CAT — Logical Caselet

Step 1:

The quoted price (in lakhs of rupees) is calculated as:

Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)

Thus,

QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)

Using conditions (1) and (2),

• 24 = 10 + 5 × 1 + 3 × 3
• 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1

In Block XX, the highest quoted price is Rs. 24 lakh.

Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).

Step 2:

From condition (3), E1 and E2 are vacant.

Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.

For E1,

Quoted Price = 15

= 12 + 5 × 0 + 3 × 1

Therefore, by condition (4), E1 is the only house in Block YY with a parking space.

Further, from condition (3), each of Column D and Column F contains at least one occupied house.

This gives two possibilities:

• If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
• If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.

Three houses A1, C1 and B2 are vacant in Block XX.

Step 1:

The quoted price (in lakhs of rupees) is calculated as:

Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)

Thus,

QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)

Using conditions (1) and (2),

• 24 = 10 + 5 × 1 + 3 × 3
• 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1

In Block XX, the highest quoted price is Rs. 24 lakh.

Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).

Step 2:

From condition (3), E1 and E2 are vacant.

Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.

For E1,

Quoted Price = 15

= 12 + 5 × 0 + 3 × 1

Therefore, by condition (4), E1 is the only house in Block YY with a parking space.

Further, from condition (3), each of Column D and Column F contains at least one occupied house.

This gives two possibilities:

• If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
• If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.

House B1 is definitely occupied.

Step 1:

The quoted price (in lakhs of rupees) is calculated as:

Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)

Thus,

QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)

Using conditions (1) and (2),

• 24 = 10 + 5 × 1 + 3 × 3
• 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1

In Block XX, the highest quoted price is Rs. 24 lakh.

Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).

Step 2:

From condition (3), E1 and E2 are vacant.

Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.

For E1,

Quoted Price = 15

= 12 + 5 × 0 + 3 × 1

Therefore, by condition (4), E1 is the only house in Block YY with a parking space.

Further, from condition (3), each of Column D and Column F contains at least one occupied house.

This gives two possibilities:

• If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
• If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.

The number of vacant houses in Row-2 is either 2 or 3.

Step 1:

The quoted price (in lakhs of rupees) is calculated as:

Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)

Thus,

QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)

Using conditions (1) and (2),

• 24 = 10 + 5 × 1 + 3 × 3
• 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1

In Block XX, the highest quoted price is Rs. 24 lakh.

Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).

Step 2:

From condition (3), E1 and E2 are vacant.

Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.

For E1,

Quoted Price = 15

= 12 + 5 × 0 + 3 × 1

Therefore, by condition (4), E1 is the only house in Block YY with a parking space.

Further, from condition (3), each of Column D and Column F contains at least one occupied house.

This gives two possibilities:

• If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
• If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.

The maximum possible quoted price for a vacant house i.e., E2 in Column-E = 10 + 5 × 1 + 3 × 2 = Rs.21 lakhs.

Step 1:

The quoted price (in lakhs of rupees) is calculated as:

Quoted Price = Base Price + 5 × (Number of adjacent roads) + 3 × (Number of neighbouring occupied houses)

Thus,

QP = 10/12 + 5(0, 1, 2) + 3(0, 1, 2, 3)

Using conditions (1) and (2),

• 24 = 10 + 5 × 1 + 3 × 3
• 15 = 10 + 5 × 1 + 3 × 0 or 15 = 12 + 5 × 0 + 3 × 1

In Block XX, the highest quoted price is Rs. 24 lakh.

Hence, B1, A2, and C2 are occupied, while A1, B2, and C1 are vacant (i.e., available for sale).

Step 2:

From condition (3), E1 and E2 are vacant.

Also, from conditions (2) and (3), exactly one of D1 and F1 is occupied.

For E1,

Quoted Price = 15

= 12 + 5 × 0 + 3 × 1

Therefore, by condition (4), E1 is the only house in Block YY with a parking space.

Further, from condition (3), each of Column D and Column F contains at least one occupied house.

This gives two possibilities:

• If D1 is occupied, then F2 must also be occupied, while D2 may be either occupied or vacant.
• If F1 is occupied, then D2 must be occupied, while F2 may be either occupied or vacant.

House E1 in Block YY has parking space.

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:

At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.

At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.

Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).

Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.

As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).

From the line graph, we obtain the following values:

• 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
• 23:30 — Inside: 32°C, Outside: 37°C
• 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
• 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
• 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
• 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
• 02:00 — Inside: 28°C, Outside: 32°C

The AC was turned off twice between 11:00 and 1:59.

Step 1:

At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.

At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.

Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).

Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.

As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).

