CAT — Distribution, Assignment & Selection

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their total imports be Ip, Ix, Ic, and Ir respectively.

Using the normalized trade balance formula:

For P:

(Ep − Ip)/(Ep + Ip) = 0

⇒ Ep = Ip

For X:

(Ex − Ix)/(Ex + Ix) = 10%

⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C:

(Ec − Ic)/(Ec + Ic) = −20%

⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

According to condition (5), P is the only country that exports to C.

Hence, C receives all its imports from P, and the import volume from P to C is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are distributed as follows:

• 90% to P ⇒ 0.9 × Ec = 720
• 4% to ROW ⇒ 0.04 × Ec = 32
• The remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

• 90% to P ⇒ 0.9Ec = 720
• 4% to ROW ⇒ 0.04Ec = 32
• Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

• From X = 0.22Ip
• From C = 720
• From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

• From P = 600
• From C = 48
• From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

Thus, 200 units are exported from P to ROW.

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

• 90% to P ⇒ 0.9Ec = 720
• 4% to ROW ⇒ 0.04Ec = 32
• Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

• From X = 0.22Ip
• From C = 720
• From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

• From P = 600
• From C = 48
• From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

1008 IC is exported from ROW to ROW.

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

• 90% to P ⇒ 0.9Ec = 720
• 4% to ROW ⇒ 0.04Ec = 32
• Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

• From X = 0.22Ip
• From C = 720
• From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

• From P = 600
• From C = 48
• From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

The trade balance of ROW = Er – Ir = 2100 – 1900 = 200 IC.

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

• 90% to P ⇒ 0.9Ec = 720
• 4% to ROW ⇒ 0.04Ec = 32
• Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

• From X = 0.22Ip
• From C = 720
• From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

• From P = 600
• From C = 48
• From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

Total trade:

P = Ep + Ip = 4000

X = Ex + Ix = (1100 + 900) = 2000

C = Ec + Ic = (800 + 1200) = 2000

Hence, the least total trade = X and C.

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Observe that 2 four-author papers contribute 2 × 4 = 8 author counts, since all four authors are involved in each paper. The same counting principle applies to the three-author and two-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

Since Aman has authored 5 papers in total, and 2 of them are four-author papers, the remaining 3 papers must be distributed among the other three categories. As every category must have at least one paper, Aman must have 1 single-author, 1 two-author, and 1 three-author paper.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Using Condition (4), Brajen cannot have 4 single-author and two-author papers combined, because the overall total for these categories is 8. He also cannot have 3, as that would leave him with 0 three-author papers, violating the given conditions.

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

Now, applying Condition (3), both Chintan and Devon must have more than 2 three-author papers, and the total number of three-author papers contributed by all authors is 9. The only feasible allocation is 3 each for Chintan and Devon.

The deductions obtained so far are summarized in the table below.

Therefore, the total number of two-author and three-author papers written by Brajen is:

2 + 2 = 4.

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Observe that 2 four-author papers contribute 2 × 4 = 8 author counts, since all four authors are involved in each paper. The same counting principle applies to the three-author and two-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

Since Aman has authored 5 papers in total, and 2 of them are four-author papers, the remaining 3 papers must be distributed among the other three categories. As every category must have at least one paper, Aman must have 1 single-author, 1 two-author, and 1 three-author paper.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Using Condition (4), Brajen cannot have 4 single-author and two-author papers combined, because the overall total for these categories is 8. He also cannot have 3, as that would leave him with 0 three-author papers, violating the given conditions.

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

Now, applying Condition (3), both Chintan and Devon must have more than 2 three-author papers, and the total number of three-author papers contributed by all authors is 9. The only feasible allocation is 3 each for Chintan and Devon.

The deductions obtained so far are summarized in the table below.

Let us check each statement:

i. The statement, Chintan wrote exactly three two-author papers may not necessarily be true.

ii. The statement, Chintan wrote more single-author papers than Devon may not necessarily be true.

Hence, neither i nor ii is definitely true.

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Observe that 2 four-author papers contribute 2 × 4 = 8 author counts, since all four authors are involved in each paper. The same counting principle applies to the three-author and two-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

Since Aman has authored 5 papers in total, and 2 of them are four-author papers, the remaining 3 papers must be distributed among the other three categories. As every category must have at least one paper, Aman must have 1 single-author, 1 two-author, and 1 three-author paper.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Using Condition (4), Brajen cannot have 4 single-author and two-author papers combined, because the overall total for these categories is 8. He also cannot have 3, as that would leave him with 0 three-author papers, violating the given conditions.

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

Now, applying Condition (3), both Chintan and Devon must have more than 2 three-author papers, and the total number of three-author papers contributed by all authors is 9. The only feasible allocation is 3 each for Chintan and Devon.

The deductions obtained so far are summarized in the table below.

There are 3 three author papers and both Chintan and Devon wrote 3 three author papers whereas Aman wrote 1 and Brajen wrote 2. So the only possible combination will be {(Chintan, Devon, Aman), (Chintan, Devon, Brajen), (Chintan, Devon, Brajen)}. Hence, the statement, Arman wrote three-author papers only with Chintan and Devon, is true. Brajen wrote three-author papers only with Chintan and Devon is also true. Hence, both (i) and (ii) are true.

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Observe that 2 four-author papers contribute 2 × 4 = 8 author counts, since all four authors are involved in each paper. The same counting principle applies to the three-author and two-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

Since Aman has authored 5 papers in total, and 2 of them are four-author papers, the remaining 3 papers must be distributed among the other three categories. As every category must have at least one paper, Aman must have 1 single-author, 1 two-author, and 1 three-author paper.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Using Condition (4), Brajen cannot have 4 single-author and two-author papers combined, because the overall total for these categories is 8. He also cannot have 3, as that would leave him with 0 three-author papers, violating the given conditions.

