CAT 2019 Question Paper DILR Slot 2

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:
1. Gordon(G) did not score consecutively in any two rounds.
2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^nd, 4^th and 6^th rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option B is correct.

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:
1. Gordon(G) did not score consecutively in any two rounds.
2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^nd, 4^th and 6^th rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option D is correct.

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:
1. Gordon(G) did not score consecutively in any two rounds.
2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^nd, 4^th and 6^th rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option D is correct.

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:
1. Gordon(G) did not score consecutively in any two rounds.
2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^nd, 4^th and 6^th rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and  TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.  

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option D is correct.

The states which satisfy the condition given in the question:

Maharashtra, Mizoram, Sikkim, Goa, Arunachal, Kerla, Meghalaya.....7 states

The ‘Heavy Monsoon States’ have a negative deviation: Arunachal, Kerla, Meghalaya

= 3/7x100=42.86%

Option A

All the states which satisfy the condition for ' Low monsoon state' are Gujrat (+25%), Karnataka (+20%), Rajasthan (+15), MP (+10%), Assam (-10%), WB (-30%), Jharkhand (-35%), Delhi (-40%) and Manipur (-60%).

The median of all the deviation is -10% Assam.

The states Assam, WB, Jharkhand, Delhi and Manipur satisfy the conditions given in the question.

The actual rainfall of all these states in 2019 are 600,600,400,300,400

Average of these states= 2300/5=460mm 

The actual rainfall in Gujarat in 2019 is 20% more than the rainfall in 2009.

So If the actual rainfall in 2009 = x mm

Then the actual rainfall in 2019= 1.2x mm

Actual rainfall in 2019= 600mm

Then, actual rainfall in 2009 = 500mm
As deviation is +25% so average 2009 - 2018 is 600/1.25 = 480
LPA 2019 = (480×10 - 500 + 600 )/10 = 490mm

Answer C

All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each.

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.

5x4=20, 10x3=30, 15x2=30

The total possible with considering the minimum number of questions of each type = 20+30+30=80 marks.

Rest 20 marks are possible by the following cases: {5,5,5,5} {5,5,10} {10,10} {5,15}

ET contained more questions than MT.

Thus MT cannot consider the case {5,5,5,5}

The number of questions in each case:

1) {5,5,5,5} = 9+4 =13 questions

2) {5,5,10} = 9+3 =12 questions

3) {10,10} = 9+2 =11 questions

4) {5,15} = 9+2 =11 questions

Considering MT and ET together, each faculty member prepared the same number of questions. The total number of questions should be multiple of 6, thus the total number of questions will be 24.

For ET and MT, there are 2 cases :

{5,5,5,5}{5,15}

{5,5,5,5}{10,10}

According to the statement (i), Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks. Thus {10,10} case is not possible.

MT {5,5,5,5,5,10,10,10,15,15,15}

ET {5,5,5,5,5,5,5,5,10,10,10,15,15}

From statement (i),(ii),(iv),(v), every other faculty member prepared two questions for MT.
we can create the following table:

{ Annie(A), Beti(B), Chetan(C), Dave (D), Fakir(F) }

There are 24 questions in total so each faculty will make 4 questions.

We can create the following table for ET.

Hence the correct option is D

All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each.

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.

5x4=20, 10x3=30, 15x2=30

The total possible with considering the minimum number of questions of each type = 20+30+30=80 marks.

Rest 20 marks are possible by the following cases: {5,5,5,5} {5,5,10} {10,10} {5,15}

ET contained more questions than MT.

Thus MT cannot consider the case {5,5,5,5}

The number of questions in each case:

1) {5,5,5,5} = 9+4 =13 questions

2) {5,5,10} = 9+3 =12 questions

3) {10,10} = 9+2 =11 questions

4) {5,15} = 9+2 =11 questions

Considering MT and ET together, each faculty member prepared the same number of questions. The total number of questions should be multiple of 6, thus the total number of questions will be 24.

For ET and MT, there are 2 cases :

{5,5,5,5}{5,15}

{5,5,5,5}{10,10}

According to the statement (i), Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks. Thus {10,10} case is not possible.

MT {5,5,5,5,5,10,10,10,15,15,15}

ET {5,5,5,5,5,5,5,5,10,10,10,15,15}

From statement (i),(ii),(iv),(v), every other faculty member prepared two questions for MT.
we can create the following table:

{ Annie(A), Beti(B), Chetan(C), Dave (D), Fakir(F) }

There are 24 questions in total so each faculty will make 4 questions.

We can create the following table for ET.

Hence the correct option is A

All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each.

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.

5x4=20, 10x3=30, 15x2=30

The total possible with considering the minimum number of questions of each type = 20+30+30=80 marks.

Rest 20 marks are possible by the following cases: {5,5,5,5} {5,5,10} {10,10} {5,15}

ET contained more questions than MT.