From the line graph, we obtain the following values:

• 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
• 23:30 — Inside: 32°C, Outside: 37°C
• 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
• 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
• 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
• 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
• 02:00 — Inside: 28°C, Outside: 32°C

The temperature outside at 1 am was 34°C.

Step 1:

At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.

At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.

Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).

Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.

As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).

From the line graph, we obtain the following values:

• 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
• 23:30 — Inside: 32°C, Outside: 37°C
• 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
• 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
• 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
• 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
• 02:00 — Inside: 28°C, Outside: 32°C

The temperature outside at 9 pm was 38 + 2 + 2 = 42°C.

Step 1:

At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.

At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.

Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).

Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.

As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).

From the line graph, we obtain the following values:

• 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
• 23:30 — Inside: 32°C, Outside: 37°C
• 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
• 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
• 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
• 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
• 02:00 — Inside: 28°C, Outside: 32°C

The AC settings could be changed only at the half-hour marks.

From the given information, we know that the AC was switched on at 00:30 hours and 01:30 hours, accounting for two setting changes.

It is also given that the Power mode was used for a longer duration than the Regular mode.

This leaves the following possible scenarios:

Case 1:

• At 23:00, the AC was set to Power mode with a target temperature of 32°C.
• At 23:30, the mode was changed from Power to Regular.

Case 2:

• At 23:00, the AC was set to Power mode with a target temperature of 32°C.
• At 23:30, the AC continued in Power mode, but the target temperature was changed to 26°C.

Case 3:

• At 23:00, the AC was set to Regular mode with a target temperature of 26°C.
• At 23:30, the mode was switched to Power, while the target temperature remained at 26°C.

In each of the above cases, the AC settings were changed exactly three times.

Step 1:

At 00:00 hours, the AC is switched off. In the next 30 minutes, the room temperature increases by 5°C. If the AC had remained off for one full hour, the temperature would have reached 36°C. Therefore, the outside temperature at 00:00 hours must have been 36°C.

At 01:00 hours, the AC is again switched off. During the next 30 minutes, the room temperature rises by 4°C. Had the AC remained off for the entire hour, the temperature would have become 34°C. Hence, the outside temperature at 01:00 hours was 34°C.

Thus, the outside temperature falls by 2°C between 00:00 and 01:00 (from 36°C to 34°C).

Since the outside temperature decreases at a constant rate between 7:00 PM and 3:00 AM, it drops by 1°C every 30 minutes. Using this, the outside temperatures at the remaining time points can be determined.

As the Power mode was active for a longer duration than the Regular mode, the only feasible distributions over the four half-hour intervals are (3, 1) or (4, 0).

From the line graph, we obtain the following values:

• 23:00 — Inside: 38°C, Outside: 38°C, AC Status: On
• 23:30 — Inside: 32°C, Outside: 37°C
• 00:00 — Inside: 26°C, Outside: 36°C, AC Status: Off
• 00:30 — Inside: 31°C, Outside: 35°C, AC Status: On
• 01:00 — Inside: 26°C, Outside: 34°C, AC Status: Off
• 01:30 — Inside: 30°C, Outside: 33°C, AC Status: On
• 02:00 — Inside: 28°C, Outside: 32°C

The maximum difference between temperature outside and inside between 11:01 pm and 1:59 am is 10°C at 00:00 hours.

Step 1:

The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.

From the given information and condition (1), the total number of applications scheduled is:

10 × 20 = 200

According to condition (3), there are 100 US visa applications.

Hence, the remaining 100 applications belong to the other three visa categories.

Using condition (2), the number of US visa applications per slot is:

100 ÷ 20 = 5

Similarly, the combined number of applications for the remaining three categories per slot is also:

100 ÷ 20 = 5

The processing time for each application is fixed within a category, although it may differ across categories.

Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.

The available information is:

• US: 4 counters, Processing time = 10 minutes
• UK: 2 counters, Processing time = 10 minutes
• Schengen: 2 counters, Processing time = Unknown
• Others: 2 counters, Processing time = Unknown

Step 2:

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.

This implies that at least three applicants were scheduled for the Schengen category in each time slot.

Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.

Therefore, at least two applicants were scheduled for the Others category in every time slot.

The total number of applications scheduled that day was 200.

Out of these, 100 were US visa applications.

Therefore, each 15-minute time slot had:

100 ÷ 20 = 5 US applications

From the previous deductions, each time slot also had:

• 3 Schengen applications
• 2 Others applications

Thus, all 10 application slots in each time slot are accounted for:

• US = 5
• Schengen = 3
• Others = 2

This leaves no applications for the UK category in any time slot.

Hence, the number of UK applications per time slot is 0.

