CAT DILR Marathon Questions 2026 with Solutions PDF (Part 3)

We know that only Quid and Xica had lesser mileage than Ligo. Hence Ligo must be ranked 3rd in the mileage parameter. Quid and Xica must be ranked either 4 or 5.  But since we want Quid to score lowest possible points, it must be ranked fifth in mileage parameter and ranked first in turning radius parameter.  Thus Xica will have rank 4 in both the parameters.  Now from a) we know that Ciago scored two points less than a maximum number of possible points. So Ciago must’ve scored 8 points in one parameter and 10 points in another parameter.  But since Xica already scored 8 points (rank 4) in turning radius, Ciago must have second rank (8 points) in mileage and fifth rank in turning radius (10 points). Thus we have,

Since Ligo had rank 3 in mileage, it must have rank 2 in turning radius. Thus J10 had rank 1 in mileage and rank 3 in turning radius.  Thus we get our final table as shown-

Total value of rank of Xica = 8
Total points scored by Xica = 12
Difference, a = 12-8 = 4

Total value of rank of Ligo = 5
Total points scored by Ligo = 10
Difference, b = 10-5 = 5

Thus required difference = 5-4 = 1

The given frequency diagram can be tabulated as follows:

=> There are 2+1 = 3 students who have secured pass class, 2+1+2 = 5 students who have secured first class and 3+1 = 4 students who have secured distinction.

Students with prime roll numbers have not secured first class. Therefore, the roll numbers of all the 6 students who secured first class must be composite. In the given range of roll numbers, there are 6 composite roll numbers - (4,6,8,9,10 and 12). 5 among these 6 students must have secured first class.

Students with composite roll numbers have not secured distinction. Therefore, the students who secured distinction must have a roll number among (2, 3, 5, 7, 11 and 13).

Students with even roll numbers have not scored above 80 marks. Therefore, the student with roll number 2 could not have secured distinction. The student with roll number 2 must have secured pass class.

Students with odd roll numbers have not secured below 50 marks. Therefore, the student with roll number 2 must have secured 40 marks.

The sum of roll numbers of students who secured first class is even. There are 5 students who have secured first class. 5 out of the 6 possible values are even. Therefore, if the student with roll number 9 had secured first class, the sum would not have been even. Since the sum is even, the student with roll number 9 must have secured pass class. He must be one of the 2 students who have secured 50 marks.

Sum of roll numbers of students who got pass class = 18
2 + 9 + x = 18
=> x = 7

Sum of roll numbers of students who got distinction = 3 + 5 + 11 + 13 = 32.
Therefore, option B is the right answer.

The given frequency diagram can be tabulated as follows:

=> There are 2+1 = 3 students who have secured pass class, 2+1+2 = 5 students who have secured first class and 3+1 = 4 students who have secured distinction.

Students with prime roll numbers have not secured first class. Therefore, the roll numbers of all the 6 students who secured first class must be composite. In the given range of roll numbers, there are 6 composite roll numbers - (4,6,8,9,10 and 12). 5 among these 6 students must have secured first class.

Students with composite roll numbers have not secured distinction. Therefore, the students who secured distinction must have a roll number among (2, 3, 5, 7, 11 and 13).

Students with even roll numbers have not scored above 80 marks. Therefore, the student with roll number 2 could not have secured distinction. The student with roll number 2 must have secured pass class.

Students with odd roll numbers have not secured below 50 marks. Therefore, the student with roll number 2 must have secured 40 marks.

The sum of roll numbers of students who secured first class is even. There are 5 students who have secured first class. 5 out of the 6 possible values are even. Therefore, if the student with roll number 9 had secured first class, the sum would not have been even. Since the sum is even, the student with roll number 9 must have secured pass class. He must be one of the 2 students who have secured 50 marks.

Sum of roll numbers of students who got less than 70 marks = 46.

2 + 9 + rest = 46
=> Sum of the roll numbers of the remaining 3 students who got less than 70 = 35.
The only possible combination is 13 + 12 + 10.
Therefore, sum of roll numbers of students who scored above 80 marks = 3 + 5 + 7 + 11 = 26.
Therefore, option A is the right answer.

The given frequency diagram can be tabulated as follows:

=> There are 2+1 = 3 students who have secured pass class, 2+1+2 = 5 students who have secured first class and 3+1 = 4 students who have secured distinction.

Students with prime roll numbers have not secured first class. Therefore, the roll numbers of all the 6 students who secured first class must be composite. In the given range of roll numbers, there are 6 composite roll numbers - (4,6,8,9,10 and 12). 5 among these 6 students must have secured first class.

Students with composite roll numbers have not secured distinction. Therefore, the students who secured distinction must have a roll number among (2, 3, 5, 7, 11 and 13).

Students with even roll numbers have not scored above 80 marks. Therefore, the student with roll number 2 could not have secured distinction. The student with roll number 2 must have secured pass class.

Students with odd roll numbers have not secured below 50 marks. Therefore, the student with roll number 2 must have secured 40 marks.

The sum of roll numbers of students who secured first class is even. There are 5 students who have secured first class. 5 out of the 6 possible values are even. Therefore, if the student with roll number 9 had secured first class, the sum would not have been even. Since the sum is even, the student with roll number 9 must have secured pass class. He must be one of the 2 students who have secured 50 marks.

Difference between the roll numbers of students who scored the highest and the second lowest = 10.
Only 2 cases are possible - (12,2) and (13,3).
12 has secured first class and 2 has scored the lowest. Therefore, we can eliminate this pair.
Therefore, the student with roll number 3 must have scored the second lowest and the student with roll number 13 must have scored the highest.
Therefore, the roll numbers of students who must have scored 90 will be 5, 7 and 11.
Sum of the roll numbers = 5 + 7 + 11 = 23.

Therefore, option D is the right answer.

Let the total number of ice creams sold in April be $$T$$. The number of units sold of the five ice cream flavours can then be written as

$$\dfrac{3T}{20},\quad \dfrac{T}{4},\quad \dfrac{T}{10},\quad \dfrac{3T}{8},\quad \dfrac{T}{8}.$$

Since the number of ice creams sold for each flavour must be a natural number, $$T$$ must be a multiple of $$ 4, 8, 10,$$ and $$20$$. Therefore, $$T$$ must be of the form

$$T=40k,$$

where $$k$$ is a natural number.

Substituting $$T=40k$$, the number of units sold of the five flavours becomes

$$6k,\quad 10k,\quad 4k,\quad 15k,\quad 5k.$$

We can see that $$6k$$ is the median of these five numbers. We are given that the median has exactly $$9$$ factors.

Therefore, the median must be of the form $$a^2b^2$$. The only possible value of the median is $$36$$, which gives

$$6k=36$$

$$\therefore k=6.$$

Thus, the number of units sold of the five flavours in April are

$$36,\quad 60,\quad 24,\quad 90,\quad 30.$$

Since Badam Milk has the lowest number of units sold in April, its number of units sold must be

$$\boxed{24}.$$

We are given that the number of units sold of Mango in April is equal to the sum of the number of units sold of Cookies and Cream and Krispy Kit. This is possible only when

$$\text{Mango}=90,\qquad(\text{Cookies and Cream},\text{Krispy Kit})=(60,30)\text{ or }(30,60).$$

Hence, the remaining flavour, Biscoff, must have a number of units sold of

$$\boxed{36}$$

in April.

The total number of units sold in May decreased by $$25\%$$. Therefore,

$$\text{May number of units sold }=0.75(\text{April number of units sold})$$

$$=0.75(240)=180.$$

Mango had $$90$$ number of units sold in April, and its number of units sold decreased by $$\dfrac{550}{9}\%$$. Let its May sales be $$x$$.

$$\dfrac{90-x}{90}\times100=\dfrac{550}{9}$$

which gives

$$x=35.$$

Therefore, for Mango, the number of units sold in May is

$$\boxed{35}.$$

Since Badam Milk recorded the lowest number of units sold in May and the sum of its number of units sold across the two months is a perfect square, its May sales must be less than $$35$$. The only possibility satisfying these conditions is

$$\boxed{\text{Badam Milk in May}=25},$$

since

$$24+25=49=7^2.$$

Exactly one flavour recorded the same number of units sold in both April and May. Such a number must be a multiple of $$5$$ and, from the number of units sold in April, can therefore only be

$$30,\quad 60,\quad 90.$$

The value cannot be $$90$$ because Mango's number of units sold changed from April to May.