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

Now, applying Condition (3), both Chintan and Devon must have more than 2 three-author papers, and the total number of three-author papers contributed by all authors is 9. The only feasible allocation is 3 each for Chintan and Devon.

The deductions obtained so far are summarized in the table below.

If Devon wrote more than one two-author papers, then the number of two-author papers written by Chintan is 3.

Step 1:

Using the given information and conditions (1)–(3):

• P received 3 votes.
• Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
• Q received 14 votes.
• Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
• Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.

Further,

• Since S voted for P, P and S cannot belong to the same department.
• Since R voted for Q, Q and R cannot belong to the same department.
• Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.

Therefore,

• P and Q belong to the M & S department.
• Q and S belong to B & H and O & Q, in either order.

Step 2:

The vote distribution is as follows:

• P received 3 votes:
• From S
• From 2 members of B & H
• Q received 14 votes:
• From R
• From 9 members of F & A
• From 4 members of O & Q
• R received 6 votes:
• From P
• From 5 members of M & S
• S received 1 vote:
• From Q

The department-wise candidate allocation is:

• F & A: 9 faculty members, 9 members, no candidate
• M & S: 7 faculty members, 5 members, Candidates – P and Q
• O & Q: 5 faculty members, 4 members, Candidate – R or S
• B & H: 3 faculty members, 2 members, Candidate – S or R

Prof. Pakrasi and Prof. Qureshi can belong to the same department M & S.

Step 1:

Using the given information and conditions (1)–(3):

• P received 3 votes.
• Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
• Q received 14 votes.
• Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
• Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.

Further,

• Since S voted for P, P and S cannot belong to the same department.
• Since R voted for Q, Q and R cannot belong to the same department.
• Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.

Therefore,

• P and Q belong to the M & S department.
• Q and S belong to B & H and O & Q, in either order.

Step 2:

The vote distribution is as follows:

• P received 3 votes:
• From S
• From 2 members of B & H
• Q received 14 votes:
• From R
• From 9 members of F & A
• From 4 members of O & Q
• R received 6 votes:
• From P
• From 5 members of M & S
• S received 1 vote:
• From Q

The department-wise candidate allocation is:

• F & A: 9 faculty members, 9 members, no candidate
• M & S: 7 faculty members, 5 members, Candidates – P and Q
• O & Q: 5 faculty members, 4 members, Candidate – R or S
• B & H: 3 faculty members, 2 members, Candidate – S or R

Nine can be the number of votes that Prof. Qureshi received from a single department.

Step 1:

Using the given information and conditions (1)–(3):

• P received 3 votes.
• Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
• Q received 14 votes.
• Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
• Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.

Further,

• Since S voted for P, P and S cannot belong to the same department.
• Since R voted for Q, Q and R cannot belong to the same department.
• Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.

Therefore,

• P and Q belong to the M & S department.
• Q and S belong to B & H and O & Q, in either order.

Step 2:

The vote distribution is as follows:

• P received 3 votes:
• From S
• From 2 members of B & H
• Q received 14 votes:
• From R
• From 9 members of F & A
• From 4 members of O & Q
• R received 6 votes:
• From P
• From 5 members of M & S
• S received 1 vote:
• From Q

The department-wise candidate allocation is:

• F & A: 9 faculty members, 9 members, no candidate
• M & S: 7 faculty members, 5 members, Candidates – P and Q
• O & Q: 5 faculty members, 4 members, Candidate – R or S
• B & H: 3 faculty members, 2 members, Candidate – S or R

Both statements are true.

Step 1:

Using the given information and conditions (1)–(3):

• P received 3 votes.
• Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
• Q received 14 votes.
• Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
• Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.

Further,

• Since S voted for P, P and S cannot belong to the same department.
• Since R voted for Q, Q and R cannot belong to the same department.
• Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.

Therefore,

• P and Q belong to the M & S department.
• Q and S belong to B & H and O & Q, in either order.

Step 2:

The vote distribution is as follows:

• P received 3 votes:
• From S
• From 2 members of B & H
• Q received 14 votes:
• From R
• From 9 members of F & A
• From 4 members of O & Q
• R received 6 votes:
• From P
• From 5 members of M & S
• S received 1 vote:
• From Q

The department-wise candidate allocation is:

• F & A: 9 faculty members, 9 members, no candidate
• M & S: 7 faculty members, 5 members, Candidates – P and Q
• O & Q: 5 faculty members, 4 members, Candidate – R or S
• B & H: 3 faculty members, 2 members, Candidate – S or R

Either Prof. Ramaswamy or Prof. Samuel was from O & Q.

Step 1:

Using the given information and conditions (1)–(3):

• P received 3 votes.
• Since S voted for P, P received the remaining 2 votes from members of B & H. Therefore, P cannot belong to the B & H department, leaving one candidate from B & H.
• Q received 14 votes.
• Since R voted for Q, Q received 9 votes from F & A and 4 votes from O & Q.
• Hence, Q cannot belong to either F & A or O & Q. Consequently, there is no candidate from F & A, and one candidate belongs to O & Q. The remaining two candidates must come from M & S.

Further,

• Since S voted for P, P and S cannot belong to the same department.
• Since R voted for Q, Q and R cannot belong to the same department.
• Similarly, R and P cannot belong to the same department, and Q and S cannot belong to the same department.

