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Instructions:
4 Chess tournaments were held all over the world last year and in each tournament 128 chess players participated. Players who participated in the 1st tournament are same for other tournaments. At the end of these four tournaments, world championship is held that consists of 32 players. These 32 players are selected on the basis of total number of wins the 128 players got in the four tournaments. Each of the games in the tournaments (including the world championship) is a knockout game i.e. a person who loses a game will not play in that tournament again. The person who wins the last round in any tournament is called the winner of that tournament. If 31 slots of the 32 slots in the world championship tournament are filled and to fill the 32nd spot there is a tie between few players, exactly one of those players is selected based on certain criteria (like coin toss).
Which of the following is the least number of wins that the winner of the world championship can have?
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
Instructions:
4 Chess tournaments were held all over the world last year and in each tournament 128 chess players participated. Players who participated in the 1st tournament are same for other tournaments. At the end of these four tournaments, world championship is held that consists of 32 players. These 32 players are selected on the basis of total number of wins the 128 players got in the four tournaments. Each of the games in the tournaments (including the world championship) is a knockout game i.e. a person who loses a game will not play in that tournament again. The person who wins the last round in any tournament is called the winner of that tournament. If 31 slots of the 32 slots in the world championship tournament are filled and to fill the 32nd spot there is a tie between few players, exactly one of those players is selected based on certain criteria (like coin toss).
Which of the following is the maximum number of wins that a player could have had and still not be selected for the world championship tournament?
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 133 win >= n matches, 34  64 win exactly one match and 65128 win no matches.
Total number of wins in 4 tournaments = 4 * 127 = 508
Wins accounted for by players 3464 = 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 121 win 12 matches, 22  33 win 11 matches, 3464 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.
Instructions:
4 Chess tournaments were held all over the world last year and in each tournament 128 chess players participated. Players who participated in the 1st tournament are same for other tournaments. At the end of these four tournaments, world championship is held that consists of 32 players. These 32 players are selected on the basis of total number of wins the 128 players got in the four tournaments. Each of the games in the tournaments (including the world championship) is a knockout game i.e. a person who loses a game will not play in that tournament again. The person who wins the last round in any tournament is called the winner of that tournament. If 31 slots of the 32 slots in the world championship tournament are filled and to fill the 32nd spot there is a tie between few players, exactly one of those players is selected based on certain criteria (like coin toss).
Which of the following is the least number of wins one needs to enter the world championship?
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.
Instructions:
4 Chess tournaments were held all over the world last year and in each tournament 128 chess players participated. Players who participated in the 1st tournament are same for other tournaments. At the end of these four tournaments, world championship is held that consists of 32 players. These 32 players are selected on the basis of total number of wins the 128 players got in the four tournaments. Each of the games in the tournaments (including the world championship) is a knockout game i.e. a person who loses a game will not play in that tournament again. The person who wins the last round in any tournament is called the winner of that tournament. If 31 slots of the 32 slots in the world championship tournament are filled and to fill the 32nd spot there is a tie between few players, exactly one of those players is selected based on certain criteria (like coin toss).
After all the five tournaments, a table was made of the players in the descending order of their wins in all the five tournaments combined. What is the maximum number of wins the top four players could have got if no player among the 128 participants won more than one tournament?
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
Instructions:
Alex, Cane and John are three drivers who drive a passenger bus, in shifts, from Paris to Luxembourg every day. Due to the heavy traffic at peak hours, the time it takes for them to travel between the two cities depends on their starting times at the originating city. If they start at 8:00 AM, 9:00 AM or 10:00 AM, it takes 18 hours to travel between the two cities. If they start at 7:00 PM, 8:00 PM or 9:00 PM, it takes them 20 hours to travel between the two cities. If they start at any other time, it takes them 15 hours to travel between the two cities. Each time they reach their destination city, they take rest for 1 hour before proceeding to the next city.
1 trip is defined as the journey from one city to another.
If they start from Paris at 8:00 PM, what is the average time taken by them to travel from Paris to Luxembourg in their first 80 trips?
Let’s solve this set by converting the times into 24hour 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.