Now, suppose $$60$$ remains constant across both months. We already know that the combined number of units sold of Badam Milk and Mango in May is

$$25+35=60.$$

Therefore, the remaining two flavours would have to account for

$$180-60-60=60$$

number of units sold.

However, the number of units sold of both these flavours is greater than $$25$$, and their number of units sold must be distinct. The minimum possible sum considering the required multiples of $$5$$ is

$$30+40=70>60.$$

Hence, $$60$$ cannot remain constant across the two months.

Therefore, $$30$$ must remain constant across both months.

We already know that Cookies and Cream recorded the highest number of units sold in May, and hence its number of units sold must be greater than Mango's number of units sold, which is $$35$$. Since $$30$$ remains constant, the flavour with $$30$$ sales must be Krispy Kit.

Thus,

$$\boxed{\text{Krispy Kit}=30}$$

in both April and May.

The number of units sold already accounted for in May is

$$25+35+30=90.$$

Therefore, the combined number of units sold of Cookies and Cream and Biscoff must be

$$180-90=90.$$

Cookies and Cream must have the highest number of units sold in May, and its number of units sold must be greater than $$35$$. Also, all numbers of units sold in May must be distinct.

The only possible allocation satisfying these conditions is

$$\boxed{\text{Cookies and Cream}=50,\qquad\text{Biscoff}=40}.$$

Hence, the number of units sold in May is

$$\boxed{25,\ 30,\ 35,\ 40,\ 50}.$$

The highest sales made in May are $$50$$ and the lowest is $$25$$. Their difference is thus $$50-25 = 25$$

Let the total number of ice creams sold in April be $$T$$. The number of units sold of the five ice cream flavours can then be written as

$$\dfrac{3T}{20},\quad \dfrac{T}{4},\quad \dfrac{T}{10},\quad \dfrac{3T}{8},\quad \dfrac{T}{8}.$$

Since the number of ice creams sold for each flavour must be a natural number, $$T$$ must be a multiple of $$ 4, 8, 10,$$ and $$20$$. Therefore, $$T$$ must be of the form

$$T=40k,$$

where $$k$$ is a natural number.

Substituting $$T=40k$$, the number of units sold of the five flavours becomes

$$6k,\quad 10k,\quad 4k,\quad 15k,\quad 5k.$$

We can see that $$6k$$ is the median of these five numbers. We are given that the median has exactly $$9$$ factors.

Therefore, the median must be of the form $$a^2b^2$$. The only possible value of the median is $$36$$, which gives

$$6k=36$$

$$\therefore k=6.$$

Thus, the number of units sold of the five flavours in April are

$$36,\quad 60,\quad 24,\quad 90,\quad 30.$$

Since Badam Milk has the lowest number of units sold in April, its number of units sold must be

$$\boxed{24}.$$

We are given that the number of units sold of Mango in April is equal to the sum of the number of units sold of Cookies and Cream and Krispy Kit. This is possible only when

$$\text{Mango}=90,\qquad(\text{Cookies and Cream},\text{Krispy Kit})=(60,30)\text{ or }(30,60).$$

Hence, the remaining flavour, Biscoff, must have a number of units sold of

$$\boxed{36}$$

in April.

The total number of units sold in May decreased by $$25\%$$. Therefore,

$$\text{May number of units sold }=0.75(\text{April number of units sold})$$

$$=0.75(240)=180.$$

Mango had $$90$$ number of units sold in April, and its number of units sold decreased by $$\dfrac{550}{9}\%$$. Let its May sales be $$x$$.

$$\dfrac{90-x}{90}\times100=\dfrac{550}{9}$$

which gives

$$x=35.$$

Therefore, for Mango, the number of units sold in May is

$$\boxed{35}.$$

Since Badam Milk recorded the lowest number of units sold in May and the sum of its number of units sold across the two months is a perfect square, its May sales must be less than $$35$$. The only possibility satisfying these conditions is

$$\boxed{\text{Badam Milk in May}=25},$$

since

$$24+25=49=7^2.$$

Exactly one flavour recorded the same number of units sold in both April and May. Such a number must be a multiple of $$5$$ and, from the number of units sold in April, can therefore only be

$$30,\quad 60,\quad 90.$$

The value cannot be $$90$$ because Mango's number of units sold changed from April to May.

Now, suppose $$60$$ remains constant across both months. We already know that the combined number of units sold of Badam Milk and Mango in May is

$$25+35=60.$$

Therefore, the remaining two flavours would have to account for

$$180-60-60=60$$

number of units sold.

However, the number of units sold of both these flavours is greater than $$25$$, and their number of units sold must be distinct. The minimum possible sum considering the required multiples of $$5$$ is

$$30+40=70>60.$$

Hence, $$60$$ cannot remain constant across the two months.

Therefore, $$30$$ must remain constant across both months.

We already know that Cookies and Cream recorded the highest number of units sold in May, and hence its number of units sold must be greater than Mango's number of units sold, which is $$35$$. Since $$30$$ remains constant, the flavour with $$30$$ sales must be Krispy Kit.

Thus,

$$\boxed{\text{Krispy Kit}=30}$$

in both April and May.

The number of units sold already accounted for in May is

$$25+35+30=90.$$

Therefore, the combined number of units sold of Cookies and Cream and Biscoff must be

$$180-90=90.$$

Cookies and Cream must have the highest number of units sold in May, and its number of units sold must be greater than $$35$$. Also, all numbers of units sold in May must be distinct.

The only possible allocation satisfying these conditions is

$$\boxed{\text{Cookies and Cream}=50,\qquad\text{Biscoff}=40}.$$

Hence, the number of units sold in May is

$$\boxed{25,\ 30,\ 35,\ 40,\ 50}.$$

The sum of Biscoff sales made in April and May is $$36+40=76$$.

The sum of the number of coins with any 2 persons facing each other is the same.
We know that the number of coins with the 10 persons varies from 1 to 10. Sum = 1+2+3+…+10 = 55
There will be 5 pairs of persons facing each other.

=> Sum of the number of coins with any 2 persons facing each other = 55/5 = 11.

Now, it has been given that no person who faces the North has an even number of coins. Therefore, all the persons who face the North should have an odd number of coins and all the persons who face the South should have an even number of coins. In each row, the number of coins with exactly one of the 2 persons sitting at the corners is prime. We know that persons facing the South have an even number of coins. 2 is the only even prime. Therefore, the person with 2 coins should be sitting at one of the corners. The person with 2 coins is sitting third to the right of P. The person with 9 coins should be facing the person with 2 coins.

The person facing P has 9 coins less than him. Therefore, P should have 10 coins and the person facing P should have 1 coin. Q is second to the right of the person with 5 coins. Therefore, the person sitting at the rightmost end of row 1 cannot have 5 coins. The person with 5 coins should be sitting next to the person with 9 coins. Q should be sitting next to the person sitting at the corner.

As we can see, Q faces P. Now, it has been given that A is not facing the person who has 3 coins but sitting next to the person who has 8 coins. Therefore, A should be sitting on the row that faces the South. Also, the person sitting at the corner cannot have 8 coins. If the person sitting at the rightmost corner has 6 coins, then the person facing him should have 5 coins (which is impossible since we know the position of the person who has 5 coins). Therefore, the person sitting at the rightmost corner should have 4 coins and the person facing him should have 7 coins. A should have 6 coins and the person sitting opposite to A should have 5 coins.

Now, B is not a neighbour of A and C faces B. => B can have 3 or 9 coins. If B has 3 coins, the R will have 5 coins and will sit adjacent to B. Therefore, B should have 9 coins and C should have 2 coins. D has 3 coins less than the person he faces. => D faces R. The final arrangement is as follows:

Each team plays 2 matches against the other teams. Thus, the total number of matches played by any team is 6, and the total number of matches played must be $$2 \times \binom{4}{2} = 12$$

From point (2), we can see that the number of wins is $$0$$ or a prime, and the numbers are unique. Since the number of wins must be less than the number of matches played by each team, the only possible numbers of wins are 0, 2, 3, or 5.

Since the total number of matches is $$12$$ but there are only 10 decisive matches, 2 matches must have ended in a draw.

Now we are given that A lost 2 against a team, won two more matches and then won 1 and lost another. So A must have won a total of 3 matches and lost 3. Thus, A must have 0 draws. From point (3), we also know that B has 0 draws. Thus, the 2 draws must have been between C and D.