Step 1:

The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.

From the given information and condition (1), the total number of applications scheduled is:

10 × 20 = 200

According to condition (3), there are 100 US visa applications.

Hence, the remaining 100 applications belong to the other three visa categories.

Using condition (2), the number of US visa applications per slot is:

100 ÷ 20 = 5

Similarly, the combined number of applications for the remaining three categories per slot is also:

100 ÷ 20 = 5

The processing time for each application is fixed within a category, although it may differ across categories.

Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.

The available information is:

• US: 4 counters, Processing time = 10 minutes
• UK: 2 counters, Processing time = 10 minutes
• Schengen: 2 counters, Processing time = Unknown
• Others: 2 counters, Processing time = Unknown

Step 2:

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.

This implies that at least three applicants were scheduled for the Schengen category in each time slot.

Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.

Therefore, at least two applicants were scheduled for the Others category in every time slot.

The maximum processing time for the Others category is 5 minutes per application.

Since there are 2 Others applications in each time slot and 20 time slots in total, the total processing time is:

= 2 × 5 × 20

= 200 minutes.

Step 1:

The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.

From the given information and condition (1), the total number of applications scheduled is:

10 × 20 = 200

According to condition (3), there are 100 US visa applications.

Hence, the remaining 100 applications belong to the other three visa categories.

Using condition (2), the number of US visa applications per slot is:

100 ÷ 20 = 5

Similarly, the combined number of applications for the remaining three categories per slot is also:

100 ÷ 20 = 5

The processing time for each application is fixed within a category, although it may differ across categories.

Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.

The available information is:

• US: 4 counters, Processing time = 10 minutes
• UK: 2 counters, Processing time = 10 minutes
• Schengen: 2 counters, Processing time = Unknown
• Others: 2 counters, Processing time = Unknown

Step 2:

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.

This implies that at least three applicants were scheduled for the Schengen category in each time slot.

Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.

Therefore, at least two applicants were scheduled for the Others category in every time slot.

Hence, 9:45 AM is the closest time at which Nandini's visa application processing would have been completed.

Step 1:

The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.

From the given information and condition (1), the total number of applications scheduled is:

10 × 20 = 200

According to condition (3), there are 100 US visa applications.

Hence, the remaining 100 applications belong to the other three visa categories.

Using condition (2), the number of US visa applications per slot is:

100 ÷ 20 = 5

Similarly, the combined number of applications for the remaining three categories per slot is also:

100 ÷ 20 = 5

The processing time for each application is fixed within a category, although it may differ across categories.

Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.

The available information is:

• US: 4 counters, Processing time = 10 minutes
• UK: 2 counters, Processing time = 10 minutes
• Schengen: 2 counters, Processing time = Unknown
• Others: 2 counters, Processing time = Unknown

Step 2:

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.

This implies that at least three applicants were scheduled for the Schengen category in each time slot.

Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.

Therefore, at least two applicants were scheduled for the Others category in every time slot.

Step 1:

The VPO operates for 20 time slots of 15 minutes each, from 9:00 AM to 2:00 PM, and has 10 counters.

From the given information and condition (1), the total number of applications scheduled is:

10 × 20 = 200

According to condition (3), there are 100 US visa applications.

Hence, the remaining 100 applications belong to the other three visa categories.

Using condition (2), the number of US visa applications per slot is:

100 ÷ 20 = 5

Similarly, the combined number of applications for the remaining three categories per slot is also:

100 ÷ 20 = 5

The processing time for each application is fixed within a category, although it may differ across categories.

Therefore, depending on the number of applications in a category and its processing time, an application assigned to a particular slot may actually be processed in a later slot.

The available information is:

• US: 4 counters, Processing time = 10 minutes
• UK: 2 counters, Processing time = 10 minutes
• Schengen: 2 counters, Processing time = Unknown
• Others: 2 counters, Processing time = Unknown

Step 2:

On a particular day, Ira, Vijay, and Nandini were scheduled for Schengen visa processing in that order for the 9:15 AM slot. However, they entered the VPO at 9:20 AM, resulting in a 5-minute delay.

This implies that at least three applicants were scheduled for the Schengen category in each time slot.

Similarly, Mahira and Osman were scheduled for visa processing under the Others category in the 9:30 AM slot.

Therefore, at least two applicants were scheduled for the Others category in every time slot.

The first five US applications require 20 minutes to be processed.

Thereafter, each subsequent batch of five applications is completed in 15 minutes.

Hence, the total time required to process all 100 US applications is:

= 20 + (15 × 19)

= 305 minutes

Therefore, the processing of all US visa applications on that day would be completed at 2:05 PM.

Step 1:

From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.

