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The following table represents the type(s) of vehicle(s) owned by people of 5 cities.
So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.
Based on the information given above, answer the questions that follow.
If it is known that all people in Hyderabad own at least one vehicle, and each person owns either only one or all the 4 vehicles, which of the following can be a possible value of x?
I + 2II + 3III + 4IV = 61 + 131 + 707 + x = 899 + x
Since, II = III = 0
I + 4IV = 899 + x
Also,
I + IV = 897
Therefore, 3IV = 2 + x
Hence, x should be of the form of 3n  2. Hence 23 and 120 can not be the right choice.
Let us assume that x = 184,
IV = 186/3=62
But there are only 61 people with a private jet, so this can't the answer.
Let us assume that x = 64.
IV = 66/2 = 33, which is possible. Hence this is the correct option.
The following table represents the type(s) of vehicle(s) owned by people of 5 cities.
So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.
Based on the information given above, answer the questions that follow.
What is the absolute difference between the highest possible value and the lowest possible value of (x+y+z+p+q+r+s)? All people have at least one vehicle. All the unknown variables are necessarily natural numbers.
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 following table represents the type(s) of vehicle(s) owned by people of 5 cities.
So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.
Based on the information given above, answer the questions that follow.
If the total number of bikes in all 5 cities combined is 2816, what is the maximum possible number of people in Chennai who own at least 3 vehicles? It is known that all people in Chennai own at least one vehicle.
The following table represents the type(s) of vehicle(s) owned by people of 5 cities.
So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.
Based on the information given above, answer the questions that follow.
In Bengaluru, the number of people who own zero vehicles is zero, the number of people who own 2 vehicles is 163, the number of people who own 3 vehicles is 36 and the number of people who own all 4 vehicles is more than the number of people who own 3 vehicles, what is the maximum value that y can take?
I + II + III + IV = 986
I + 2 II + 3 III + 4 IV = 1265 + y
Subtracting the first from the second,
II + 2 III + 3 IV = 279 + y
163 + 72 + 3 IV = 279 + y
y = 3 IV  44
Now, IV > III, so , IV > 36
But IV <= 57
For maximum y, we have to put maximum IV,
y = 3 x 57  44 = 171  44 = 127
The following table represents the type(s) of vehicle(s) owned by people of 5 cities.
So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.
Based on the information given above, answer the questions that follow.
If it is given that all unknowns are equal to 150, what is the maximum number of people in all 5 cities combined who owns exactly 4 vehicles? Also, every person in all of the 5 cities owns at least one vehicle.
In the following table, I represents people who own only one vehicle, II represents people who own two vehicles, III represents people who own three vehicles and IV represents people who own four vehicles.
We have to calculate the additional value for every city.
Then we try to allocate the maximum to IV for each city. If for a city, this value is more than any particular vehicle value, we take that value.
Hence sum = 326
The following table represents the type(s) of vehicle(s) owned by people of 5 cities.
So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.
Based on the information given above, answer the questions that follow.
Thus, in Hyderabad, the excess value is 152. We will try to allot maximum to IV, and automatically we will get the maximum I value.
152 = 3 x 50 + 2 x 1
Hence, 50 people own 4 vehicles and 1 person owns 3 vehicles.
Number of people left with 1 vehicle = 897  51 = 846
In Bengaluru, the excess value is 429.
But a maximum of 57 people can own all 4 vehicles. We will try to allocate the rest among III.
429 = 3 X 57 + 2 X 129
But 129 people cannot own 3 vehicles, because people owning a scooter is 150, and 57 + 129 exceeds this value.
Hence, the maximum number of people who can own 3 vehicles = 150  57 = 93.
Hence we are left with... 429  3 x 57  2 x 93 = 72
Hence, 57 people own 4 vehicles and 93 people own 3 vehicles, 72 people own 2 vehicles.
Number of people left with 1 vehicle =986  57  93  72 = 764.
Hence, difference = (846 + 764)  (50 + 57)
1503 is the right answer.
TPL is an upcoming football tournament. 12 teams named A, B, C, D, E F, G, H, I, J. K, and L participated in the tournament. The structure of the Tournaments was as follows:
a) In a match if a team wins they get 3 points, 1 point is awarded on the draw and 0 points awarded for a loss.
b)In round 1, the 12 teams will be randomly divided into 3 groups of 4 team each. In this round, every team will play a match with every other team in the group. Once all the matches are done, the team will be ranked based on their points scored. In case of a tie, the team who scored more goals in that round will be given priority. The top 2 teams of each group will go to round 2
c) In round 2, 2 groups are formed. The first group contains the Rank 1 team from round 1 and the second team will consist of Rank 2 teams from round 1. In these rounds, the team will play a match against all the team in the same group. And will be ranked in a similar manner as that of round 1. The top 2 teams from each of the group will move forward to round 3
d) In round 3, the Rank1 teams from both the group play against each other and the Rank 2 teams play each other. The winner of the matches will go on to play the finals to play for rank 1 and rank 2. The losers of the match will play for rank 3 and rank 4
After TPL was over, the following this were known:
In round 1:
a) No team has all of their matches resulting in Draw
b) B had a match with A in round 2
c) Team H never had a match with Team
d) A won the match against E by 32
e) C lost the match against F by 23
f) No 2 teams in a group have the same points as well as the same number of goals.
g) L did not play any match with B
h) In one of the matches where A had a draw, it scored 2 goals
i) Every team scored at least 1 goals in every match except E and F as they scored 0 goals in exactly 1 of their matches
Which of the following is the maximum number of goals team I can score against its opponent?