Instructions:
Alex, Cane and John are three drivers who drive a passenger bus, in shifts, from Paris to Luxembourg every day. Due to the heavy traffic at peak hours, the time it takes for them to travel between the two cities depends on their starting times at the originating city. If they start at 8:00 AM, 9:00 AM or 10:00 AM, it takes 18 hours to travel between the two cities. If they start at 7:00 PM, 8:00 PM or 9:00 PM, it takes them 20 hours to travel between the two cities. If they start at any other time, it takes them 15 hours to travel between the two cities. Each time they reach their destination city, they take rest for 1 hour before proceeding to the next city.
1 trip is defined as the journey from one city to another.
If they start at 6:00 AM from Luxembourg, what is the average time per trip for their first 25 trips?
Let’s solve this set by converting the times into 24hour 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.
Instructions:
Alex, Cane and John are three drivers who drive a passenger bus, in shifts, from Paris to Luxembourg every day. Due to the heavy traffic at peak hours, the time it takes for them to travel between the two cities depends on their starting times at the originating city. If they start at 8:00 AM, 9:00 AM or 10:00 AM, it takes 18 hours to travel between the two cities. If they start at 7:00 PM, 8:00 PM or 9:00 PM, it takes them 20 hours to travel between the two cities. If they start at any other time, it takes them 15 hours to travel between the two cities. Each time they reach their destination city, they take rest for 1 hour before proceeding to the next city.
1 trip is defined as the journey from one city to another.
If they start at 4:00 AM from Paris, after a minimum of how many trips (including the first one) will they start from Luxembourg at 8:00 AM?
Let’s solve this set by converting the times into 24hour 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.
Instructions:
Alex, Cane and John are three drivers who drive a passenger bus, in shifts, from Paris to Luxembourg every day. Due to the heavy traffic at peak hours, the time it takes for them to travel between the two cities depends on their starting times at the originating city. If they start at 8:00 AM, 9:00 AM or 10:00 AM, it takes 18 hours to travel between the two cities. If they start at 7:00 PM, 8:00 PM or 9:00 PM, it takes them 20 hours to travel between the two cities. If they start at any other time, it takes them 15 hours to travel between the two cities. Each time they reach their destination city, they take rest for 1 hour before proceeding to the next city.
1 trip is defined as the journey from one city to another.
If they start at 2: 00 PM from Luxembourg, how many times will they start between 1:00 PM and 11:00 PM from Paris in their first 100 trips?
Let’s solve this set by converting the times into 24hour 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.
Instructions:
A Bschool has 4 clubs  Sports, Dramatics, Literary, and Quiz. 60 students joined these clubs in the first year. A person can join only one of these 4 clubs. After the first year was over, some students did not like the club they joined and hence, moved from one club to another.
Further the following information is known:
No student moved from the quiz club to the sports club.
The number of students who moved from dramatics club to the sports club is the same as the number of students who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well.
The number of students who moved out from the sports and quiz clubs are equal.
The number of students who moved to the literary club is one more than the number of persons who moved to the sports club.
At least one student moved from the sports club to literary club.
The number of students who left the quiz club for the dramatics club and the literary club is the same.
A total of 21 students shifted from one club to another.
4 students moved out of dramatics club and 5 students moved out of literary club.
The number of students who joined sports club is exactly half the number of students who left it.
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.
From which club did maximum number of people leave for the dramatics club?
Let us start by tabulating the data available.
We have no information about the number of persons in each club.
The sportssports 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 sportsdramatics (rowcolumn) represents the number of students from sports club who left for dramatics club. Therefore, the cell sportstotal will provide the number of students who left the sports club and the cell totalsports 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 (quizsports = 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 = 3a
g =6a
Substituting these values in (4), we get,
3a+6a+k = 5
92a+k = 5
k = 2a4
Substituting the value of 'k' in (5), we get,
j+2a4 = 2
j = 62a
(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=2a4. 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.
Instructions:
A Bschool has 4 clubs  Sports, Dramatics, Literary, and Quiz. 60 students joined these clubs in the first year. A person can join only one of these 4 clubs. After the first year was over, some students did not like the club they joined and hence, moved from one club to another.
Further the following information is known:
No student moved from the quiz club to the sports club.
The number of students who moved from dramatics club to the sports club is the same as the number of students who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well.
The number of students who moved out from the sports and quiz clubs are equal.
