CAT 2025 Slot 1

Based on the information about the initial positions of Aslam and Chhavi sitting next to each other, while Bashir and Davies have empty chairs on either side of their seats, the possible combinations are,

This has 4 possible combinations.

We are also given that Davies occupies Chair 2 after Turn 1 and Chhavi occupies Chair 7 after Turn 2. We know that the positions of Davies and Chhavi won't be changing after turn 2, as we know that in those turns, only the positions of Aslam and Bashir are being changed in the first 2 turns. Therefore, we can determine that Davies' initial position is chair 2, and Chhavi's initial position is chair 7. 

Out of the 4 possible combinations, Davies on chair 2 and Chhavi on chair 7 are only possible in one case, which is,

So, Bashir's initial position is chair 4.

Hence, the correct answer is 4.

Based on the information about the initial positions of Aslam and Chhavi sitting next to each other, while Bashir and Davies have empty chairs on either side of their seats, the possible combinations are,

This has 4 possible combinations.

We are also given that Davies occupies Chair 2 after Turn 1 and Chhavi occupies Chair 7 after Turn 2. We know that the positions of Davies and Chhavi won't be changing after turn 2, as we know that in those turns, only the positions of Aslam and Bashir are being changed in the first 2 turns. Therefore, we can determine that Davies' initial position is chair 2, and Chhavi's initial position is chair 7.

Out of the 4 possible combinations, Davies on chair 2 and Chhavi on chair 7 are only possible in one case, which is,

We now have the initial position, and we are told that all the members are adjacent only after turn 2 and turn 6.

We know that A changed his position in turn 1.

Turn 1: Aslam moves

If A changed his position anticlockwise, he would go to chair 5, and if he changed his position clockwise, he would go to chair 1.

If A ends in chair 5 after turn 1, then it is not possible for all of them to be adjacent to each other after B changes his position in turn 2.

So, A has to end in chair 1 after turn 1, so that B will have a chance in turn 2 for all of them to be adjacent.

Turn 2: Bashir moves

 For all of them to be adjacent after turn 2, B must change his position anticlockwise, and after turn 2, he would end up on chair 3.

Turn 3: Chhavi moves

We are given that Davies occupies chair 4 after turn 5, and for this to happen, chair 4 must be empty and cannot be occupied by anyone who does not move before turn 5. So, in turn 3, C must go anticlockwise and occupy chair 6, as if he goes clockwise, he would occupy chair 4, which must be empty for Davies.

Turn 4: Davies moves

Davies will not move in turn 5, and for him to occupy chair 4 after turn 5, he must occupy it on turn 4. So, D has to go clockwise in turn 4 and must occupy chair 4.

Turn 5: Aslam moves

We know that all of them are again adjacent after turn 6. If A goes clockwise in turn 5, then he will occupy the chair 2, and in that case, it is not possible for them to be adjacent after turn 6 in either of the cases of B going clockwise or anticlockwise.

So, A must go anticlockwise and occupy chair 7 for them to be adjacent after turn 6.

Turn 6: Bashir moves

Now, for them to be adjacent after turn 6, B must go clockwise and must occupy chair 5.

Turn 7: Chavvi moves

We do not have any information to determine whether C went clockwise or anticlockwise. So, both cases are possible after turn 7.

After turn 3, chair 4 is occupied by no one.

Hence, the correct answer is option D.

Based on the information about the initial positions of Aslam and Chhavi sitting next to each other, while Bashir and Davies have empty chairs on either side of their seats, the possible combinations are,

This has 4 possible combinations.

We are also given that Davies occupies Chair 2 after Turn 1 and Chhavi occupies Chair 7 after Turn 2. We know that the positions of Davies and Chhavi won't be changing after turn 2, as we know that in those turns, only the positions of Aslam and Bashir are being changed in the first 2 turns. Therefore, we can determine that Davies' initial position is chair 2, and Chhavi's initial position is chair 7.

Out of the 4 possible combinations, Davies on chair 2 and Chhavi on chair 7 are only possible in one case, which is,

We now have the initial position, and we are told that all the members are adjacent only after turn 2 and turn 6.

We know that A changed his position in turn 1.

Turn 1: Aslam moves

If A changed his position anticlockwise, he would go to chair 5, and if he changed his position clockwise, he would go to chair 1.

If A ends in chair 5 after turn 1, then it is not possible for all of them to be adjacent to each other after B changes his position in turn 2.

So, A has to end in chair 1 after turn 1, so that B will have a chance in turn 2 for all of them to be adjacent.

Turn 2: Bashir moves

For all of them to be adjacent after turn 2, B must change his position anticlockwise, and after turn 2, he would end up on chair 3.

Turn 3: Chhavi moves

We are given that Davies occupies chair 4 after turn 5, and for this to happen, chair 4 must be empty and cannot be occupied by anyone who does not move before turn 5. So, in turn 3, C must go anticlockwise and occupy chair 6, as if he goes clockwise, he would occupy chair 4, which must be empty for Davies.

Turn 4: Davies moves

Davies will not move in turn 5, and for him to occupy chair 4 after turn 5, he must occupy it on turn 4. So, D has to go clockwise in turn 4 and must occupy chair 4.

Turn 5: Aslam moves

We know that all of them are again adjacent after turn 6. If A goes clockwise in turn 5, then he will occupy the chair 2, and in that case, it is not possible for them to be adjacent after turn 6 in either of the cases of B going clockwise or anticlockwise.