Thus MT cannot consider the case {5,5,5,5}

The number of questions in each case:

1) {5,5,5,5} = 9+4 =13 questions

2) {5,5,10} = 9+3 =12 questions

3) {10,10} = 9+2 =11 questions

4) {5,15} = 9+2 =11 questions

Considering MT and ET together, each faculty member prepared the same number of questions. The total number of questions should be multiple of 6, thus the total number of questions will be 24.

For ET and MT, there are 2 cases :

{5,5,5,5}{5,15}

{5,5,5,5}{10,10}

According to the statement (i), Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks. Thus {10,10} case is not possible.

MT {5,5,5,5,5,10,10,10,15,15,15}

ET {5,5,5,5,5,5,5,5,10,10,10,15,15}

From statement (i),(ii),(iv),(v), every other faculty member prepared two questions for MT.
we can create the following table:

{ Annie(A), Beti(B), Chetan(C), Dave (D), Fakir(F) }

There are 24 questions in total so each faculty will make 4 questions.

We can create the following table for ET.

Hence the correct option is D

All six sections had a common midterm (MT) and a common end term (ET) worth 100 marks each.

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.

5x4=20, 10x3=30, 15x2=30

The total possible with considering the minimum number of questions of each type = 20+30+30=80 marks.

Rest 20 marks are possible by the following cases: {5,5,5,5} {5,5,10} {10,10} {5,15}

ET contained more questions than MT.

Thus MT cannot consider the case {5,5,5,5}

The number of questions in each case:

1) {5,5,5,5} = 9+4 =13 questions

2) {5,5,10} = 9+3 =12 questions

3) {10,10} = 9+2 =11 questions

4) {5,15} = 9+2 =11 questions

Considering MT and ET together, each faculty member prepared the same number of questions. The total number of questions should be multiple of 6, thus the total number of questions will be 24.

For ET and MT, there are 2 cases :

{5,5,5,5}{5,15}

{5,5,5,5}{10,10}

According to the statement (i), Annie prepared the fifth question for both MT and ET. For MT, this question carried 5 marks. Thus {10,10} case is not possible.

MT {5,5,5,5,5,10,10,10,15,15,15}

ET {5,5,5,5,5,5,5,5,10,10,10,15,15}

From statement (i),(ii),(iv),(v), every other faculty member prepared two questions for MT.
we can create the following table:

{ Annie(A), Beti(B), Chetan(C), Dave (D), Fakir(F) }

There are 24 questions in total so each faculty will make 4 questions.

We can create the following table for ET.

Hence the correct option is D

We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

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. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2. 

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column. 

The total amount of money (in rupees) in the three pouches kept in the first column of the second row=13

Correct answer 13

We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

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. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 8

We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

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. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 2

We can make the following table from "the total amount of money kept in the three pouches in the first column of the third row is Rs. 4."

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

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. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 ...{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 3

If all the doctors served the patients one after the other, then in 2.5 hrs, Ben will serve 15 patients, Kane will serve 10 patients and Wayne will serve 6 patients.

A total of 31 patients can be served on a particular day.

If all the doctors served the patients one after the other, then in 2.5 hrs, Ben will serve 15 patients, Kane will serve 10 patients and Wayne will serve 6 patients.

Ben will earn = 15*100=Rs 1500 

Kane will earn = 10*200=Rs 2000

Wayne will earn = 6*300=Rs 1800

So Kane will earn the maximum amount in consultation charges on that day.

Option B

Mr Singh is 13th in the sequence on all the three days. 

The following table will show the sequence for Monday, Wednesday and Friday.

He will stay the longest when the 13th guy is served by Doctor Wayne.

From the table, on Monday he had to wait at the clinic for the maximum duration: till 10:15.

Option C

On Thursday, the preference order for the patients is Wayne, Ben and Kane. 

The first two customers will be served by Wayne and Ben. While Kane will be empty for the first 15 mins. Then he and Ben will serve the next two customers and Wayne will be empty for 5 minutes as shown in the figure below.

As we can see that the cycle will repeat after every 30 mins.

So all three doctors are never simultaneously free.

Option D

From statement 1 and 2,Each team speaks exactly four languages. English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team, multiple options are possible.

In the following tables: A, B, C can be any team among Team 1,Team 2,Team 3.

From the data given in the question, the person who speaks Arabic also speaks French. Thus the only option possible is 'Table 2'.

According to statement 4, "Paula and Sally are together in exactly two teams."

Sally knows Basque, thus, she will be in group A and B, with Paula.

According to statement 5, Robert(Arabic) is in Team 1 and Quentin(Dutch) is in Team 3.

Thus, Group C is Team 1 and Group A is Team 3.

From the table, the correct option is C.

From statement 1 and 2,Each team speaks exactly four languages. English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team, multiple options are possible.

In the following tables: A, B, C can be any team among Team 1,Team 2,Team 3.

From the data given in the question, the person who speaks Arabic also speaks French. Thus the only option possible is 'Table 2'.