The number of students who moved to the literary club is one more than the number of persons who moved to the sports club.
At least one student moved from the sports club to literary club.
The number of students who left the quiz club for the dramatics club and the literary club is the same.
A total of 21 students shifted from one club to another.
4 students moved out of dramatics club and 5 students moved out of literary club.
The number of students who joined sports club is exactly half the number of students who left it.
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.
How many persons moved from the sports club to quiz club?
Let us start by tabulating the data available.
We have no information about the number of persons in each club.
The sportssports 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 sportsdramatics (rowcolumn) represents the number of students from sports club who left for dramatics club. Therefore, the cell sportstotal will provide the number of students who left the sports club and the cell totalsports 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 (quizsports = 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 = 3a
g =6a
Substituting these values in (4), we get,
3a+6a+k = 5
92a+k = 5
k = 2a4
Substituting the value of 'k' in (5), we get,
j+2a4 = 2
j = 62a
(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=2a4. 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.
Instructions:
A Bschool has 4 clubs  Sports, Dramatics, Literary, and Quiz. 60 students joined these clubs in the first year. A person can join only one of these 4 clubs. After the first year was over, some students did not like the club they joined and hence, moved from one club to another.
Further the following information is known:
No student moved from the quiz club to the sports club.
The number of students who moved from dramatics club to the sports club is the same as the number of students who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well.
The number of students who moved out from the sports and quiz clubs are equal.
The number of students who moved to the literary club is one more than the number of persons who moved to the sports club.
At least one student moved from the sports club to literary club.
The number of students who left the quiz club for the dramatics club and the literary club is the same.
A total of 21 students shifted from one club to another.
4 students moved out of dramatics club and 5 students moved out of literary club.
The number of students who joined sports club is exactly half the number of students who left it.
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.
What is the absolute difference between the number of students who joined literary club and the number of students who left it?
Let us start by tabulating the data available.
We have no information about the number of persons in each club.
The sportssports 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 sportsdramatics (rowcolumn) represents the number of students from sports club who left for dramatics club. Therefore, the cell sportstotal will provide the number of students who left the sports club and the cell totalsports 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 (quizsports = 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 = 3a
g =6a
Substituting these values in (4), we get,
3a+6a+k = 5
92a+k = 5
k = 2a4
Substituting the value of 'k' in (5), we get,
j+2a4 = 2
j = 62a
(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=2a4. 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 54 = 1.
Instructions:
A Bschool has 4 clubs  Sports, Dramatics, Literary, and Quiz. 60 students joined these clubs in the first year. A person can join only one of these 4 clubs. After the first year was over, some students did not like the club they joined and hence, moved from one club to another.
Further the following information is known:
No student moved from the quiz club to the sports club.
The number of students who moved from dramatics club to the sports club is the same as the number of students who moved from the sports club to the dramatics club. The same is the case with sports club and literary club as well.
The number of students who moved out from the sports and quiz clubs are equal.
The number of students who moved to the literary club is one more than the number of persons who moved to the sports club.
At least one student moved from the sports club to literary club.
The number of students who left the quiz club for the dramatics club and the literary club is the same.
A total of 21 students shifted from one club to another.
4 students moved out of dramatics club and 5 students moved out of literary club.
The number of students who joined sports club is exactly half the number of students who left it.
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.
The strength of which club decreased the most after the change?
Let us start by tabulating the data available.
We have no information about the number of persons in each club.
The sportssports 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 sportsdramatics (rowcolumn) represents the number of students from sports club who left for dramatics club. Therefore, the cell sportstotal will provide the number of students who left the sports club and the cell totalsports 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 (quizsports = 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 = 3a
g =6a
Substituting these values in (4), we get,
3a+6a+k = 5
92a+k = 5
k = 2a4
Substituting the value of 'k' in (5), we get,
j+2a4 = 2
j = 62a
(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=2a4. 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.
Read the information carefully and answer the questions that follow.