Now, neither C nor D could have won 5 matches, because their total number of matches would then exceed 6. Thus, B wins 5 matches.

Also given that C wins 2 of its matches, D must have been the team that won 0 matches.

Our final table looks like this

Now we need to find the final scores of the matches. We are given that the total number of goals scored by B is 11. Since no team scores more than 2 in any match, the maximum number of goals that a team can score is 12 if it scores 2 in every match.

Thus, B must have scored 2 in every match that it won and scored 1 in the match that it lost.

Also, C won both its home matches. Since C drew both its matches against D, it must have won one match against A and another against B. Since B scored 1 goal in the match that it lost, the score in the match that C won against B must be 2-1.

We can make our table based on the information so far as follows.

Now, it's given that the sum of the goals scored by A and the opponent is the same in all the matches that A played. Based on the information so far, we can infer that the score must be either 2-0 or 1-1 if the number of goals scored is 2 or 2-1 if the number of goals scored is 3. But since none of A's matches ended in a draw, it can either be 2-0 or 2-1.

It can't be 2-0, as then the total number of goals in the matches that it won would be 3*2 = 6 and the matches that it lost would be 0*3=0. But we know that the total number of goals scored by A is 9. Hence, the scoreline in all of A's matches must be 2-1.

We also know that A won two matches against the same team, lost two matches against another team and won one and lost another match against the third team. Since we know that D won 0 matches, A must have won both against D. Since the only match that B lost is against C, A must have lost both matches against B and won one and lost another against C. We can thus fill the table further

Now we know that both the draws had unique scores, which could be any two of 0-0,1-1, 2-2.

But we know that D has only scored 3 goals in the tournament, of which 2 have already been accounted for. Thus, 2-2 cannot be the score for one of the draws. The only possible scores are 1-1 and 0-0.

Now, using the information that B has conceded 4 goals in the tournament, we can fill the table.

We have 6 restaurants, which are being rated in 5 different categories such that no two restaurants got the same rating in any category and no restaurant got the same rating in multiple categories. So, we will get a 6 x 5 table. We will include one more column, called as Dummy Rating, which will contain the rating that any restaurant will not get in any of the categories.

So, the required table will look like:

From statement 1, we know that the ratings of E in Food, B in Value and F in Ambience are forming an increasing geometric progression, in the same order. The only increasing geometric progression possible is 1, 2 and 4, with a common ratio of 2. So, the ratings of E in Food, B in Value and F in Ambience are 1, 2 and 4, respectively.

From statement 2, we have two distinct arithmetic progression, whose common differences are equal to the common ratio of the geometric progression. So, the possible values of the ratings will be (1, 3, 5) and (2, 4, 6).

Now, two cases will be possible.

Case 1: A's rating in Hygiene, Food and Value is (1, 3 and 5) and ratings which are not assigned to C, E and A are (2, 4 and 6).

But from statement 5, we know that the Popularity Rating of F is the lowest possible. We know that the Popularity Rating is calculated by the sum of the ratings received by any restaurant in all 5 categories. Since, all the 5 ratings are distinct, the lowest possible PR of restaurant F will be 15 (when it didn't receive 6 as its rating). So, the dummy rating for both F and A will be 6, which is not possible. Hence, this case is rejected.

Case 2: A's rating in Hygiene, Food and Value is (2, 4 and 6) and ratings which are not assigned to C, E and A are (1, 3 and 5).

From statement 5, we know that the Popularity Rating of F is the lowest possible. Since, all the 5 ratings are distinct, the lowest possible Popularity Rating of restaurant F will be 15 (when it didn't receive 6 as its rating).

We know that B received a rating of 2 in Value and the ratings which not assigned to B and D can either 2 or 4. So, B has not been assigned a rating of 4 and D will not been assigned a rating of 2.

D's rating in Value is twice the common difference of the arithmetic progression, which is 4.

The ratings of C, E and F in Value can be 1, 3 and 5, in any order. But E got a rating of 1 in Food and didn't receive a rating of 3 in any of the categories. So, E got a rating of 5 in Value. Also, C didn't get a rating of 1 in any of the categories. So, F and C will get a rating of 1 and 3, respectively in Value.

Now, from statement 4, A's rating in Ambience is equal to the rating of F in Hygiene and A has only two ratings, 1 and 3, which has not been assigned in any of the categories. So, if A's rating in Ambience is 1, F's rating in Hygiene will be 1. But it is not possible as F has already got a rating of 1 in Value. So, A's rating in Ambience will be 3 and F's rating in Hygiene will also be 3. So, A's rating in Service will be 1.

Also, D's rating in Value can be one of 1, 3, 5 and 6. We know that D's Food rating matches F's Service rating. F has already received a rating of 1, 3, 4 and 6 in Value, Hygiene, Ambience and Dummy Rating. So, D's rating in Value can only be 5 and F's rating in Services will also be 5. So, F's rating in Food will be 2.

So, the updated table for the ratings will be:

Now, Food has ratings of 3 and 6 left to be assigned. C has already got a rating of 3 in Value. So, C will get a rating of 6 in Food and B will get 3 in Food.

Ratings of 1, 5 and 6 are left to be assigned to B. A and F has already got a rating of 1 and 5 in Service. So, B will get a rating of 6 in Service. Similarly, D will get a rating of 3 in Service.

From statement 3, we know that E's Hygiene rating is the highest possible, which is 6.

Now, E has ratings of 2 and 4, which is needed to be distributed to Ambience and Service. But, F has already received a rating of 4 in Ambience. So, E will get a rating of 2 and 4 in Ambience and Service, respectively. Hence, C will get a rating of 2 in Service.

Now, C will get a rating of 5 in Ambience and 4 in Hygiene.

Similarly, in Hygiene, B and D will get a rating of 5 and 1, respectively. In Ambience, B and D will get a rating of 1 and 6, respectively.

Finally, the Popularity Rating of A, B, C, D, E and F is 16, 17, 20, 19, 18 and 15, respectively.

So, the final table of the ratings will be:

Now, for the weights, from statement 1, we know that, all the weights are going to be in an arithmetic progression and the sum is 100.

Let the weights be (a-2d), (a-d), a, (a+d), (a+2d).

$$a-2d+a-d+a+a+d+a+2d=5a=100$$

$$a=20$$

Now, from statement 2, we know that Food will have the highest weightage and Ambience will be second lowest. So, Food will have a weightage of (a+2d) and Ambience will have (a-d). Difference of these weights is 15. So, $$(a+2d)-(a-d)=3d=15$$. So, the value of $$d$$ is 5.

So, the value of the weights will be 10, 15, 20, 25 and 30. Out of these, 30 annd 15 has already been assigned to Food and Ambience.

Out of the remaining weights, weight of Value is twice of the difference between the weights of Hygiene and Service.

It is only possible when the weight of Value is 10 and the weights of Hygiene and Service will be one of the 20 and 25, in any order.

So, the final table, including the weights and the ratings is:

Now, let's answer the questions:

The weight of Service is 25.

Compliance Rating of C = $$\left(30\times6\right)+\left(15\times5\right)+\left(25\times2\right)+\left(10\times3\right)+\left(20\times4\right)=415$$

Popularity Rating of C = 20

Difference = $$\left(415-20\right)=395$$

Hence, the required answer is 395.

All the friends were given 5 targets each and 6 attempts on each target. The average number of attempts per target for the 6 friends is given as 4.4, 5, 5.2, 5.4, 5.8 and 6.

Sum of the attempts taken to hit all five targets by each friend = 22, 25, 26, 27, 29 and 30 (in some order)

All the friends were ranked by the average number of total attempts, with the highest average receiving the 1st rank and the lowest rank receiving the 6th rank. So, the friend who is taking a total of 30 attempts is the first person to attend the training, and the friend who is taking a total of 22 attempts is the last person to attend the training.

From statement 1, exactly two friends used the same number of attempts per target. Since all the targets were shot, the total number of attempts taken by both the friends should be a multiple of 5. The only possibilities are 25 and 30.

So, the total number of attempts taken by these two friends is 25 and 30, where the number of attempts for each target is (5,5,5,5,5) and (6,6,6,6,6), respectively.

From statements 1 and 4, we know that Chandu took a total of 22 attempts (as he is the last person to attend the training), and he never took the maximum possible attempts to shoot any target. It is also given that his attempts for 3 targets formed a geometric progression and the attempts for the other two targets were equal.

