All questions of Logical Reasoning (LR) and Data Interpretation (DI): Section-wise Tests for CAT Exam

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.

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

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.

There are 5 ladies whose weights are 50 or above
There ages are 50,50, 70,60 and 80

Average = 310/5 = 62

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. 

Number of students from college Y and Z are same. H is from college Y. E is from college Z. According to this information the following table is obtained.

D won medal in 200 m race event and he is not from college Y. No student from college Z has won medal in 200 m race event. Student from each college has won medal in 100 m race event. C is not from college Z nor he has won medal in 100 m race event. According to this the following table is obtained.

Number of students who won medals in 100 m race event is more than the number of students who won medals in 400 m race event. D and G are not from same college. All the medals of 400 m race event are won by students of same college. A and G won medals in same race event but they are not from same college. C is not from college Z nor he has won medal in 100 m race event. According to this the final table is shown below.

Two students from college Y won 200 m race.

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.

Bike owners in Chennai = 2816 – 707 – 971 – 432 – 342 = 364.

To own at least 3 vehicles, one can own either 3 or 4 vehicles.

I + II + III + IV = 1034

I + 2 II + 3 III + 4 IV = 1987

The difference has to be adjusted among people owning 2, 3 and 4 vehicles. To maximise the sum of people owning 3 and 4 vehicles, we will try to allocate the maximum possible to 3 and the remaining to 4.

1987 – 1034 = 953

2III + 3IV = 953.

III = 475

IV = 1

But, if we observe the values for Chennai, the number of people having a bike is 364 and the number of people having a private jet is 24. Hence, even if we consider that people who own a bike also own a 4-wheeler and a scooter(but not a private jet) and people who own a private jet also own a 4-wheeler and a scooter(but not a bike), we won’t be able to reach the above numbers. We would be able to achieve a maximum value of 24 + 364 = 388.
Let us verify if we can represent the above condition in a 4-set Venn Diagram.

Now, we need to arrange the remaining people who own a 4-wheeler and those who own a scooter.

I + II + III + IV = 1034

IV = 0, because we have already assigned all people who own a Private jet and a bike into III.

III = 388.

I + II = 1034 – 388 = 646

I + 2II + 3III + 4IV = 1987

I + 2II + 3 X 388 = 1987

I + 2II = 823

II = 823 – 646 = 177

I = 469

Hence, we get the following Venn Diagram:

Hence, maximum people who own 3 vehicles = 388.

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

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.

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. 

Number of volunteers involved in both TR and ER but not FR = b
a + b + 4 = 10
a + b = 6
Let number of volunteers involved in both FR and ER but not TR = x
Total = 17 + x + 8 + 8 = 37
x = 4
Consider the Venn diagram shown below:

Let m volunteers be added to the TR project and n be added to each of the FR and ER projects.
Then, 7 + m = 8 + n
m = n + 1
Also, b + 2 = 5
∴ b = 3 and a = 3
Number of volunteers working on TR = 7 + n + 1 + 4 + 5 = 17 + n
Number of volunteers working on FR = 17 + n
Number of volunteers working on ER = 18 + n
Thus, ER has the highest number of volunteers.
Hence, option 4 is the correct choice.

Given,
Richard earned a total of 22.5 points and he was not the first in any level. The only way he can win 22.5 points without being first in any level is if he earned 7.5 in two levels, 5 in one level and 2.5 in another level.
From (2), Connor and Poppy did not win the highest points in the game.
From (6), Sarah also did not win the highest points in the game.
From (4), one of the persons was the first in two levels and this person did not win the highest points. This person must have earned a minimum of 10 + 10 + 2.5 + 2.5 = 25 points in the game. The person who earned the highest points must have earned more than this. From this, we can infer that Richard also cannot be the person who earned the highest points.
Hence, only Mason can be the person who earned the highest points.
From (5), Mason was the first in one level. He is not the third in any level. Also, Richard was the second in two levels (since he earned 7.5 in two levels). Hence, the maximum points that Mason can win = 10 + 7.5 + 7.5 + 2.5 = 27.5 points.
Since Mason has to win more than 25 points, he must have earned 27.5 points. The person who was first in two levels must have earned 25 points (this is the only way for him to have earned less points than Mason).
From (2) and (6), Poppy and Sarah cannot be the person who earned 25 points.
Hence, Connor must be the person who earned 25 points. This is possible if he was the first in two levels and fourth/fifth in two other levels.
From (2), Poppy must have earned 20 points. This is possible only if Poppy was the first in one level, the third in another level and fourth/fifth in two levels (Poppy cannot be the second in any level because Richard and Mason are second in two levels each). The points that Poppy wins in this case = 10 + 5 + 2.5 + 2.5 = 20 points.
Sarah must have been the third in two levels and fourth/fifth in two levels. The total points that Sarah earned = 5 + 5 + 2.5 + 2.5 = 15 points.
The following table presents the positions of the five persons in the four levels and the total points earned by them in the game.

