CAT 2023 Slot 3 - DILR

In a course, there are only three female students named Amala, Koli, and Rini, and only three male students named Biman, Mathew, and Shyamal.
It is known that the total score in the course is calculated as a weighted average of two components, with both weights being positive and summing up to 1.
Let's assume that the project score component be x, while the test score is represented by (1-x).Projects are completed in pairs, with each pair consisting of one female and one male student, totaling three pairs.Both members of each pair receive the same score for the project. So, the scores achieved in the project are 40, 60, and 80, respectively.
Hence, it can be concluded that each female student will belong to a unique group, and no two male or female students will be assigned to the same group.
Regarding the test scores, there are six scores provided for six students, with four being unique and the remaining two being average scores, both of which are 60. Additionally, it is understood that the highest possible score is 80, while the lowest is 40.
Therefore, the unique scores are 80, 70, 50, and 40 (as all test scores are multiples of 10), while the remaining two scores are both 60.
Based on point 3, we deduce that Amala's project score was twice that of Koli's, while Koli scored 20 points higher than Amala in the test. Therefore, Amala's project score is 80, and Koli's is 40, resulting in Rini's project score being 60. Koli's test score, being 20 points higher than Amala's, could be either 80, 70, or 60.
So, the score obtained by them is as follows:

It is given that Amala attained the highest overall score, while Shyamal achieved the second highest on the test. His score surpassed Koli's by two points, yet fell short of Amala's aggregate by two points.
Therefore, Shyamal's test score is 70, which means Koli cannot score 70 in the test, leading to the inference that Amala cannot score 50 in the test.

As stated, Shyamal's aggregate score surpassed Koli's by two points but fell short of Amala's by two points. Consequently, Amala's aggregate score is four points higher than Koli's, and she holds the highest aggregate score.
Case (i): The test of Amala is 40

Hence, 40(1 – x) + 80x = 60(1 – x) + 40x + 4
60x = 24
x = 0.4
Therefore, Amala's aggregate score is calculated as:
= 40(1 – 0.4) + 80 × 0.4
24 + 32 = 56
Shyamal's minimum aggregate score, calculated as 70(1 – 0.4) + 40 × 0.4, equals 58, which surpasses Amala's.
Therefore, Case 1 is not possible.
So, the below table is as follows:

Hence, 60(1 – x) + 80x = 80(1 – x) + 40x + 4
60 + 20x = 84 – 40x
60x = 24
x = 0.24
Therefore, Amala's aggregate score, calculated as 60(1-0.4) + 80 × 0.4, amounts to 68, indicating that Shyamal's aggregate score is (68 – 2) = 66
Thus, Shyamal's project score is calculated as
(66-70×0.6)/0.4 = 60
It is further understood that Biman achieved the second lowest score in the test, indicating his test score to be 50, and he attained the lowest aggregate score. Additionally, Mathew's project score exceeded Rini's but fell short of her test score. Consequently, Mathew's project score is 80 (as Rini scored 60 in the project), while Biman's project score is 40.
Likewise, Rini outperformed Mathew on the test, indicating Rini's score to be 60 and Mathew's to be 40.
Therefore, the final table will be as follows:

From the table, we can see that the score obtained by Rini in the project is 60.
So, the correct answer is 60.

In a course, there are only three female students named Amala, Koli, and Rini, and only three male students named Biman, Mathew, and Shyamal.
It is known that the total score in the course is calculated as a weighted average of two components, with both weights being positive and summing up to 1.
Let's assume that the project score component be x, while the test score is represented by (1-x).Projects are completed in pairs, with each pair consisting of one female and one male student, totaling three pairs.Both members of each pair receive the same score for the project. So, the scores achieved in the project are 40, 60, and 80, respectively.
Hence, it can be concluded that each female student will belong to a unique group, and no two male or female students will be assigned to the same group.
Regarding the test scores, there are six scores provided for six students, with four being unique and the remaining two being average scores, both of which are 60. Additionally, it is understood that the highest possible score is 80, while the lowest is 40.
Therefore, the unique scores are 80, 70, 50, and 40 (as all test scores are multiples of 10), while the remaining two scores are both 60.
Based on point 3, we deduce that Amala's project score was twice that of Koli's, while Koli scored 20 points higher than Amala in the test. Therefore, Amala's project score is 80, and Koli's is 40, resulting in Rini's project score being 60. Koli's test score, being 20 points higher than Amala's, could be either 80, 70, or 60.
So, the score obtained by them is as follows:

It is given that Amala attained the highest overall score, while Shyamal achieved the second highest on the test. His score surpassed Koli's by two points, yet fell short of Amala's aggregate by two points.
Therefore, Shyamal's test score is 70, which means Koli cannot score 70 in the test, leading to the inference that Amala cannot score 50 in the test.

As stated, Shyamal's aggregate score surpassed Koli's by two points but fell short of Amala's by two points. Consequently, Amala's aggregate score is four points higher than Koli's, and she holds the highest aggregate score.
Case (i): The test of Amala is 40

Hence, 40(1 – x) + 80x = 60(1 – x) + 40x + 4
60x = 24
x = 0.4
Therefore, Amala's aggregate score is calculated as:
= 40(1 – 0.4) + 80 × 0.4
24 + 32 = 56
Shyamal's minimum aggregate score, calculated as 70(1 – 0.4) + 40 × 0.4, equals 58, which surpasses Amala's.
Therefore, Case 1 is not possible.
So, the below table is as follows:

Hence, 60(1 – x) + 80x = 80(1 – x) + 40x + 4
60 + 20x = 84 – 40x
60x = 24
x = 0.24
Therefore, Amala's aggregate score, calculated as 60(1-0.4) + 80 × 0.4, amounts to 68, indicating that Shyamal's aggregate score is (68 – 2) = 66
Thus, Shyamal's project score is calculated as
(66-70×0.6)/0.4 = 60
It is further understood that Biman achieved the second lowest score in the test, indicating his test score to be 50, and he attained the lowest aggregate score. Additionally, Mathew's project score exceeded Rini's but fell short of her test score. Consequently, Mathew's project score is 80 (as Rini scored 60 in the project), while Biman's project score is 40.
Likewise, Rini outperformed Mathew on the test, indicating Rini's score to be 60 and Mathew's to be 40.
Therefore, the final table will be as follows:

From the solution, we get that the weight of the test component is 0.6
So, the correct answer is 0.60

In a course, there are only three female students named Amala, Koli, and Rini, and only three male students named Biman, Mathew, and Shyamal.
It is known that the total score in the course is calculated as a weighted average of two components, with both weights being positive and summing up to 1.
Let's assume that the project score component be x, while the test score is represented by (1-x).Projects are completed in pairs, with each pair consisting of one female and one male student, totaling three pairs.Both members of each pair receive the same score for the project. So, the scores achieved in the project are 40, 60, and 80, respectively.
Hence, it can be concluded that each female student will belong to a unique group, and no two male or female students will be assigned to the same group.
Regarding the test scores, there are six scores provided for six students, with four being unique and the remaining two being average scores, both of which are 60. Additionally, it is understood that the highest possible score is 80, while the lowest is 40.
Therefore, the unique scores are 80, 70, 50, and 40 (as all test scores are multiples of 10), while the remaining two scores are both 60.
Based on point 3, we deduce that Amala's project score was twice that of Koli's, while Koli scored 20 points higher than Amala in the test. Therefore, Amala's project score is 80, and Koli's is 40, resulting in Rini's project score being 60. Koli's test score, being 20 points higher than Amala's, could be either 80, 70, or 60.
So, the score obtained by them is as follows:

It is given that Amala attained the highest overall score, while Shyamal achieved the second highest on the test. His score surpassed Koli's by two points, yet fell short of Amala's aggregate by two points.
Therefore, Shyamal's test score is 70, which means Koli cannot score 70 in the test, leading to the inference that Amala cannot score 50 in the test.