Eight friends, Abhi, Anupam, Deepak, Vinod, Prakash, Neha, Sukriti, and Jyoti are the alumni from different colleges viz. IIM A, IIM B, IIM C, IIM L, IIM K, IIM I, IIM R, IIM S, not necessarily in that order are sitting around a round table with eight seats.. Further, they are working in different domains viz. HR, Marketing, Finance, Operations, Consulting, Strategy, Sales, and General Management not in any particular order. It is also known that:
1. The friend, who is sitting opposite Vinod, works in General management and is an alumnus of IIM I.
2. Neha is sitting opposite the friend working in Marketing while the friends working in HR and Strategy are sitting adjacent to each other.
3. The alumnus from IIM K, who works in Strategy is sitting opposite the alumnus from IIM C, who works in Operations.
4. Anupam works in Finance and he sits second to the left of Prakash, who sits to the immediate right of Jyoti.
5. The alumnus of IIM S sits second to the right of Abhi, who sits opposite the alumnus from IIM B.
6. Vinod, who works in Consulting, sits adjacent to the friend working in Sales. Deepak sits opposite the friend who is an alumnus of IIM R.
7. The friends working in Sales and Marketing sit opposite each other and Prakash does not work in Marketing.
8. Anupam is an alumnus of IIM B. The alumnus of IIM A sits second to the right of Jyoti.
Who is sitting opposite Neha?
Anupam works in Finance and he sits second to the left of Prakash, who sits to the immediate right of Jyoti. The alumnus of IIM A sits second to the right of Jyoti. Anupam is an alumnus of IIM Bangalore. The alumnus of IIM S sits second to the right of Abhi, who sits opposite the alumnus from IIM B.
Neha is sitting opposite the friend working in Marketing. The friends working in sales and marketing sit opposite each other and Prakash does not work in Marketing. So, Neha must be working in Sales. Vinod, who works in Consulting, sits adjacent to the friend working in Sales. So, Vinod sits adjacent to Neha. So, Neha must be sitting either to the immediate right of Abhi or to the immediate left of Anupam. In either case, Vinod must be sitting opposite Prakash. The friend, who is sitting opposite Vinod, works in General management and is an alumnus of IIM I. Thus, Prakash must be from IIM I and he must be working in General Management.
The alumnus from IIM K, who works in Strategy is sitting opposite the alumnus from IIM C, who works in Operations. The friends working in HR and Strategy are sitting adjacent to each other. There is only one possibility  the friends working in HR and Strategy sit between Abhi and Vinod. Therefore, Abhi must be working in HR.
Neha sits opposite the friend who works in Marketing. So, she must be sitting between Vinod and Anupam. Also, the friend who is an alumnus of IIM A must be working in Marketing.
Deepak sits opposite the friend who is an alumnus of IIM R. So, Neha must be an alumnus from IIM R and Deepak must be sitting between Abhi and Prakash. Thus, Sukriti must be sitting between Vinod and Abhi and Abhi must be an alumnus of IIM L. Thus, we get the final arrangement as:
From the arrangement, we can see that Deepak sits opposite Neha.
Hence, option A is the correct answer.
Read the information carefully and answer the questions that follow.
Eight friends, Abhi, Anupam, Deepak, Vinod, Prakash, Neha, Sukriti, and Jyoti are the alumni from different colleges viz. IIM A, IIM B, IIM C, IIM L, IIM K, IIM I, IIM R, IIM S, not necessarily in that order are sitting around a round table with eight seats.. Further, they are working in different domains viz. HR, Marketing, Finance, Operations, Consulting, Strategy, Sales, and General Management not in any particular order. It is also known that:
1. The friend, who is sitting opposite Vinod, works in General management and is an alumnus of IIM I.
2. Neha is sitting opposite the friend working in Marketing while the friends working in HR and Strategy are sitting adjacent to each other.
3. The alumnus from IIM K, who works in Strategy is sitting opposite the alumnus from IIM C, who works in Operations.
4. Anupam works in Finance and he sits second to the left of Prakash, who sits to the immediate right of Jyoti.
5. The alumnus of IIM S sits second to the right of Abhi, who sits opposite the alumnus from IIM B.
6. Vinod, who works in Consulting, sits adjacent to the friend working in Sales. Deepak sits opposite the friend who is an alumnus of IIM R.
7. The friends working in Sales and Marketing sit opposite each other and Prakash does not work in Marketing.
8. Anupam is an alumnus of IIM B. The alumnus of IIM A sits second to the right of Jyoti.
Who is an alumnus of IIM C?