This leads to the following two possibilities:

Case 1:

• R-A, V1: Rs. 15 lakh
• R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

• R-A, V1 = 15 lakh
• R-A, V3 = 7 lakh
• R-B, V2 = 12 lakh
• R-B, V3 = 8 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Case 2:

• R-A, V1: Rs. 7 lakh
• R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

• R-A, V1 = 7 lakh
• R-A, V3 = 15 lakh
• R-B, V1 = 8 lakh
• R-B, V2 = 12 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Let us examine each of the given statements:

1. The ATM at the intersection of R-C and V2 has a cash requirement of Rs. 8 lakh. False
2. The cash requirement of the ATM at the intersection of R-C and V2 cannot be determined uniquely. False
3. There is no ATM at the intersection of R-C and V2. False
4. The ATM at the intersection of R-C and V2 has a cash requirement of Rs. 9 lakh. True

Hence, only Statement 4 is correct.

Step 1:

From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.

This leads to the following two possibilities:

Case 1:

• R-A, V1: Rs. 15 lakh
• R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

• R-A, V1 = 15 lakh
• R-A, V3 = 7 lakh
• R-B, V2 = 12 lakh
• R-B, V3 = 8 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Case 2:

• R-A, V1: Rs. 7 lakh
• R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

• R-A, V1 = 7 lakh
• R-A, V3 = 15 lakh
• R-B, V1 = 8 lakh
• R-B, V2 = 12 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

There are three ATMs that have cash requirements of Rs. 10 Lakhs or more.

Step 1:

From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.

This leads to the following two possibilities:

Case 1:

• R-A, V1: Rs. 15 lakh
• R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

• R-A, V1 = 15 lakh
• R-A, V3 = 7 lakh
• R-B, V2 = 12 lakh
• R-B, V3 = 8 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Case 2:

• R-A, V1: Rs. 7 lakh
• R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

• R-A, V1 = 7 lakh
• R-A, V3 = 15 lakh
• R-B, V1 = 8 lakh
• R-B, V2 = 12 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

The second-highest and second-lowest cash requirements are Rs. 12 lakh and Rs. 8 lakh, respectively.

• In Case 1, the distance between these two ATMs is 7 km.
• In Case 2, the corresponding distance is 4 km.

Hence, the required distance cannot be determined uniquely. It can be either 4 km or 7 km.

Step 1:

From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.

This leads to the following two possibilities:

Case 1:

• R-A, V1: Rs. 15 lakh
• R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

• R-A, V1 = 15 lakh
• R-A, V3 = 7 lakh
• R-B, V2 = 12 lakh
• R-B, V3 = 8 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Case 2:

• R-A, V1: Rs. 7 lakh
• R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

• R-A, V1 = 7 lakh
• R-A, V3 = 15 lakh
• R-B, V1 = 8 lakh
• R-B, V2 = 12 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Let us evaluate each statement individually:

Statement A: Each of R-A, R-B, and R-C contains exactly two ATMs.

This condition is satisfied in both possible cases. Hence, Statement A is true.

Statement B: Each of V1, V2, and V3 contains exactly two ATMs.

This condition is not satisfied in Case 1. Hence, Statement B is false.

Therefore, only Statement A is true.

Step 1:

From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.

This leads to the following two possibilities:

Case 1:

• R-A, V1: Rs. 15 lakh
• R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

• R-A, V1 = 15 lakh
• R-A, V3 = 7 lakh
• R-B, V2 = 12 lakh
• R-B, V3 = 8 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

Case 2:

• R-A, V1: Rs. 7 lakh
• R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

• R-A, V1 = 7 lakh
• R-A, V3 = 15 lakh
• R-B, V1 = 8 lakh
• R-B, V2 = 12 lakh
• R-C, V2 = 9 lakh
• R-C, V3 = 11 lakh

From the two possible arrangements, the positions of the ATMs with cash requirements of Rs. 12 lakh, Rs. 11 lakh, and Rs. 9 lakh remain unchanged in both cases.

However, the positions of the ATMs with cash requirements of Rs. 7 lakh, Rs. 8 lakh, and Rs. 15 lakh vary between the two arrangements.

Therefore, the positions of 3 ATMs cannot be determined uniquely.

Hence, the correct answer is 3.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

From the completed rating assignment, every individual rating can be determined uniquely using the given information.

Therefore, no individual rating remains undetermined.

Hence, the correct answer is 0.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

R2 did not give a rating of 4 to any workers.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

R1 gave a rating of 3 to Xavier.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

The median of the ratings given by R3 to the five workers is 4.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

Restaurant R4 gave its median rating to exactly one of the workers.

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.