So, A must go anticlockwise and occupy chair 7 for them to be adjacent after turn 6.

Turn 6: Bashir moves

Now, for them to be adjacent after turn 6, B must go clockwise and must occupy chair 5.

Turn 7: Chavvi moves

We do not have any information to determine whether C went clockwise or anticlockwise. So, both cases are possible after turn 7.

The chairs occupied after turn 6 are 4, 5, 6 and 7.

Hence, the correct answer is option A.

Based on the information about the initial positions of Aslam and Chhavi sitting next to each other, while Bashir and Davies have empty chairs on either side of their seats, the possible combinations are,

This has 4 possible combinations.

We are also given that Davies occupies Chair 2 after Turn 1 and Chhavi occupies Chair 7 after Turn 2. We know that the positions of Davies and Chhavi won't be changing after turn 2, as we know that in those turns, only the positions of Aslam and Bashir are being changed in the first 2 turns. Therefore, we can determine that Davies' initial position is chair 2, and Chhavi's initial position is chair 7.

Out of the 4 possible combinations, Davies on chair 2 and Chhavi on chair 7 are only possible in one case, which is,

We now have the initial position, and we are told that all the members are adjacent only after turn 2 and turn 6.

We know that A changed his position in turn 1.

Turn 1: Aslam moves

If A changed his position anticlockwise, he would go to chair 5, and if he changed his position clockwise, he would go to chair 1.

If A ends in chair 5 after turn 1, then it is not possible for all of them to be adjacent to each other after B changes his position in turn 2.

So, A has to end in chair 1 after turn 1, so that B will have a chance in turn 2 for all of them to be adjacent.

Turn 2: Bashir moves

For all of them to be adjacent after turn 2, B must change his position anticlockwise, and after turn 2, he would end up on chair 3.

Turn 3: Chhavi moves

We are given that Davies occupies chair 4 after turn 5, and for this to happen, chair 4 must be empty and cannot be occupied by anyone who does not move before turn 5. So, in turn 3, C must go anticlockwise and occupy chair 6, as if he goes clockwise, he would occupy chair 4, which must be empty for Davies.

Turn 4: Davies moves

Davies will not move in turn 5, and for him to occupy chair 4 after turn 5, he must occupy it on turn 4. So, D has to go clockwise in turn 4 and must occupy chair 4.

Turn 5: Aslam moves

We know that all of them are again adjacent after turn 6. If A goes clockwise in turn 5, then he will occupy the chair 2, and in that case, it is not possible for them to be adjacent after turn 6 in either of the cases of B going clockwise or anticlockwise.

So, A must go anticlockwise and occupy chair 7 for them to be adjacent after turn 6.

Turn 6: Bashir moves

Now, for them to be adjacent after turn 6, B must go clockwise and must occupy chair 5.

Turn 7: Chavvi moves

We do not have any information to determine whether C went clockwise or anticlockwise. So, both cases are possible after turn 7.

In either of the cases after turn 7, the seat adjacent to Bashir is occupied only by Davies, and the other one is empty.

Hence, the correct answer is option B.

Denoting Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.

Putting all the given information in the table, we get,

B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.

We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.

Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,

Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.

We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.

If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.

If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.

After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.

Putting all the calculated values, we get the final table as,

Ekalavya's score after quarter 2 is 4.

Hence, the correct answer is 4.

Denoting Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.

Putting all the given information in the table, we get,

B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.

We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.

Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,

Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.

We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.

If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.

If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.

After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.

Putting all the calculated values, we get the final table as,

As we can see in the table, no employee has changed their group more than once.

Hence, the correct answer is 0.

Denoting Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.

Putting all the given information in the table, we get,

B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.

We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.

Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,

Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.

We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.

If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.

If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.

After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.

Putting all the calculated values, we get the final table as,

Bunty's score after quarter 3 is 5.

Hence, the correct answer is 5.

Denoting Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.

Putting all the given information in the table, we get,

B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.

We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.

Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,

Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.

We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.

If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.

If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.

After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.

Putting all the calculated values, we get the final table as,

The cumulative score of 4 employees can be determined with certainty at the end of quarter 3.

Hence, the correct answer is 4.

Denoting Asha, Bunty, Chintu, Dolly, Eklavya, and Falguni as A, B, C, D, E and F for easy usage. The values in the brackets are the ratings obtained in that quarter, and the values outside are the cumulative ratings after that Quarter.

Putting all the given information in the table, we get,

B came from Elite to Novice after Q1, and B cannot go back to Elite after Q2 and come back to Novice after Q3 because B came to Novice in Q1, and A came to Novice in Q3, so C has to come to Novice after Q3 to be present in Novice at the start of Q4 as per the table. Therefore, we can conclude that C came to Novice after Q3, and B stayed in Novice in Q3 as well, receiving a rating of 3 in Q3. Similarly, D cannot go to Elite after Q1 and come back after Q2 because we already know that A is coming to Novice after Q2. So, we can conclude that D stayed in Novice during Q2 as well. We are also given that A and F received the same rating during all the quarters, and we know that the rating of D is 2 in Q3, so we can conclude its rating to be 2 in all the quarters. For B to come to Novice after Q1, it must receive a rating of 1 in Q1.

We are also given that F has the same rating across all the quarters, and it has to be either 1 or 3, as D already has a rating of 2 in Q1 in that group. If the rating of F is 1, then it will stay in Novice in quarter 2 and also will get a rating of 1 in Q2, which makes the cumulative score 2. But we know that both E and F are going to elite after quarters 1 and 2 in some order, and if the cumulative rating of F is 2 at the end of Q2, then it is not possible for it to go to elite after Q2, as there are members with ratings higher than 2 in Novice after Q2.