As stated, Shyamal's aggregate score surpassed Koli's by two points but fell short of Amala's by two points. Consequently, Amala's aggregate score is four points higher than Koli's, and she holds the highest aggregate score.
Case (i): The test of Amala is 40

Hence, 40(1 – x) + 80x = 60(1 – x) + 40x + 4
60x = 24
x = 0.4
Therefore, Amala's aggregate score is calculated as:
= 40(1 – 0.4) + 80 × 0.4
24 + 32 = 56
Shyamal's minimum aggregate score, calculated as 70(1 – 0.4) + 40 × 0.4, equals 58, which surpasses Amala's.
Therefore, Case 1 is not possible.
So, the below table is as follows:

Hence, 60(1 – x) + 80x = 80(1 – x) + 40x + 4
60 + 20x = 84 – 40x
60x = 24
x = 0.24
Therefore, Amala's aggregate score, calculated as 60(1-0.4) + 80 × 0.4, amounts to 68, indicating that Shyamal's aggregate score is (68 – 2) = 66
Thus, Shyamal's project score is calculated as
(66-70×0.6)/0.4 = 60
It is further understood that Biman achieved the second lowest score in the test, indicating his test score to be 50, and he attained the lowest aggregate score. Additionally, Mathew's project score exceeded Rini's but fell short of her test score. Consequently, Mathew's project score is 80 (as Rini scored 60 in the project), while Biman's project score is 40.
Likewise, Rini outperformed Mathew on the test, indicating Rini's score to be 60 and Mathew's to be 40.
Therefore, the final table will be as follows:

From the table, we can see that 68 is the maximum aggregrate score.
So, the correct answer is 68.

In a course, there are only three female students named Amala, Koli, and Rini, and only three male students named Biman, Mathew, and Shyamal.
It is known that the total score in the course is calculated as a weighted average of two components, with both weights being positive and summing up to 1.
Let's assume that the project score component be x, while the test score is represented by (1-x).Projects are completed in pairs, with each pair consisting of one female and one male student, totaling three pairs.Both members of each pair receive the same score for the project. So, the scores achieved in the project are 40, 60, and 80, respectively.
Hence, it can be concluded that each female student will belong to a unique group, and no two male or female students will be assigned to the same group.
Regarding the test scores, there are six scores provided for six students, with four being unique and the remaining two being average scores, both of which are 60. Additionally, it is understood that the highest possible score is 80, while the lowest is 40.
Therefore, the unique scores are 80, 70, 50, and 40 (as all test scores are multiples of 10), while the remaining two scores are both 60.
Based on point 3, we deduce that Amala's project score was twice that of Koli's, while Koli scored 20 points higher than Amala in the test. Therefore, Amala's project score is 80, and Koli's is 40, resulting in Rini's project score being 60. Koli's test score, being 20 points higher than Amala's, could be either 80, 70, or 60.
So, the score obtained by them is as follows:

It is given that Amala attained the highest overall score, while Shyamal achieved the second highest on the test. His score surpassed Koli's by two points, yet fell short of Amala's aggregate by two points.
Therefore, Shyamal's test score is 70, which means Koli cannot score 70 in the test, leading to the inference that Amala cannot score 50 in the test.

