[SLOT 2]
INSTRUCTIONTen players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2,..., players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10, 1 and 2 in Round 7 and so on. The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.
The following information is known about Rounds 1 through 6:
 Gordon did not score consecutively in any two rounds.
 Eric and Fatima both scored in a round.
The following information is known about Rounds 7 through 10:
 Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.
 Joshin scored in Round 7, while Amita scored in Round 10.
 No player scored in all the four rounds.
Question 1:
What were the scores of Chen, David, and Eric respectively after Round 3?
 $3,3,3$
 $3,0,3$
 $3,6,3$
 $3,3,0$
Question 2:
Which three players were in the last three positions after Round 4?
 Bala, Chen, Gordon
 Bala, Hansa, lkea
 Bala, Ikea, Joshin
 Hansa, Ikea, Joshin
Question 3:
Which player scored points in maximum number of rounds?
 Ikea
 Amita
 Chen
 Joshin
Question 4:
Which players scored points in the last round?
 Amita, Chen, Eric
 Amita, Chen, David
 Amita, Bala, Chen
 Amita, Eric, Joshin
From the given information, we can see that each player participated in only a certain number of rounds. The following table provides the rounds that each person participated in (split by Rounds 1  6 and Rounds 7  10):
Person 
Rounds participated From 1 to 6 
Rounds participated From 7 to 10 
Amita 
1, 6 
7, 8, 9, 10 
Bala 
1, 2 
7, 8, 9, 10 
Chen 
1, 2, 3 
8, 9, 10 
David 
1,2, 3,4 
9, 10 
Eric 
1,2, 3, 4,5 
10 
Fatima 
1,2, 3, 4, 5,6 
 
Gordon 
2, 3, 4, 5, 6 
7 
Hansa 
3, 4, 5, 6 
7,8 
Ikea 
4, 5,6 
7, 8, 9 
Joshin 
5, 6 
7, 8, 9, 10 
Rounds 1 to 6:
The points that Amita scored till Round 6 is 8. Among the rounds 1 to 6. he participated only in Round 1 and Round 6. Hence, he must have scored 7 and 1 points in Round 1 and Round 6 in any order.
Since Bala scored 2 points till Round 6, he must have scored 1 point in Round 1 and 1 point in Round 2.
Since Bala scored 1 point in Round 1. Amita could not have scored 1 point in Round 1. Hence. Amita must have scored 7 points in Round 1 and 1 point in Round 6.
Gordon scored 17 points in Rounds 2 to 6. It is given that from Round 1 to Round 6. Gordon did not score consecutively in any two rounds. Since it is not possible to score 17 points in two rounds. Gordon must have scored 17 points in three rounds  Round 2, Round 4 and Round 6.
In these three rounds, he must have scored 7, 7 and 3 points, in any order.
Joshin scored 14 points in Rounds 5 and 6. Hence, he must have scored 7 points in each of these two rounds. Since Joshin scored 7 points in Round 6, Gordon cannot score 7 points in Round 6. Hence, Gordon must have scored 3 points in Round 6 and 7 points in each of Round 2 and Round 4.
Ikea scored 2 points till Round 6. In the first 6 rounds, he played in only Rounds 4. 5 and 6, of which he scored 0 points in Round 6. Hence, he must have scored 1 point each in Round 4 and Round 5.
Hansa scored 1 point in the first 6 rounds. He played in only Rounds 3 to 6. Of these rounds, he did not score any point in Round 6. In Rounds 4 and 5. Ikea scored 1 point. Hence. Hansa could not have scored 1 point in Rounds 4 and 5.
Therefore, Hansa must have scored 1 point in Round 3.
Eric scored 3 points in the first 6 rounds. He could have scored 1 point in each of three rounds OR 3 points in one round. He participated only in Rounds 1 to 5. Of these rounds, Bala scored 1 point each in Round 1 and 2. Hansa scored 1 point in Round 3, while Ikea scored 1 point in Round 4 and Round 5. Hence. Eric could not have scored 1 point each in three rounds. Hence. Eric must have scored 3 points in one round.
Similarly. Fatima cannot score 7. 1. 1, 1 point in four rounds (among the first 6 rounds) and must have scored 7 points and 3 points in two rounds.
Given that Eric and Fatima both scored in one round. In the round that both of them scored. Eric must have scored 3 points, while Fatima must have score 7 points (since both of them cannot score the same number of points in a round).
Among the rounds that both Eric and Fatima played (i.e.. Rounds 1 to 5), the only round in which Fatima could have scored 7 points is Round 3. Hence, in round 3. Fatima must have scored 7 points and Eric must have scored 3 points. Eric should have scored 0 in all the other rounds from rounds 1 to 6.