Therefore, for all conditions to be satisfied, the rating of F must be 3 in all quarters, and F advances to Elite after Q1 and remains there. Also, we can conclude that E goes to Elite after Q2. Putting all the information in the table, we get,

Now, if we look at the Elite table, we know that A and C received 2 and 3 in some order in Q1 and 1 and 2 in Q2 in some order.

We know that A goes to Novice from Elite after Q2, so the cumulative rating of A must either be less than C after Q2 or equal to C , and the rating of A in Q2 is less than C. These are the two possibilities.

If A received a rating of 3 in Q1, then C gets 2 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 4 after 2 quarters which is same as the cumulative rating of C after 2 quarters and because C gets a rating of 2 and A gets a rating of 1 in Q2, A goes to Novice and C stays in elite even with the same cumulative rating.

If A received a rating of 2 in Q1, then C gets 3 in Q1. For A to get to Novice after Q2, it must obtain a rating of 1 in Q2, so that it will have a cumulative rating of 3 after 2 quarters, which is less than the cumulative rating of C after 2 quarters. If it gets a rating of 2 in Q2, then its cumulative becomes 4, which would be the same as C in that case, and because the rating in Q2 would be less for C in that case, it gets demoted to Novice, which is not the case. So, in the case of A getting 2 in Q1, it must get 1 in Q2.

After Q2, E and C will either have the same cumulative score or C will have 1 point more than E's cumulative score. For E to remain in Elite even after Q3, it must get a rating of 2 points in Q3, and C must get a rating of 1 point.

Putting all the calculated values, we get the final table as,

I. Asha received a rating of 2 in Quarter 1 - This is not necessarily true, as it can also be 3 in Quarter 1.
II. Asha received a rating of 1 in Quarter 2. - This is definitely true, as Asha received a rating of 1 in Quarter 1.

Only II is definitely correct.

Hence, the correct answer is option D.

The values given in both charts together are represented in the table below, with the import tariff percentages charged by each of the five countries on the others represented as a percentage, and the import tariffs levied by each country on other countries are represented in brackets(in billion USD).

Japan's export to India would be the same as India's import from Japan.

India is charging a 50% tariff on Japan, and the tariff by India on Japan equals 3.5 billion USD.

So, 50% of the imports equals 3.5 billion USD.

$$\dfrac{50}{100}\ \times\ $$ Imports $$=\ 3.5$$

Imports $$=\ 3.5\ \times\ 2\ =\ 7$$ billion USD.

The value of imports by India from Japan = Japan's export to India = 7 billion USD.

Hence, the correct answer is option C.

The values given in both charts together are represented in the table below, with the import tariff percentages charged by each of the five countries on the others represented as a percentage, and the import tariffs levied by each country on other countries are represented in brackets(in billion USD).

Option A) Exports by Japan to the UK would be the same as the UK's imports from Japan.

The UK is charging a 40% tariff on Japan, and the tariff imposed by the UK on Japan equals 6 billion USD.

So, 40% of the imports equals 6 billion USD.

$$\dfrac{40}{100}\ \times\ $$ Imports $$=\ 6$$

Imports $$=\ 3\ \times\ 5\ =\ 15$$ billion USD.

The value of imports by the UK from Japan = Japan's export to the UK = 15 billion USD.

Option B) Imports by the US from France.

The US is charging a 20% tariff on France, and the tariff by the US on France equals 6 billion USD.

So, 20% of the imports equals 6 billion USD.

$$\dfrac{20}{100}\ \times\ $$ Imports $$=\ 6$$

Imports $$=\ 6\ \times\ 5\ =\ 30$$ billion USD.

The value of imports by the US from France = 30 billion USD.

Option C) Exports by France to Japan would be the same as Japan's imports from France.

Japan is charging a 30% tariff on France, and the tariff by Japan on France equals 3 billion USD.

So, 30% of the imports equals 3 billion USD.

$$\dfrac{30}{100}\ \times\ $$ Imports $$=\ 3$$

Imports $$=\ 1\ \times\ 10\ =\ 10$$ billion USD.

The value of imports by Japan from France = France's export to Japan = 10 billion USD.

Option D) Imports by France from India.

France is charging a 40% tariff on India, and the tariff by France on India equals 6.5 billion USD.

So, 40% of the imports equals 6.5 billion USD.

$$\dfrac{40}{100}\ \times\ $$ Imports $$=\ 6.5$$

Imports $$=\ 6.5\ \times\ 2.5\ =\ 16.25$$ billion USD.

The value of imports by France from India = 16.25 billion USD.

Out of all the options, the value of imports by the US from France is the highest.

Hence, the correct answer is option B.

The values given in both charts together are represented in the table below, with the import tariff percentages charged by each of the five countries on the others represented as a percentage, and the import tariffs levied by each country on other countries are represented in brackets(in billion USD).

Trade surplus/trade deficit of India with the UK = Exports from India to the UK - Imports from India to the UK

Exports by India to the UK would be the same as the UK's imports from India.

The UK is charging a 30% tariff on India, and the tariff imposed by the UK on India equals 3 billion USD.

So, 30% of the imports equals 3 billion USD.

$$\dfrac{30}{100}\ \times\ $$ Imports $$=\ 3$$

Imports $$=\ 1\ \times\ 10\ =\ 10$$ billion USD.

