of students who left the sports club and the cell total-sports will provide the number of students who left other clubs for the sports club. 
The number of persons who moved from dramatics club to the sports club is the same as the number of persons who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well. Let us use ‘a’ to denote the number of persons who moved from the dramatics club to the sports club and ‘b’ to denote the number of persons who moved from sports club to literary club. The number of persons who moved out of sports and quiz clubs is the same. Let us denote it by ‘c’. 
 A total of 21 students left one club for another. No person moved from the quiz club to sports club (quiz-sports = 0).

The number of persons who moved to the literary club is one more than the number of persons who moved to the sports club. Let the number of persons who moved to the literary club be ‘d+1’ and the number of persons who moved to the sports club be ‘d’.
The number of persons who left the quiz club for the dramatics club and the literary club is the same. Let us denote the number of persons who left the quiz club for dramatics club by ‘e’.  4 students moved out of dramatics club and 5 students moved out of literary club.

As we can see from the table, c+c+4+5 =21
⇒ c = 6 
The number of students who left other clubs for dramatics club is 4 more than the number of students who left other clubs for the quiz club.
Let the number of students who left other clubs for quiz club be ‘f’.
⇒ Number of students who left other clubs for dramatics club = f + 4.
The number of students who joined sports club is exactly half the number of students who left it. We know that 6 students left the sports club. Therefore, 3 students should have joined the sports club.  ⇒ d = 3
We can see from the table that e + e = 6 
⇒ e = 3 
Let us fill the vacant cells with variables from g to k. We get the following table. 

f + f + 4 + 3 + 4 = 21
⇒ 2f = 10
f = 5
a+b = 3
a+b+i = 6
⇒ i = 3
a+h+j = 4 --------------(1)
b+h = 1 ----------------(2)
a+b = 3 ----------------(3)
b+g+k = 5 -------------(4)
j + k = 2 --------------(5)
a+g = 6 ---------------(6)
Let us rewrite every variable in terms of 'a'.
b = 3-a
g =6-a
Substituting these values in (4), we get,
3-a+6-a+k = 5
9-2a+k = 5
k = 2a-4
Substituting the value of 'k' in (5), we get,
j+2a-4 = 2
j = 6-2a
(1)⇒ a+h+j = 4
a + h + 6 - 2a = 4
⇒ h = a - 2
It has been given that at least one student moved from Sports to Literary club. Therefore, the value of 'a' cannot be 3.
We know that k=2a-4. Therefore, the value of 'a' should be at least 2. 
2 is the only value that falls within the range.
Solving the equations using a=2, we get the following table:

3 persons have moved from sports club to quiz club. Therefore, option D is the right answer. 

Let us start by tabulating the data available.
We have no information about the number of persons in each club. 
The sports-sports cell in the table represents the number of persons from sports club who stayed in sports club. Since we do not have this information (we have information only regarding the movement from one club to another), let us mark all such cells with X. 
The cell sports-dramatics (row-column) represents the number of students from sports club who left for dramatics club. Therefore, the cell sports-total will provide the number of students who left the sports club and the cell total-sports will provide the number of students who left other clubs for the sports club. 
The number of persons who moved from dramatics club to the sports club is the same as the number of persons who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well. Let us use ‘a’ to denote the number of persons who moved from the dramatics club to the sports club and ‘b’ to denote the number of persons who moved from sports club to literary club. The number of persons who moved out of sports and quiz clubs is the same. Let us denote it by ‘c’. 
 A total of 21 students left one club for another. No person moved from the quiz club to sports club (quiz-sports = 0).

The number of persons who moved to the literary club is one more than the number of persons who moved to the sports club. Let the number of persons who moved to the literary club be ‘d+1’ and the number of persons who moved to the sports club be ‘d’.
The number of persons who left the quiz club for the dramatics club and the literary club is the same. Let us denote the number of persons who left the quiz club for dramatics club by ‘e’.  4 students moved out of dramatics club and 5 students moved out of literary club.