Since each target has a limited attempt of 6, the only possible GP with a common ratio of more than 1 is (1, 2, 4). But if we take 1, 2 and 4, the maximum possible number of attempts will be (1+2+4+6+6)=19, which is less than the lowest number of attempts taken by any one of the friends (22).

So, the other possible GPs will be (4,4,4), (5,5,5), and (6,6,6), as any GP less than (4,4,4) will not reach upto 22 attempts.

Now, for (4,4,4), the maximum possible number of attempts is (4+4+4+6+6)=24. So, the only possible number of attempts for Chandu will be (4,4,4,5,5), as the remaining 10 attempts were equally divided among two targets.

Now, for (5,5,5), the maximum value of the possible number of attempts will be (5+5+5+6+6)=27. But 27 will not be included, as Chandu never took the maximum possible number of attempts for any target. Now, the other possibility can be 25 but the distribution of the number of attempts in this case will be (5,5,5,5,5), which is not possible as the number of attempts of exactly 3 targets are in geometric progression.

Now, for (6,6,6), it is not possible.

So, for Chandu, the total number of attempts is equal to 22, and the distribution is (5,5,4,4,4).

Now, from statement 2, Eshaan and Bipro attended the training first and second, respectively. So, Eshaan and Bipro took 30 and 29 attempts, respectively. For 29 attempts, only one distribution is possible, which is (6,6,6,6,5).

So, for Eshaan and Bipro, the distributions of attempts are (6,6,6,6,6) and (6,6,6,6,5).

Now, for a total of 26 attempts, the possible distributions are (6,6,6,6,2), (6,6,6,5,3), (6,6,6,4,4), (6,6,5,5,4), (6,5,5,5,5).

Similarly, for 27 attempts, the possible distributions are (6,6,6,6,3), (6,6,6,5,4), (6,6,5,5,5).

From statement 3, we know that the median number of attempts for Akash was 5 and the range of attempts was 2. The remaining attempts, which are yet to be assigned, are 25, 26 and 27.

For 25, the median is 5, and the range is 0 $$(5-5=0)$$. Not possible.

For 26, the medians are 6, 6, 6, 5 and 5, with the range of $$(6-2=4)$$, $$(6-3=3)$$, $$(6-4=2)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively.

The only possibility matching the conditions given in the 3rd statement is (6,6,5,5,4).

For 27, the medians are 6, 6 and 5, with the ranges of $$(6-3=3)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively. It is not possible.

So, for Akash, the number of attempts are 26 and the possible distributions are (6,6,5,5,4).

Now, from Statement 2, the mode of the number of attempts of Farah exceeded the mean of her attempts. It is not possible if Farah's number of attempts is 25. So, it is only possible if the number of attempts is 27.

From the possible distributions of Farah, the valid ones are (6,6,6,6,3) and (6,6,6,5,4), as (6,6,5,5,5) has a mode of 5, which is less than the mean of the attempts (5.4), and hence not possible.

Hence, Dipanshu will have a total number of attempts of 25, with a distribution of (5,5,5,5,5).

Case 1: When the distribution of the attempts of Farah is (6,6,6,6,3).

It is given that each friend's training hours are equal to the median number of attempts across all targets. It was also given that, if there are two modes, the training hours will be equal to the smaller mode.

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be 3 hours (Farah's range of attempts is 3). With the break, the Wednesday slot will be over.

Now, Akash will train on Thursday (till 3 pm), the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 5 pm on Thursday. He will train on Friday (till 2 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 4 pm) and will train for 2 hours (till 12 pm) on Saturday.

The training course will be completed at 12 pm on Saturday.

Case 2: When the distribution of the attempts of Farah is (6,6,6,5,4).

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be of 2 hours (Range of attempts for Farah is 2). The break will be over at 5 pm on Wednesday, and Akash will start his training at the same time and will train till 2 pm on Thursday and the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 4 pm on Thursday. He will train on Friday (till 1 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 3 pm) and will train for 1 hour (till 11 am) on Saturday.

The training course will be completed at 11 am on Saturday.

Now, let's answer the question:

Since the training is ending at 12 pm, we are talking about case 1. In this case, Farah, Dipanshu and Chandu attended the training on 2 different days.

So, the required answer is Option C.

All the friends were given 5 targets each and 6 attempts on each target. The average number of attempts per target for the 6 friends is given as 4.4, 5, 5.2, 5.4, 5.8 and 6.

Sum of the attempts taken to hit all five targets by each friend = 22, 25, 26, 27, 29 and 30 (in some order)

All the friends were ranked by the average number of total attempts, with the highest average receiving the 1st rank and the lowest rank receiving the 6th rank. So, the friend who is taking a total of 30 attempts is the first person to attend the training, and the friend who is taking a total of 22 attempts is the last person to attend the training.

From statement 1, exactly two friends used the same number of attempts per target. Since all the targets were shot, the total number of attempts taken by both the friends should be a multiple of 5. The only possibilities are 25 and 30.

So, the total number of attempts taken by these two friends is 25 and 30, where the number of attempts for each target is (5,5,5,5,5) and (6,6,6,6,6), respectively.

From statements 1 and 4, we know that Chandu took a total of 22 attempts (as he is the last person to attend the training), and he never took the maximum possible attempts to shoot any target. It is also given that his attempts for 3 targets formed a geometric progression and the attempts for the other two targets were equal.

Since each target has a limited attempt of 6, the only possible GP with a common ratio of more than 1 is (1, 2, 4). But if we take 1, 2 and 4, the maximum possible number of attempts will be (1+2+4+6+6)=19, which is less than the lowest number of attempts taken by any one of the friends (22).

So, the other possible GPs will be (4,4,4), (5,5,5), and (6,6,6), as any GP less than (4,4,4) will not reach upto 22 attempts.

Now, for (4,4,4), the maximum possible number of attempts is (4+4+4+6+6)=24. So, the only possible number of attempts for Chandu will be (4,4,4,5,5), as the remaining 10 attempts were equally divided among two targets.

Now, for (5,5,5), the maximum value of the possible number of attempts will be (5+5+5+6+6)=27. But 27 will not be included, as Chandu never took the maximum possible number of attempts for any target. Now, the other possibility can be 25 but the distribution of the number of attempts in this case will be (5,5,5,5,5), which is not possible as the number of attempts of exactly 3 targets are in geometric progression.

Now, for (6,6,6), it is not possible.

So, for Chandu, the total number of attempts is equal to 22, and the distribution is (5,5,4,4,4).

Now, from statement 2, Eshaan and Bipro attended the training first and second, respectively. So, Eshaan and Bipro took 30 and 29 attempts, respectively. For 29 attempts, only one distribution is possible, which is (6,6,6,6,5).

So, for Eshaan and Bipro, the distributions of attempts are (6,6,6,6,6) and (6,6,6,6,5).

Now, for a total of 26 attempts, the possible distributions are (6,6,6,6,2), (6,6,6,5,3), (6,6,6,4,4), (6,6,5,5,4), (6,5,5,5,5).

Similarly, for 27 attempts, the possible distributions are (6,6,6,6,3), (6,6,6,5,4), (6,6,5,5,5).

From statement 3, we know that the median number of attempts for Akash was 5 and the range of attempts was 2. The remaining attempts, which are yet to be assigned, are 25, 26 and 27.

For 25, the median is 5, and the range is 0 $$(5-5=0)$$. Not possible.

For 26, the medians are 6, 6, 6, 5 and 5, with the range of $$(6-2=4)$$, $$(6-3=3)$$, $$(6-4=2)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively.

The only possibility matching the conditions given in the 3rd statement is (6,6,5,5,4).

For 27, the medians are 6, 6 and 5, with the ranges of $$(6-3=3)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively. It is not possible.

So, for Akash, the number of attempts are 26 and the possible distributions are (6,6,5,5,4).

Now, from Statement 2, the mode of the number of attempts of Farah exceeded the mean of her attempts. It is not possible if Farah's number of attempts is 25. So, it is only possible if the number of attempts is 27.

From the possible distributions of Farah, the valid ones are (6,6,6,6,3) and (6,6,6,5,4), as (6,6,5,5,5) has a mode of 5, which is less than the mean of the attempts (5.4), and hence not possible.

Hence, Dipanshu will have a total number of attempts of 25, with a distribution of (5,5,5,5,5).

Case 1: When the distribution of the attempts of Farah is (6,6,6,6,3).