The value of imports by the UK from India = India's export to the UK = 10 billion USD. 

Imports by India from the UK can be calculated as,

India is charging a 20% tariff on the UK, and the tariff imposed by India on the UK equals 5 billion USD.

So, 20% of the imports equals 5 billion USD.

$$\dfrac{20}{100}\ \times\ $$ Imports $$=\ 5$$

Imports $$=\ 5\ \times\ 5\ =\ 25$$ billion USD.

The value of imports by India from the UK = 25 billion USD.

We can clearly see that the Imports are greater than the exports for India from the UK. So, the trade deficit can be calculated as,

Trade deficit = Exports from India to the UK - Imports from India to the UK = 10 billion USD - 25 billion USD = -15 billion USD.

So, there is a deficit of 15 billion USD.

Hence, the correct answer is option B. 

The values given in both charts together are represented in the table below, with the import tariff percentages charged by each of the five countries on the others represented as a percentage, and the import tariffs levied by each country on other countries are represented in brackets(in billion USD).

For France - Trade surplus/trade deficit of France with the US = Exports from France to the US - Imports from the US to France

Exports by France to the US would be the same as the US's imports from France.

The US is charging a 20% tariff on India, and the tariff imposed by the US on France equals 6 billion USD.

So, 20% of the imports equals 6 billion USD.

$$\dfrac{20}{100}\ \times\ $$ Imports $$=\ 6$$

Imports $$=\ 6\ \times\ 5\ =\ 30$$ billion USD.

The value of imports by the US from France = France's export to the US = 30 billion USD.

Imports by France from the US can be calculated as,

France is charging a 30% tariff on the US, and the tariff imposed by France on the US equals 5.5 billion USD.

So, 30% of the imports equals 5.5 billion USD.

$$\dfrac{30}{100}\ \times\ $$ Imports $$=\ 5.5$$

Imports $$=\ 5.5\ \times\ 3.34\ =\ 18.34$$ billion USD.

The value of imports by France from the US = 18.34 billion USD.

We can clearly see that the Imports are less than the exports for France from the US. So, the trade surplus can be calculated as,

Trade surplus = Exports from France to the US - Imports from France to the US = 30 billion USD - 18.34 billion USD = 11.67 billion USD.

So, there is a surplus of 11.67 billion USD.

For the UK - The Trade surplus/trade deficit of the UK with the US = Exports from the UK to the US - Imports from the US to the UK

Exports by the UK to the US would be the same as the US's imports from the UK.

The US is charging a 30% tariff on the UK, and the tariff imposed by the US on the UK equals 3 billion USD.

So, 30% of the imports equals 3 billion USD.

$$\dfrac{30}{100}\ \times\ $$ Imports $$=\ 3$$

Imports $$=\ 1\ \times\ 10\ =\ 10$$ billion USD.

The value of imports by the US from the UK = The UK's export to the US = 10 billion USD.

Imports by the UK from the US can be calculated as,

The UK is charging a 20% tariff on the US, and the tariff imposed by the UK on the US equals 2.5 billion USD.

So, 20% of the imports equals 2.5 billion USD.

$$\dfrac{20}{100}\ \times\ $$ Imports $$=\ 2.5$$

Imports $$=\ 2.5\ \times\ 5\ =\ 12.5$$ billion USD.

The value of imports by the UK from the US = 12.5 billion USD.

We can clearly see that the Imports are greater than the exports for the UK from the US. So, the trade deficit can be calculated as,

Trade deficit = Exports from the UK to the US - Imports from the UK to the US = 10 billion USD - 12.5 billion USD = -2.5 billion USD.

So, there is a deficit of 2.5 billion USD.

We can see that France has a surplus and the UK has a deficit with the UK.

Hence, the correct answer is option D.

Assuming the tickets from A - C, A - D, and A - E to be $$a,b$$ and $$c$$, respectively. Also, assuming the tickets from C - D and C - E to be $$x$$ and $$y$$, respectively. We are also given that the number of tickets from A - C is equal to the number of tickets booked from A - E, so the value of $$a=c$$. We are also given that 40 tickets were booked from B to C, and 30 tickets were booked from B to E. Other information provided is that no tickets were booked from A to B, from B to D, or from D to E.

Putting all the given information in the table, we get,

From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,

A - B $$=a+b+c=2a+b$$

B - C $$=a+b+c+30+40=2a+b+70$$

C - D $$=a+b+x+y+30$$

D - E $$=c+y+30=a+y+30$$

We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,

Seats occupied in segment C - D $$=\dfrac{95}{100}\times\ 200\ =\ 190$$

We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.

We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.

Equating it with the above equation, we get,

$$2a+b+70=200$$

$$2a+b=130$$ --(1)

$$a+b+x+y+30 = 190$$

$$a+b+x+y=160$$ --(2)

We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.

Equating the expressions, we get,

$$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$$

$$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$  ---(3)

So, the seats occupied during D - E $$=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$

We know that this value has to be an integer.

We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.

We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.

The only value of a at which the equation (3) is an integer is when $$a+30$$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $$a=50$$, as at $$a=40$$ and 60, the value of $$a+30$$ is not a multiple of 4.

We can conclude that the value of $$a=50$$, and substituting in (1), we get the value of $$b=30$$.