The only persons who were second in any level were Richard and Mason. In the level that Mason was first, Richard must be second. Hence, the statement given in option (1) is definitely true. The other statements need not necessarily be true.

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.
 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.

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word, i.e. either 'Silence' or 'Rosebud'. The word next to Titli's word must be a 6-letter word, which is 'Burden'. It cannot be 'Barely' because it has been given that 'Barely' was the first word to be entered. Also, the word preceding Titli's word must be a 7-letter word.
Thus, both the 7-letter words and 'Burden' must be at consecutive places. They can be at either the second, third and fourth positions or the third, fourth and fifth positions. But, in the second case, 'Baadshah' would be at the second position, which is not possible as given in the question. So, both the 7-letter words and 'Burden' must be at the second, third and fourth positions, respectively, and 'Baadshah' would be at the fifth position. Also, 'Silence' is not at the third position. So, it must be at the second position and 'Rosebud' must be at the third position.

15 across and 15 down must start from the same letter and their length must be unequal. Also, 'Silence' is the answer to an across clue which can be either 15 across or 21 across. However, if 'Silence' is the answer to 15 across, the answer to 15 down must also start with 'S'. But there is no other word starting with 'S'; so 'Silence' must be the answer to the 21 across clue. The fifth one to be worked out was the answer to an across clue. So, 'Baadshah' should be the answer to an across clue.
Elsie's word was longer than Bineet's. So, Elsie's word must be either 'Baadshah' or 'Silence' and Bineet's word should be either 'Barely' or 'Silence'. In both the cases, 'Burden' would be Easwar's word and it would be the answer to the 15 down clue because 'Burden' has been produced by one of the men but it is not Bineet according to instruction 2 and Easwar did not solve 4 down. Sheela was neither the first nor the last to come up with an answer. So, her answer must be 'Silence'. Bineet's word must be 'Barely' and Elsie's word must be 'Baadshah'.

Easwar produced the answer 'Burden'.

It is given that there were 6 attempts in the game and all the five players got different number of dollars. The maximum number of attempts a single player could have rolled the slot machine has to be 2 (if one player rolled the slot machine in 3 attempts, then two players would end up with the same number of dollars).
Jack was the only player to get 2 attempts consecutively and he did not get the 1^st attempt. The number of maximum possible attempts by a player is 2, which means Jack got $10 in the first attempt, and in the second attempt he paid a penalty amount of $5, but from (iv), the player who rolled the slot machine in the last attempt won the game.
Jack couldn't be at 6^th.
Possible positions for Jack are: 2, 3, or 3, 4, or 4, 5, but there is change in sequence (from (ii)) at the beginning of the 4^th attempt and the 5^th attempt. So, Jack couldn't go at the 4^th and 5^th attempt.
Sequence[attempt number(name, amount)]: 1, 2(Jack, $10), 3(Jack, -$5), 4, 5, 6(xx, $10)
This means, the 1^st, 4^th and 5^th attempts didn't hit correctly and they would pay the penalty.
Sequence[attempt number(name, amount)]: 1(xx, -$5), 2(Jack, $10), 3(Jack, -$5), 4(xx, -$5), 5(xx, -$5), 6(xx, $10)
There has to be another person who rolled the slot machine in 2 attempts. This person also could not have rolled the slot machine in the 5^th and 6^th attempts.
Hence, he must have rolled the slot machine in the 1^st and 5^th attempts.
From (ii), this person has to be Luke because only by getting the 5^th attempt could Luke move to front of Jack.
Sequence[attempt number(name, amount)]: 1(Luke, -$5), 2(Jack, $10), 3(Jack, -$5), 4(xx, -$5), 5(Luke, -$5), 6(xx, $10)
Will must have rolled the slot machine in one attempt without hitting the jackpot and Franklin must have rolled the slot machine in one attempt and hit the jackpot.
Sequence[attempt number(name, amount)]: 1(Luke, -$5), 2(Jack, $10), 3(Jack, -$5), 4(Will, -$5), 5(Luke, -$5), 6(Franklin, $10)

From the above table, it is clear that Franklin got the highest amount, i.e. $10, so he stood first in the game.