Therefore,

• P and Q belong to the M & S department.
• Q and S belong to B & H and O & Q, in either order.

Step 2:

The vote distribution is as follows:

• P received 3 votes:
• From S
• From 2 members of B & H
• Q received 14 votes:
• From R
• From 9 members of F & A
• From 4 members of O & Q
• R received 6 votes:
• From P
• From 5 members of M & S
• S received 1 vote:
• From Q

The department-wise candidate allocation is:

• F & A: 9 faculty members, 9 members, no candidate
• M & S: 7 faculty members, 5 members, Candidates – P and Q
• O & Q: 5 faculty members, 4 members, Candidate – R or S
• B & H: 3 faculty members, 2 members, Candidate – S or R

Only statement B is true.

Step 1:

From Condition (1), Ustad Samiran never trains more than one of the four students in the same year. Therefore, each student must have trained under Ustad Samiran for 3 years.

Using Condition (2), Acharya Raghunath does not train any student during 2015–2018 and 2021–2024, creating two gaps of four years each. Hence, Acharya Raghunath trains every student for exactly 2 years.

From Condition (4), Ananya and Bhaskar began their training under Pandit Meghnath and Ustad Samiran, respectively, in 2013. Since no student can train under more than one guru simultaneously, Charu and Devendra must have started under Acharya Raghunath in 2013.

It also follows that Ananya and Bhaskar trained under Acharya Raghunath during 2019–2020, completing their two-year training period. As the training durations under the three gurus are 2, 3, and 4 years, and we have already determined that Acharya Raghunath trains for 2 years while Pandit Meghnath trains for 3 years, Ustad Samiran must train each student for 4 years.

Since Ananya started her training in 2013, and a student cannot train under two gurus at the same time, Bhaskar can train under Pandit Meghnath only during 2021–2024.

Now, applying Condition (3), among the six possible student pairs (AB, AC, AD, BC, BD, and CD), the pairs AD and BC cannot be gurubhais. Each of the remaining four pairs must be gurubhais for exactly 2 years.

Accordingly, Ananya and Charu train together under Pandit Meghnath during 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.

Step 2:

Under Ustad Samiran, each student has a distinct 3-year training period with no overlap. Therefore, Ananya must train during 2022–2024, as any other three-year period would overlap with her training under the other two gurus.

The remaining schedules can now be determined uniquely. Devendra trains under Ustad Samiran from 2016–2018, while Charu trains under Ustad Samiran from 2019–2021.

The complete schedule is shown in the table below.

Hence, Ananya and Bhaskar were gurubhai in 2020.

Step 1:

From Condition (1), Ustad Samiran never trains more than one of the four students in the same year. Therefore, each student must have trained under Ustad Samiran for 3 years.

Using Condition (2), Acharya Raghunath does not train any student during 2015–2018 and 2021–2024, creating two gaps of four years each. Hence, Acharya Raghunath trains every student for exactly 2 years.

From Condition (4), Ananya and Bhaskar began their training under Pandit Meghnath and Ustad Samiran, respectively, in 2013. Since no student can train under more than one guru simultaneously, Charu and Devendra must have started under Acharya Raghunath in 2013.

It also follows that Ananya and Bhaskar trained under Acharya Raghunath during 2019–2020, completing their two-year training period. As the training durations under the three gurus are 2, 3, and 4 years, and we have already determined that Acharya Raghunath trains for 2 years while Pandit Meghnath trains for 3 years, Ustad Samiran must train each student for 4 years.

Since Ananya started her training in 2013, and a student cannot train under two gurus at the same time, Bhaskar can train under Pandit Meghnath only during 2021–2024.

Now, applying Condition (3), among the six possible student pairs (AB, AC, AD, BC, BD, and CD), the pairs AD and BC cannot be gurubhais. Each of the remaining four pairs must be gurubhais for exactly 2 years.

Accordingly, Ananya and Charu train together under Pandit Meghnath during 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.

Step 2:

Under Ustad Samiran, each student has a distinct 3-year training period with no overlap. Therefore, Ananya must train during 2022–2024, as any other three-year period would overlap with her training under the other two gurus.

The remaining schedules can now be determined uniquely. Devendra trains under Ustad Samiran from 2016–2018, while Charu trains under Ustad Samiran from 2019–2021.

The complete schedule is shown in the table below.

Charu began her training under Pandit Meghnath in 2015.

Step 1:

From Condition (1), Ustad Samiran never trains more than one of the four students in the same year. Therefore, each student must have trained under Ustad Samiran for 3 years.

Using Condition (2), Acharya Raghunath does not train any student during 2015–2018 and 2021–2024, creating two gaps of four years each. Hence, Acharya Raghunath trains every student for exactly 2 years.

From Condition (4), Ananya and Bhaskar began their training under Pandit Meghnath and Ustad Samiran, respectively, in 2013. Since no student can train under more than one guru simultaneously, Charu and Devendra must have started under Acharya Raghunath in 2013.

It also follows that Ananya and Bhaskar trained under Acharya Raghunath during 2019–2020, completing their two-year training period. As the training durations under the three gurus are 2, 3, and 4 years, and we have already determined that Acharya Raghunath trains for 2 years while Pandit Meghnath trains for 3 years, Ustad Samiran must train each student for 4 years.

Since Ananya started her training in 2013, and a student cannot train under two gurus at the same time, Bhaskar can train under Pandit Meghnath only during 2021–2024.

Now, applying Condition (3), among the six possible student pairs (AB, AC, AD, BC, BD, and CD), the pairs AD and BC cannot be gurubhais. Each of the remaining four pairs must be gurubhais for exactly 2 years.