It is given that each friend's training hours are equal to the median number of attempts across all targets. It was also given that, if there are two modes, the training hours will be equal to the smaller mode.

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be 3 hours (Farah's range of attempts is 3). With the break, the Wednesday slot will be over.

Now, Akash will train on Thursday (till 3 pm), the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 5 pm on Thursday. He will train on Friday (till 2 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 4 pm) and will train for 2 hours (till 12 pm) on Saturday.

The training course will be completed at 12 pm on Saturday.

Case 2: When the distribution of the attempts of Farah is (6,6,6,5,4).

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be of 2 hours (Range of attempts for Farah is 2). The break will be over at 5 pm on Wednesday, and Akash will start his training at the same time and will train till 2 pm on Thursday and the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 4 pm on Thursday. He will train on Friday (till 1 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 3 pm) and will train for 1 hour (till 11 am) on Saturday.

The training course will be completed at 11 am on Saturday.

Now, let's answer the question:

Akash attended the training on 2 different days in Case 2. The course ended at 11 am in Case 2.

So, the required answer is Option C.

All the friends were given 5 targets each and 6 attempts on each target. The average number of attempts per target for the 6 friends is given as 4.4, 5, 5.2, 5.4, 5.8 and 6.

Sum of the attempts taken to hit all five targets by each friend = 22, 25, 26, 27, 29 and 30 (in some order)

All the friends were ranked by the average number of total attempts, with the highest average receiving the 1st rank and the lowest rank receiving the 6th rank. So, the friend who is taking a total of 30 attempts is the first person to attend the training, and the friend who is taking a total of 22 attempts is the last person to attend the training.

From statement 1, exactly two friends used the same number of attempts per target. Since all the targets were shot, the total number of attempts taken by both the friends should be a multiple of 5. The only possibilities are 25 and 30.

So, the total number of attempts taken by these two friends is 25 and 30, where the number of attempts for each target is (5,5,5,5,5) and (6,6,6,6,6), respectively.

From statements 1 and 4, we know that Chandu took a total of 22 attempts (as he is the last person to attend the training), and he never took the maximum possible attempts to shoot any target. It is also given that his attempts for 3 targets formed a geometric progression and the attempts for the other two targets were equal.

Since each target has a limited attempt of 6, the only possible GP with a common ratio of more than 1 is (1, 2, 4). But if we take 1, 2 and 4, the maximum possible number of attempts will be (1+2+4+6+6)=19, which is less than the lowest number of attempts taken by any one of the friends (22).

So, the other possible GPs will be (4,4,4), (5,5,5), and (6,6,6), as any GP less than (4,4,4) will not reach upto 22 attempts.

Now, for (4,4,4), the maximum possible number of attempts is (4+4+4+6+6)=24. So, the only possible number of attempts for Chandu will be (4,4,4,5,5), as the remaining 10 attempts were equally divided among two targets.

Now, for (5,5,5), the maximum value of the possible number of attempts will be (5+5+5+6+6)=27. But 27 will not be included, as Chandu never took the maximum possible number of attempts for any target. Now, the other possibility can be 25 but the distribution of the number of attempts in this case will be (5,5,5,5,5), which is not possible as the number of attempts of exactly 3 targets are in geometric progression.

Now, for (6,6,6), it is not possible.

So, for Chandu, the total number of attempts is equal to 22, and the distribution is (5,5,4,4,4).

Now, from statement 2, Eshaan and Bipro attended the training first and second, respectively. So, Eshaan and Bipro took 30 and 29 attempts, respectively. For 29 attempts, only one distribution is possible, which is (6,6,6,6,5).

So, for Eshaan and Bipro, the distributions of attempts are (6,6,6,6,6) and (6,6,6,6,5).

Now, for a total of 26 attempts, the possible distributions are (6,6,6,6,2), (6,6,6,5,3), (6,6,6,4,4), (6,6,5,5,4), (6,5,5,5,5).

Similarly, for 27 attempts, the possible distributions are (6,6,6,6,3), (6,6,6,5,4), (6,6,5,5,5).

From statement 3, we know that the median number of attempts for Akash was 5 and the range of attempts was 2. The remaining attempts, which are yet to be assigned, are 25, 26 and 27.

For 25, the median is 5, and the range is 0 $$(5-5=0)$$. Not possible.

For 26, the medians are 6, 6, 6, 5 and 5, with the range of $$(6-2=4)$$, $$(6-3=3)$$, $$(6-4=2)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively.

The only possibility matching the conditions given in the 3rd statement is (6,6,5,5,4).

For 27, the medians are 6, 6 and 5, with the ranges of $$(6-3=3)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively. It is not possible.

So, for Akash, the number of attempts are 26 and the possible distributions are (6,6,5,5,4).

Now, from Statement 2, the mode of the number of attempts of Farah exceeded the mean of her attempts. It is not possible if Farah's number of attempts is 25. So, it is only possible if the number of attempts is 27.

From the possible distributions of Farah, the valid ones are (6,6,6,6,3) and (6,6,6,5,4), as (6,6,5,5,5) has a mode of 5, which is less than the mean of the attempts (5.4), and hence not possible.

Hence, Dipanshu will have a total number of attempts of 25, with a distribution of (5,5,5,5,5).

Case 1: When the distribution of the attempts of Farah is (6,6,6,6,3).

It is given that each friend's training hours are equal to the median number of attempts across all targets. It was also given that, if there are two modes, the training hours will be equal to the smaller mode.

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be 3 hours (Farah's range of attempts is 3). With the break, the Wednesday slot will be over.

Now, Akash will train on Thursday (till 3 pm), the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 5 pm on Thursday. He will train on Friday (till 2 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 4 pm) and will train for 2 hours (till 12 pm) on Saturday.

The training course will be completed at 12 pm on Saturday.

Case 2: When the distribution of the attempts of Farah is (6,6,6,5,4).

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be of 2 hours (Range of attempts for Farah is 2). The break will be over at 5 pm on Wednesday, and Akash will start his training at the same time and will train till 2 pm on Thursday and the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 4 pm on Thursday. He will train on Friday (till 1 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 3 pm) and will train for 1 hour (till 11 am) on Saturday.

The training course will be completed at 11 am on Saturday.

Now, let's answer the question:

On Thursday, Akash and Dipanshu are the only ones who can attend the training in both the cases.

So, the required answer is Option B.

All the friends were given 5 targets each and 6 attempts on each target. The average number of attempts per target for the 6 friends is given as 4.4, 5, 5.2, 5.4, 5.8 and 6.

Sum of the attempts taken to hit all five targets by each friend = 22, 25, 26, 27, 29 and 30 (in some order)

All the friends were ranked by the average number of total attempts, with the highest average receiving the 1st rank and the lowest rank receiving the 6th rank. So, the friend who is taking a total of 30 attempts is the first person to attend the training, and the friend who is taking a total of 22 attempts is the last person to attend the training.

From statement 1, exactly two friends used the same number of attempts per target. Since all the targets were shot, the total number of attempts taken by both the friends should be a multiple of 5. The only possibilities are 25 and 30.

So, the total number of attempts taken by these two friends is 25 and 30, where the number of attempts for each target is (5,5,5,5,5) and (6,6,6,6,6), respectively.

From statements 1 and 4, we know that Chandu took a total of 22 attempts (as he is the last person to attend the training), and he never took the maximum possible attempts to shoot any target. It is also given that his attempts for 3 targets formed a geometric progression and the attempts for the other two targets were equal.

Since each target has a limited attempt of 6, the only possible GP with a common ratio of more than 1 is (1, 2, 4). But if we take 1, 2 and 4, the maximum possible number of attempts will be (1+2+4+6+6)=19, which is less than the lowest number of attempts taken by any one of the friends (22).

So, the other possible GPs will be (4,4,4), (5,5,5), and (6,6,6), as any GP less than (4,4,4) will not reach upto 22 attempts.

Now, for (4,4,4), the maximum possible number of attempts is (4+4+4+6+6)=24. So, the only possible number of attempts for Chandu will be (4,4,4,5,5), as the remaining 10 attempts were equally divided among two targets.

Now, for (5,5,5), the maximum value of the possible number of attempts will be (5+5+5+6+6)=27. But 27 will not be included, as Chandu never took the maximum possible number of attempts for any target. Now, the other possibility can be 25 but the distribution of the number of attempts in this case will be (5,5,5,5,5), which is not possible as the number of attempts of exactly 3 targets are in geometric progression.