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.

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.

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.

It is given that there were 6 attempts in the game and all the five players got different number of dollars. The maximum number of attempts a single player could have rolled the slot machine has to be 2 (if one player rolled the slot machine in 3 attempts, then two players would end up with the same number of dollars).
Jack was the only player to get 2 attempts consecutively and he did not get the 1^st attempt. The number of maximum possible attempts by a player is 2, which means Jack got $10 in the first attempt, and in the second attempt he paid a penalty amount of $5, but from (iv), the player who rolled the slot machine in the last attempt won the game.
Jack couldn't be at 6^th.
Possible positions for Jack are: 2, 3, or 3, 4, or 4, 5, but there is change in sequence (from (ii)) at the beginning of the 4^th attempt and the 5^th attempt. So, Jack couldn't go at the 4^th and 5^th attempt.
Sequence[attempt number(name, amount)]: 1, 2(Jack, $10), 3(Jack, -$5), 4, 5, 6(xx, $10)
This means, the 1^st, 4^th and 5^th attempts didn't hit correctly and they would pay the penalty.
Sequence[attempt number(name, amount)]: 1(xx, -$5), 2(Jack, $10), 3(Jack, -$5), 4(xx, -$5), 5(xx, -$5), 6(xx, $10)
There has to be another person who rolled the slot machine in 2 attempts. This person also could not have rolled the slot machine in the 5^th and 6^th attempts.
Hence, he must have rolled the slot machine in the 1^st and 5^th attempts.
From (ii), this person has to be Luke because only by getting the 5^th attempt could Luke move to front of Jack.
Sequence[attempt number(name, amount)]: 1(Luke, -$5), 2(Jack, $10), 3(Jack, -$5), 4(xx, -$5), 5(Luke, -$5), 6(xx, $10)
Will must have rolled the slot machine in one attempt without hitting the jackpot and Franklin must have rolled the slot machine in one attempt and hit the jackpot.
Sequence[attempt number(name, amount)]:
1(Luke, -$5), 2(Jack, $10), 3(Jack, -$5), 4(Will, -$5), 5(Luke, -$5), 6(Franklin, $10)
OR
1(Luke, -$5), 2(Jack, -$5), 3(Jack, $10), 4(Will, -$5), 5(Luke, -$5), 6(Franklin, $10)

Thus, we can see that only 2nd or 3rd attempt and the last attempt hit the jackpot.
Since the question asks for a 'definite' hit, the answer is 6th attempt.