In a group, there are 4 teams. Thus each team will play 3 matches in Round 1.
A team can score 7 points only if it wins 2 matches and draws the third match.
A team can score 6 points only if it wins 2 matches and loses the third match.
A team can score 5 points only if it wins 1 matches and draws in 2 matches.
A team can score 4 points only if it wins 1 matches and draws in 2 matches.
A team can score 3 points only if it wins 1 matches and loses in 2 matches or draws in all 3 matches.
A team can score 2 points only if it loses 1 matches and draws in 2 matches.
A team can score1 points only if it draws 1 match and loses in 2 matches.
A team can score 0points only if it loses all 3 matches.
Thus the table can be updated as follows
From above we can see that K has a score of 7 which is the greatest among all the teams and it also scored 12 goals which are the greatest among all the team. Thus K will play in round 2. Furthermore, we are given that A plays against B in round 2. It implies A and B got rank 1 in their groups. Thus A, B and K are from different groups and they were the top ranking team in their groups. Let the groups be called I, II and III.
Further, A won the match against E, thus A and E are in the same group
Since B scored 6 points and was able to go to round 2 implies that H was not in the group with B. We are also given that H and K are not in the same team. Thus H has to be in the same group as A
For any match, if it has a result then 3 points are given and if it is a draw then 2 points are given(1 to each team). If all the matches of group 1 have a result then the total sum of points scored by the team will be 6 \times 3 =186×3=18. But we know that A has a match which was resulted draw. Thus maximum possible
For group 1 we know that at least 1 match was draw. Thus the sum of points of Group I has to be less than 18.
Points of A+H+E = 7+7+3 = 17. Thus the fourth team has to be with 0 points i.e. team I and exactly 1 draw match was played in group 1. Thus we can deduce that E won a match and lost 2 matches to score 3 points.
L never played a match with B, thus L had to be in group III.
Team L and Team C have the same points and same goals. Thus, C has to be in group II. F played a match with C thus F is in the same group as C
If we combine the results of B, C and F. We get a total of 4 wins, 4 Loss and 1 Draw. Since for any group number of wins has to be equal to the loss. The fourth team must have 3 draws( not possible as only C has 1 draw) or 1 Draw, 1 Win and 1 Loss. J is the only such team. Thus J is in group II. The remaining D and G have to be in group III
Upon arranging them based on their points scored and goal scored we get
For Group I we can see that
A had a Draw match with H.
A won against E (32)
A won against I
H won against E
H won against I
E won against I.
We are given that in the Draw match A scored 2 goals, Thus H also scored 2 goals in that Match. In its third match, A has to score 3 goals against I to have total of 8 goals in total.
We know that E scored exactly 0 goals once. It has to happen in a Match against E. Thus E scored 5 goals against I.
H scored a total of 5 goals out of which, 2 were against A. Remaining 3 had to be against E and I. We know that in both the matches H won the game. Thus H had to score at least 1 more goal than E and I in respective matches. We also know that I scored at least 1 goal in each match. Thus H has to score at least 1 goal against E and 2 goals against I. This equal 3 thus, it is the only possible case
I have to score either 1 or 2 goals against team A. Thus I have to score 2 or 3 goals against E.
TPL is an upcoming football tournament. 12 teams named A, B, C, D, E F, G, H, I, J. K, and L participated in the tournament. The structure of the Tournaments was as follows:
a) In a match if a team wins they get 3 points, 1 point is awarded on the draw and 0 points awarded for a loss.
b)In round 1, the 12 teams will be randomly divided into 3 groups of 4 team each. In this round, every team will play a match with every other team in the group. Once all the matches are done, the team will be ranked based on their points scored. In case of a tie, the team who scored more goals in that round will be given priority. The top 2 teams of each group will go to round 2
c) In round 2, 2 groups are formed. The first group contains the Rank 1 team from round 1 and the second team will consist of Rank 2 teams from round 1. In these rounds, the team will play a match against all the team in the same group. And will be ranked in a similar manner as that of round 1. The top 2 teams from each of the group will move forward to round 3
d) In round 3, the Rank1 teams from both the group play against each other and the Rank 2 teams play each other. The winner of the matches will go on to play the finals to play for rank 1 and rank 2. The losers of the match will play for rank 3 and rank 4
After TPL was over, the following this were known:
In round 1:
a) No team has all of their matches resulting in Draw
b) B had a match with A in round 2
c) Team H never had a match with Team
d) A won the match against E by 32
e) C lost the match against F by 23
f) No 2 teams in a group have the same points as well as the same number of goals.
g) L did not play any match with B
h) In one of the matches where A had a draw, it scored 2 goals
i) Every team scored at least 1 goals in every match except E and F as they scored 0 goals in exactly 1 of their matches
Exact score of how many matches can be known by the given information?