Now, for (6,6,6), it is not possible.

So, for Chandu, the total number of attempts is equal to 22, and the distribution is (5,5,4,4,4).

Now, from statement 2, Eshaan and Bipro attended the training first and second, respectively. So, Eshaan and Bipro took 30 and 29 attempts, respectively. For 29 attempts, only one distribution is possible, which is (6,6,6,6,5).

So, for Eshaan and Bipro, the distributions of attempts are (6,6,6,6,6) and (6,6,6,6,5).

Now, for a total of 26 attempts, the possible distributions are (6,6,6,6,2), (6,6,6,5,3), (6,6,6,4,4), (6,6,5,5,4), (6,5,5,5,5).

Similarly, for 27 attempts, the possible distributions are (6,6,6,6,3), (6,6,6,5,4), (6,6,5,5,5).

From statement 3, we know that the median number of attempts for Akash was 5 and the range of attempts was 2. The remaining attempts, which are yet to be assigned, are 25, 26 and 27.

For 25, the median is 5, and the range is 0 $$(5-5=0)$$. Not possible.

For 26, the medians are 6, 6, 6, 5 and 5, with the range of $$(6-2=4)$$, $$(6-3=3)$$, $$(6-4=2)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively.

The only possibility matching the conditions given in the 3rd statement is (6,6,5,5,4).

For 27, the medians are 6, 6 and 5, with the ranges of $$(6-3=3)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively. It is not possible.

So, for Akash, the number of attempts are 26 and the possible distributions are (6,6,5,5,4).

Now, from Statement 2, the mode of the number of attempts of Farah exceeded the mean of her attempts. It is not possible if Farah's number of attempts is 25. So, it is only possible if the number of attempts is 27.

From the possible distributions of Farah, the valid ones are (6,6,6,6,3) and (6,6,6,5,4), as (6,6,5,5,5) has a mode of 5, which is less than the mean of the attempts (5.4), and hence not possible.

Hence, Dipanshu will have a total number of attempts of 25, with a distribution of (5,5,5,5,5).

Case 1: When the distribution of the attempts of Farah is (6,6,6,6,3).

It is given that each friend's training hours are equal to the median number of attempts across all targets. It was also given that, if there are two modes, the training hours will be equal to the smaller mode.

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be 3 hours (Farah's range of attempts is 3). With the break, the Wednesday slot will be over.

Now, Akash will train on Thursday (till 3 pm), the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 5 pm on Thursday. He will train on Friday (till 2 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 4 pm) and will train for 2 hours (till 12 pm) on Saturday.

The training course will be completed at 12 pm on Saturday.

Case 2: When the distribution of the attempts of Farah is (6,6,6,5,4).

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be of 2 hours (Range of attempts for Farah is 2). The break will be over at 5 pm on Wednesday, and Akash will start his training at the same time and will train till 2 pm on Thursday and the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 4 pm on Thursday. He will train on Friday (till 1 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 3 pm) and will train for 1 hour (till 11 am) on Saturday.

The training course will be completed at 11 am on Saturday.

Now, let's answer the question:

We can see that Farah attended a total of 6 hours of training in both cases.

So, the required answer is Option A.

All the friends were given 5 targets each and 6 attempts on each target. The average number of attempts per target for the 6 friends is given as 4.4, 5, 5.2, 5.4, 5.8 and 6.

Sum of the attempts taken to hit all five targets by each friend = 22, 25, 26, 27, 29 and 30 (in some order)

All the friends were ranked by the average number of total attempts, with the highest average receiving the 1st rank and the lowest rank receiving the 6th rank. So, the friend who is taking a total of 30 attempts is the first person to attend the training, and the friend who is taking a total of 22 attempts is the last person to attend the training.

From statement 1, exactly two friends used the same number of attempts per target. Since all the targets were shot, the total number of attempts taken by both the friends should be a multiple of 5. The only possibilities are 25 and 30.

So, the total number of attempts taken by these two friends is 25 and 30, where the number of attempts for each target is (5,5,5,5,5) and (6,6,6,6,6), respectively.

From statements 1 and 4, we know that Chandu took a total of 22 attempts (as he is the last person to attend the training), and he never took the maximum possible attempts to shoot any target. It is also given that his attempts for 3 targets formed a geometric progression and the attempts for the other two targets were equal.

Since each target has a limited attempt of 6, the only possible GP with a common ratio of more than 1 is (1, 2, 4). But if we take 1, 2 and 4, the maximum possible number of attempts will be (1+2+4+6+6)=19, which is less than the lowest number of attempts taken by any one of the friends (22).

So, the other possible GPs will be (4,4,4), (5,5,5), and (6,6,6), as any GP less than (4,4,4) will not reach upto 22 attempts.

Now, for (4,4,4), the maximum possible number of attempts is (4+4+4+6+6)=24. So, the only possible number of attempts for Chandu will be (4,4,4,5,5), as the remaining 10 attempts were equally divided among two targets.

Now, for (5,5,5), the maximum value of the possible number of attempts will be (5+5+5+6+6)=27. But 27 will not be included, as Chandu never took the maximum possible number of attempts for any target. Now, the other possibility can be 25 but the distribution of the number of attempts in this case will be (5,5,5,5,5), which is not possible as the number of attempts of exactly 3 targets are in geometric progression.

Now, for (6,6,6), it is not possible.

So, for Chandu, the total number of attempts is equal to 22, and the distribution is (5,5,4,4,4).

Now, from statement 2, Eshaan and Bipro attended the training first and second, respectively. So, Eshaan and Bipro took 30 and 29 attempts, respectively. For 29 attempts, only one distribution is possible, which is (6,6,6,6,5).

So, for Eshaan and Bipro, the distributions of attempts are (6,6,6,6,6) and (6,6,6,6,5).

Now, for a total of 26 attempts, the possible distributions are (6,6,6,6,2), (6,6,6,5,3), (6,6,6,4,4), (6,6,5,5,4), (6,5,5,5,5).

Similarly, for 27 attempts, the possible distributions are (6,6,6,6,3), (6,6,6,5,4), (6,6,5,5,5).

From statement 3, we know that the median number of attempts for Akash was 5 and the range of attempts was 2. The remaining attempts, which are yet to be assigned, are 25, 26 and 27.

For 25, the median is 5, and the range is 0 $$(5-5=0)$$. Not possible.

For 26, the medians are 6, 6, 6, 5 and 5, with the range of $$(6-2=4)$$, $$(6-3=3)$$, $$(6-4=2)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively.

The only possibility matching the conditions given in the 3rd statement is (6,6,5,5,4).

For 27, the medians are 6, 6 and 5, with the ranges of $$(6-3=3)$$, $$(6-4=2)$$ and $$(6-5=1)$$, respectively. It is not possible.

So, for Akash, the number of attempts are 26 and the possible distributions are (6,6,5,5,4).

Now, from Statement 2, the mode of the number of attempts of Farah exceeded the mean of her attempts. It is not possible if Farah's number of attempts is 25. So, it is only possible if the number of attempts is 27.

From the possible distributions of Farah, the valid ones are (6,6,6,6,3) and (6,6,6,5,4), as (6,6,5,5,5) has a mode of 5, which is less than the mean of the attempts (5.4), and hence not possible.

Hence, Dipanshu will have a total number of attempts of 25, with a distribution of (5,5,5,5,5).

Case 1: When the distribution of the attempts of Farah is (6,6,6,6,3).

It is given that each friend's training hours are equal to the median number of attempts across all targets. It was also given that, if there are two modes, the training hours will be equal to the smaller mode.

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be 3 hours (Farah's range of attempts is 3). With the break, the Wednesday slot will be over.

Now, Akash will train on Thursday (till 3 pm), the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 5 pm on Thursday. He will train on Friday (till 2 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 4 pm) and will train for 2 hours (till 12 pm) on Saturday.

The training course will be completed at 12 pm on Saturday.

Case 2: When the distribution of the attempts of Farah is (6,6,6,5,4).

Now, the training slot is from 10:00 am to 6:00 pm.

Eshaan is ranked 1. So, he will start with the training and will train for 6 hours. So, his training will be till 4 pm on Monday.

Now, it was given that, between any two candidates, the trainer took a break during the slots, which was equal to the previous candidate's range of attempts, or 2 hours if the range was zero.

Eshaan's range of attempts is 0. So, the trainer will take a 2-hour break.

So, after 4 pm on Monday, the trainer took a 2-hour break. So, the slot will be over as it ends at 6 pm.