According to statement 4, "Paula and Sally are together in exactly two teams."

Sally knows Basque, thus, she will be in group A and B, with Paula.

According to statement 5, Robert(Arabic) is in Team 1 and Quentin(Dutch) is in Team 3.

Thus, Group C is Team 1 and Group A is Team 3.

From the table, the correct option is C.

From statement 1 and 2,Each team speaks exactly four languages. English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team, multiple options are possible.

In the following tables: A, B, C can be any team among Team 1,Team 2,Team 3.

From the data given in the question, the person who speaks Arabic also speaks French. Thus the only option possible is 'Table 2'.

According to statement 4, "Paula and Sally are together in exactly two teams."

Sally knows Basque, thus, she will be in group A and B, with Paula.

According to statement 5, Robert(Arabic) is in Team 1 and Quentin(Dutch) is in Team 3.

Thus, Group C is Team 1 and Group A is Team 3.

From the table, the correct option is C.

From statement 1 and 2,Each team speaks exactly four languages. English and Chinese are spoken by all three teams, Basque and French by exactly two teams and the other languages by exactly one team, multiple options are possible.

In the following tables: A, B, C can be any team among Team 1,Team 2,Team 3.

From the data given in the question, the person who speaks Arabic also speaks French. Thus the only option possible is 'Table 2'.

According to statement 4, "Paula and Sally are together in exactly two teams."

Sally knows Basque, thus, she will be in group A and B, with Paula.

According to statement 5, Robert(Arabic) is in Team 1 and Quentin(Dutch) is in Team 3.

Thus, Group C is Team 1 and Group A is Team 3.

From the table, the correct option is A.

We can make the following table from the web chart given in the question:

If we consider the total cost in the year

2016 as 100, then according to Statement 1, the total revenue in 2016 must be 200.

The store’s revenue doubled from 2016 to 2017, thus the total revenue in the year 2017 = 400.

We can find the revenue for the individual department in the year 2017, from the table.

There was no profit from the Electronics department in 2017, thus, we can find the total cost in 2017= 300

Considering the statement 4, we can find the total revenue in 2018 and tabulate the following table.

The percentage profit of the store in 2018= (250-200)/200= 25%

We can make the following table from the web chart given in the question:

If we consider the total cost in the year

2016 as 100, then according to Statement 1, the total revenue in 2016 must be 200.

The store’s revenue doubled from 2016 to 2017, thus the total revenue in the year 2017 = 400.

We can find the revenue for the individual department in the year 2017, from the table.

There was no profit from the Electronics department in 2017, thus, we can find the total cost in 2017= 300

Considering the statement 4, we can find the total revenue in 2018 and tabulate the following table.

The ratio of revenue generated from the Produce department in 2017 to that in 2018 = 160:100= 8:5

We can make the following table from the web chart given in the question:

If we consider the total cost in the year

2016 as 100, then according to Statement 1, the total revenue in 2016 must be 200.

The store’s revenue doubled from 2016 to 2017, thus the total revenue in the year 2017 = 400.

We can find the revenue for the individual department in the year 2017, from the table.

There was no profit from the Electronics department in 2017, thus, we can find the total cost in 2017= 300

Considering the statement 4, we can find the total revenue in 2018 and tabulate the following table.

Profit in 2016 = 200-100= 100
Profit in the electronic department in 2016= 100-30=70
The total profits= for the store in 2016 were from the Electronics department 70%

We can make the following table from the web chart given in the question:

If we consider the total cost in the year

2016 as 100, then according to Statement 1, the total revenue in 2016 must be 200.

The store’s revenue doubled from 2016 to 2017, thus the total revenue in the year 2017 = 400.

We can find the revenue for the individual department in the year 2017, from the table.

There was no profit from the Electronics department in 2017, thus, we can find the total cost in 2017= 300

Considering the statement 4, we can find the total revenue in 2018 and tabulate the following table.

Profit percentage in 2017= (400-300)/300 %= 33.33%

Profit percentage in 2018= (250-200)/200 %= 25%

The approximate difference in profit percentages of the store in 2017 and 2018= (33.33-25)%= 8.33%

Option C.

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, "Among the 200 students who preferred Sunita as student union president, 80% supported proposal A."

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

The number of students who preferred Sunita and the proposal A=160

=160/250= 64%

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, "Among the 200 students who preferred Sunita as student union president, 80% supported proposal A."

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

The percentage of the students surveyed who did not support proposal A preferred Ragini as student union president = 210/250=84%

Answer 84 

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, "Among the 200 students who preferred Sunita as student union president, 80% supported proposal A."

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

According to the Venn diagram,  the students surveyed who supported both proposals A and B preferred Sunita as student union president $$\frac{50}{50+50}$$ % =50%

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, "Among the 200 students who preferred Sunita as student union president, 80% supported proposal A."

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

From the diagram, we can understand that option A is correct.