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.
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.
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 54 -16 = 38(6 +12 +20)
We have the following figure for the number of coins in the pouches in each slot

. From 9-11: 30, we have 150 minutes. Doctors Ben, Kane and Wayne take 10,15 and 20
minutes respectively for each patient. Therefore Ben, Kane and Wayne can see 150/10=15,
150/15=10 and 150/25=6 respectively. Therefore Ben, Kane and Wayne can see a maximum
of 15,10 and 6 patients respectively every day. Sum =31

Given, on Saturday, the queue is not empty.
=> Each doctor sees the maximum number of patients on a day.
Given Ben, Kane and Wayne charge Rs.100, 200 and 300 respectively.
=> Ben earns 15 (100)= Rs.1500,Kane earns 10(200)
= Rs .2000, Wayne earns 6(300)= Rs .1800
Therefore, Kane earns the maximum amount.
Ans: (Dr. Kane)

Mr. Singh takes maximum duration when he enters Dr. Wayne’s room, who sees each
patient for 25 minutes

He was at the clinic for 85 minutes.
Similarly On Wednesday, he would meet Ben and he would be at the clinic for 70 minutes.
On Friday, he would meet Ben and he would be at the clinic for 70 minutes.
Therefore, Singh stays at the clinic for the maximum duration on Monday
Ans: (Monday)

The above pattern continues.
Hence, there is no time where all the doctors are simultaneously free.
Ans: (0)

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.

The students who supported proposal B but not A are b (Sunita) and b (Ragini). Among
them those supported Ragini are b (Ragini) 150. Ans: (150)

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.

The required value is 160/250*100 = 64

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.

The required answer is 210/250*100 = 84

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.

The required answer is 50/250*100 = 50

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

 The Low monsoon states are Gujarat, Karnataka, Rajasthan, MP, Assam, WB, Jharkhand,
Delhi and Manipur. The deviation from LPA for these states are 25,20,15, 10,-10,-30,-35,-40
and -60. The median value is -10

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

The states which have a negative deviation from LPA and have an actual rainfall of 600 mm
or less are Assam, WB, Jharkhand, Delhi and Manipur. The average rainfall in these states is
2300/5 = 460 mm

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

The actual rainfall in Gujarat in 2019 is 600 mm. The rainfall in Gujarat in 2009 was 500 mm.
As the value of 500 is replaced by 600 in calculating the LPA, the LPA would increase by 10
as it is the average of 10 years.
Ans: (490 mm)

The actual rainfall in 2019 and the Long Period

Average (LPA) for the different states are as follows.

The heavy monsoon states are Maharashtra, Sikkim, Mizoram, Goa, Arunachal, Kerala and
Meghalaya. Among these, Arunachal, Kerala and Meghalaya have a negative deviation from
respective LPAs in 2019. The required percentage = 3/7*100 = 42.86

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.

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.

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.

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.