Anupam works in Finance and he sits second to the left of Prakash, who sits to the immediate right of Jyoti. The alumnus of IIM A sits second to the right of Jyoti. Anupam is an alumnus of IIM Bangalore. The alumnus of IIM S sits second to the right of Abhi, who sits opposite the alumnus from IIM B.
Neha is sitting opposite the friend working in Marketing. The friends working in sales and marketing sit opposite each other and Prakash does not work in Marketing. So, Neha must be working in Sales. Vinod, who works in Consulting, sits adjacent to the friend working in Sales. So, Vinod sits adjacent to Neha. So, Neha must be sitting either to the immediate right of Abhi or to the immediate left of Anupam. In either case, Vinod must be sitting opposite Prakash. The friend, who is sitting opposite Vinod, works in General management and is an alumnus of IIM I. Thus, Prakash must be from IIM I and he must be working in General Management.
The alumnus from IIM K, who works in Strategy is sitting opposite the alumnus from IIM C, who works in Operations. The friends working in HR and Strategy are sitting adjacent to each other. There is only one possibility  the friends working in HR and Strategy sit between Abhi and Vinod. Therefore, Abhi must be working in HR.
Neha sits opposite the friend who works in Marketing. So, she must be sitting between Vinod and Anupam. Also, the friend who is an alumnus of IIM A must be working in Marketing.
Deepak sits opposite the friend who is an alumnus of IIM R. So, Neha must be an alumnus from IIM R and Deepak must be sitting between Abhi and Prakash. Thus, Sukriti must be sitting between Vinod and Abhi and Abhi must be an alumnus of IIM L. Thus, we get the final arrangement as:
From the arrangement, we can see that Jyoti is an alumnus of IIM C.
Hence, option B is the correct answer.
Read the information carefully and answer the questions that follow.
Eight friends, Abhi, Anupam, Deepak, Vinod, Prakash, Neha, Sukriti, and Jyoti are the alumni from different colleges viz. IIM A, IIM B, IIM C, IIM L, IIM K, IIM I, IIM R, IIM S, not necessarily in that order are sitting around a round table with eight seats.. Further, they are working in different domains viz. HR, Marketing, Finance, Operations, Consulting, Strategy, Sales, and General Management not in any particular order. It is also known that:
1. The friend, who is sitting opposite Vinod, works in General management and is an alumnus of IIM I.
2. Neha is sitting opposite the friend working in Marketing while the friends working in HR and Strategy are sitting adjacent to each other.
3. The alumnus from IIM K, who works in Strategy is sitting opposite the alumnus from IIM C, who works in Operations.
4. Anupam works in Finance and he sits second to the left of Prakash, who sits to the immediate right of Jyoti.
5. The alumnus of IIM S sits second to the right of Abhi, who sits opposite the alumnus from IIM B.
6. Vinod, who works in Consulting, sits adjacent to the friend working in Sales. Deepak sits opposite the friend who is an alumnus of IIM R.
7. The friends working in Sales and Marketing sit opposite each other and Prakash does not work in Marketing.
8. Anupam is an alumnus of IIM B. The alumnus of IIM A sits second to the right of Jyoti.
What is the position of the friend who is an alumnus of IIM A with respect to the friend who works in HR?
Anupam works in Finance and he sits second to the left of Prakash, who sits to the immediate right of Jyoti. The alumnus of IIM A sits second to the right of Jyoti. Anupam is an alumnus of IIM Bangalore. The alumnus of IIM S sits second to the right of Abhi, who sits opposite the alumnus from IIM B.
Neha is sitting opposite the friend working in Marketing. The friends working in sales and marketing sit opposite each other and Prakash does not work in Marketing. So, Neha must be working in Sales. Vinod, who works in Consulting, sits adjacent to the friend working in Sales. So, Vinod sits adjacent to Neha. So, Neha must be sitting either to the immediate right of Abhi or to the immediate left of Anupam. In either case, Vinod must be sitting opposite Prakash. The friend, who is sitting opposite Vinod, works in General management and is an alumnus of IIM I. Thus, Prakash must be from IIM I and he must be working in General Management.