As stated, Shyamal's aggregate score surpassed Koli's by two points but fell short of Amala's by two points. Consequently, Amala's aggregate score is four points higher than Koli's, and she holds the highest aggregate score.
Case (i): The test of Amala is 40

Hence, 40(1 – x) + 80x = 60(1 – x) + 40x + 4
60x = 24
x = 0.4
Therefore, Amala's aggregate score is calculated as:
= 40(1 – 0.4) + 80 × 0.4
24 + 32 = 56
Shyamal's minimum aggregate score, calculated as 70(1 – 0.4) + 40 × 0.4, equals 58, which surpasses Amala's.
Therefore, Case 1 is not possible.
So, the below table is as follows:

Hence, 60(1 – x) + 80x = 80(1 – x) + 40x + 4
60 + 20x = 84 – 40x
60x = 24
x = 0.24
Therefore, Amala's aggregate score, calculated as 60(1-0.4) + 80 × 0.4, amounts to 68, indicating that Shyamal's aggregate score is (68 – 2) = 66
Thus, Shyamal's project score is calculated as
(66-70×0.6)/0.4 = 60
It is further understood that Biman achieved the second lowest score in the test, indicating his test score to be 50, and he attained the lowest aggregate score. Additionally, Mathew's project score exceeded Rini's but fell short of her test score. Consequently, Mathew's project score is 80 (as Rini scored 60 in the project), while Biman's project score is 40.
Likewise, Rini outperformed Mathew on the test, indicating Rini's score to be 60 and Mathew's to be 40.
Therefore, the final table will be as follows:

From the table, we can see that Mathew has got a score of 40 in the test.
So, the correct answer is 40.

In a course, there are only three female students named Amala, Koli, and Rini, and only three male students named Biman, Mathew, and Shyamal.
It is known that the total score in the course is calculated as a weighted average of two components, with both weights being positive and summing up to 1.
Let's assume that the project score component be x, while the test score is represented by (1-x).Projects are completed in pairs, with each pair consisting of one female and one male student, totaling three pairs.Both members of each pair receive the same score for the project. So, the scores achieved in the project are 40, 60, and 80, respectively.
Hence, it can be concluded that each female student will belong to a unique group, and no two male or female students will be assigned to the same group.
Regarding the test scores, there are six scores provided for six students, with four being unique and the remaining two being average scores, both of which are 60. Additionally, it is understood that the highest possible score is 80, while the lowest is 40.
Therefore, the unique scores are 80, 70, 50, and 40 (as all test scores are multiples of 10), while the remaining two scores are both 60.
Based on point 3, we deduce that Amala's project score was twice that of Koli's, while Koli scored 20 points higher than Amala in the test. Therefore, Amala's project score is 80, and Koli's is 40, resulting in Rini's project score being 60. Koli's test score, being 20 points higher than Amala's, could be either 80, 70, or 60.
So, the score obtained by them is as follows:

It is given that Amala attained the highest overall score, while Shyamal achieved the second highest on the test. His score surpassed Koli's by two points, yet fell short of Amala's aggregate by two points.
Therefore, Shyamal's test score is 70, which means Koli cannot score 70 in the test, leading to the inference that Amala cannot score 50 in the test.

As stated, Shyamal's aggregate score surpassed Koli's by two points but fell short of Amala's by two points. Consequently, Amala's aggregate score is four points higher than Koli's, and she holds the highest aggregate score.
Case (i): The test of Amala is 40

Hence, 40(1 – x) + 80x = 60(1 – x) + 40x + 4
60x = 24
x = 0.4
Therefore, Amala's aggregate score is calculated as:
= 40(1 – 0.4) + 80 × 0.4
24 + 32 = 56
Shyamal's minimum aggregate score, calculated as 70(1 – 0.4) + 40 × 0.4, equals 58, which surpasses Amala's.
Therefore, Case 1 is not possible.
So, the below table is as follows:

Hence, 60(1 – x) + 80x = 80(1 – x) + 40x + 4
60 + 20x = 84 – 40x
60x = 24
x = 0.24
Therefore, Amala's aggregate score, calculated as 60(1-0.4) + 80 × 0.4, amounts to 68, indicating that Shyamal's aggregate score is (68 – 2) = 66
Thus, Shyamal's project score is calculated as
(66-70×0.6)/0.4 = 60
It is further understood that Biman achieved the second lowest score in the test, indicating his test score to be 50, and he attained the lowest aggregate score. Additionally, Mathew's project score exceeded Rini's but fell short of her test score. Consequently, Mathew's project score is 80 (as Rini scored 60 in the project), while Biman's project score is 40.
Likewise, Rini outperformed Mathew on the test, indicating Rini's score to be 60 and Mathew's to be 40.
Therefore, the final table will be as follows:

From the table, we can see that both the pairs Amala is with Mathew, Koli is with Biman and Shyamal is with Rini. So, the pairs given in the questions are not there.
So, the correct answer is Neither (i) nor (ii).

In every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was double that of online registrations.
Let x be the number of online registrations
2x be the number of offline registrations
Total number of registrations be x + 2x = 3x
According to the table data:
3x should lie between the minimum and maximum total number of registrations.
x = 40 (as x should also be a multiple of 10)
In January, 40 and 80 are the online and offline registrations respectively.
From 5, we can say that the number of online registrations is highest in may.
In may, there are total 100 online registrations.
Maximum possible total registrations is 130
Lowest possible number of offline registrations is 30
Let x be the number of offline registrations in May which is equal to number of online registrations in March.
Let's arrange the data in table:

From the table mentioned in question, 50 is the median for Offline data
Now x should lie between 50 and 80
For 80 to be the median for the online data
y should lie between 80 and 100
Now, let's consider the following:
Feb ⇒ Minimum value of y + x
= 80 + 50 = 130
Therefore, x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March:
50 + z = 110
z = 60
Let's complete the table:

So, total number of April's registrations is 120.

In every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was double that of online registrations.
Let x be the number of online registrations
2x be the number of offline registrations
Total number of registrations be x + 2x = 3x
According to the table data:
3x should lie between the minimum and maximum total number of registrations.
x = 40 (as x should also be a multiple of 10)
In January, 40 and 80 are the online and offline registrations respectively.
From 5, we can say that the number of online registrations is highest in may.
In may, there are total 100 online registrations.
Maximum possible total registrations is 130
Lowest possible number of offline registrations is 30
Let x be the number of offline registrations in May which is equal to number of online registrations in March.
Let's arrange the data in table:

From the table mentioned in question, 50 is the median for Offline data
Now x should lie between 50 and 80
For 80 to be the median for the online data
y should lie between 80 and 100
Now, let's consider the following:
Feb ⇒ Minimum value of y + x
= 80 + 50 = 130
Therefore, x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March:
50 + z = 110
z = 60
Let's complete the table:

The correct answer is 40.

In every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was double that of online registrations.
Let x be the number of online registrations
2x be the number of offline registrations
Total number of registrations be x + 2x = 3x
According to the table data:
3x should lie between the minimum and maximum total number of registrations.
x = 40 (as x should also be a multiple of 10)
In January, 40 and 80 are the online and offline registrations respectively.
From 5, we can say that the number of online registrations is highest in may.
In may, there are total 100 online registrations.
Maximum possible total registrations is 130
Lowest possible number of offline registrations is 30
Let x be the number of offline registrations in May which is equal to number of online registrations in March.
Let's arrange the data in table:

From the table mentioned in question, 50 is the median for Offline data
Now x should lie between 50 and 80
For 80 to be the median for the online data
y should lie between 80 and 100
Now, let's consider the following:
Feb ⇒ Minimum value of y + x
= 80 + 50 = 130
Therefore, x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March:
50 + z = 110
z = 60
Let's complete the table:

The correct answer is Only I.