Based on Condition II, we understand that the student who missed the Mathematics examination did not miss any other examination. This indicates that the Maths score is bound to be the average of the best 3 out of the 4 exam scores obtained by this candidate. Based on this inference, we can proceed with identifying the math score that can be represented as an average of the rest of the scores. We can straightaway eliminate Deep and Esha as potential candidates, given that their Mathematics score is greater than the rest of the exam scores. After estimating the average scores for the rest of the candidates, we observe that only Carl has missed his Mathematics examination.
For Carl: best 3 out of 4 - 80(Hindi), 90(Social Science), 100(Science)
Avg. = 270/3 = 90 which matches the given value
 ∴  Carl missed his Mathematics examination.
Further, based on Condition III, we can surmise that the student who missed Hindi and Science should have similar average scores in these two subjects. We notice that Alva has the same score of 75 in both Hindi and Science. The same can be said about Deep, who has a score of 90 in both these subjects. Thus, one out of Alva and Deep missed out on Hindi and Science examination, while the second individual missed out only on the Hindi examination.
Since we know that Carl, Alva and Deep are unlikely to have missed out on the English exam, we can divert our attention to determining which individual out of Bithi, Esha and Foni failed to appear for this subject. However, we notice that Bithi's English score is greater than the rest of her scores, thereby helping us eliminate her as the potential candidate.
For Esha: best 3 out of 4 - 85(Hindi), 95(Mathematics), 60(Science)
Avg. = 240/3 = 80 which matches the given value
∴  Esha most likely missed her English examination.
For Foni: best 3 out of 4 - 78(Mathematics), 83(Social Science), 88(Science)
Avg. = 249/3 = 83 which matches the given value
∴  Foni most likely missed her English examination.
Based on Condition I, we know that exactly two candidates missed the examinations for English, Hindi, Science, and Social Science.
For English, we determined these individuals to be Esha and Foni. For Hindi, we determined these individuals to be Alva and Deep. For Science, we know one of the individuals is either Alva or Deep. Given that Carl, Alva and Deep cannot be a part of the group that missed Science or Social Science exam, we can proceed by carefully scrutinizing the rest of the group that includes Bithi, Esha and Foni.
We notice that Bithi has a similar score in both Science and Social Science examination. Assuming that she did miss these exams, let us proceed to check if this was actually the case.
For Bithi: Best 2 out 3 - 90(English), 80(Hindi)
Avg = 170/2 = 85 which matches the given value
∴  Bithi is likely to have missed her Science and Social Science examinations.
We additionally notice that Foni has a similar score in English and Social Science. On considering the best 2 out of 3 scores, the average value of the score for both the subject holds (equal to 83). Thus, we can conclude that Bithi and Foni missed their Social Science examination.
Thus, the students who missed just one exam were: Carl (Mathematics); Esha (English) and one out of Alva and Deep (Hindi).
Hence of the six students, we can correctly determine the missed subjects for four of them (except Alva and Deep):
Mathematics: Carl ; English: Esha & Foni ; Hindi: Alva & Deep; Science: Bithi & one out of Alva and Deep ; Social Science: Foni & Bithi
Hence, the correct answer to this question is Option B: Alva and Deep.

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

Given,
Richard earned a total of 22.5 points and he was not the first in any level. The only way he can win 22.5 points without being first in any level is if he earned 7.5 in two levels, 5 in one level and 2.5 in another level.
From (2), Connor and Poppy did not win the highest points in the game.
From (6), Sarah also did not win the highest points in the game.
From (4), one of the persons was the first in two levels and this person did not win the highest points. This person must have earned a minimum of 10 + 10 + 2.5 + 2.5 = 25 points in the game. The person who earned the highest points must have earned more than this. From this, we can infer that Richard also cannot be the person who earned the highest points.
Hence, only Mason can be the person who earned the highest points.
From (5), Mason was the first in one level. He is not the third in any level. Also, Richard was the second in two levels (since he earned 7.5 in two levels). Hence, the maximum points that Mason can win = 10 + 7.5 + 7.5 + 2.5 = 27.5 points.
Since Mason has to win more than 25 points, he must have earned 27.5 points. The person who was first in two levels must have earned 25 points (this is the only way for him to have earned less points than Mason).
From (2) and (6), Poppy and Sarah cannot be the person who earned 25 points.
Hence, Connor must be the person who earned 25 points. This is possible if he was the first in two levels and fourth/fifth in two other levels.
From (2), Poppy must have earned 20 points. This is possible only if Poppy was the first in one level, the third in another level and fourth/fifth in two levels (Poppy cannot be the second in any level because Richard and Mason are second in two levels each). The points that Poppy wins in this case = 10 + 5 + 2.5 + 2.5 = 20 points.
Sarah must have been the third in two levels and fourth/fifth in two levels. The total points that Sarah earned = 5 + 5 + 2.5 + 2.5 = 15 points.
The following table presents the positions of the five persons in the four levels and the total points earned by them in the game.

Connor earned 2.5 points in two levels. In these two levels, Sarah could have got 2.5 points. Hence, we cannot say that Connor definitely earned more than Sarah in at least three levels.
Richard could have earned 2.5 points in the same level that Sarah earned 2.5 points. In the level that Richard earned 5 points, Sarah could have earned only 2.5 points (since two persons could not have earned 5 points). In the other two levels that Richard earned 7.5 points, Sarah would have got less points than Richard. Hence, Richard would have definitely earned higher points than Sarah in three levels. Poppy earned 2.5 points in two levels. In these two levels, Sarah could have earned 2.5 points. Hence, we cannot say that Poppy definitely earned more than Sarah in at least three levels. Mason earned 10 points in one level and 7.5 points in two levels. In all the three levels, Mason would have earned higher points than Sarah. Hence, the given condition is satisfied for two persons, Richard and Mason.

Only one goal was scored in every odd-numbered game.