Accordingly, Ananya and Charu train together under Pandit Meghnath during 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.

Step 2:

Under Ustad Samiran, each student has a distinct 3-year training period with no overlap. Therefore, Ananya must train during 2022–2024, as any other three-year period would overlap with her training under the other two gurus.

The remaining schedules can now be determined uniquely. Devendra trains under Ustad Samiran from 2016–2018, while Charu trains under Ustad Samiran from 2019–2021.

The complete schedule is shown in the table below.

Bhaskar and Devendra were Gurubhai in 2022.

Step 1:

From Condition (1), Ustad Samiran never trains more than one of the four students in the same year. Therefore, each student must have trained under Ustad Samiran for 3 years.

Using Condition (2), Acharya Raghunath does not train any student during 2015–2018 and 2021–2024, creating two gaps of four years each. Hence, Acharya Raghunath trains every student for exactly 2 years.

From Condition (4), Ananya and Bhaskar began their training under Pandit Meghnath and Ustad Samiran, respectively, in 2013. Since no student can train under more than one guru simultaneously, Charu and Devendra must have started under Acharya Raghunath in 2013.

It also follows that Ananya and Bhaskar trained under Acharya Raghunath during 2019–2020, completing their two-year training period. As the training durations under the three gurus are 2, 3, and 4 years, and we have already determined that Acharya Raghunath trains for 2 years while Pandit Meghnath trains for 3 years, Ustad Samiran must train each student for 4 years.

Since Ananya started her training in 2013, and a student cannot train under two gurus at the same time, Bhaskar can train under Pandit Meghnath only during 2021–2024.

Now, applying Condition (3), among the six possible student pairs (AB, AC, AD, BC, BD, and CD), the pairs AD and BC cannot be gurubhais. Each of the remaining four pairs must be gurubhais for exactly 2 years.

Accordingly, Ananya and Charu train together under Pandit Meghnath during 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.

Step 2:

Under Ustad Samiran, each student has a distinct 3-year training period with no overlap. Therefore, Ananya must train during 2022–2024, as any other three-year period would overlap with her training under the other two gurus.

The remaining schedules can now be determined uniquely. Devendra trains under Ustad Samiran from 2016–2018, while Charu trains under Ustad Samiran from 2019–2021.

The complete schedule is shown in the table below.

Let us check each statement:

1. Ananya was training under Ustad Samiran in 2015, is not true.

2. Charu was training under Ustad Samiran in 2019, is true.

3. Ananya was training under Ustad Samiran in 2018, is not true

4. Charu was training under Ustad Samiran in 2018, is not true.

Step 1:

From Condition (1), Ustad Samiran never trains more than one of the four students in the same year. Therefore, each student must have trained under Ustad Samiran for 3 years.

Using Condition (2), Acharya Raghunath does not train any student during 2015–2018 and 2021–2024, creating two gaps of four years each. Hence, Acharya Raghunath trains every student for exactly 2 years.

From Condition (4), Ananya and Bhaskar began their training under Pandit Meghnath and Ustad Samiran, respectively, in 2013. Since no student can train under more than one guru simultaneously, Charu and Devendra must have started under Acharya Raghunath in 2013.

It also follows that Ananya and Bhaskar trained under Acharya Raghunath during 2019–2020, completing their two-year training period. As the training durations under the three gurus are 2, 3, and 4 years, and we have already determined that Acharya Raghunath trains for 2 years while Pandit Meghnath trains for 3 years, Ustad Samiran must train each student for 4 years.

Since Ananya started her training in 2013, and a student cannot train under two gurus at the same time, Bhaskar can train under Pandit Meghnath only during 2021–2024.

Now, applying Condition (3), among the six possible student pairs (AB, AC, AD, BC, BD, and CD), the pairs AD and BC cannot be gurubhais. Each of the remaining four pairs must be gurubhais for exactly 2 years.

Accordingly, Ananya and Charu train together under Pandit Meghnath during 2015 and 2016. Similarly, Bhaskar and Devendra train together during 2021 and 2022.

Step 2:

Under Ustad Samiran, each student has a distinct 3-year training period with no overlap. Therefore, Ananya must train during 2022–2024, as any other three-year period would overlap with her training under the other two gurus.

The remaining schedules can now be determined uniquely. Devendra trains under Ustad Samiran from 2016–2018, while Charu trains under Ustad Samiran from 2019–2021.

The complete schedule is shown in the table below.

Between 2013-24, there were 4 years when only two of these four musicians were training under these three Gurus. These years were 2017, 2018, 2023 and 2024.

Step 1:

From conditions (3) and (4), exactly two players received the same rating.

Let the repeated rating be N.

Then,

(1 + 2 + 3 + 4 + 5 + 6 + 7 + N) / 8 = 4

⇒ N = 4

Hence, Players 5 and 7 both received a rating of 4.

Since Xena trained more players than Yuki, there are two possible distributions:

• Possibility 1:
• Xena trains 3 players.
• Yuki trains 2 players.
• Zara trains 3 players.
• Possibility 2:
• Xena trains 4 players.
• Yuki trains 2 players.
• Zara trains 2 players.

For Possibility 1,

X = 38/13, which is not an integer and is therefore not feasible.

For Possibility 2,

X = 3, which satisfies the conditions.

Thus,

• Xena trained 4 players with an average score of 3, giving a total score of 12.
• Yuki trained 2 players with an average score of 6, giving a total score of 12.
• Zara trained 2 players with an average score of 4, giving a total score of 8.

Step 2:

From condition (5), Player 2 received a rating of 7.