Now, on Tuesday, Bipro will start his training and will train till 4 pm. The trainer has 2 hours left, and he will take a 1-hour break (which is the range of attempts for Bipro: $$(6-5-1)$$). So, the trainer's break will be over at 5 pm on Tuesday. It is given that if the trainer completes a candidate's training and the break within a given time slot, he immediately begins training the next candidate in the remaining time. So, he will start training Farah from 5 pm on Tuesday till 6 pm.

Farah will train for 6 hours, and her training will be over at 3 pm on Wednesday. Now, the break will be of 2 hours (Range of attempts for Farah is 2). The break will be over at 5 pm on Wednesday, and Akash will start his training at the same time and will train till 2 pm on Thursday and the trainer will take a 2-hour break (Akash's range of attempts), and Dipanshu will start training at 4 pm on Thursday. He will train on Friday (till 1 pm), and the trainer will take a 2-hour break.

Now, Chandu will start his training on Friday (from 3 pm) and will train for 1 hour (till 11 am) on Saturday.

The training course will be completed at 11 am on Saturday.

Now, let's answer the question:

We can see that the ranks of Bipro and Dipanshu were 2 and 5, respectively. So, Farah and Akash can train between Bipro and Dipanshu.

So, the required answer is Option A.

Based on the drop and increase in attendance, we can conclude that the next mock attendance can be found by multiplying the current mock attendance by 1.25 (for a 25% increase), 0.8 (for a 20% decrease), or 1 (for no change).

M1 attendance was 10,000 (clue 1). Now, total cumulative attendance is 40,500. If we assume each mock had the same attendance, it would total up upto 40,000. We need 500 more. This is only possible if one mock has attendance of 8,000 and another has attendance of 12,500. Thus, there are two paths possible for the attendance change from M1 to M4:

Path A: 10,000 (M1) $$\rightarrow$$ 8,000 (M2) $$\rightarrow$$ 10,000 (M3) $$\rightarrow$$ 12,500 (M4)

Path B: 10,000 (M1) $$\rightarrow$$ 12,500 (M2) $$\rightarrow$$ 10,000 (M3) $$\rightarrow$$ 8,000 (M4)

The average QA rating is 3.5, so the sum is 14. Also, all 4 QA ratings are distinct (clue 2). To have 4 distinct integers in a 1-5 scale summing to 14, the ratings must be {2, 3, 4, 5} in some order. Also, M4 had the highest QA rating. So it must be equal to 5.

The average LA rating across the 4 mocks was 2. The LA ratings for M2, M3, and M4 were the same. Thus, the total LA rating will be 8. Let the LA rating be $$a$$ for each of M2, M3, and M4. Let the LA rating of M1 be $$b$$.

$$\Rightarrow$$ $$b$$ + 3$$a$$ = 8.

From this, we can get two possible solutions for $$a$$ and $$b$$. $$\Rightarrow$$ ($$a$$, $$b$$) = (1, 5) OR (2, 2)

The sum of VA ratings across the 4 mocks was 6, and no VA section received a rating higher than 2. Thus, only possibility of ratings is {2, 2, 1, 1} in some order.

Now, assume for any mock that the QA, LA, and VA ratings are $$x$$, $$y$$, and $$z$$, respectively. The weightages of QA, LA and VA are in the ratio 2 : 1 : 1. Hence, the ODI of that particular mock will be:

ODI = (2$$x$$ + $$y$$ + $$z$$)/4

For drop in attendance: If a mock's ODI is greater than 3.5, the attendance for the next mock drops.

$$\Rightarrow$$ (2$$x$$ + $$y$$ + $$z$$)/4 > 3.5.

$$\Rightarrow$$ (2$$x$$ + $$y$$ + $$z$$) > 14

For an increase in attendance: If a mock's ODI is less than 2.5, the attendance for the next mock increases.

$$\Rightarrow$$ (2$$x$$ + $$y$$ + $$z$$)/4 < 2.5

$$\Rightarrow$$ (2$$x$$ + $$y$$ + $$z$$) < 10

For no change in attendance: If a mock's ODI is between 2.5 and 3.5 (both inclusive), attendance for the next mock remains unchanged.

$$\Rightarrow$$ 2.5 $$\leq$$ (2$$x$$ + $$y$$ + $$z$$)/4 $$\leq$$ 3.5

$$\Rightarrow$$ 10 $$\leq$$ (2$$x$$ + $$y$$ + $$z$$) $$\leq$$ 14

Now, if we consider Path B, after both M2 and M3, there is a drop in attendance. Hence, the ODI of both M2 and M3 must be more than 3.5.

$$\Rightarrow$$ (2$$x$$ + $$y$$ + $$z$$) > 14

The maximum values that $$x$$, $$y$$, and $$z$$ can take are 4, 2, and 2, respectively. In this case, the LHS equals 12, which is less than 14. Hence, even at the maximum values, we cannot get the ODI above 3.5. Hence, Path B is not correct. Hence, Path A must be the valid one.

Let the values of (2$$x$$ + $$y$$ + $$z$$) for M1, M2, and M3 be S1, S2, and S3. In Path A, M1 triggers a drop, whereas M2 and M3 trigger an increase.

Hence, for M1 $$\Rightarrow$$ S1 > 14

$$\Rightarrow$$ (2$$x$$ + $$y$$ + $$z$$) > 14.

Maximum $$x$$ available is 4, and maximum $$z$$ available is 2. If we put these values, we get $$\Rightarrow$$ 10 + $$y$$ > 14. Thus, the only possible value that $$y$$ can take in M1 is 5. Hence, the LA ratings are {5, 1, 1, 1} for M1 to M4. This also locks the QA and VARC ratings at 4 and 2, respectively. Thus, we can make a table as follows:

Now, the remaining QA ratings for M2 and M3 are {2, 3}. The remaining VARC ratings for M2, M3, and M4 are {1, 1, 2}. Now, for M2 and M3, if we put the maximum values of $$x$$ and $$z$$, we get,

S2 = 2 $$\times$$ 3 + 1 + 2 = 9, which is less than 10 and hence satisfies the drop condition. It will be true for S3 as well. Hence, we can make the final table as follows:

The ODI of M4 is 3 or 3.25, which is greater than 2.5 but less than 3.5. Thus, it falls in the band where the attendance for the next mock remains unchanged. Hence, the attendance for M5 will be equal to M4, which is 12,500 and hence option C is the correct choice.

We know only an employee of lower seniority designation can be promoted to a higher seniority designation and there is no demotion.
So, we can create a table like one below:

Here, any entry represents the number of employees promoted from the designation in that row to the designation in that column.
For example, $$a$$ represents the number of employees promoted from Analyst to Associate.
From point (6), $$a=12$$.
From point (2), $$c=0$$.
Also, the number of analysts in 2020 and 2025 were 80 and 28 respectively.
So, the change in number of analysts$$=80-28=52$$.
$$a+b+c=52$$
$$b=52-(12+0)=40$$

From point (3), $$f=2e$$.
Also, $$f=7k$$.
So, $$f$$ can be 14,28,42,.... and $$e$$ can be 7,14,21,....
Also, the change in number of associates$$=a-d-e=12-d-e$$ is non-negative.
So, $$12-d-e\ge\ 0$$
$$d+e\le12$$
So, $$e\le12$$.
Only one possibility exists: $$e=7$$ and $$f=14$$.
Now, $$d\le5$$
Also, from point (5), $$b+d-f$$ is a multiple of 10.
So, $$26+d$$ is a multiple of 10 and $$d\le5$$.
Only one possibility exist: $$d=4$$.

So, we can make the final table:

Here the diagonal entries represent the no. of people in each designation who were not promoted from 2020 to 2025.

Number of people promoted to Partner$$=7+14=21$$
Number of people promoted from Associate to any senior designation$$=4+7=11$$
Required Difference$$=10$$
Answer is 10.