Substituting in (3), we get,

$$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$$

$$\ y\ +\ 80\ =\ 140$$

$$\ y\ =\ 60$$

Substituting all the values in (2), we get,

$$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$$

$$x\ =\ 20$$

Putting all the values in the table, we get,

A - B $$=a+b+c=2a+b=130$$

B - C $$=a+b+c+30+40=2a+b+70=200$$

C - D $$=a+b+x+y+30=190$$

D - E $$=c+y+30=a+y+30=140$$

Occupancy factor of segment D - E can be calculated as,

Occupancy factor $$=\dfrac{140}{200}\times\ 100\ =\ 70\%$$

Hence, the correct answer is option B.

Assuming the tickets from A - C, A - D, and A - E to be $$a,b$$ and $$c$$, respectively. Also, assuming the tickets from C - D and C - E to be $$x$$ and $$y$$, respectively. We are also given that the number of tickets from A - C is equal to the number of tickets booked from A - E, so the value of $$a=c$$. We are also given that 40 tickets were booked from B to C, and 30 tickets were booked from B to E. Other information provided is that no tickets were booked from A to B, from B to D, or from D to E.

Putting all the given information in the table, we get,

From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,

A - B $$=a+b+c=2a+b$$

B - C $$=a+b+c+30+40=2a+b+70$$

C - D $$=a+b+x+y+30$$

D - E $$=c+y+30=a+y+30$$

We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,

Seats occupied in segment C - D $$=\dfrac{95}{100}\times\ 200\ =\ 190$$

We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.

We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.

Equating it with the above equation, we get,

$$2a+b+70=200$$

$$2a+b=130$$ --(1)

$$a+b+x+y+30 = 190$$

$$a+b+x+y=160$$ --(2)

We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.

Equating the expressions, we get,

$$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$$

$$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$ ---(3)

So, the seats occupied during D - E $$=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$

We know that this value has to be an integer.

We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.

We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.

The only value of a at which the equation (3) is an integer is when $$a+30$$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $$a=50$$, as at $$a=40$$ and 60, the value of $$a+30$$ is not a multiple of 4.

We can conclude that the value of $$a=50$$, and substituting in (1), we get the value of $$b=30$$.

Substituting in (3), we get,

$$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$$

$$\ y\ +\ 80\ =\ 140$$

$$\ y\ =\ 60$$

Substituting all the values in (2), we get,

$$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$$

$$x\ =\ 20$$

Putting all the values in the table, we get,

A - B $$=a+b+c=2a+b=130$$

B - C $$=a+b+c+30+40=2a+b+70=200$$

C - D $$=a+b+x+y+30=190$$

D - E $$=c+y+30=a+y+30=140$$

The number of tickets booked from station A to E from the above table is 50.

Hence, the correct answer is 50.

Assuming the tickets from A - C, A - D, and A - E to be $$a,b$$ and $$c$$, respectively. Also, assuming the tickets from C - D and C - E to be $$x$$ and $$y$$, respectively. We are also given that the number of tickets from A - C is equal to the number of tickets booked from A - E, so the value of $$a=c$$. We are also given that 40 tickets were booked from B to C, and 30 tickets were booked from B to E. Other information provided is that no tickets were booked from A to B, from B to D, or from D to E.

Putting all the given information in the table, we get,

From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,

A - B $$=a+b+c=2a+b$$

B - C $$=a+b+c+30+40=2a+b+70$$

C - D $$=a+b+x+y+30$$

D - E $$=c+y+30=a+y+30$$

We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,

Seats occupied in segment C - D $$=\dfrac{95}{100}\times\ 200\ =\ 190$$

We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.

We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.

Equating it with the above equation, we get,

$$2a+b+70=200$$

$$2a+b=130$$ --(1)

$$a+b+x+y+30 = 190$$

$$a+b+x+y=160$$ --(2)

We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.

Equating the expressions, we get,

$$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$$

$$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$ ---(3)

So, the seats occupied during D - E $$=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$

We know that this value has to be an integer.

We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.

We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.

The only value of a at which the equation (3) is an integer is when $$a+30$$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $$a=50$$, as at $$a=40$$ and 60, the value of $$a+30$$ is not a multiple of 4.

We can conclude that the value of $$a=50$$, and substituting in (1), we get the value of $$b=30$$.

Substituting in (3), we get,

$$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$$

$$\ y\ +\ 80\ =\ 140$$

$$\ y\ =\ 60$$

Substituting all the values in (2), we get,

$$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$$

$$x\ =\ 20$$

Putting all the values in the table, we get,

A - B $$=a+b+c=2a+b=130$$

B - C $$=a+b+c+30+40=2a+b+70=200$$

C - D $$=a+b+x+y+30=190$$

D - E $$=c+y+30=a+y+30=140$$

Number of tickets booked from station C = 20 + 60 = 80.

Hence, the correct answer is 80.

Assuming the tickets from A - C, A - D, and A - E to be $$a,b$$ and $$c$$, respectively. Also, assuming the tickets from C - D and C - E to be $$x$$ and $$y$$, respectively. We are also given that the number of tickets from A - C is equal to the number of tickets booked from A - E, so the value of $$a=c$$. We are also given that 40 tickets were booked from B to C, and 30 tickets were booked from B to E. Other information provided is that no tickets were booked from A to B, from B to D, or from D to E.

Putting all the given information in the table, we get,

From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,

A - B $$=a+b+c=2a+b$$

B - C $$=a+b+c+30+40=2a+b+70$$

C - D $$=a+b+x+y+30$$

D - E $$=c+y+30=a+y+30$$

We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,

Seats occupied in segment C - D $$=\dfrac{95}{100}\times\ 200\ =\ 190$$

We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.

We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.