In a group, there are 4 teams. Thus each team will play 3 matches in Round 1.
A team can score 7 points only if it wins 2 matches and draws the third match.
A team can score 6 points only if it wins 2 matches and loses the third match.
A team can score 5 points only if it wins 1 matches and draws in 2 matches.
A team can score 4 points only if it wins 1 matches and draws in 2 matches.
A team can score 3 points only if it wins 1 matches and loses in 2 matches or draws in all 3 matches.
A team can score 2 points only if it loses 1 matches and draws in 2 matches.
A team can score1 points only if it draws 1 match and loses in 2 matches.
A team can score 0points only if it loses all 3 matches.
Thus the table can be updated as follows
From above we can see that K has a score of 7 which is the greatest among all the teams and it also scored 12 goals which are the greatest among all the team. Thus K will play in round 2. Furthermore, we are given that A plays against B in round 2. It implies A and B got rank 1 in their groups. Thus A, B and K are from different groups and they were the top ranking team in their groups. Let the groups be called I, II and III.
Further, A won the match against E, thus A and E are in the same group
Since B scored 6 points and was able to go to round 2 implies that H was not in the group with B. We are also given that H and K are not in the same team. Thus H has to be in the same group as A
For any match, if it has a result then 3 points are given and if it is a draw then 2 points are given(1 to each team). If all the matches of group 1 have a result then the total sum of points scored by the team will be 6 \times 3 =186×3=18. But we know that A has a match which was resulted draw. Thus maximum possible
For group 1 we know that at least 1 match was draw. Thus the sum of points of Group I has to be less than 18.
Points of A+H+E = 7+7+3 = 17. Thus the fourth team has to be with 0 points i.e. team I and exactly 1 draw match was played in group 1. Thus we can deduce that E won a match and lost 2 matches to score 3 points.
L never played a match with B, thus L had to be in group III.
Team L and Team C have the same points and same goals. Thus, C has to be in group II. F played a match with C thus F is in the same group as C
If we combine the results of B, C and F. We get a total of 4 wins, 4 Loss and 1 Draw. Since for any group number of wins has to be equal to the loss. The fourth team must have 3 draws( not possible as only C has 1 draw) or 1 Draw, 1 Win and 1 Loss. J is the only such team. Thus J is in group II. The remaining D and G have to be in group III
For Group I we can see that
A had a Draw match with H.
A won against E (32)
A won against I
H won against E
H won against I
E won against I.
We are given that in the Draw match A scored 2 goals, Thus H also scored 2 goals in that Match. In its third match, A has to score 3 goals against I to have total of 8 goals in total.
We know that E scored exactly 0 goals once. It has to happen in a Match against E. Thus E scored 5 goals against I.
H scored a total of 5 goals out of which, 2 were against A. Remaining 3 had to be against E and I. We know that in both the matches H won the game.
Thus H had to score at least 1 more goal than E and I in respective matches. We also know that I scored at least 1 goal in each match. Thus H has to score at least 1 goal against E and 2 goals against I. This equal 3 thus, it is the only possible case
We have determined a score of 4 matches in group I.
For group II we know there was a match between F and which had a score 32
Total 5 matches' scores are known
TPL is an upcoming football tournament. 12 teams named A, B, C, D, E F, G, H, I, J. K, and L participated in the tournament. The structure of the Tournaments was as follows:
a) In a match if a team wins they get 3 points, 1 point is awarded on the draw and 0 points awarded for a loss.
b)In round 1, the 12 teams will be randomly divided into 3 groups of 4 team each. In this round, every team will play a match with every other team in the group. Once all the matches are done, the team will be ranked based on their points scored. In case of a tie, the team who scored more goals in that round will be given priority. The top 2 teams of each group will go to round 2
c) In round 2, 2 groups are formed. The first group contains the Rank 1 team from round 1 and the second team will consist of Rank 2 teams from round 1. In these rounds, the team will play a match against all the team in the same group. And will be ranked in a similar manner as that of round 1. The top 2 teams from each of the group will move forward to round 3
d) In round 3, the Rank1 teams from both the group play against each other and the Rank 2 teams play each other. The winner of the matches will go on to play the finals to play for rank 1 and rank 2. The losers of the match will play for rank 3 and rank 4
After TPL was over, the following this were known:
In round 1:
a) No team has all of their matches resulting in Draw
b) B had a match with A in round 2
c) Team H never had a match with Team
d) A won the match against E by 32
e) C lost the match against F by 23
f) No 2 teams in a group have the same points as well as the same number of goals.
g) L did not play any match with B
h) In one of the matches where A had a draw, it scored 2 goals
i) Every team scored at least 1 goals in every match except E and F as they scored 0 goals in exactly 1 of their matches
In round 1, How many matches ended in a draw?
In a group, there are 4 teams. Thus each team will play 3 matches in Round 1.