In every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was double that of online registrations.
Let x be the number of online registrations
2x be the number of offline registrations
Total number of registrations be x + 2x = 3x
According to the table data:
3x should lie between the minimum and maximum total number of registrations.
x = 40 (as x should also be a multiple of 10)
In January, 40 and 80 are the online and offline registrations respectively.
From 5, we can say that the number of online registrations is highest in may.
In may, there are total 100 online registrations.
Maximum possible total registrations is 130
Lowest possible number of offline registrations is 30
Let x be the number of offline registrations in May which is equal to number of online registrations in March.
Let's arrange the data in table:

From the table mentioned in question, 50 is the median for Offline data
Now x should lie between 50 and 80
For 80 to be the median for the online data
y should lie between 80 and 100
Now, let's consider the following:
Feb ⇒ Minimum value of y + x
= 80 + 50 = 130
Therefore, x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March:
50 + z = 110
z = 60
Let's complete the table:

The correct answer is 50

In every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was double that of online registrations.
Let x be the number of online registrations
2x be the number of offline registrations
Total number of registrations be x + 2x = 3x
According to the table data:
3x should lie between the minimum and maximum total number of registrations.
x = 40 (as x should also be a multiple of 10)
In January, 40 and 80 are the online and offline registrations respectively.
From 5, we can say that the number of online registrations is highest in may.
In may, there are total 100 online registrations.
Maximum possible total registrations is 130
Lowest possible number of offline registrations is 30
Let x be the number of offline registrations in May which is equal to number of online registrations in March.
Let's arrange the data in table:

From the table mentioned in question, 50 is the median for Offline data
Now x should lie between 50 and 80
For 80 to be the median for the online data
y should lie between 80 and 100
Now, let's consider the following:
Feb ⇒ Minimum value of y + x
= 80 + 50 = 130
Therefore, x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March:
50 + z = 110
z = 60
Let's complete the table:

The correct answer is Both I and II.

Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = A/4  and Split AC's = 3A/4
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5
From the given condition 3, we get
A/4 - B/5 = 6 & 4B/5 = 36
So, B = 46 and A = 60

So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table:

As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table:

So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36
4z = 28
z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2 × 5 = 10

Now, by looking at the table, we get that, 14 split ACs are sold by D2.

Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = A/4  and Split AC's = 3A/4
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5
From the given condition 3, we get
A/4 - B/5 = 6 & 4B/5 = 36
So, B = 46 and A = 60

So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table:

As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table:

So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36
4z = 28
z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2 × 5 = 10

Now, by looking at the table, we get that total number of non-inverter ACs is 9 + 6 = 15.
So, needed percentage:
15 out of 60 i.e.   15/60 × 10 = 25%

Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = A/4  and Split AC's = 3A/4
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5
From the given condition 3, we get
A/4 - B/5 = 6 & 4B/5 = 36
So, B = 46 and A = 60

So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table:

As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table:

So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36
4z = 28
z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2 × 5 = 10

Now, the total number of ACs sold by D2 and D4: = 60 – D1 – D3
= 60 – 15 – 12
= 33
Therefore, the correct answer is 33.

Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = A/4  and Split AC's = 3A/4
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5
From the given condition 3, we get
A/4 - B/5 = 6 & 4B/5 = 36
So, B = 46 and A = 60

So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table:

As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table:

So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36
4z = 28
z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2 × 5 = 10

From the table, we get that D1 and D3 sold 27 ACs together which is less than 60 – 27 = 33 which is sold by D2 and D4 together.

Suppose, A is the total number of AC's sold.
From the given information, that the total number of ACs sold in the city, 25% were of Window variant.
Window AC's = A/4  and Split AC's = 3A/4
Now, Suppose B is the total Number of inverter ACs
From the given information, that among the Inverter ACs sold, 20% were of Window variant.
Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5
From the given condition 3, we get
A/4 - B/5 = 6 & 4B/5 = 36
So, B = 46 and A = 60