Equating it with the above equation, we get,

$$2a+b+70=200$$

$$2a+b=130$$ --(1)

$$a+b+x+y+30 = 190$$

$$a+b+x+y=160$$ --(2)

We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.

Equating the expressions, we get,

$$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$$

$$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$ ---(3)

So, the seats occupied during D - E $$=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$

We know that this value has to be an integer.

We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.

We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.

The only value of a at which the equation (3) is an integer is when $$a+30$$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $$a=50$$, as at $$a=40$$ and 60, the value of $$a+30$$ is not a multiple of 4.

We can conclude that the value of $$a=50$$, and substituting in (1), we get the value of $$b=30$$.

Substituting in (3), we get,

$$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$$

$$\ y\ +\ 80\ =\ 140$$

$$\ y\ =\ 60$$

Substituting all the values in (2), we get,

$$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$$

$$x\ =\ 20$$

Putting all the values in the table, we get,

A - B $$=a+b+c=2a+b=130$$

B - C $$=a+b+c+30+40=2a+b+70=200$$

C - D $$=a+b+x+y+30=190$$

D - E $$=c+y+30=a+y+30=140$$

The number of tickets booked to station C = 50 + 40 = 90

The number of tickets booked to station D = 30 + 20 = 50

Difference = 90 - 50 = 40.

Hence, the correct answer is 40.

Assuming the tickets from A - C, A - D, and A - E to be $$a,b$$ and $$c$$, respectively. Also, assuming the tickets from C - D and C - E to be $$x$$ and $$y$$, respectively. We are also given that the number of tickets from A - C is equal to the number of tickets booked from A - E, so the value of $$a=c$$. We are also given that 40 tickets were booked from B to C, and 30 tickets were booked from B to E. Other information provided is that no tickets were booked from A to B, from B to D, or from D to E.

Putting all the given information in the table, we get,

From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,

A - B $$=a+b+c=2a+b$$

B - C $$=a+b+c+30+40=2a+b+70$$

C - D $$=a+b+x+y+30$$

D - E $$=c+y+30=a+y+30$$

We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,

Seats occupied in segment C - D $$=\dfrac{95}{100}\times\ 200\ =\ 190$$

We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.

We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.

Equating it with the above equation, we get,

$$2a+b+70=200$$

$$2a+b=130$$ --(1)

$$a+b+x+y+30 = 190$$

$$a+b+x+y=160$$ --(2)

We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.

Equating the expressions, we get,

$$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$$

$$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$ ---(3)

So, the seats occupied during D - E $$=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$$

We know that this value has to be an integer.

We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.

We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.

The only value of a at which the equation (3) is an integer is when $$a+30$$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $$a=50$$, as at $$a=40$$ and 60, the value of $$a+30$$ is not a multiple of 4.

We can conclude that the value of $$a=50$$, and substituting in (1), we get the value of $$b=30$$.

Substituting in (3), we get,

$$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$$

$$\ y\ +\ 80\ =\ 140$$

$$\ y\ =\ 60$$

Substituting all the values in (2), we get,

$$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$$

$$x\ =\ 20$$

Putting all the values in the table, we get,

A - B $$=a+b+c=2a+b=130$$

B - C $$=a+b+c+30+40=2a+b+70=200$$

C - D $$=a+b+x+y+30=190$$

D - E $$=c+y+30=a+y+30=140$$

The number of tickets booked for exactly one segment can be calculated using values from the above table as,

Tickets booked for exactly one segment = A - B + B - C + C - D + D - E = 0 + 40 + 20 + 0 = 60.

Hence, the correct answer is 60.

We are given that Alia tapped 6 times, Dilshan tapped 11 times, and Ehsan tapped 9 times. We also know that the total number of taps is 40. So, the sum of taps by Badal and Clive can be calculated as,

6 + Badal + Clive + 11 + 9 = 40

Badal + Clive = 14

We are given that taps by Clive are more than taps by Badal.

So, the possible pairs for taps by Clive and Badal such that the sum is 14 are (8, 6), (9, 5), (10, 4)...(14, 0).

We are given that no one gave more than two 'Yes' answers. We know that yes means one tap. So, the maximum number of 1 taps out of the 4 questions answered can be 2. The minimum possible number of taps for any person for the four questions would be 1 + 1 + 2 + 2 = 6. It is not possible for any person to have fewer than 6 taps as the maximum number of yes is 2.

So, out of all the possibilities, the only possible case for Badal is 6, and Clive is 8, as in all the other cases, the number by Badal was less than 6.

We are given that Alia answered Yes to Clive and Dilshan's questions. We know that the total number of taps by Alia is 6, and she tapped once for both those questions. The number of taps by Alia to the questions by Badal and Ehsan must be greater than 1, as we know that a person must have a maximum of two single taps. The sum of taps by Alia to Badal and Ehsan can be calculated as, 

Badal + 1 + 1 + Ehsan = 6

Badal + Ehsan = 4

We know that both the values are greater than 1 and their sum is 4, so the only possibility is for both of them to be equal to 2.

So, Alia tapped twice for both Badal and Ehsan.

We are given that Dilshan answered no to Badal's question which means he tapped twice. The total number of taps by Dilshan is 11. So, the sum of taps from Alia, Clive and Ehsan's questions can be calculated as,

Alia + 2 + Clive + Ehsan = 11

Alia + Clive + Ehsan = 9

We know that the maximum possibility for each of the values is 3, and for the sum to be 9, all the values must be equal to 3.