A team can score 7 points only if it wins 2 matches and draws the third match.
A team can score 6 points only if it wins 2 matches and loses the third match.
A team can score 5 points only if it wins 1 matches and draws in 2 matches.
A team can score 4 points only if it wins 1 matches and draws in 2 matches.
A team can score 3 points only if it wins 1 matches and loses in 2 matches or draws in all 3 matches.
A team can score 2 points only if it loses 1 matches and draws in 2 matches.
A team can score1 points only if it draws 1 match and loses in 2 matches.
A team can score 0points only if it loses all 3 matches.
Thus the table can be updated as follows
From above we can see that K has a score of 7 which is the greatest among all the teams and it also scored 12 goals which are the greatest among all the team. Thus K will play in round 2. Furthermore, we are given that A plays against B in round 2. It implies A and B got rank 1 in their groups. Thus A, B and K are from different groups and they were the top ranking team in their groups. Let the groups be called I, II and III.
Further, A won the match against E, thus A and E are in the same group
Since B scored 6 points and was able to go to round 2 implies that H was not in the group with B. We are also given that H and K are not in the same team. Thus H has to be in the same group as A
For any match, if it has a result then 3 points are given and if it is a draw then 2 points are given(1 to each team). If all the matches of group 1 have a result then the total sum of points scored by the team will be 6 \times 3 =186×3=18. But we know that A has a match which was resulted draw. Thus maximum possible
For group 1 we know that at least 1 match was draw. Thus the sum of points of Group I has to be less than 18.
Points of A+H+E = 7+7+3 = 17. Thus the fourth team has to be with 0 points i.e. team I and exactly 1 draw match was played in group 1. Thus we can deduce that E won a match and lost 2 matches to score 3 points.
L never played a match with B, thus L had to be in group III.
Team L and Team C have the same points and same goals. Thus, C has to be in group II. F played a match with C thus F is in the same group as C
If we combine the results of B, C and F. We get a total of 4 wins, 4 Loss and 1 Draw. Since for any group number of wins has to be equal to the loss. The fourth team must have 3 draws( not possible as only C has 1 draw) or 1 Draw, 1 Win and 1 Loss. J is the only such team. Thus J is in group II. The remaining D and G have to be in group III
Upon arranging them based on their points scored and goal scored we get
From here we can see that group 1 had 1 draw matches. Group II has 1 Draw and Group III has 3 draws. Total draws = 5
TPL is an upcoming football tournament. 12 teams named A, B, C, D, E F, G, H, I, J. K, and L participated in the tournament. The structure of the Tournaments was as follows:
a) In a match if a team wins they get 3 points, 1 point is awarded on the draw and 0 points awarded for a loss.
b)In round 1, the 12 teams will be randomly divided into 3 groups of 4 team each. In this round, every team will play a match with every other team in the group. Once all the matches are done, the team will be ranked based on their points scored. In case of a tie, the team who scored more goals in that round will be given priority. The top 2 teams of each group will go to round 2
c) In round 2, 2 groups are formed. The first group contains the Rank 1 team from round 1 and the second team will consist of Rank 2 teams from round 1. In these rounds, the team will play a match against all the team in the same group. And will be ranked in a similar manner as that of round 1. The top 2 teams from each of the group will move forward to round 3
d) In round 3, the Rank1 teams from both the group play against each other and the Rank 2 teams play each other. The winner of the matches will go on to play the finals to play for rank 1 and rank 2. The losers of the match will play for rank 3 and rank 4
After TPL was over, the following this were known:
In round 1:
a) No team has all of their matches resulting in Draw
b) B had a match with A in round 2
c) Team H never had a match with Team
d) A won the match against E by 32
e) C lost the match against F by 23
f) No 2 teams in a group have the same points as well as the same number of goals.
g) L did not play any match with B
h) In one of the matches where A had a draw, it scored 2 goals
i) Every team scored at least 1 goals in every match except E and F as they scored 0 goals in exactly 1 of their matches
What is the rank of team J in its group in round 1?
In a group, there are 4 teams. Thus each team will play 3 matches in Round 1.
A team can score 7 points only if it wins 2 matches and draws the third match.
A team can score 6 points only if it wins 2 matches and loses the third match.
A team can score 5 points only if it wins 1 matches and draws in 2 matches.
A team can score 4 points only if it wins 1 matches and draws in 2 matches.
A team can score 3 points only if it wins 1 matches and loses in 2 matches or draws in all 3 matches.
A team can score 2 points only if it loses 1 matches and draws in 2 matches.
A team can score1 points only if it draws 1 match and loses in 2 matches.
A team can score 0points only if it loses all 3 matches.
Thus the table can be updated as follows
From above we can see that K has a score of 7 which is the greatest among all the teams and it also scored 12 goals which are the greatest among all the team. Thus K will play in round 2. Furthermore, we are given that A plays against B in round 2. It implies A and B got rank 1 in their groups. Thus A, B and K are from different groups and they were the top ranking team in their groups. Let the groups be called I, II and III.