So, From condition-6
(i) If D1 and D4 sold no window Non-inverter ACs, then D2 sold twice as many as D3, meaning D2 sold 4 and D3 sold 2 of this type.
From condition-2
(ii) Let's say D1 sold x window inverter ACs, then the number of split inverter ACs sold would be 13 minus x.
From condition-5
(iii) Let's suppose y represents the number of window ACs sold by D3 and D4. In that case, D2 sold 3y ACs of this type.
From condition-4
(iv) Number of split ACs sold by D1 will be 2x
From condition-7
(v) Let's say z represents the number of split inverter ACs sold by D3 and D4. In that case, D2 sold twice as many, totaling 2z ACs of this type.
Now, let use the (i), (ii), (iii), (iv) and (v) to create a table:

As per the question, the total number of window ACs is 15.
So, x + 3y + y + y = 15
x + 5y = 15 (where x and y should be greater than or equal to 2 from condition 1)
Hence, the only solution is x = 5 and y = 2
Now, use the values to fill the table:

So, the Number of split Inverter ACs is 36, which means that 8 + 2z + z + z = 36
4z = 28
z = 7
Now, use the value of z and using the (5), the number of split AC's sold by D1 is 2 × 5 = 10

So, by looking at the table, we can say that:
The number of non-inverter ACs sold by D2 is 
4 + 1 = 5

At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already in route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).

Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.

At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.

So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows:

From the table, we can see that among the options given, Station E is the station visited the largest number of times.

At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already in route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).

Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.

At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.

So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows:

From the data available in the table, we can see that the teams have crossed the B Station 2 times in the given time period.
So, the correct answer is 2.

At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already in route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).

Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.

At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.

So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows:

From the above table, we can infer that at 10:15 am Team 3 is travelling from Station D to Station E.
So, the correct answer is Team 3.

At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already in route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).

Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.

At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.

So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows:

From the table, we can easily see that only the Team 4 is passing through the Station E twice a day i.e 2 times in a day.
So, the correct answer is 2.

At 9:00 a.m., it's understood that all four teams have selected distinct paths from the starting point since no street sees more than one team traveling in any direction simultaneously.
At 10:00 a.m., we know Team 2 is at station E, and Team 3 is at station D. Moreover, only Team 1 and Team 4 patrol the street connecting stations A and E.
This scenario is only possible if Team 2 traveled from A to E through F, and Team 3 arrived at station D via station C.
Also, it's confirmed that only Teams 1 and 3 are at Station E by 10:30 a.m., and Team 4 avoids passing through Stations B, D, or F. Consequently, Team 1 likely opted for the (A-B) route initially, while Team 4 likely selected the (A-E) route at 9:00 a.m.
So, Team 1 is expected to arrive at B by 9:30 a.m., return to A by 10:00 a.m., and then proceed to E by 10:30 a.m.
Given that Team 4 avoids stations B, D, and F, they can only travel through stations A, E, and C. Therefore, Team 4's routes to reach station E by 11:30 could either be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Given that Team 1 is already in route from A to E at 10:00 a.m., Team 4 cannot opt for the same route at that time. Therefore, the definitive route for Team 4 to reach E by 11:30 a.m. is (A-E-A-C-A-E), and by 12:00 p.m., Team 4 will return to station A.
Therefore, the complete route map for Team 4 is (A-E-A-C-A-E-A).

Observing that Team 1 arrives at station E by 10:30 a.m., we note that they will reach station B by 11:30 a.m., indicating they must travel to B via A.
Therefore, the complete route plan for Team 1 is (A-B-A-E-A-B-A). Additionally, it's confirmed that Teams 1 and 3 are the sole occupants of station E at 10:30 a.m.

At 10:00 a.m., Team 2 has only one option: they must go from E to F since Team 3 is already on the E to D route. For Team 3 to reach A by 12:00 p.m., their only feasible route is E-D-C-A.

So, At 10:30 a.m., Team 2's possible routes back to A are either (F-A-F-A) or (F-E-F-A).
Therefore, the final table is as follows:

From the table, we can infer that Team 3 and Team 4 are passing through the Station C on the mentioned day.
So, the correct answer is 2.