We are given that taps received for the questions by Badal, Dilshan and Ehsan are equal, so let us assume the value to be a.

Putting all the calculated values in the table, we get,

We know that each question received atleast one yes, one no and one maybe as an answer. If we examine the question asked by B, we already have two 'No's, so out of the other two, one must be 'Yes', and the other must be 'Maybe'. The taps received by B can be calculated as 2 + 2 + 1 + 3 = 8. We calculated the value of a to be 8, and the value of taps received by C can be calculated as,

9 + 8 + c + 8 + 8  = 40

c = 7

So, the number of taps received by Clive is 7.

Hence, the correct answer is 7.

We are given that Alia tapped 6 times, Dilshan tapped 11 times, and Ehsan tapped 9 times. We also know that the total number of taps is 40. So, the sum of taps by Badal and Clive can be calculated as,

6 + Badal + Clive + 11 + 9 = 40

Badal + Clive = 14

We are given that taps by Clive are more than taps by Badal.

So, the possible pairs for taps by Clive and Badal such that the sum is 14 are (8, 6), (9, 5), (10, 4)...(14, 0).

We are given that no one gave more than two 'Yes' answers. We know that yes means one tap. So, the maximum number of 1 taps out of the 4 questions answered can be 2. The minimum possible number of taps for any person for the four questions would be 1 + 1 + 2 + 2 = 6. It is not possible for any person to have fewer than 6 taps as the maximum number of yes is 2.

So, out of all the possibilities, the only possible case for Badal is 6, and Clive is 8, as in all the other cases, the number by Badal was less than 6.

We are given that Alia answered Yes to Clive and Dilshan's questions. We know that the total number of taps by Alia is 6, and she tapped once for both those questions. The number of taps by Alia to the questions by Badal and Ehsan must be greater than 1, as we know that a person must have a maximum of two single taps. The sum of taps by Alia to Badal and Ehsan can be calculated as,

Badal + 1 + 1 + Ehsan = 6

Badal + Ehsan = 4

We know that both the values are greater than 1 and their sum is 4, so the only possibility is for both of them to be equal to 2.

So, Alia tapped twice for both Badal and Ehsan.

We are given that Dilshan answered no to Badal's question which means he tapped twice. The total number of taps by Dilshan is 11. So, the sum of taps from Alia, Clive and Ehsan's questions can be calculated as,

Alia + 2 + Clive + Ehsan = 11

Alia + Clive + Ehsan = 9

We know that the maximum possibility for each of the values is 3, and for the sum to be 9, all the values must be equal to 3.

We are given that taps received for the questions by Badal, Dilshan and Ehsan are equal, so let us assume the value to be a.

Putting all the calculated values in the table, we get,

Alia and Badal are the only people with an equal number of taps.

Hence, the correct answer is option D.

We are given that Alia tapped 6 times, Dilshan tapped 11 times, and Ehsan tapped 9 times. We also know that the total number of taps is 40. So, the sum of taps by Badal and Clive can be calculated as,

6 + Badal + Clive + 11 + 9 = 40

Badal + Clive = 14

We are given that taps by Clive are more than taps by Badal.

So, the possible pairs for taps by Clive and Badal such that the sum is 14 are (8, 6), (9, 5), (10, 4)...(14, 0).

We are given that no one gave more than two 'Yes' answers. We know that yes means one tap. So, the maximum number of 1 taps out of the 4 questions answered can be 2. The minimum possible number of taps for any person for the four questions would be 1 + 1 + 2 + 2 = 6. It is not possible for any person to have fewer than 6 taps as the maximum number of yes is 2.

So, out of all the possibilities, the only possible case for Badal is 6, and Clive is 8, as in all the other cases, the number by Badal was less than 6.

We are given that Alia answered Yes to Clive and Dilshan's questions. We know that the total number of taps by Alia is 6, and she tapped once for both those questions. The number of taps by Alia to the questions by Badal and Ehsan must be greater than 1, as we know that a person must have a maximum of two single taps. The sum of taps by Alia to Badal and Ehsan can be calculated as,

Badal + 1 + 1 + Ehsan = 6

Badal + Ehsan = 4

We know that both the values are greater than 1 and their sum is 4, so the only possibility is for both of them to be equal to 2.

So, Alia tapped twice for both Badal and Ehsan.

We are given that Dilshan answered no to Badal's question which means he tapped twice. The total number of taps by Dilshan is 11. So, the sum of taps from Alia, Clive and Ehsan's questions can be calculated as,

Alia + 2 + Clive + Ehsan = 11

Alia + Clive + Ehsan = 9

We know that the maximum possibility for each of the values is 3, and for the sum to be 9, all the values must be equal to 3.

We are given that taps received for the questions by Badal, Dilshan and Ehsan are equal, so let us assume the value to be a.

Putting all the calculated values in the table, we get,

We know that each question received atleast one yes, one no and one maybe as an answer. If we examine the question asked by B, we already have two 'No's, so out of the other two, one must be 'Yes', and the other must be 'Maybe'. The taps received by B can be calculated as 2 + 2 + 1 + 3 = 8. We calculated the value of a to be 8, and the value of taps received by C can be calculated as,

9 + 8 + c + 8 + 8 = 40

c = 7

So, the number of taps received by Clive is 7.