Further, A won the match against E, thus A and E are in the same group
Since B scored 6 points and was able to go to round 2 implies that H was not in the group with B. We are also given that H and K are not in the same team. Thus H has to be in the same group as A
For any match, if it has a result then 3 points are given and if it is a draw then 2 points are given(1 to each team). If all the matches of group 1 have a result then the total sum of points scored by the team will be 6 \times 3 =186×3=18. But we know that A has a match which was resulted draw. Thus maximum possible
For group 1 we know that at least 1 match was draw. Thus the sum of points of Group I has to be less than 18.
Points of A+H+E = 7+7+3 = 17. Thus the fourth team has to be with 0 points i.e. team I and exactly 1 draw match was played in group 1. Thus we can deduce that E won a match and lost 2 matches to score 3 points.
L never played a match with B, thus L had to be in group III.
Team L and Team C have the same points and same goals. Thus, C has to be in group II. F played a match with C thus F is in the same group as C
If we combine the results of B, C and F. We get a total of 4 wins, 4 Loss and 1 Draw. Since for any group number of wins has to be equal to the loss. The fourth team must have 3 draws( not possible as only C has 1 draw) or 1 Draw, 1 Win and 1 Loss. J is the only such team. Thus J is in group II. The remaining D and G have to be in group III
Upon arranging them based on their points scored and goal scored we get
Instructions: Krishna is a black market gold coins dealer. He buys gold coins from various traders. He checks the quality of gold coins by weighing them. He either uses a Spring Balance or a Beam balance for the same. Ananth, Siddhartha, Pramit and Prasoon are the 4 traders who tried to sell gold coins to Krishna. Because of his prior experience Krishna know that all 4 of these traders will try to sneak in exactly 1 fake gold coin among the real gold coins
Ananth tried to sell 213 gold coins to Krishna. How many minimum number of weighings does Krishna need to do if he chooses to use Balance Beam. He knows that Ananth has added a fake coin with lesser weight than the actual gold coins
Here the Balance Beam is used. Therefore to find out the minimum number of trials to find the fake coin we will divide it into 3 parts
Let A, B, C be 3 groups each having 71 coins each.
We will compare A and B first.
In the first weighing, We can have three cases:
a) A > B, thus B has the fake coin
b) A < B thus A has the fake coin
c) A = B, thus C has the fake coin
In any of the scenario, We would know that among the 71 coins, there is a fake lighter coin and the other 2 sets have all the good coins
For second weighing we will try to divide 71 coins in 3 equal sets, It is possible when the 23,24,24 set are made. In again comparing 2 sets of 24 coins each we will know exactly which set has the fake coin. In the worst possible scenario, there will be 24 coins, among which 1 is a fake coin
Next, we divide 24 in 8,8 and 8 and thus in a total of 3 weighings we would have narrowed it down to 8 coins, out of which 1 is fake.
In the fourth weighing, we will divide the 8 coins into 3,3,2 coins. On comparing 3 and 3 we will figure out which set has the fake coin. In the worstcase set of 3 coins will have a fake coin
In the final weighing, we will compare the two coins, If they are equal, then the third coin is fake. If they unequal then the lighter one will be fake.
Thus a total of 5 trials is needed.
Instructions: Krishna is a black market gold coins dealer. He buys gold coins from various traders. He checks the quality of gold coins by weighing them. He either uses a Spring Balance or a Beam balance for the same. Ananth, Siddhartha, Pramit and Prasoon are the 4 traders who tried to sell gold coins to Krishna. Because of his prior experience Krishna know that all 4 of these traders will try to sneak in exactly 1 fake gold coin among the real gold coins
Prasoon wants to sell 120 coins and Krishna knows the weight of the real coins. What is the minimum number of weighings does Krishna need to do to figure out the fake coins if he uses a Spring Balance.
The weight of a real coin is known.
120 coins will be divided into 2 parts of 60 each. And one of them will be weighed. If the weight equals the calculated weight of 60 coins then that set has all real coins and others will have the fake coin.
Thus after the 1{st}1st step we will be left with 60 coins out of which 1 is fake. We will follow the similar process
Thus after the 2{nd}2nd step we will be left with 30 coins out of which 1 is fake.
Thus after the 3{nd}3nd step we will be left with 15 coins out of which 1 is fake.
Thus after the 4{th}4th step, we will be left with at most 8 coins out of which 1 is fake.
Thus after the 5{th}5th step, we will be left with 4 coins out of which 1 is fake.
Thus after the 6{th}6th step, we will be left with 2 coins out of which 1 is fake.
We need 7 trials
Instructions: In the virtual game of 'Shoot'em all', M number of virtual people are seated in a circular arrangement, all of them facing the centre as follows. Also, they are number from 1 to M.
There is a shooter at the centre of the circle, who can point to and shoot any particular person among the M people.