Chen must have scored 3 points in one of the rounds, while David must have scored 3 points in two rounds each. Irrespective of which round Chen scored 3 points, David must have scored 3 points in Round 4. In the first two rounds, Chen and David must have scored 3 points in any order. Hence, Fatima could have scored 3 points only in Round 5.
Rounds 7 to 10:
In rounds 710, Chen must have scored 3 points (since he scored 6 points in total, after Round 10). Since Chen played 3 rounds from 710 and he scored in three consecutive rounds among these rounds, he must have scored 1 point each in Round 8, Round 9 and Round 10.
Ikea scored 15 points in Rounds 7 to 10. Since he played only in Rounds 7, 8 and 9. he must have scored 7, 7 and 1 points in these three rounds, in any order. Since Chen scored 1 point each in Rounds 8 and 9. Ikea could have
scored 1 point only in Round 7. In Rounds 8 and 9, he must have scored 7 points each.
Joshin scored 3 points in Rounds 7 to 10. He could not have scored 1 point in each of three rounds (since Ikea and Chen are the only persons who scored in three consecutive rounds and no one else scored in any two consecutive rounds).
Hence, Joshin must have scored 3 points in one round. Since it is given that Joshin scored in Round 7, he must
have scored 3 points in Round 7 and no points in the other rounds.
In Round 7, Ikea scored 1 point and Joshin scored 3 points. The only person who could have scored 7 points in Round 7 is Amita.
Hansa would have scored 3 points in Round 8 (since he scored 0 in Round 7).
Eric scored 7 points in Rounds 7 to 10. Among the rounds that he played (Rounds 9, 10), he must have scored 7 points in Round 10 (since Ikea scored 7 points in Round 9).
David did not score any points in Rounds 7 through 10.
Amita must have scored 3 points in Round 10 (since Amita scored in Round 10) and Bala must have scored 3 points in Round 9.
The table below provides the points scored by all the players in all the rounds. An X' indicates that the player did not participate in that round.
A 
B 
C 
D 
E 
F 
G 
H 
1 
J 

1 
7 
1 
3/0 
0/3 
0 
0 
X 
X 
X 
X 
2 
X 
1 
0/3 
3/0 
0 
0 
7 
X 
X 
X 
3 
X 
X 
0 
0 
3 
7 
0 
1 
X 
X 
4 
X 
X 
X 
3 
0 
0 
7 
0 
1 
X 
5 
X 
X 
X 
X 
0 
3 
0 
0 
1 
7 
6 
1 
X 
X 
X 
X 
0 
3 
0 
0 
7 
7 
7 
0 
X 
X 
X 
X 
0 
0 
1 
3 
8 
0 
0 
1 
X 
X 
X 
X 
3 
7 
0 
9 
0 
3 
1 
0 
X 
X 
X 
X 
7 
0 
10 
3 
0 
1 
0 
7 
X 
X 
X 
X 
0 
INSTRUCTION
In the table below the check marks indicate all languages spoken by five people: Paula, Quentin, Robert, Sally and Terence. For example, Paula speaks only Chinese and English.
These five people form three teams, Team 1, Team 2 and Team 3. Each team has either 2 or 3 members. A team is said to speak a particular language if at least one of its members speak that language.
The following facts are known.
 Each team speaks exactly four languages and has the same number of members.
 English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team.
 None of the teams include both Quentin and Robert.
 Paula and Sally are together in exactly two teams.
 Robert is in Team 1 and Quentin is in Team 3.
Question 5:
Who among the following four is not a member of Team 2?
 Quentin
 Paula
 Sally
 Terence
Question 6:
Who among the following four people is a part of exactly two teams?
 Sally
 Robert
 Paula
 Quentin
Question 7:
Who among the five people is a member of all teams?
 Sally
 No one
 Terence
 Paula
Question 8:
Apart from Chinese and English, which languages are spoken by Team 1?
 Arabic and Basque
 Arabic and French
 Basque and Dutch
 Basque and French
From (1) and (5), the persons in Team 1 speak English. Chinese. Arabic and French. (Robert speaks both Arabic and French).
From (1) and (5), the persons in Team 3 speak. English, Chinese and Dutch. (Quentin speaks Dutch and English). Since each person speaks two languages and each team speaks exactly four languages, we need to find one person for Team 3. who speaks one language among English, Chinese and Dutch and a different language apart from these three.
Since, Paula and Sally together speak Basque, Chinese and English and they are together in exactly two teams, they cannot be in Team 1. They must be in Teams 2 and 3.
