The question is " If the last location visited is Ahmednagar, then what is the total distance covered in the route (in km)? "

Hence, the answer is '35'

The question is " If the total number of widgets delivered in a day is 250 units, then what is the total distance covered in the route (in km)? "

Hence, the answer is '38'

The question is " What is the chance that the total number of widgets delivered in a day is 260 units and the route ends at Bikrampore? "

Hence, the answer is '7.56%'

Choice C is the correct answer.

The question is " If the first location visited from the warehouse is Ahmednagar, then what is the chance that the total distance covered in the route is 40 km? "

Hence, the answer is '18%'

Choice D is the correct answer.

The question is " If Ahmednagar is not the first location to be visited in a route and the total route distance is 29 km, then which of the following is a possible number of widgets delivered on that day? "

Hence, the answer is '210'

Choice A is the correct answer.

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.





2. On both days, Lahur met the same number of households and sold the same number of items.



3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.



4. Tohri met 30 more households on the second day than on the first day.



5. Tohri's success rate was twice that of Lahur's on the first day, and it was 75% of Lahur's on the second day.



yx−15yx−15 = 50x50x*2
100−yx+15100−yx+15 = 50x50x*3434
xy = 100(x–15).
4x(100–y) = 150(x+15)
400x – 4xy = 150(x+15)
400x – 4*100(x–15) = 150(x+15) 400 * 15 = 150 (x+15). x = 25



y10y10 = 50255025*2
y = 40

There are 20 slots of 15 minutes each from 9:00 AM to 2:00 PM. There are ten counters in the VPO.
From the given information and condition (1), Total number of applications scheduled = 10 × 20 = 200
From condition (3), Number of US applications = 100. Number of application for rest 3 categories = 100.
From condition (2), Number of US applications for each slot = 100/20 = 5. Similarly, number of applications of rest 3 categories for each slot = 100/20 = 5.
The processing time for an application is the same within each category. But it may vary across the categories.
Depending on the number of applications in a category and time required to process an application for that category,
it is possible that an applicant for a slot may be processed later.
The information can be shown in the table below. 

Step 2:

On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15
AM slot but entered the VPO at 9:20 AM i.e., they had delayed by 5 minutes.
So at least 3 people were scheduled for each slot for a Schengen visa.
Mahira and Osman were scheduled in the 9:30 AM slot on that day for visa processing in the Others category

So at least 2 people were scheduled for each slot for an Others category visa.
Total applications on that day = 200
US applications = 100
So there are 5 applications of US out of total 10 in one time slot.
There were 3 applications of Schengen and 2 applications of Others in 1 time slot.
Hence, there are 0 applications of UK in one time slot. 

There are 20 slots of 15 minutes each from 9:00 AM to 2:00 PM. There are ten counters in the VPO.
From the given information and condition (1), Total number of applications scheduled = 10 × 20 = 200
From condition (3), Number of US applications = 100. Number of application for rest 3 categories = 100.
From condition (2), Number of US applications for each slot = 100/20 = 5. Similarly, number of applications of rest 3 categories for each slot = 100/20 = 5.
The processing time for an application is the same within each category. But it may vary across the categories.
Depending on the number of applications in a category and time required to process an application for that category,
it is possible that an applicant for a slot may be processed later.
The information can be shown in the table below. 

Step 2:

On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15
AM slot but entered the VPO at 9:20 AM i.e., they had delayed by 5 minutes.
So at least 3 people were scheduled for each slot for a Schengen visa.
Mahira and Osman were scheduled in the 9:30 AM slot on that day for visa processing in the Others category.
So at least 2 people were scheduled for each slot for an Others category visa. 

There are 20 slots of 15 minutes each from 9:00 AM to 2:00 PM. There are ten counters in the VPO.
From the given information and condition (1), Total number of applications scheduled = 10 × 20 = 200
From condition (3), Number of US applications = 100. Number of application for rest 3 categories = 100.
From condition (2), Number of US applications for each slot = 100/20 = 5. Similarly, number of applications of rest 3 categories for each slot = 100/20 = 5.
The processing time for an application is the same within each category. But it may vary across the categories.
Depending on the number of applications in a category and time required to process an application for that category,
it is possible that an applicant for a slot may be processed later.
The information can be shown in the table below. 

Step 2:
On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15
AM slot but entered the VPO at 9:20 AM i.e., they had delayed by 5 minutes. 

So at least 3 people were scheduled for each slot for a Schengen visa.
Mahira and Osman were scheduled in the 9:30 AM slot on that day for visa processing in the Others category.
So at least 2 people were scheduled for each slot for an Others category visa. 

There are 20 slots of 15 minutes each from 9:00 AM to 2:00 PM. There are ten counters in the VPO.
From the given information and condition (1), Total number of applications scheduled = 10 × 20 = 200
From condition (3), Number of US applications = 100. Number of application for rest 3 categories = 100.
From condition (2), Number of US applications for each slot = 100/20 = 5. Similarly, number of applications of rest 3 categories for each slot = 100/20 = 5.
The processing time for an application is the same within each category. But it may vary across the categories.
Depending on the number of applications in a category and time required to process an application for that category,
it is possible that an applicant for a slot may be processed later.
The information can be shown in the table below. 

Step 2:
On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15
AM slot but entered the VPO at 9:20 AM i.e., they had delayed by 5 minutes.
So at least 3 people were scheduled for each slot for a Schengen visa.
Mahira and Osman were scheduled in the 9:30 AM slot on that day for visa processing in the Others category.
So at least 2 people were scheduled for each slot for an Others category visa. 