There are N rounds to this game. Initially, the shooter is pointing the gun towards the Kth numbered person, where 1 <= K <= N, but he does not shoot him yet. After the first round, he turns to the first person to the left of the person he was facing and shoots him. After the second round, he turns to the second person to the left of the person he was facing and shoots him, similarly, after the Nth round, he turns to the Nth person to the left of the person he was facing and shoots him. While counting, we will also count the vacant spots. For example, if the person numbered 3 is already shot dead, the 2nd person to the left of 2 is 4, not 5, since we have to count 3 as well. Also, while counting, if the count exceeds the total count of the table(i.e M), we go around the circle again. For example, if there are 8 people, the 10th person to the left of 1 is 3, and the 8th person to the left of 1 is 1 himself.
If at the end of any round, the shooter is pointing to a person who has already been shot in one of the previous rounds, he does not shoot anyone in that round and continues to the next round.
Also, suppose, all the persons are already shot in the nth round, where n < N, then in the (n+1)th round, the shooter shoots himself, and the game ends, it does not move to the (n+2)th round.
For example, suppose M = 8, N = 4, and K = 3, after the first round, the shooter shoots 4, after the second round, the shooter shoots 6, after the 3rd round, he shoots 1, and after the 4th and final round, he shoots 5.
Based on the information given above, answer the questions that follow.
Suppose M = 12 and N = 12, persons numbered 6, 12, 3, and 9 are alive at the end of all rounds. How many values can K take?
Since M = 12, we can draw the arrangement as
Now, let us assume that K = 1, so, after the first round, 2 is killed, after the second, 4 is killed, and if we continue, we get
Hence, the remaining ones are 3, 6, 9, 12.
Thus, when K =1, the arrangement is as above, on observation, we can say that if K = 4, 7, 10, we get the above arrangement as well.
Hence, the total number of values that K can take is 4.
Instructions: In the virtual game of 'Shoot'em all', M number of virtual people are seated in a circular arrangement, all of them facing the centre as follows. Also, they are number from 1 to M.
There is a shooter at the centre of the circle, who can point to and shoot any particular person among the M people.
There are N rounds to this game. Initially, the shooter is pointing the gun towards the Kth numbered person, where 1 <= K <= N, but he does not shoot him yet. After the first round, he turns to the first person to the left of the person he was facing and shoots him. After the second round, he turns to the second person to the left of the person he was facing and shoots him, similarly, after the Nth round, he turns to the Nth person to the left of the person he was facing and shoots him. While counting, we will also count the vacant spots. For example, if the person numbered 3 is already shot dead, the 2nd person to the left of 2 is 4, not 5, since we have to count 3 as well. Also, while counting, if the count exceeds the total count of the table(i.e M), we go around the circle again. For example, if there are 8 people, the 10th person to the left of 1 is 3, and the 8th person to the left of 1 is 1 himself.
If at the end of any round, the shooter is pointing to a person who has already been shot in one of the previous rounds, he does not shoot anyone in that round and continues to the next round.
Also, suppose, all the persons are already shot in the nth round, where n < N, then in the (n+1)th round, the shooter shoots himself, and the game ends, it does not move to the (n+2)th round.
For example, suppose M = 8, N = 4, and K = 3, after the first round, the shooter shoots 4, after the second round, the shooter shoots 6, after the 3rd round, he shoots 1, and after the 4th and final round, he shoots 5.
Based on the information given above, answer the questions that follow.
Suppose M = 25, in the first 20 rounds, in how many rounds does the shooter not shoot anyone?
Let us assume K =1, so, we can tabulate the persons killed in each round and also rounds in which no one was killed.
Hence, there were 10 rounds after which no one was killed.
Instructions: In the virtual game of 'Shoot'em all', M number of virtual people are seated in a circular arrangement, all of them facing the centre as follows. Also, they are number from 1 to M.
There is a shooter at the centre of the circle, who can point to and shoot any particular person among the M people.
There are N rounds to this game. Initially, the shooter is pointing the gun towards the Kth numbered person, where 1 <= K <= N, but he does not shoot him yet. After the first round, he turns to the first person to the left of the person he was facing and shoots him. After the second round, he turns to the second person to the left of the person he was facing and shoots him, similarly, after the Nth round, he turns to the Nth person to the left of the person he was facing and shoots him. While counting, we will also count the vacant spots. For example, if the person numbered 3 is already shot dead, the 2nd person to the left of 2 is 4, not 5, since we have to count 3 as well. Also, while counting, if the count exceeds the total count of the table(i.e M), we go around the circle again. For example, if there are 8 people, the 10th person to the left of 1 is 3, and the 8th person to the left of 1 is 1 himself.
If at the end of any round, the shooter is pointing to a person who has already been shot in one of the previous rounds, he does not shoot anyone in that round and continues to the next round.
Also, suppose, all the persons are already shot in the nth round, where n < N, then in the (n+1)th round, the shooter shoots himself, and the game ends, it does not move to the (n+2)th round.
For example, suppose M = 8, N = 4, and K = 3, after the first round, the shooter shoots 4, after the second round, the shooter shoots 6, after the 3rd round, he shoots 1, and after the 4th and final round, he shoots 5.
Based on the information given above, answer the questions that follow.
Suppose M = 25, in the first 20 rounds, in how many rounds does the shooter not shoot anyone?