As we can see from the table, c+c+4+5 =21
⇒ c = 6 
The number of students who left other clubs for dramatics club is 4 more than the number of students who left other clubs for the quiz club.
Let the number of students who left other clubs for quiz club be ‘f’.
⇒ Number of students who left other clubs for dramatics club = f + 4.
The number of students who joined sports club is exactly half the number of students who left it. We know that 6 students left the sports club. Therefore, 3 students should have joined the sports club.  ⇒ d = 3
We can see from the table that e + e = 6 
⇒ e = 3 
Let us fill the vacant cells with variables from g to k. We get the following table. 

f + f + 4 + 3 + 4 = 21
⇒ 2f = 10
f = 5
a+b = 3
a+b+i = 6
⇒ i = 3
a+h+j = 4 --------------(1)
b+h = 1 ----------------(2)
a+b = 3 ----------------(3)
b+g+k = 5 -------------(4)
j + k = 2 --------------(5)
a+g = 6 ---------------(6)
Let us rewrite every variable in terms of 'a'.
b = 3-a
g =6-a
Substituting these values in (4), we get,
3-a+6-a+k = 5
9-2a+k = 5
k = 2a-4
Substituting the value of 'k' in (5), we get,
j+2a-4 = 2
j = 6-2a
(1)⇒ a+h+j = 4
a + h + 6 - 2a = 4
⇒ h = a - 2
It has been given that at least one student moved from Sports to Literary club. Therefore, the value of 'a' cannot be 3.
We know that k=2a-4. Therefore, the value of 'a' should be at least 2. 
2 is the only value that falls within the range.
Solving the equations using a=2, we get the following table:

Number of persons who joined literary club = 4
Number of persons who left literary club = 5
Therefore, the required difference is 5-4 = 1.

Let us start by tabulating the data available.
We have no information about the number of persons in each club. 
The sports-sports cell in the table represents the number of persons from sports club who stayed in sports club. Since we do not have this information (we have information only regarding the movement from one club to another), let us mark all such cells with X. 
The cell sports-dramatics (row-column) represents the number of students from sports club who left for dramatics club. Therefore, the cell sports-total will provide the number of students who left the sports club and the cell total-sports will provide the number of students who left other clubs for the sports club. 
The number of persons who moved from dramatics club to the sports club is the same as the number of persons who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well. Let us use ‘a’ to denote the number of persons who moved from the dramatics club to the sports club and ‘b’ to denote the number of persons who moved from sports club to literary club. The number of persons who moved out of sports and quiz clubs is the same. Let us denote it by ‘c’. 
 A total of 21 students left one club for another. No person moved from the quiz club to sports club (quiz-sports = 0).

The number of persons who moved to the literary club is one more than the number of persons who moved to the sports club. Let the number of persons who moved to the literary club be ‘d+1’ and the number of persons who moved to the sports club be ‘d’.
The number of persons who left the quiz club for the dramatics club and the literary club is the same. Let us denote the number of persons who left the quiz club for dramatics club by ‘e’.  4 students moved out of dramatics club and 5 students moved out of literary club.

As we can see from the table, c+c+4+5 =21
⇒ c = 6 
The number of students who left other clubs for dramatics club is 4 more than the number of students who left other clubs for the quiz club.
Let the number of students who left other clubs for quiz club be ‘f’.
⇒ Number of students who left other clubs for dramatics club = f + 4.
The number of students who joined sports club is exactly half the number of students who left it. We know that 6 students left the sports club. Therefore, 3 students should have joined the sports club.  ⇒ d = 3
We can see from the table that e + e = 6 
⇒ e = 3 
Let us fill the vacant cells with variables from g to k. We get the following table. 

f + f + 4 + 3 + 4 = 21
⇒ 2f = 10
f = 5
a+b = 3
a+b+i = 6
⇒ i = 3
a+h+j = 4 --------------(1)
b+h = 1 ----------------(2)
a+b = 3 ----------------(3)
b+g+k = 5 -------------(4)
j + k = 2 --------------(5)
a+g = 6 ---------------(6)
Let us rewrite every variable in terms of 'a'.
b = 3-a
g =6-a
Substituting these values in (4), we get,
3-a+6-a+k = 5
9-2a+k = 5
k = 2a-4
Substituting the value of 'k' in (5), we get,
j+2a-4 = 2
j = 6-2a
(1)⇒ a+h+j = 4
a + h + 6 - 2a = 4
⇒ h = a - 2
It has been given that at least one student moved from Sports to Literary club. Therefore, the value of 'a' cannot be 3.
We know that k=2a-4. Therefore, the value of 'a' should be at least 2. 
2 is the only value that falls within the range.
Solving the equations using a=2, we get the following table:

Number of members lost by sports club = 6 - 3 = 3.
Number of members gained by dramatics club = 9 - 4 = 5.
Number of members lost by literary club = 5 - 4 =1
Number of members lost by quiz club = 6 - 5 = 1.
As we can see, sports club lost the highest number of members.
Therefore, option C is the right answer. 