The alumnus from IIM K, who works in Strategy is sitting opposite the alumnus from IIM C, who works in Operations. The friends working in HR and Strategy are sitting adjacent to each other. There is only one possibility  the friends working in HR and Strategy sit between Abhi and Vinod. Therefore, Abhi must be working in HR.
Neha sits opposite the friend who works in Marketing. So, she must be sitting between Vinod and Anupam. Also, the friend who is an alumnus of IIM A must be working in Marketing.
Deepak sits opposite the friend who is an alumnus of IIM R. So, Neha must be an alumnus from IIM R and Deepak must be sitting between Abhi and Prakash. Thus, Sukriti must be sitting between Vinod and Abhi and Abhi must be an alumnus of IIM L. Thus, we get the final arrangement as:
From the arrangement, we can see that the friend who is an alumnus of IIM A sits to the immediate left of the friend who works in HR.
Hence, option B is the correct answer.
Instructions:
A code ABCDEFGH where A, B, C, D, E, F, G, H are all digits, guards the secret formula of a high demand soft drink. The code should follow the following constraints:
1. The 2 digit number AB should be a multiple of 11.
2. The digit C should be a multiple of B.
3. The digit D should be a multiple of B.
4. The 2digit number EF should be a multiple of the 2digit number CD.
5. G and H can be any 2 consecutive digits from 1 to 9 such that G<H.
6. None of the digits should be zero or one.
Based on the information given above, answer the questions that follow.
How many codes are possible if all the digits from A to H are greater than 4?
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.
Instructions:
A code ABCDEFGH where A, B, C, D, E, F, G, H are all digits, guards the secret formula of a high demand soft drink. The code should follow the following constraints:
1. The 2 digit number AB should be a multiple of 11.
2. The digit C should be a multiple of B.
3. The digit D should be a multiple of B.
4. The 2digit number EF should be a multiple of the 2digit number CD.
5. G and H can be any 2 consecutive digits from 1 to 9 such that G<H.
6. None of the digits should be zero or one.
Based on the information given above, answer the questions that follow.
What is the absolute difference between the number of valid codes and invalid codes among the ones given below?22666689
33996656
99999912
22469234
55555556
44848434
22489667
33333332
44888845
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.
Instructions:
A code ABCDEFGH where A, B, C, D, E, F, G, H are all digits, guards the secret formula of a high demand soft drink. The code should follow the following constraints:
1. The 2 digit number AB should be a multiple of 11.
2. The digit C should be a multiple of B.
3. The digit D should be a multiple of B.
4. The 2digit number EF should be a multiple of the 2digit number CD.
5. G and H can be any 2 consecutive digits from 1 to 9 such that G<H.
6. None of the digits should be zero or one.
Based on the information given above, answer the questions that follow.
How many codes are possible such that each digit from A to H is necessarily prime?
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.
Instructions:
A code ABCDEFGH where A, B, C, D, E, F, G, H are all digits, guards the secret formula of a high demand soft drink. The code should follow the following constraints:
1. The 2 digit number AB should be a multiple of 11.
2. The digit C should be a multiple of B.
3. The digit D should be a multiple of B.
4. The 2digit number EF should be a multiple of the 2digit number CD.
5. G and H can be any 2 consecutive digits from 1 to 9 such that G<H.
6. None of the digits should be zero or one.
Based on the information given above, answer the questions that follow.
How many codes are possible?
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
The government had not declared many parts of the state as droughthit to help farmers get loans.
A. The farmers want their area to be declared droughthit.
B. Farmers in the state need loans.
C. Farmers in droughthit area are not given loans.
D. The farmers in the areas not hit by drought do not need loans.
Which assumption will be perfect for statement
The statement does not refer to what the farmers want. It only states the action taken by the government. Hence, choice (1) is out of context.
The reason behind the government's decision to not declare many areas drought hit is to let the farmers avail loans. From this it is clear that the government is assuming that there is a necessity for the farmers to avail loans.Hence, choice (2) is an implicit assumption.
The government does not want the farmers to be denied loans. As a result the government is not declaring many areas as drought hit. From this it can be inferred that the farmers do not get loans, once their area is declared drought hit.
Hence, choice (3) is an inference.
The statement has no reference to the farmers who are in the areas that are not drought hit. Hence, choice (4) is out of context.
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