We are given that the answer by Alia does not match the answers people gave to her question. We know that the taps by Badal have to be 1, 1, 2 and 2 in some order. We know that Alia tapped twice to Badal's question, so Badal cannot tap twice to Alia's question, which means Badal tapped once for Alia's question. Now the sum of Clive and Ehsan's taps for Alia's question has to be 9 - 3 - 1 = 5. So, they have to be 2 and 3 in some order. We also know that Alia's answer to Ehsan's question is 2, so Ehsan has only one option to answer Alia, which is 3, and Clive's answer to Alia's question is 2.

Badal and Ehsan's answer to Clive's question has to be 1 and 2 in some order. Similarly, Badal and Clive's answers to Ehsan's question have to be 1 and 3 in some order.

We can also calculate the sum of taps by Badal, Clive and Ehsan to Dilshan's question is 8 - 1 = 7. So, their taps have to be 2, 2 and 3 in some order as this is the only possibility satisfying the condition of at least one Yes, one No and one Maybe for every question. We know that a tap Badal has to be either one or two, so the only possibility for Badal's answer to Dilshan's question is 2 and the other two has to be two and three in some order.

We are also given that the answer by Ehsan does not match the answers people gave to his question. Dilshan's answer to Ehsan's question is 3, so the answer by Ehsan to Dilshan's question has to be 2.

The answer by Clive to Badal's question has to be 1, as if it is 3, then the answer to Ehsan's question by Clive becomes 0, which is not possible. With that value, we can determine all the other values and putting the values in the table, we get,

Clive's answer to Ehsan's question is No, which is denoted by 2.

Hence, the correct answer is option A.

We are given that Alia tapped 6 times, Dilshan tapped 11 times, and Ehsan tapped 9 times. We also know that the total number of taps is 40. So, the sum of taps by Badal and Clive can be calculated as,

6 + Badal + Clive + 11 + 9 = 40

Badal + Clive = 14

We are given that taps by Clive are more than taps by Badal.

So, the possible pairs for taps by Clive and Badal such that the sum is 14 are (8, 6), (9, 5), (10, 4)...(14, 0).

We are given that no one gave more than two 'Yes' answers. We know that yes means one tap. So, the maximum number of 1 taps out of the 4 questions answered can be 2. The minimum possible number of taps for any person for the four questions would be 1 + 1 + 2 + 2 = 6. It is not possible for any person to have fewer than 6 taps as the maximum number of yes is 2.

So, out of all the possibilities, the only possible case for Badal is 6, and Clive is 8, as in all the other cases, the number by Badal was less than 6.

We are given that Alia answered Yes to Clive and Dilshan's questions. We know that the total number of taps by Alia is 6, and she tapped once for both those questions. The number of taps by Alia to the questions by Badal and Ehsan must be greater than 1, as we know that a person must have a maximum of two single taps. The sum of taps by Alia to Badal and Ehsan can be calculated as,

Badal + 1 + 1 + Ehsan = 6

Badal + Ehsan = 4

We know that both the values are greater than 1 and their sum is 4, so the only possibility is for both of them to be equal to 2.

So, Alia tapped twice for both Badal and Ehsan.

We are given that Dilshan answered no to Badal's question which means he tapped twice. The total number of taps by Dilshan is 11. So, the sum of taps from Alia, Clive and Ehsan's questions can be calculated as,

Alia + 2 + Clive + Ehsan = 11

Alia + Clive + Ehsan = 9

We know that the maximum possibility for each of the values is 3, and for the sum to be 9, all the values must be equal to 3.

We are given that taps received for the questions by Badal, Dilshan and Ehsan are equal, so let us assume the value to be a.

Putting all the calculated values in the table, we get,

We know that each question received atleast one yes, one no and one maybe as an answer. If we examine the question asked by B, we already have two 'No's, so out of the other two, one must be 'Yes', and the other must be 'Maybe'. The taps received by B can be calculated as 2 + 2 + 1 + 3 = 8. We calculated the value of a to be 8, and the value of taps received by C can be calculated as,

9 + 8 + c + 8 + 8 = 40

c = 7

So, the number of taps received by Clive is 7.

We are given that the answer by Alia does not match the answers people gave to her question. We know that the taps by Badal have to be 1, 1, 2 and 2 in some order. We know that Alia tapped twice to Badal's question, so Badal cannot tap twice to Alia's question, which means Badal tapped once for Alia's question. Now the sum of Clive and Ehsan's taps for Alia's question has to be 9 - 3 - 1 = 5. So, they have to be 2 and 3 in some order. We also know that Alia's answer to Ehsan's question is 2, so Ehsan has only one option to answer Alia, which is 3, and Clive's answer to Alia's question is 2.

Badal and Ehsan's answer to Clive's question has to be 1 and 2 in some order. Similarly, Badal and Clive's answers to Ehsan's question have to be 1 and 3 in some order.

We can also calculate the sum of taps by Badal, Clive and Ehsan to Dilshan's question is 8 - 1 = 7. So, their taps have to be 2, 2 and 3 in some order as this is the only possibility satisfying the condition of at least one Yes, one No and one Maybe for every question. We know that a tap Badal has to be either one or two, so the only possibility for Badal's answer to Dilshan's question is 2 and the other two has to be two and three in some order.

We are also given that the answer by Ehsan does not match the answers people gave to his question. Dilshan's answer to Ehsan's question is 3, so the answer by Ehsan to Dilshan's question has to be 2.

The answer by Clive to Badal's question has to be 1, as if it is 3, then the answer to Ehsan's question by Clive becomes 0, which is not possible. With that value, we can determine all the other values and putting the values in the table, we get,

The total number of Yes responses is equal to the number of 1's in the table, which is 6.

Hence, the correct answer is 6.