Since AB is a multiple of 11, and both A and B are greater than 4, they must be at least 5. So, the possible values for AB are:
55
66
77
88
99
Now, for each of the given values of AB, let us find out all possible values of C and D. Since any other multiple of AB in each case is a number having more than 2 digits, the only multiple is AB itself:
5555
6666
7777
8888
9999
The same is true for EF,
555555
666666
777777
888888
999999
Now for each of the values of ABCDEF, we can have 4 values of GH, 56, 67, 78, 89.
Hence, total number of possible cases = 4 x 5 = 20.

The festival began on the Monday evening and ended on the Sunday evening. Each day, only one band performed. Each band performed only once. The seven bands are - Cactus, Axis, Enigma, Boom, Fish, Dhoom and Bodhi Tree.
Bodhi Tree didn't precede any performance; so it performed on Sunday. Sid, the lead vocalist of the rock band Cactus, would perform only on Monday. Rupam of a fusion band with Fish performed on Wednesday. Angelina with Enigma refused to perform after Thursday and she didn't perform before Meet. So, Meet of a rock band performed on Tuesday and Angelina performed on Thursday. We can tabulate the results as follows :

Rupam is the only male among the lead vocalists of the fusion band, so the other two are females; hence, they should be Angelina and Bony.
Fusion band was performing on Wednesday and Thursday [Rupam, Angelina], so definitely it should be followed by a rock band [since all the three bands of the same genre were not allowed to perform consecutively]. So, rock band performed on Friday and fusion band performed on Saturday [by Bony]. So, Ali performed on Friday. Boom (a rock band) refused to perform immediately before or after Fish. So, it had to perform on Friday. The final arrangement is as follows:

If Bony was the lead vocalist of Axis, then Meet was the lead vocalist of Dhoom but not Boom.

22666689 - right
33996656 - wrong - EF is not a multiple of CD
99999912 - wrong - 1 is not allowed
22469234 - right
55555556 - right
44848434 - right
22489667 - right
33333332 - wrong - G should be less than H
44888845 - right
|3 - 6| = 3.

The festival began on the Monday evening and ended on the Sunday evening. Each day, only one band performed. Each band performed only once. The seven bands are - Cactus, Axis, Enigma, Boom, Fish, Dhoom and Bodhi Tree.
Bodhi Tree didn't precede any performance; so it performed on Sunday. Sid, the lead vocalist of the rock band Cactus, would perform only on Monday. Rupam of a fusion band with Fish performed on Wednesday. Angelina with Enigma refused to perform after Thursday and she didn't perform before Meet. So, Meet of a rock band performed on Tuesday and Angelina performed on Thursday. We can tabulate the results as follows :

Rupam is the only male among the lead vocalists of the fusion band, so the other two are females; hence, they should be Angelina and Bony.
Fusion band was performing on Wednesday and Thursday [Rupam, Angelina], so definitely it should be followed by a rock band [since all the three bands of the same genre were not allowed to perform consecutively]. So, rock band performed on Friday and fusion band performed on Saturday [by Bony]. So, Ali performed on Friday. Boom (a rock band) refused to perform immediately before or after Fish. So, it had to perform on Friday. The final arrangement is as follows:

Boom performed on Friday and Meet performed on Tuesday.

Since we can use only prime digits, GH can only be 23.
AB can be 22, 33, 55, 77.
When AB is 22, ABCDEFGH can be 22222223
When AB is 33, ABCDEFGH can be 33333323
When AB is 55, ABCDEFGH can be 55555523
When AB is 77, ABCDEFGH can be 77777723
Hence, only 4 codes.