From point 3, we can say that Arun scored 2, 3 or 4 on all five days.
From point 4, we see Bhavesh and Dhruv's total scores across five days were 10 and 21, respectively.
From point 7, we see that no athlete recorded any particular score more than twice during the five-day period.
So, the only possible score for Dhruv across five days is (5, 5, 4, 4, 3) in some order.
From point 1, Dhruv's score on Day 2 is 3, and Bhavesh's score on Day 4 is 4.
Also, Bhavesh's total score is 10.
So, the only possible score for Bhavesh across five days is (1, 1, 2, 2, 4) in some order, with Day 4 score as 4.
From point 7, Bhavesh scored 2 on day 5.
Also, Arun's total score is 15, and he scored 3 on day 5.
Now, since Arun scores between 2 and 4. Only one possibility for Arun exists: (2, 2, 3, 4, 4) in any order.
From point 6, Chetan recorded his lowest score on Day 4, and we know his total score is 12.
Chetan's lowest score can be 1 or 2.
If his lowest score is 2, then his minimum score on other four days would be (3,3,4,4) and the sum would be $$2+3+3+4+4=16$$, which is not possible.
So, Chetan's lowest score is 1, and it is recorded on Day 4.
From point 3, Chetan and Dhruv had equal scores on Day 5. It can be 4 or 5.
We can make the following table:

From point 5, we can say the following about the total scores on each day:
Day 1 < Day 2 < Day 3 < Day 4 and Day 4 > Day 5
From the table above, we can see that the maximum total score possible on Day 4 is 14, and the minimum total score possible on Day 5 is 13.
So, Day 4 total score $$=14$$ and Day 5 total score $$=13$$.
Also, the minimum possible score on Day 1 $$=2+1+2+4=9$$.
So, the minimum score on Day 2 would be 10, and from point 2, the minimum score on Day 3 is 12.
But Day 3 score can't be 13, as the scores have to be distinct.
So, the total scores of each day are below:
Day 1 - 9
Day 2 - 10
Day 3 - 12
Day 4 - 14
Day 5 - 13

We can make the following table:

There are only two cases possible now:

Case 1:

Case 2:

So, there are two possible combinations.
Option A is the answer.

From point 3, we can say that Arun scored 2, 3 or 4 on all five days.
From point 4, we see Bhavesh and Dhruv's total scores across five days were 10 and 21, respectively.
From point 7, we see that no athlete recorded any particular score more than twice during the five-day period.
So, the only possible score for Dhruv across five days is (5, 5, 4, 4, 3) in some order.
From point 1, Dhruv's score on Day 2 is 3, and Bhavesh's score on Day 4 is 4.
Also, Bhavesh's total score is 10.
So, the only possible score for Bhavesh across five days is (1, 1, 2, 2, 4) in some order, with Day 4 score as 4.
From point 7, Bhavesh scored 2 on day 5.
Also, Arun's total score is 15, and he scored 3 on day 5.
Now, since Arun scores between 2 and 4. Only one possibility for Arun exists: (2, 2, 3, 4, 4) in any order.
From point 6, Chetan recorded his lowest score on Day 4, and we know his total score is 12.
Chetan's lowest score can be 1 or 2.
If his lowest score is 2, then his minimum score on other four days would be (3,3,4,4) and the sum would be $$2+3+3+4+4=16$$, which is not possible.
So, Chetan's lowest score is 1, and it is recorded on Day 4.
From point 3, Chetan and Dhruv had equal scores on Day 5. It can be 4 or 5.
We can make the following table:

From point 5, we can say the following about the total scores on each day:
Day 1 < Day 2 < Day 3 < Day 4 and Day 4 > Day 5
From the table above, we can see that the maximum total score possible on Day 4 is 14, and the minimum total score possible on Day 5 is 13.
So, Day 4 total score $$=14$$ and Day 5 total score $$=13$$.
Also, the minimum possible score on Day 1 $$=2+1+2+4=9$$.
So, the minimum score on Day 2 would be 10, and from point 2, the minimum score on Day 3 is 12.
But Day 3 score can't be 13, as the scores have to be distinct.
So, the total scores of each day are below:
Day 1 - 9
Day 2 - 10
Day 3 - 12
Day 4 - 14
Day 5 - 13

We can make the following table:

There are only two cases possible now:

Case 1:

Case 2:

So, Arun scored 4 on Day 4, and Dhruv scored 5 on Day 3.
Total$$=4+5=9$$
Option C.

From point 3, we can say that Arun scored 2, 3 or 4 on all five days.
From point 4, we see Bhavesh and Dhruv's total scores across five days were 10 and 21, respectively.
From point 7, we see that no athlete recorded any particular score more than twice during the five-day period.
So, the only possible score for Dhruv across five days is (5, 5, 4, 4, 3) in some order.
From point 1, Dhruv's score on Day 2 is 3, and Bhavesh's score on Day 4 is 4.
Also, Bhavesh's total score is 10.
So, the only possible score for Bhavesh across five days is (1, 1, 2, 2, 4) in some order, with Day 4 score as 4.
From point 7, Bhavesh scored 2 on day 5.
Also, Arun's total score is 15, and he scored 3 on day 5.
Now, since Arun scores between 2 and 4. Only one possibility for Arun exists: (2, 2, 3, 4, 4) in any order.
From point 6, Chetan recorded his lowest score on Day 4, and we know his total score is 12.
Chetan's lowest score can be 1 or 2.
If his lowest score is 2, then his minimum score on other four days would be (3,3,4,4) and the sum would be $$2+3+3+4+4=16$$, which is not possible.
So, Chetan's lowest score is 1, and it is recorded on Day 4.
From point 3, Chetan and Dhruv had equal scores on Day 5. It can be 4 or 5.
We can make the following table:

From point 5, we can say the following about the total scores on each day:
Day 1 < Day 2 < Day 3 < Day 4 and Day 4 > Day 5
From the table above, we can see that the maximum total score possible on Day 4 is 14, and the minimum total score possible on Day 5 is 13.
So, Day 4 total score $$=14$$ and Day 5 total score $$=13$$.
Also, the minimum possible score on Day 1 $$=2+1+2+4=9$$.
So, the minimum score on Day 2 would be 10, and from point 2, the minimum score on Day 3 is 12.
But Day 3 score can't be 13, as the scores have to be distinct.
So, the total scores of each day are below:
Day 1 - 9
Day 2 - 10
Day 3 - 12
Day 4 - 14
Day 5 - 13

We can make the following table:

There are only two cases possible now:

Case 1:

Case 2:

In case 1, Chetan scored 2 on Day 1 and Day 2. So, the question is asking about Case 2.
In case 2, Arun's score on Day 2 and Bhavesh's score on Day 3 are 2 and 1, respectively.
So, sum $$=2+1=3$$
The answer is 3.

From point 3, we can say that Arun scored 2, 3 or 4 on all five days.
From point 4, we see Bhavesh and Dhruv's total scores across five days were 10 and 21, respectively.
From point 7, we see that no athlete recorded any particular score more than twice during the five-day period.
So, the only possible score for Dhruv across five days is (5, 5, 4, 4, 3) in some order.
From point 1, Dhruv's score on Day 2 is 3, and Bhavesh's score on Day 4 is 4.
Also, Bhavesh's total score is 10.
So, the only possible score for Bhavesh across five days is (1, 1, 2, 2, 4) in some order, with Day 4 score as 4.
From point 7, Bhavesh scored 2 on day 5.
Also, Arun's total score is 15, and he scored 3 on day 5.
Now, since Arun scores between 2 and 4. Only one possibility for Arun exists: (2, 2, 3, 4, 4) in any order.
From point 6, Chetan recorded his lowest score on Day 4, and we know his total score is 12.
Chetan's lowest score can be 1 or 2.
If his lowest score is 2, then his minimum score on other four days would be (3,3,4,4) and the sum would be $$2+3+3+4+4=16$$, which is not possible.
So, Chetan's lowest score is 1, and it is recorded on Day 4.
From point 3, Chetan and Dhruv had equal scores on Day 5. It can be 4 or 5.
We can make the following table:

From point 5, we can say the following about the total scores on each day:
Day 1 < Day 2 < Day 3 < Day 4 and Day 4 > Day 5
From the table above, we can see that the maximum total score possible on Day 4 is 14, and the minimum total score possible on Day 5 is 13.
So, Day 4 total score $$=14$$ and Day 5 total score $$=13$$.
Also, the minimum possible score on Day 1 $$=2+1+2+4=9$$.
So, the minimum score on Day 2 would be 10, and from point 2, the minimum score on Day 3 is 12.
But Day 3 score can't be 13, as the scores have to be distinct.
So, the total scores of each day are below:
Day 1 - 9
Day 2 - 10
Day 3 - 12
Day 4 - 14
Day 5 - 13

We can make the following table:

There are only two cases possible now:

Case 1:

Case 2:

Two is recorded 6 times across the whole table.