Emma recorded two goals in Game 4.

Goals scored in Game 1, Game 2 and Game 4 are in GP.
Let the common ratio be r.

Goals scored should be a real number and total goals should be 14, so the only possible value of r is 2.

The game with the second highest number of goals scored is Game 6.
As the sum of scores is 14 and already in the above table 10 goals have been scored; in Game 6, the score will be 3 and the Game 8 score will be 1.

No two players had the same total number of goals.
So, their individuals score will be 2, 3, 4, and 5 as this is the only possible way for the sum to be 14.
Emma emerged as a dominant scorer – it means Emma scored 5 goals.
Hannah is known for her lowest scoring capabilities, and scored in Game 2 and Game 8.
The player with the highest number of goals achieved this feat scoring in exactly four games, including Game 3 and Game 7.

Olivia and Grace both scored 2 goals each in Game 4 and Game 6, respectively.

Either Olivia's score is 3 and Grace score's is 4 or Olivia's score is 4 and Grace's score is 3.
Based on the information provided:

For the highest possible value, the individual values must be equal to their maximum possible values.
x = 897 [number of people in the city]
y = 986 [number of people in the city]
z = 1034 [number of people in the city]
p = q = 564 [number of people in the city]
r = s = 1067 [number of people in the city]
x + y + z + p + q + r + s = 6179.
For the lowest possible value, the individual values must be equal to their minimum possible values.
x = 1 [total number of vehicles already exceeds the total number of people]
y = 1 [total number of vehicles already exceeds the total number of people]
z = 1 [total number of vehicles already exceeds the total number of people]
p and q are interdependent. p + q = 87 [If p+q = 87, only then can at least one person own one vehicle, that is, the number of vehicles = number of people]
r and s are interdependent. r + s = 1 + 1 = 2 [total number of vehicles already exceeds the total number of people]
Hence, x + y + z + p + q + r + s = 1 + 1 + 1 + 87 + 2 = 92
Difference = 6179 - 92 = 6087.

The percentage increase in the USA market index is the maximum among the three markets in a year.

From the table, we can see that the % decline in the rupee was maximum in the month of July among the options and also the stock index of UK was lowest in July among the given months so the maximum decline in the portfolio was in July.

Overall rank of only three countries ( USA, China and India) is higher than Japan.

But their nuclear rank also is higher.

Hence, option 1.

The total number of goals scored by all the players combined is 19. From (i), the winning team (Fierce Werewolves) must have scored 10 goals and the losing team must have scored 9 goals. From (ii), the winning team had the ball for 40 minutes and the losing team had the ball for 50 minutes.
Sean cannot be in Fierce Werewolves because Fierce Werewolves (winning team) had the ball for 40 minutes and Bonnie (possession time: 20 minutes) is in Fierce Werewolves.
If Sean and Casey were in Deadly Sharks with 9 goals and 38 minutes between them, the other three players in the team must have scored 1 goal and a possession time of 12 minutes. In this case, for any combination of three players, the possession time cannot be 50 minutes. Hence, Sean and Casey must be in different teams.
Therefore, Bonnie and Casey must be in Fierce Werewolves and Sean in Deadly Sharks. Bonnie and Casey together scored 8 goals and have a possession time of 33 minutes. The remaining players must have scored 2 goals and a possession time of 7 minutes. The 2 goals must have been scored by a single player (Alex, William or Rjay) since two players could not have scored 1 goal each. Between Alex, William and Rjay, only William can be a part of Fierce Werewolves because the other two have a higher possession time. The remaining two players must not have scored any goals and have a possession time of 5 minutes. From the table we can see that Charlie must be a part of Fierce Werewolves. One among Michael and Harry must be a player of Fierce Werewolves.
Deadly Sharks must comprise Sean, Alex, Rjay, Jamie, and one among Michael and Harry.
The following table represents the team.

The registration number of white car is 1115.

Given that the registration number of white car has less than 3 different digits. It cannot have only one digit (since no digit occurs 4 times). Hence, the registration number of white car must be a multiple of 5 (i.e. end in a 0 or a 5) and have 2 different digits, i.e. be of the form (aaab/abbb) OR (aabb). Hence, 3a + b = 8 (for which the only possibility is a = 1 and b = 5, since 0 appears only twice) or 2a + 2b = 8 (for which there is no possibility, since 4 does not appear even once and at least one of a and b must be 0 or 5). Hence. only 1115 is possible.