There are 20 slots of 15 minutes each from 9:00 AM to 2:00 PM. There are ten counters in the VPO.
From the given information and condition (1), Total number of applications scheduled = 10 × 20 = 200
From condition (3), Number of US applications = 100. Number of application for rest 3 categories = 100.
From condition (2), Number of US applications for each slot = 100/20 = 5. Similarly, number of applications of rest 3 categories for each slot = 100/20 = 5.
The processing time for an application is the same within each category. But it may vary across the categories.
Depending on the number of applications in a category and time required to process an application for that category,
it is possible that an applicant for a slot may be processed later.
The information can be shown in the table below. 

On a particular day, Ira, Vijay and Nandini were scheduled for Schengen visa processing in that order. They had a 9:15
AM slot but entered the VPO at 9:20 AM i.e., they had delayed by 5 minutes.
So at least 3 people were scheduled for each slot for a Schengen visa.
Mahira and Osman were scheduled in the 9:30 AM slot on that day for visa processing in the Others category.
So at least 2 people were scheduled for each slot for an Others category visa. 

Step 1:
Mean rating of Ullas is 2.2. So his total rating = 2.2 × 5 = 11.
Since his median rating is 2.
So his middle most rating is 2.
Since his mode rating is 2. So his one more rating is 2.
Since his range of rating is 3. So his two ratings are 1 and 5 or 2 and 5 or 1 and 4.
Hence, only possible ratings of Ullas is (1, 2, 2, 2, 4).
Mean rating of Vasu is 3.8. So his total rating is 3.8 × 5 = 19. His median rating is 4. So his middle most rating is 4.
His mode rating is 4. So his one more rating is 4.
Since range of rating is 3. So only possible ratings of Vasu is (2, 4, 4, 4, 5).
Similarly, possible ratings of Waman, Xavier and Yusuf are (1, 2, 4, 5, 5), (1, 3, 4, 5, 5) and (1, 1, 3,
4, 4) respectively

Step 1:
Mean rating of Ullas is 2.2. So his total rating = 2.2 × 5 = 11.
Since his median rating is 2.
So his middle most rating is 2.
Since his mode rating is 2. So his one more rating is 2.
Since his range of rating is 3. So his two ratings are 1 and 5 or 2 and 5 or 1 and 4.
Hence, only possible ratings of Ullas is (1, 2, 2, 2, 4).
Mean rating of Vasu is 3.8. So his total rating is 3.8 × 5 = 19. His median rating is 4. So his middle most rating is 4.
His mode rating is 4. So his one more rating is 4.
Since range of rating is 3. So only possible ratings of Vasu is (2, 4, 4, 4, 5).
Similarly, possible ratings of Waman, Xavier and Yusuf are (1, 2, 4, 5, 5), (1, 3, 4, 5, 5) and (1, 1, 3,
4, 4) respectively

Step 1:
Mean rating of Ullas is 2.2. So his total rating = 2.2 × 5 = 11.
Since his median rating is 2.
So his middle most rating is 2.
Since his mode rating is 2. So his one more rating is 2.
Since his range of rating is 3. So his two ratings are 1 and 5 or 2 and 5 or 1 and 4.
Hence, only possible ratings of Ullas is (1, 2, 2, 2, 4).
Mean rating of Vasu is 3.8. So his total rating is 3.8 × 5 = 19. His median rating is 4. So his middle most rating is 4.
His mode rating is 4. So his one more rating is 4.
Since range of rating is 3. So only possible ratings of Vasu is (2, 4, 4, 4, 5).
Similarly, possible ratings of Waman, Xavier and Yusuf are (1, 2, 4, 5, 5), (1, 3, 4, 5, 5) and (1, 1, 3,
4, 4) respectively

Step 1:
Mean rating of Ullas is 2.2. So his total rating = 2.2 × 5 = 11.
Since his median rating is 2.
So his middle most rating is 2.
Since his mode rating is 2. So his one more rating is 2.
Since his range of rating is 3. So his two ratings are 1 and 5 or 2 and 5 or 1 and 4.
Hence, only possible ratings of Ullas is (1, 2, 2, 2, 4).
Mean rating of Vasu is 3.8. So his total rating is 3.8 × 5 = 19. His median rating is 4. So his middle most rating is 4.
His mode rating is 4. So his one more rating is 4.
Since range of rating is 3. So only possible ratings of Vasu is (2, 4, 4, 4, 5).
Similarly, possible ratings of Waman, Xavier and Yusuf are (1, 2, 4, 5, 5), (1, 3, 4, 5, 5) and (1, 1, 3,
4, 4) respectively

Step 1:
Mean rating of Ullas is 2.2. So his total rating = 2.2 × 5 = 11.
Since his median rating is 2.
So his middle most rating is 2.
Since his mode rating is 2. So his one more rating is 2.
Since his range of rating is 3. So his two ratings are 1 and 5 or 2 and 5 or 1 and 4.
Hence, only possible ratings of Ullas is (1, 2, 2, 2, 4).
Mean rating of Vasu is 3.8. So his total rating is 3.8 × 5 = 19. His median rating is 4. So his middle most rating is 4.
His mode rating is 4. So his one more rating is 4.
Since range of rating is 3. So only possible ratings of Vasu is (2, 4, 4, 4, 5).
Similarly, possible ratings of Waman, Xavier and Yusuf are (1, 2, 4, 5, 5), (1, 3, 4, 5, 5) and (1, 1, 3,
4, 4) respectively