Also, Player 4 received a rating of 2, and Player 8 received a rating of 1.

The known ratings are:

• Player 2 → 7
• Player 4 → 2
• Player 5 → 4
• Player 7 → 4
• Player 8 → 1

Yuki trained only two players, whose ratings add up to 12.

The only possible combination is 7 and 5.

Since 7 belongs to Player 2, the rating 5 must belong to Player 6.

As Yuki trained only even-numbered players and only Player 6 remains, Players 2 and 6 must have been trained by Yuki.

Zara's trainees have a total score of 8.

This total can be obtained either as 2 + 6 or 4 + 4.

Since the rating 2 belongs to Player 4, who is even-numbered, and Zara trained only odd-numbered players, the 2 + 6 combination is not possible.

Therefore, Zara must have trained Players 5 and 7, both of whom have a rating of 4.

The remaining players—1, 3, 4, and 8—were trained by Xena.

Thus, the final assignments are:

• Player 1 → Rating 3/6, Trained by Xena
• Player 2 → Rating 7, Trained by Yuki
• Player 3 → Rating 6/3, Trained by Xena
• Player 4 → Rating 2, Trained by Xena
• Player 5 → Rating 4, Trained by Zara
• Player 6 → Rating 5, Trained by Yuki
• Player 7 → Rating 4, Trained by Zara
• Player 8 → Rating 1, Trained by Xena

Step 1:

From conditions (3) and (4), exactly two players received the same rating.

Let the repeated rating be N.

Then,

(1 + 2 + 3 + 4 + 5 + 6 + 7 + N) / 8 = 4

⇒ N = 4

Hence, Players 5 and 7 both received a rating of 4.

Since Xena trained more players than Yuki, there are two possible distributions:

• Possibility 1:
• Xena trains 3 players.
• Yuki trains 2 players.
• Zara trains 3 players.
• Possibility 2:
• Xena trains 4 players.
• Yuki trains 2 players.
• Zara trains 2 players.

For Possibility 1,

X = 38/13, which is not an integer and is therefore not feasible.

For Possibility 2,

X = 3, which satisfies the conditions.

Thus,

• Xena trained 4 players with an average score of 3, giving a total score of 12.
• Yuki trained 2 players with an average score of 6, giving a total score of 12.
• Zara trained 2 players with an average score of 4, giving a total score of 8.

Step 2:

From condition (5), Player 2 received a rating of 7.

Also, Player 4 received a rating of 2, and Player 8 received a rating of 1.

The known ratings are:

• Player 2 → 7
• Player 4 → 2
• Player 5 → 4
• Player 7 → 4
• Player 8 → 1

Yuki trained only two players, whose ratings add up to 12.

The only possible combination is 7 and 5.

Since 7 belongs to Player 2, the rating 5 must belong to Player 6.

As Yuki trained only even-numbered players and only Player 6 remains, Players 2 and 6 must have been trained by Yuki.

Zara's trainees have a total score of 8.

This total can be obtained either as 2 + 6 or 4 + 4.

Since the rating 2 belongs to Player 4, who is even-numbered, and Zara trained only odd-numbered players, the 2 + 6 combination is not possible.

Therefore, Zara must have trained Players 5 and 7, both of whom have a rating of 4.

The remaining players—1, 3, 4, and 8—were trained by Xena.

Thus, the final assignments are:

• Player 1 → Rating 3/6, Trained by Xena
• Player 2 → Rating 7, Trained by Yuki
• Player 3 → Rating 6/3, Trained by Xena
• Player 4 → Rating 2, Trained by Xena
• Player 5 → Rating 4, Trained by Zara
• Player 6 → Rating 5, Trained by Yuki
• Player 7 → Rating 4, Trained by Zara
• Player 8 → Rating 1, Trained by Xena

Step 1:

From conditions (3) and (4), exactly two players received the same rating.

Let the repeated rating be N.

Then,

(1 + 2 + 3 + 4 + 5 + 6 + 7 + N) / 8 = 4

⇒ N = 4

Hence, Players 5 and 7 both received a rating of 4.

Since Xena trained more players than Yuki, there are two possible distributions:

• Possibility 1:
• Xena trains 3 players.
• Yuki trains 2 players.
• Zara trains 3 players.
• Possibility 2:
• Xena trains 4 players.
• Yuki trains 2 players.
• Zara trains 2 players.

For Possibility 1,

X = 38/13, which is not an integer and is therefore not feasible.

For Possibility 2,

X = 3, which satisfies the conditions.

Thus,

• Xena trained 4 players with an average score of 3, giving a total score of 12.
• Yuki trained 2 players with an average score of 6, giving a total score of 12.
• Zara trained 2 players with an average score of 4, giving a total score of 8.

Step 2:

From condition (5), Player 2 received a rating of 7.

Also, Player 4 received a rating of 2, and Player 8 received a rating of 1.

The known ratings are:

• Player 2 → 7
• Player 4 → 2
• Player 5 → 4
• Player 7 → 4
• Player 8 → 1

Yuki trained only two players, whose ratings add up to 12.

The only possible combination is 7 and 5.

Since 7 belongs to Player 2, the rating 5 must belong to Player 6.

As Yuki trained only even-numbered players and only Player 6 remains, Players 2 and 6 must have been trained by Yuki.

Zara's trainees have a total score of 8.

This total can be obtained either as 2 + 6 or 4 + 4.

Since the rating 2 belongs to Player 4, who is even-numbered, and Zara trained only odd-numbered players, the 2 + 6 combination is not possible.

Therefore, Zara must have trained Players 5 and 7, both of whom have a rating of 4.