Hence, from (5) and the above, Paula, Quentin and Sally, (Basque. Chinese. Dutch and English) are in Team 3. Since there are three persons in Team 3. Teams 1 and 2 should also have three persons each. Team 1 speaks, English, Chinese, Arabic and French. Robert (Arabic and French) is one of the team members. Now, two more persons, who speak languages among the above four are to be selected. It is possible only with Paula and Terence.
From (2) Basque and French are spoken by two teams. Hence, Team 2 speaks these two languages. Paula and Sally are there in Team 2 (Basque, Chinese and English). We need to find one more person, who speaks one of these three languages and French. It is possible with only Terence.
Team 
1 
2 
3 
Persons 
Robert, Paula, Terence 
Paula, Sally, Terence 
Quentin, Paula, Sally 
Languages 
Arabic, Chinese, English, French 
Basque, Chinese, English, French 
Basque, Chinese, Dutch, English 
INSTRUCTION
Students in a college are discussing two proposals 
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.
A student does not necessarily support either of the two proposals.
In an upcoming election for student union president, there are two candidates in fray: Sunita and Ragini. Every student prefers one of the two candidates.
A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.
 250 students supported proposal A and 250 students supported proposal B.
 Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.
 Among those who preferred Ragini, 30% supported proposal A.
 20% of those who supported proposal B preferred Sunita.
 40% of those who did not support proposal B preferred Ragini.
 Every student who preferred Sunita and supported proposal B also supported proposal A.
 Among those who preferred Ragini, 20% did not support any of the proposals.
Question 9:
Among the students surveyed who supported proposal A, what percentage preferred Sunita for student union president?
Question 10:
What percentage of the students surveyed who did not support proposal A preferred Ragini as student union president?
Question 11:
What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?
 50
 40
 20
 25
Question 12:
How many of the students surveyed supported proposal B, did not support proposal A and preferred Ragini as student union president?
 200
 150
 210
 40
The set of students who like Sunita and Ragini are disjoint sets.
Hence, the Venn diagram can be drawn as follows
There are 500 students in all.
From statement (2)
Sunita = 200. Hence, Ragini = 300.
From statement (1) A (Sunita) + A (Ragini) = 250 and B (Sunita) + B (Ragini) = 250.
From (2), A (Sunita) = 160. Hence, A (Ragini) = 90.
From (4), B (Sunita) = 20 % of 250 = 50. Hence, B (Ragini) = 200.
From (6), g (Sunita) = 50 and hence, b (Sunita) = 0 and a (Sunita) = 110. Hence, n (Sunita) = 40.
From (7), n (Ragini) = 60
It is given that 250 support B, hence the other 250 do not support B.
From (5), (a + n) of Ragini = 40 % of 250 = 100. Hence, a (Ragini) = 40.
Thus, the final solution is as follows.
INSTRUCTION
The first year students in a business school are split into six sections. In 2019 the Business Statistics course was taught in these six sections by Annie, Beti, Chetan, Dave, Esha, and Fakir. All six sections had a common midterm (MT) and a common endterm (ET) worth 100 marks each. ET contained more questions than MT. Questions for MT and ET were prepared collectively by the six faculty members. Considering MT and ET together, each faculty member prepared the same number of questions. Each of MT and ET had at least four questions that were worth 5 marks, at least three questions that were worth 10 marks, and at least two questions that were worth 15 marks. In both MT and ET, all the 5mark questions preceded the 10mark questions, and all the 15 mark questions followed the 10mark questions.
The following additional facts are known.
 Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks.
 Annie prepared one question for MT. Every other faculty member prepared more than one questions for MT.
 All questions prepared by a faculty member appeared consecutively in MT as well as ET.
 Chetan prepared the third question in both MT and ET; and Esha prepared the eighth question in both.
 Fakir prepared the first question of MT and the last one in ET. Dave prepared the last question of MT and the first one in ET.
Question 13:
The second question in ET was prepared by:
 Dave
 Beti
 Chetan
 Esha
Question 14:
How many 5mark questions were there in MT and ET combined?
 12
 10
 13
 Cannot be determined
Question 15:
Who prepared 15mark questions for MT and ET?
 Only Dave and Fakir
 Only Beti, Dave, Esha and Fakir
 Only Esha and Fakir
 Only Dave, Esha and Fakir
Question 16:
Which of the following questions did Beti prepare in ET?
 Seventh question
 Fourth question
 Ninth question
 Tenth question
In the following the names of the faculties are referred by first letter of their names.