The festival began on the Monday evening and ended on the Sunday evening. Each day, only one band performed. Each band performed only once. The seven bands are - Cactus, Axis, Enigma, Boom, Fish, Dhoom and Bodhi Tree.
Bodhi Tree didn't precede any performance; so it performed on Sunday. Sid, the lead vocalist of the rock band Cactus, would perform only on Monday. Rupam of a fusion band with Fish performed on Wednesday. Angelina with Enigma refused to perform after Thursday and she didn't perform before Meet. So, Meet of a rock band performed on Tuesday and Angelina performed on Thursday. We can tabulate the results as follows :

Rupam is the only male among the lead vocalists of the fusion band, so the other two are females; hence, they should be Angelina and Bony.
Fusion band was performing on Wednesday and Thursday [Rupam, Angelina], so definitely it should be followed by a rock band [since all the three bands of the same genre were not allowed to perform consecutively]. So, rock band performed on Friday and fusion band performed on Saturday [by Bony]. So, Ali performed on Friday. Boom (a rock band) refused to perform immediately before or after Fish. So, it had to perform on Friday. The final arrangement is as follows:

'Cactus, Axis/Dhoom, Fish, Enigma, Boom, Dhoom/Axis' is a possible performance sequence.

AB = 22, Possible values of CD are 22, 24, 26, 28, 42, 44, 46, 48, 62, 64, 66, 68, 82, 84, 86, 88.
AB = 33, Possible values of CD are 33, 36, 39, 63, 66, 69, 93, 96, 99.
AB = 44, Possible values of CD are 44, 48, 84, 88.
AB = 55, Possible values of CD are 55.
AB = 66, Possible values of CD are 66.
AB = 77, Possible values of CD are 77.
AB = 88, Possible values of CD are 88.
AB = 99, Possible values of CD are 99.
For the following values of ABCD, we get these values of EF.
ABCD = 2222, EF = 22, 44, 66, 88
ABCD = 2224, EF = 24, 48, 72, 96
ABCD = 2226, EF = 26, 52, 78
ABCD = 2228, EF = 28, 56, 84
ABCD = 2242, EF = 42, 84
ABCD = 2244, EF = 44, 88
ABCD = 2246, EF = 46, 92
ABCD = 2248, EF = 48, 96
ABCD = 2262, EF = 62
ABCD = 2264, EF = 64
ABCD = 2266, EF = 66
ABCD = 2268, EF = 68
ABCD = 2282, EF = 82
ABCD = 2284, EF = 84
ABCD = 2286, EF = 86
ABCD = 2288, EF = 88
ABCD = 3333, EF = 33, 66, 99
ABCD = 3336, EF = 36, 72
ABCD = 3339, EF = 39, 78
ABCD = 3363, EF = 63
ABCD = 3366, EF = 66
ABCD = 3369, EF = 69
ABCD = 3393, EF = 93
ABCD = 3396, EF = 96
ABCD = 3399, EF = 99
ABCD = 4444, EF = 44, 88
ABCD = 4448, EF = 48, 96
ABCD = 4484, EF = 84
ABCD = 4488, EF = 88
ABCD = 5555, EF = 55
ABCD = 6666, EF = 66
ABCD = 7777, EF = 77
ABCD = 8888, EF = 88
ABCD = 9999, EF = 99
Total cases = 54.
Each one will have 23, 34, 45, 56, 67, 78, 89 i.e 7 cases with it.
54 x 7 = 378

1. Assign Fixed Performances:

• Monday: Cactus (Rock) - Sid
• Wednesday: Fish (Fusion) - Rupam
• Sunday: Bodhi Tree (Home band)

2. Determine Possible Days for Remaining Bands:

• Angelina (Enigma) must perform on or before Thursday.
• Meet must perform before Angelina.
• Boom cannot perform on Tuesday or Thursday (immediately before or after Fish).
• Ali (with Boom) cannot perform on Saturday.

3. Schedule the Bands:

• Tuesday: Meet's band (Rock)
• Thursday: Enigma (Fusion) - Angelina
• Friday: Boom (Rock) - Ali
• Saturday: Dhoom (Fusion) - Bony
  

• Counting Fusion and Rock Band Consecutive Performances:

• Monday (Rock) → Tuesday (Rock): Same genre.
• Tuesday (Rock) → Wednesday (Fusion): 1st Rock-Fusion transition.
• Wednesday (Fusion) → Thursday (Fusion): Same genre.
• Thursday (Fusion) → Friday (Rock): 2nd Fusion-Rock transition.
• Friday (Rock) → Saturday (Fusion): 3rd Rock-Fusion transition.

• Total Transitions between Fusion and Rock bands from Monday to Saturday: 3