Let us assume K =1, so, we can tabulate the persons killed in each round and also rounds in which no one was killed.
Hence, there were 10 rounds after which no one was killed.
Instructions: In the virtual game of 'Shoot'em all', M number of virtual people are seated in a circular arrangement, all of them facing the centre as follows. Also, they are number from 1 to M.
There is a shooter at the centre of the circle, who can point to and shoot any particular person among the M people.
There are N rounds to this game. Initially, the shooter is pointing the gun towards the Kth numbered person, where 1 <= K <= N, but he does not shoot him yet. After the first round, he turns to the first person to the left of the person he was facing and shoots him. After the second round, he turns to the second person to the left of the person he was facing and shoots him, similarly, after the Nth round, he turns to the Nth person to the left of the person he was facing and shoots him. While counting, we will also count the vacant spots. For example, if the person numbered 3 is already shot dead, the 2nd person to the left of 2 is 4, not 5, since we have to count 3 as well. Also, while counting, if the count exceeds the total count of the table(i.e M), we go around the circle again. For example, if there are 8 people, the 10th person to the left of 1 is 3, and the 8th person to the left of 1 is 1 himself.
If at the end of any round, the shooter is pointing to a person who has already been shot in one of the previous rounds, he does not shoot anyone in that round and continues to the next round.
Also, suppose, all the persons are already shot in the nth round, where n < N, then in the (n+1)th round, the shooter shoots himself, and the game ends, it does not move to the (n+2)th round.
For example, suppose M = 8, N = 4, and K = 3, after the first round, the shooter shoots 4, after the second round, the shooter shoots 6, after the 3rd round, he shoots 1, and after the 4th and final round, he shoots 5.
Based on the information given above, answer the questions that follow.
Suppose M = 12, K =6, who is the last person to be killed if N = 5000?
If M = 12, we can denote them as follows:
Also, K = 6, so after round 1, 7 is killed, after round 2, 9 is killed and so on as follows:
Hence, after Round 24, we see the pattern repeating, hence, all those who have been killed till the 24th round comprise the persons killed. No one is killed post the 24th round.
The last person to be killed is 1 in the 10th round.
Directions. Read the following information carefully and answer the questions that follow.
Eight friends  A, B, C, D, E, F, G and H  speak one languages each viz, Bengali, Telugu, Marathi, Tamil, Urdu, Gujarati, Kannada and Punjabi, not necessarily in the same order. Each of them also owns a car from among Volkswagen, Jaguar, BMW, Skoda, Audi, Maruti, Mercedes and Scorpio, not necessarily in the same order. It is also known that:
1) A speaks Bengali and does not own the Jaguar while D does not speak Marathi and owns the Skoda.
2) E neither speaks Marathi nor Gujarati but owns either the Audi or BMW.
3) The Jaguar is not owned by C or F while H owns the Scorpio.
4) The person who speaks Gujarati owns the Maruti but is not C. The person who speaks Marathi owns neither the Mercedes nor the Scorpio.
5) Gujarati is not the language which G, who owns the Mercedes, speaks.
6) C does not speak Telugu and does not own the Volkswagen.
7) The person who owns the Jaguar speaks neither Marathi nor Urdu and the person who speaks Punjabi owns neither the Mercedes nor the Jaguar.
8) D speaks Tamil while G speaks neither Urdu nor Telugu.
If C owns the Audi, which car does E own?
Consider the solution to the first question.
If C owns the Audi, E owns the BMW.
Hence, option B is correct.
A block in Queens has 25 blocks arranged in a 5x5 grid as shown below. Friendly Neighborhood Spiderman patrols the area and fights crime in this region only.
The rows of the block are denoted by Roman numbers and the columns are denoted by Alphabets. The numbers in the cells denote the height of the building. For eg, block CIII has 6 units in height.
Spiderman moves from the terrace of one building to another either horizontally or vertically. He has a certain amount of webfluid in his webshooters. If Spiderman has to move from a building of low height to a building of high height, he has to use exactly 2ml of fluid to reach up 1 unit height. For eg: If Spiderman moves from DI(4units) to EI(6units) he has to climb 2 units and thus has to spend 4ml of fluid to do so.
However, while moving from a high building to a low building, Spiderman does not have to use any of the webfluid.
Unless stated otherwise, SpiderMan uses web fluid to move from one building to another
What is the minimum amount of webfluid(in mL) required for SpiderMan to reach from BII to DV?
Both the building are of the height 9 units. Thus in order to ensure that SpiderMan uses least amount of the webfluid, he has to ensure he doesn't move towards the building which has less height, as Spiderman has to use the fluid to regain the height of 9 units.
If Spiderman Moves from BII to BIII to BIV to BV to CV to DV, it has to climb only 3 units overall. Thus 6ml of fluid will be consumed as 2ml is required for 1 unit floor climbing
A block in Queens has 25 blocks arranged in a 5x5 grid as shown below. Friendly Neighborhood Spiderman patrols the area and fights crime in this region only.
The rows of the block are denoted by Roman numbers and the columns are denoted by Alphabets. The numbers in the cells denote the height of the building. For eg, block CIII has 6 units in height.