The remaining players—1, 3, 4, and 8—were trained by Xena.

Thus, the final assignments are:

• Player 1 → Rating 3/6, Trained by Xena
• Player 2 → Rating 7, Trained by Yuki
• Player 3 → Rating 6/3, Trained by Xena
• Player 4 → Rating 2, Trained by Xena
• Player 5 → Rating 4, Trained by Zara
• Player 6 → Rating 5, Trained by Yuki
• Player 7 → Rating 4, Trained by Zara
• Player 8 → Rating 1, Trained by Xena

Step 1:

From conditions (3) and (4), exactly two players received the same rating.

Let the repeated rating be N.

Then,

(1 + 2 + 3 + 4 + 5 + 6 + 7 + N) / 8 = 4

⇒ N = 4

Hence, Players 5 and 7 both received a rating of 4.

Since Xena trained more players than Yuki, there are two possible distributions:

• Possibility 1:
• Xena trains 3 players.
• Yuki trains 2 players.
• Zara trains 3 players.
• Possibility 2:
• Xena trains 4 players.
• Yuki trains 2 players.
• Zara trains 2 players.

For Possibility 1,

X = 38/13, which is not an integer and is therefore not feasible.

For Possibility 2,

X = 3, which satisfies the conditions.

Thus,

• Xena trained 4 players with an average score of 3, giving a total score of 12.
• Yuki trained 2 players with an average score of 6, giving a total score of 12.
• Zara trained 2 players with an average score of 4, giving a total score of 8.

Step 2:

From condition (5), Player 2 received a rating of 7.

Also, Player 4 received a rating of 2, and Player 8 received a rating of 1.

The known ratings are:

• Player 2 → 7
• Player 4 → 2
• Player 5 → 4
• Player 7 → 4
• Player 8 → 1

Yuki trained only two players, whose ratings add up to 12.

The only possible combination is 7 and 5.

Since 7 belongs to Player 2, the rating 5 must belong to Player 6.

As Yuki trained only even-numbered players and only Player 6 remains, Players 2 and 6 must have been trained by Yuki.

Zara's trainees have a total score of 8.

This total can be obtained either as 2 + 6 or 4 + 4.

Since the rating 2 belongs to Player 4, who is even-numbered, and Zara trained only odd-numbered players, the 2 + 6 combination is not possible.

Therefore, Zara must have trained Players 5 and 7, both of whom have a rating of 4.

The remaining players—1, 3, 4, and 8—were trained by Xena.

Thus, the final assignments are:

• Player 1 → Rating 3/6, Trained by Xena
• Player 2 → Rating 7, Trained by Yuki
• Player 3 → Rating 6/3, Trained by Xena
• Player 4 → Rating 2, Trained by Xena
• Player 5 → Rating 4, Trained by Zara
• Player 6 → Rating 5, Trained by Yuki
• Player 7 → Rating 4, Trained by Zara
• Player 8 → Rating 1, Trained by Xena

Step 1:

From conditions (3) and (4), exactly two players received the same rating.

Let the repeated rating be N.

Then,

(1 + 2 + 3 + 4 + 5 + 6 + 7 + N) / 8 = 4

⇒ N = 4

Hence, Players 5 and 7 both received a rating of 4.

Since Xena trained more players than Yuki, there are two possible distributions:

• Possibility 1:
• Xena trains 3 players.
• Yuki trains 2 players.
• Zara trains 3 players.
• Possibility 2:
• Xena trains 4 players.
• Yuki trains 2 players.
• Zara trains 2 players.

For Possibility 1,

X = 38/13, which is not an integer and is therefore not feasible.

For Possibility 2,

X = 3, which satisfies the conditions.

Thus,

• Xena trained 4 players with an average score of 3, giving a total score of 12.
• Yuki trained 2 players with an average score of 6, giving a total score of 12.
• Zara trained 2 players with an average score of 4, giving a total score of 8.

Step 2:

From condition (5), Player 2 received a rating of 7.

Also, Player 4 received a rating of 2, and Player 8 received a rating of 1.

The known ratings are:

• Player 2 → 7
• Player 4 → 2
• Player 5 → 4
• Player 7 → 4
• Player 8 → 1

Yuki trained only two players, whose ratings add up to 12.

The only possible combination is 7 and 5.

Since 7 belongs to Player 2, the rating 5 must belong to Player 6.

As Yuki trained only even-numbered players and only Player 6 remains, Players 2 and 6 must have been trained by Yuki.

Zara's trainees have a total score of 8.

This total can be obtained either as 2 + 6 or 4 + 4.

Since the rating 2 belongs to Player 4, who is even-numbered, and Zara trained only odd-numbered players, the 2 + 6 combination is not possible.

Therefore, Zara must have trained Players 5 and 7, both of whom have a rating of 4.

The remaining players—1, 3, 4, and 8—were trained by Xena.

Thus, the final assignments are:

• Player 1 → Rating 3/6, Trained by Xena
• Player 2 → Rating 7, Trained by Yuki
• Player 3 → Rating 6/3, Trained by Xena
• Player 4 → Rating 2, Trained by Xena
• Player 5 → Rating 4, Trained by Zara
• Player 6 → Rating 5, Trained by Yuki
• Player 7 → Rating 4, Trained by Zara
• Player 8 → Rating 1, Trained by Xena

Step 1:

From conditions (1) and (2), it can be inferred that Chitra completed Ride 3 from 9:00 AM to 10:00 AM and Ride 1 from 10:00 AM to 11:00 AM. This follows because Anjali had to wait for 30 minutes to complete Ride 1, which finished at 11:00 AM.