Given that each of MT and ET carries 100 marks. Each of MT and ET has at least four questions of 5 marks (4 x 5 = 20), at least three questions of 10 marks (3 x 10 =30) and at least two questions of 15 marks (2 x 15 = 30). These together add up to 80 marks. The remaining 20 marks can be of the following possible combinations. (Four questions of 5 marks) or (two questions of 10 marks) or (one question of 5 marks and 1question of 15 marks)or (two questions of five marks and one question of ten marks). Hence, the total number of questions in MT or ET can be 11 or 12 or 13. It is given that ET has more number of questions than in MT. Hence, MT has 11 or 12 questions.
It is given that the number of questions given by any faculty in both MT and ET together is the same. If MT has 12 questions and ET has 13 questions, or if MT has 11 questions and ET has 12 questions, this condition cannot be satisfied. Hence, MT has 11 (Five 5 marks, three 10 marks and three 15 marks) questions and ET has 13 questions (eight 5 marks, three ten marks and two 15 marks). This implies, each faculty has given four questions in MT and ET together. Since it is given that faculty A has given only one question in MT and each of the other faculties has given more than one question, each of the faculties B, C, D, E and F has given two questions in MT. This implies faculty A has given three questions in ET and all other faculties have given two questions each in ET. From the given data we get the following.
Except A, every other faculty gave at least two questions for MT and all the questions of a faculty appeared consecutively. Hence, 2^{nd} question in MT is given by F, 4^{th} by C, 10^{th} by D. B also has given two questions and both appeared consecutively. Hence, 6^{th} and 7^{th} questions are given by B and the 9^{th} question is given by E. In each test first all 5 marks questions appeared followed by 10 marks questions and then 15 marks questions. It can be understood that MT has five 5 marks, three 10 marks and three 15 marks. Hence, in MT questions 1 to 5 carry 5 marks each, 6 to 8 carry 10 marks each and 9 to 11 carry 15 marks each.
It can be understood that A has given three questions for ET and each of the others has given two questions. Hence, the 2^{nd} question of ET is given by D, the 4^{th} question by C, 6^{th} and 7^{th} questions by A, 9^{th} by E, 10^{th} and 11^{th} by B and 12^{th} by F. Since ET has eight 5 marks questions, three ten marks questions and two 15 marks questions, questions 1 to 8 of ET carry 8 marks each, 9 to 11 carry ten marks each, 12 and 13 carry 15 marks each. Thus, we get the following:
INSTRUCTION
Three doctors, Dr. Ben, Dr. Kane and Dr. Wayne visit a particular clinic Monday to Saturday to see patients. Dr. Ben sees each patient for 10 minutes and charges Rs. 100/. Dr. Kane sees each patient for 15 minutes and charges Rs. 200/, while Dr. Wayne sees each patient for 25 minutes and charges Rs. 300/.
The clinic has three rooms numbered 1, 2 and 3 which are assigned to the three doctors as per the following table.
The clinic is open from 9 a.m. to 11.30 a.m. every Monday to Saturday.
On arrival each patient is handed a numbered token indicating their position in the queue, starting with token number 1 every day. As soon as any doctor becomes free, the next patient in the queue enters that emptied room for consultation. If at any time, more than one room is free then the waiting patient enters the room with the smallest number. For example, if the next two patients in the queue have token numbers 7 and 8 and if rooms numbered 1 and 3 are free, then patient with token number 7 enters room number 1 and patient with token number 8 enters room number 3.
Question 17:
What is the maximum number of patients that the clinic can cater to on any single day?
 30
 12
 31
 15
Question 18:
The queue is never empty on one particular Saturday. Which of the three doctors would earn the maximum amount in consultation charges on that day?
 Dr. Wayne
 Dr. Kane
 Both Dr. Wayne and Dr. Kane
 Dr. Ben
Question 19:
Mr. Singh visited the clinic on Monday, Wednesday, and Friday of a particular week, arriving at 8:50 a.m. on each of the three days. His token number was 13 on all three days. On which day was he at the clinic for the maximum duration?
 Monday
 Friday
 Wednesday
 Same duration on all three days
Question 20:
On a slow Thursday, only two patients are waiting at 9 a.m. After that two patients keep arriving at exact 15 minute intervals starting at 9:15 a.m.  i.e. at 9:15 a.m., 9:30 a.m., 9:45 a.m. etc. Then the total duration in minutes when all three doctors are simultaneously free is
 15
 30
 10
 0
INSTRUCTION
To compare the rainfall data, India Meteorological Department (IMD) calculated the Long Period Average (LPA) of rainfall during period JuneAugust for each of the 16 states. The figure given below shows the actual rainfall (measured in mm) during JuneAugust, 2019 and the percentage deviations from LPA of respective states in 2018. Each state along with its actual rainfall is presented in the figure.