Spiderman moves from the terrace of one building to another either horizontally or vertically. He has a certain amount of webfluid in his webshooters. If Spiderman has to move from a building of low height to a building of high height, he has to use exactly 2ml of fluid to reach up 1 unit height. For eg: If Spiderman moves from DI(4units) to EI(6units) he has to climb 2 units and thus has to spend 4ml of fluid to do so.
However, while moving from a high building to a low building, Spiderman does not have to use any of the webfluid.
Unless stated otherwise, SpiderMan uses web fluid to move from one building to another
If SpiderMan has exactly 7ml of fluid, Then from which of the following buildings he can reach the Building AI?
Spiderman has total number of 7ml of the fluids. Thus he can maximum climb 3 units of height in his journey to other buildings.
Let us start with rowI,
BI to AI will be possible as it requires 1 unit climbing. Thus it can also be reached from CI. From DI it will take minimum of 4 units, hence DI is not possible. Similarly E1 is also not possible in 3 units climb.
In row II: Spiderman can reach AI from AII and BII without using any webfluids. Spiderman will have to climb 1 unit up from CII to BII after which he'll reach AI. DII will require more than 3 units of climbing, hence it is not possible. We see that from EII spider man can visit EIII then DIII, from which it can go to CIII without spending any of the webfluid. From here it can go 1 unit up and reach BIII and 2 unit up to BII. From this, it can easily reach AI. From AIII, it will take more than 3 units to reach AI
Thus till now we have found out that BI, CI, AII, BII, CII, EII, BIII, CIII, DIII and EIII are possible buildings
Now coming at rowIV
From AIV \rightarrow\→ BIV \rightarrow\→ BII \rightarrow\→ BII \rightarrow\→ AII \rightarrow\→ AI is possible, thus AIV and BIV are the buildings which satisfy the criteria. CIV,DIV and EIV will require more than 3 units of upward travel and hence rejected.
For row V, AV is not possible.
From BV \rightarrow\→ BIV which requires 1 step up. And from BIV it takes 2 units of upward movement to reach AI. Since height of BV < CV < DV, CV and DV are also possible answer.
Total 15 such buildings are there from which Spiderman can reach AI in 7ml of webfluids
The buildings are BI, CI, AII,BII,CII, EII, BIII,CIII, DIII, EIII, AIV, BIV, BV, CV and DV.
BV is one of the buildings
A block in Queens has 25 blocks arranged in a 5x5 grid as shown below. Friendly Neighborhood Spiderman patrols the area and fights crime in this region only.
The rows of the block are denoted by Roman numbers and the columns are denoted by Alphabets. The numbers in the cells denote the height of the building. For eg, block CIII has 6 units in height.
Spiderman moves from the terrace of one building to another either horizontally or vertically. He has a certain amount of webfluid in his webshooters. If Spiderman has to move from a building of low height to a building of high height, he has to use exactly 2ml of fluid to reach up 1 unit height. For eg: If Spiderman moves from DI(4units) to EI(6units) he has to climb 2 units and thus has to spend 4ml of fluid to do so.
However, while moving from a high building to a low building, Spiderman does not have to use any of the webfluid.
Unless stated otherwise, SpiderMan uses web fluid to move from one building to another
If SpiderMan has exactly 7ml of fluid, Then from which of the following buildings he cannot reach the Building AI?
Spiderman has total number of 7ml of the fluids. Thus he can maximum climb 3 units of height in his journey to other buildings.
Let us start with rowI,
BI to AI will be possible as it requires 1 unit climbing. Thus it can also be reached from CI. From DI it will take minimum of 4 units, hence DI is not possible. Similarly E1 is also not possible in 3 units climb.
In row II: Spiderman can reach AI from AII and BII without using any webfluids. Spiderman will have to climb 1 unit up from CII to BII after which he'll reach AI. DII will require more than 3 units of climbing, hence it is not possible. We see that from EII spider man can visit EIII then DIII, from which it can go to CIII without spending any of the webfluid. From here it can go 1 unit up and reach BIII and 2 unit up to BII. From this, it can easily reach AI. From AIII, it will take more than 3 units to reach AI
Thus till now we have found out that BI, CI, AII, BII, CII, EII, BIII, CIII, DIII and EIII are possible buildings
Now coming at rowIV
From AIV \rightarrow\→ BIV \rightarrow\→ BII \rightarrow\→ BII \rightarrow\→ AII \rightarrow\→ AI is possible, thus AIV and BIV are the buildings which satisfy the criteria. CIV,DIV and EIV will require more than 3 units of upward travel and hence rejected.
For row V, AV is not possible.
From BV \rightarrow\→ BIV which requires 1 step up. And from BIV it takes 2 units of upward movement to reach AI. Since height of BV < CV < DV, CV and DV are also possible answer.
Total 15 such buildings are there from which Spiderman can reach AI in 7ml of webfluids
The buildings are BI, CI, AII,BII,CII, EII, BIII,CII, DIII, EIII, AIV, BIV, BV, CV and DV.
CIV is not one of the buildings
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