From condition (3), Bipasha paid for only one ride until 12:15 PM, i.e., Ride 2, which ran from 11:30 AM to 12:30 PM. Since Bipasha completed three rides in total, she could not have taken Ride 3, as it ended at 1:00 PM.

The schedule at this stage is:

• Anjali: Ride 1 – Unknown, Ride 2 – Unknown, Ride 3 – Unknown, Ride 4 – Unknown
• Bipasha: Ride 2 – 11:30 AM to 12:30 PM, Ride 3 – Not taken
• Chitra: Ride 1 – 10:00 AM to 11:00 AM, Ride 3 – 9:00 AM to 10:00 AM, Ride 2 and Ride 4 – Not taken

Step 2:

Using condition (4), Bipasha's schedule can be determined as follows:

• 11:30 AM – 12:30 PM: Ride 2
• 12:30 PM – 1:30 PM: Free
• 1:30 PM – 2:30 PM: Coffee break
• 2:30 PM – 3:00 PM: Waiting
• 3:00 PM – 4:00 PM: Ride 4

Since Bipasha waited from 2:30 PM to 3:00 PM for Anjali to complete the previous ride, her final ride started at 3:00 PM and ended at 4:00 PM.

Hence, the final schedule is:

• Anjali
• Ride 1: 11:00 AM – 12:00 PM
• Ride 2: 1:00 PM – 2:00 PM
• Ride 3: 12:00 PM – 1:00 PM
• Ride 4: 2:00 PM – 3:00 PM
• Bipasha
• Ride 2: 11:30 AM – 12:30 PM
• Ride 4: 3:00 PM – 4:00 PM
• Chitra
• Ride 1: 10:00 AM – 11:00 AM
• Ride 3: 9:00 AM – 10:00 AM



Total amount spent on tickets (in Rs.) by Bipasha is Rs. 110.

Step 1:

From conditions (1) and (2), it can be inferred that Chitra completed Ride 3 from 9:00 AM to 10:00 AM and Ride 1 from 10:00 AM to 11:00 AM. This follows because Anjali had to wait for 30 minutes to complete Ride 1, which finished at 11:00 AM.

From condition (3), Bipasha paid for only one ride until 12:15 PM, i.e., Ride 2, which ran from 11:30 AM to 12:30 PM. Since Bipasha completed three rides in total, she could not have taken Ride 3, as it ended at 1:00 PM.

The schedule at this stage is:

• Anjali: Ride 1 – Unknown, Ride 2 – Unknown, Ride 3 – Unknown, Ride 4 – Unknown
• Bipasha: Ride 2 – 11:30 AM to 12:30 PM, Ride 3 – Not taken
• Chitra: Ride 1 – 10:00 AM to 11:00 AM, Ride 3 – 9:00 AM to 10:00 AM, Ride 2 and Ride 4 – Not taken

Step 2:

Using condition (4), Bipasha's schedule can be determined as follows:

• 11:30 AM – 12:30 PM: Ride 2
• 12:30 PM – 1:30 PM: Free
• 1:30 PM – 2:30 PM: Coffee break
• 2:30 PM – 3:00 PM: Waiting
• 3:00 PM – 4:00 PM: Ride 4

Since Bipasha waited from 2:30 PM to 3:00 PM for Anjali to complete the previous ride, her final ride started at 3:00 PM and ended at 4:00 PM.

Hence, the final schedule is:

• Anjali
• Ride 1: 11:00 AM – 12:00 PM
• Ride 2: 1:00 PM – 2:00 PM
• Ride 3: 12:00 PM – 1:00 PM
• Ride 4: 2:00 PM – 3:00 PM
• Bipasha
• Ride 2: 11:30 AM – 12:30 PM
• Ride 4: 3:00 PM – 4:00 PM
• Chitra
• Ride 1: 10:00 AM – 11:00 AM
• Ride 3: 9:00 AM – 10:00 AM



Ride 1 was taken by all three visitors.

Step 1:

From conditions (1) and (2), it can be inferred that Chitra completed Ride 3 from 9:00 AM to 10:00 AM and Ride 1 from 10:00 AM to 11:00 AM. This follows because Anjali had to wait for 30 minutes to complete Ride 1, which finished at 11:00 AM.

From condition (3), Bipasha paid for only one ride until 12:15 PM, i.e., Ride 2, which ran from 11:30 AM to 12:30 PM. Since Bipasha completed three rides in total, she could not have taken Ride 3, as it ended at 1:00 PM.

The schedule at this stage is:

• Anjali: Ride 1 – Unknown, Ride 2 – Unknown, Ride 3 – Unknown, Ride 4 – Unknown
• Bipasha: Ride 2 – 11:30 AM to 12:30 PM, Ride 3 – Not taken
• Chitra: Ride 1 – 10:00 AM to 11:00 AM, Ride 3 – 9:00 AM to 10:00 AM, Ride 2 and Ride 4 – Not taken

Step 2:

Using condition (4), Bipasha's schedule can be determined as follows:

• 11:30 AM – 12:30 PM: Ride 2
• 12:30 PM – 1:30 PM: Free
• 1:30 PM – 2:30 PM: Coffee break
• 2:30 PM – 3:00 PM: Waiting
• 3:00 PM – 4:00 PM: Ride 4

Since Bipasha waited from 2:30 PM to 3:00 PM for Anjali to complete the previous ride, her final ride started at 3:00 PM and ended at 4:00 PM.