Question 21:
If a ‘Heavy Monsoon State’ is defined as a state with actual rainfall from JuneAugust, 2019 of 900 mm or more, then approximately what percentage of ‘Heavy Monsoon States’ have a negative deviation from respective LPAs in 2019?
 14.29
 57.14
 42.86
 75.00
Question 22:
If a ‘Low Monsoon State’ is defined as a state with actual rainfall from JuneAugust,2019 of 750 mm or less, then what is the median ‘deviation from LPA’ (as defined in the Yaxis of the figure) of ‘Low Monsoon States’?
 $10 \%$
 $30 \%$
 $20 \%$
 $10 \%$
Question 23:
What is the average rainfall of all states that have actual rainfall of 600 mm or less in 2019 and have a negative deviation from LPA?
 500 mm
 460 mm
 367 mm
 450 mm
Question 24:
The LPA of a state for a year is defined as the average rainfall in the preceding 10 years considering the period of JuneAugust. For example, LPA in 2018 is the average rainfall during 20092018 and LPA in 2019 is the average rainfall during 20102019. It is also observed that the actual rainfall in Gujarat in 2019 is 20% more than the rainfall in 2009. The LPA of Gujarat in 2019 is closest to
 $525 \mathrm{mm}$
 $490 \mathrm{mm}$
 $475 \mathrm{mm}$
 $505 \mathrm{mm}$
The actual rainfall in 2019 and the Long Period
Average (LPA) for the different states are as follows.
INSTRUCTION
Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3× 3 grid, as shown in the figure. Every pouch has a certain number of onerupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.
There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.
Question 25:
What is the total amount of money (in rupees) in the three pouches kept in the first column of the second row?
Question 26:
How many pouches contain exactly one coin?
Question 27:
What is the number of slots for which the average amount (in rupees) of its three pouches is an integer?
Question 28:
The number of slots for which the total amount in its three pouches strictly exceeds Rs. 10 is
The minimum and maximum and possible number of coins (overall) in each slot would be as follows.
It is given that the average amount of money kept in the nine pouches in any column or any row is an integer (a multiple of nine).
The total amount of money in the first column must be either 18 or 27 . The minimum value of the sum of money in the three slots is \(8+11+4=23\) and the maximum value is \(10+13+4=27\).
\(\therefore\) The number of coins in the first column of the three rows are \(10(2+4+4), 13(3+5+5)\) and \(4(1+2+1)\) Similarly in the third row, the sum must be 18 and in the second column, the sum must be 27 .
\(\therefore\) The number of coins in the second column is \(20(6+\) \(6+8)+3(1+1+1)\) and \(4(1+1+2)\)
The third column in the first row would be \(6(1+2+3)\) and the third column in the third row would be \(10(2+3\) +5)
In the last column, the value in the second row would be \(5416=38(6+12+20)\)
We have the following figure for the number of coins in the pouches in each slot.
INSTRUCTION
A large store has only three departments, Clothing, Produce, and Electronics. The following figure shows the percentages of revenue and cost from the three departments for the years 2016, 2017 and 2018. The dotted lines depict percentage levels. So for example, in 2016, 50% of store's revenue came from its Electronics department while 40% of its costs were incurred in the Produce department.
In this setup, Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost × 100%.
It is known that
 The percentage profit for the store in 2016 was 100%.
 The store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.
 There was no profit from the Electronics department in 2017.
 In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department.
Question 29:
What was the percentage profit of the store in 2018?
Question 30:
What was the ratio of revenue generated from the Produce department in 2017 to that in 2018?
 16: 9
 9: 16
 8: 5
 4: 3
Question 31:
What percentage of the total profits for the store in 2016 was from the Electronics department?
Question 32:
What was the approximate difference in profit percentages of the store in 2017 and 2018?
 33.3
 15.5
 8.3
 25.0
The percentage share in Revenue and cost in the different years are as follows.
Assume the cost of the store in 2016 to be 100.
As the profit percentage that year was 100, the revenue of the store in that year would be 200 .
\(\therefore\) Revenue in 2017 would be 400 and cost of the store in 2018 would be 200 . Given that in \(2017,30 \%\) of \(400=40 \%\) of cost.
\(\therefore 120=40 \%\) of cost or cost in 2017=300
In \(2018,50 \%\) of \(200=40 \%\) of Revenue
\(\therefore\) Revenue in \(2018=\frac{100}{0.4}=250\)
\(\therefore\) We have the following values for Revenue and cost for the different years.