The registration number of white car is exactly a factor of registration number of black car or we can say that the registration number of black car is a multiple of registration number of white car. Multiples of 1115 with four digits are 2230, 3345, 4460, 5575, 6690, 7805 and 8920. Of these, only 6690 is possible (8920 and 7805 are not possible because the registration number of red car should have three digits, which are not in anyone else's registration number, i.e. the registration number of red car should have three digits out of 2, 7, 8 and 9).

Since 9 is present in the registration number of black car, red car's registration number should contain 2, 7, 8 and 5/0. The registration number of grey car will have the digits 3, 3, 5 and 5/0. But, if there are two 5s in the registration number of grey car, it will not have at least three distinct digits. Hence, the registration number of grey car will have the digits 3, 3, 5 and 0 and the registration number of red car will be either 2785 or 2875.

From (5), 2785 + 500 = 3285 and 2875 + 500 = 3375. The registration number of grey car should lie between 3285 and 3475. The only possible values of registration number of grey car are 3350 and 3305. Since both these values are less than 3375, the registration number of red car cannot be 2875. Therefore, the registration number of red car must be 2785 and the registration number of grey car can be 3350 or 3305.

From the given information, we can make the following empty table:

From (ii) and (iii), we get that Q got F grade in match II and match IV.
From (v) and (vi), P got same grade in match II and match III. Now, P could obtain grade B in match II and match III (from (vi)).
Now, in every match, at least one century was scored; thus, R obtained grade A in match II and match IV (from (iv)).
Thus, grade C would be obtained by Q in match I, R in match III and P in match IV (from (iii) and (v)).
Now, remaining rows and columns can be filled giving the final table as follows:

Since on 1st October, the value of the rupee has gotten stronger than its value on 1st February so the investor will be able to draw more dollars for a given amount of rupee and also the Indian stock market has reached the second-highest value among the options given so the value investment is maximum on 1st October in dollars term.

Total CP of magazines of week 1 = 54 × 7.5 + 88 × 8 = 1,109
Total CP of magazines of week 2 = 34 × 8 + 88 × 6.75 + 2 × 6.5 = 879
Total CP of magazines of week 3 = 72 × 6.5 + 36 × 7.75 = 747
Total CP of magazines of week 4 = 84 × 7.75 + 30 × 6 = 831
Total CP of magazines of week 5 = 66 × 6 + 42 × 6.52 = 669
Total CP of magazines of week 6 = 22 × 6.52 + 82 × 8.50 = 840

In week 3, Mr. Brown made the third highest percent of profit.

Since G₁ has 10 points, it must have won 3 rounds and drew 1 round (winning 2 and drawing 4 is not possible). G₂ could have won 1 round and drew 2 rounds or drew all 5 rounds. From A, G₂ must have won 1 round and drew 2 rounds. Since G₃ has two losses, it must have won the remaining 3 rounds for 9 points. G₄ must have won 4 rounds and lost 1 round, and G₅ must have won 1 round and lost 4 rounds.

In Zed Academy. K₁ could have 2 wins or 1 win and 3 draws. K₂ could have 2 wins and 1 draw or 1 win and 4 draws. K₃ must have 2 wins and 2 draws. K₄ and K₅ each can have 2 wins or 1 win and 3 draws.

The total number of draws that the groups in Zed Academy can have is 3 (since the total number of draws in Alex Academy is 3). Since K₃ already has 2 draws, the only possibility is K₂ having 2 wins and 1 draw.

Therefore, K₁, K₄ and K₅ each has 2 wins and 3 losses.

The table below presents this information:


Since G₁ and G₂ drew 3 rounds and K₂ and K₃ also drew three rounds, G₂ must have drawn against both K₂ and K₃ while G₁ must have drawn against K₃. Since G₄ lost against K₁, it must have won all the remaining rounds. Since G₁ lost against K₅, it must have won against K₁, K₂ and K₄.

K₂ lost to G₁ and G₄. Hence, it must have won against G₃ and G₅. K₃ lost to G₄. Hence, it must have won against G₃ and G₅. Since G₃ lost two rounds, it must have won the rounds against K₁, K₄ and K₅. K₄ must have won against G₂ and G₅. The table below gives the results of the rounds (with the group that won the round in each cell and '-' representing a draw).

K₄ lost three rounds.