Hence, the final schedule is:

• Anjali
• Ride 1: 11:00 AM – 12:00 PM
• Ride 2: 1:00 PM – 2:00 PM
• Ride 3: 12:00 PM – 1:00 PM
• Ride 4: 2:00 PM – 3:00 PM
• Bipasha
• Ride 2: 11:30 AM – 12:30 PM
• Ride 4: 3:00 PM – 4:00 PM
• Chitra
• Ride 1: 10:00 AM – 11:00 AM
• Ride 3: 9:00 AM – 10:00 AM



The rides that Anjali completed by 2 pm are Ride 1, Ride 2 and Ride 3.

Step 1:

From conditions (1) and (2), it can be inferred that Chitra completed Ride 3 from 9:00 AM to 10:00 AM and Ride 1 from 10:00 AM to 11:00 AM. This follows because Anjali had to wait for 30 minutes to complete Ride 1, which finished at 11:00 AM.

From condition (3), Bipasha paid for only one ride until 12:15 PM, i.e., Ride 2, which ran from 11:30 AM to 12:30 PM. Since Bipasha completed three rides in total, she could not have taken Ride 3, as it ended at 1:00 PM.

The schedule at this stage is:

• Anjali: Ride 1 – Unknown, Ride 2 – Unknown, Ride 3 – Unknown, Ride 4 – Unknown
• Bipasha: Ride 2 – 11:30 AM to 12:30 PM, Ride 3 – Not taken
• Chitra: Ride 1 – 10:00 AM to 11:00 AM, Ride 3 – 9:00 AM to 10:00 AM, Ride 2 and Ride 4 – Not taken

Step 2:

Using condition (4), Bipasha's schedule can be determined as follows:

• 11:30 AM – 12:30 PM: Ride 2
• 12:30 PM – 1:30 PM: Free
• 1:30 PM – 2:30 PM: Coffee break
• 2:30 PM – 3:00 PM: Waiting
• 3:00 PM – 4:00 PM: Ride 4

Since Bipasha waited from 2:30 PM to 3:00 PM for Anjali to complete the previous ride, her final ride started at 3:00 PM and ended at 4:00 PM.

Hence, the final schedule is:

• Anjali
• Ride 1: 11:00 AM – 12:00 PM
• Ride 2: 1:00 PM – 2:00 PM
• Ride 3: 12:00 PM – 1:00 PM
• Ride 4: 2:00 PM – 3:00 PM
• Bipasha
• Ride 2: 11:30 AM – 12:30 PM
• Ride 4: 3:00 PM – 4:00 PM
• Chitra
• Ride 1: 10:00 AM – 11:00 AM
• Ride 3: 9:00 AM – 10:00 AM



Anjali and Chitra take 6 Rides in total.

Step 1:

From conditions (1) and (2), it can be inferred that Chitra completed Ride 3 from 9:00 AM to 10:00 AM and Ride 1 from 10:00 AM to 11:00 AM. This follows because Anjali had to wait for 30 minutes to complete Ride 1, which finished at 11:00 AM.

From condition (3), Bipasha paid for only one ride until 12:15 PM, i.e., Ride 2, which ran from 11:30 AM to 12:30 PM. Since Bipasha completed three rides in total, she could not have taken Ride 3, as it ended at 1:00 PM.

The schedule at this stage is:

• Anjali: Ride 1 – Unknown, Ride 2 – Unknown, Ride 3 – Unknown, Ride 4 – Unknown
• Bipasha: Ride 2 – 11:30 AM to 12:30 PM, Ride 3 – Not taken
• Chitra: Ride 1 – 10:00 AM to 11:00 AM, Ride 3 – 9:00 AM to 10:00 AM, Ride 2 and Ride 4 – Not taken

Step 2:

Using condition (4), Bipasha's schedule can be determined as follows:

• 11:30 AM – 12:30 PM: Ride 2
• 12:30 PM – 1:30 PM: Free
• 1:30 PM – 2:30 PM: Coffee break
• 2:30 PM – 3:00 PM: Waiting
• 3:00 PM – 4:00 PM: Ride 4

Since Bipasha waited from 2:30 PM to 3:00 PM for Anjali to complete the previous ride, her final ride started at 3:00 PM and ended at 4:00 PM.

Hence, the final schedule is:

• Anjali
• Ride 1: 11:00 AM – 12:00 PM
• Ride 2: 1:00 PM – 2:00 PM
• Ride 3: 12:00 PM – 1:00 PM
• Ride 4: 2:00 PM – 3:00 PM
• Bipasha
• Ride 2: 11:30 AM – 12:30 PM
• Ride 4: 3:00 PM – 4:00 PM
• Chitra
• Ride 1: 10:00 AM – 11:00 AM
• Ride 3: 9:00 AM – 10:00 AM



The total amount spent on tickets by Anjali is Rs. 140.

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

Hence, the PI of Whimshire is 45.

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The PI of Fogglia = 35

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The PI of Humbleset = 50

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The pair of cities that definitely belong to the same states Noodleton and Quackford.

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

We can identify the pIs of all cities and NURs and also identify the state they belong to.

Common Solution:

The available information can initially be represented as follows:

• A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C receives: P = 0 (Total = 45)
• D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

• C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
• D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

• A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
• D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

The available information can initially be represented as follows:

• A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C receives: P = 0 (Total = 45)
• D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

• C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
• D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

• A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
• D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

The available information can initially be represented as follows:

• A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C receives: P = 0 (Total = 45)
• D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

• C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
• D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

• A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
• D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

The available information can initially be represented as follows:

• A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C receives: P = 0 (Total = 45)
• D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

• C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
• D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

• A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
• B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
• C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
• D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.