TISSNET 2023

Let the two-digit number be of the form ab.

Number = 10a+b     ----- (1)

On reversing, the number will be ba, i.e. 10b+a    ----- (2)

It is given that sum of the two-digit number and the number obtained by reversing the digits is 66.

=> 10a+b+10b+a = 66

=> 11a + 11b = 66

=> a + b = 6

Also, |a - b| = 2 

There are two cases possible

Case 1: a > b

=> a - b = 2 and a + b = 6

=> Solving, we get a = 4 and b = 2

Case 2: b > a

=> b - a = 2 and a + b = 6

=> Solving, we get a = 2 and b = 4

Therefore, a total of 2 numbers are possible 24 and 42.

The answer is option D.

 Maximum difference(at most):

Total engineers are 8, and managers are 15. To find the maximum difference for average appraisal score, if we take minimum values for managers, it will eventually be less compared to when we take minimum values for engineers. Therefore, we take the minimum possible values for managers and maximum possible values for engineers. (As any number divided by 15 will be very less, compared to 8)

Engineers

For engineers with 3 years of experience from 3 engineers, one will have an appraisal score of 12; the other two will have a score equal to 16.

So, a total score = 12+16+16 = 44

For engineers with 4 years of experience, from 5 engineers, one will have an appraisal score of 9, other four will have a score equal to 15.

So, a total score = 5+15+15+15+15 = 65

Therefore, the average appraisal score is $$\ \frac{44+65}{3+5}$$ = 14.125

Managers

For managers with 3 years of experience from 10 managers, one will have an appraisal score of 12; the other nine will have a score equal to 10.

So, a total score = 12+10*9 = 102

For managers with 4 years of experience from 5 managers, one will have an appraisal score of 13; the other four will have a score equal to 10.

So, a total score = 13+10*4 = 53

Therefore, the average appraisal score is $$\ \frac{102+53}{10+5}$$ = 10.2

The average performance appraisal score of managers differs from that of engineers by at most (14.125-10.2) points

=> 3.79 points

Consider only cases where the performance appraisal score lies in the range of 15 or above.

 For engineers with 3 years of experience from 3 engineers, at least one will have an appraisal score of 12 other two can have a score above or equal to 15. Therefore, there are at most two engineers with 3 years of experience.

For engineers with 4 years of experience from 5 engineers, at least one will have an appraisal score of 9, the other four can have a score above or equal to 15. Therefore, there are at most four engineers with 4 years of experience.

For Managers with 1 year of experience from 6 managers, all will have a score above or equal to 15.

Therefore, a total of 12 employees can have an appraisal score equal to or more than 15

Required percentage = $$\frac{12}{40}\times100=30\%$$

The answer is option A.

With 1 year of experience from 6 managers, one will have an appraisal score of 17 other five can have a score equal to 15.

=> Total score = 15*5+17 = 92

With 2 years of experience from 6 managers, one will have an appraisal score of 14 other five can have a score equal to 12.

=> Total score = 12*5+14 = 74

With 3 years of experience from 10 managers, one will have an appraisal score of 12 other nine can have a score equal to 10.

=> Total score = 10*9+12 = 102

With 4 years of experience from 5 managers, one will have an appraisal score of 13 other four can have a score equal to 10.

=> Total score = 10*4+13 = 53

Total score of all managers = 92+74+102+53 = 321

Total number of managers = 27

Therefore, the minimum average appraisal score for all the managers in the company = $$\ \frac{321}{27}$$ = 11.89

The answer is option A.

Converting the data given in bar chart into a tabular form -

The percentage decline (approximate) in the average value an UPI transaction in 2026-27 compared to 2022-23 = $$\left(\dfrac{1573-861\ }{1573}\times100\right)$$ = $$\left(\dfrac{712\ }{1573}\times100\right)$$ = 45.2%

The answer is option B.

Converting the data given in bar chart into a tabular form -

The average value of an UPI transaction in 2023-24 equals Rs. 1354

The answer is option D.

Converting the data given in bar chart into a tabular form -

The decline in the average value of an UPI transaction in 2025-26 compared to 2023-24

=> Rs. 1354 - Rs. 1000

=> Rs. 354

Converting the data given in the chart into a tabular form:

1. Monday - Location(1) - 8 vehicles and Location(2) - 5 vehicles 
Total vehicles on all locations on Monday exceeding the speed limit = 8+5+0+1 = 14
Total vehicles from location(1) and (2) exceeding the speed limit = 13(more than half of total)

2. Tuesday - Location(1) - 2 vehicles and Location(2) - 4 vehicles 
Total vehicles on all locations on Tuesday exceeding the speed limit = 2+4+7+10 = 23
Total vehicles from locations (1) and (2) exceeding the speed limit = 6(not More than half of the total)

3. Wednesday - Location(1) - 2 vehicles and Location(2) - 6 vehicles
Total vehicles on all locations on Wednesday exceeding the speed limit = 2+6+5+1 = 14
Total vehicles from location(1) and (2) exceeding the speed limit = 8(more than half of total)

4. Thursday - Location(1) - 7 vehicles and Location(2) - 4 vehicles
Total vehicles on all locations on Thursday exceeding the speed limit = 7+4+2+8 = 21
Total vehicles from location(1) and (2) exceeding the speed limit = 11(more than half of total)

5. Friday - Location(1) - 4 vehicles and Location(2) - 5 vehicles
Total vehicles on all locations on Friday exceeding the speed limit = 4+5+3+4 = 16
Total vehicles from location(1) and (2) exceeding the speed limit = 9(more than half of total)

Therefore, in a total of 4 days, more than half of the total number of vehicles exceeded the speed limit in locations 1 and 2.
The answer is option C.

Converting the data given in the chart into a tabular form -

 Monday and Wednesday had the least number of vehicles recorded exceeding the speed limit.
The answer is option B.

1. Monday - Total vehicles on all locations on Monday exceeding the speed limit = 8+5+0+1 = 14

2. Tuesday - Total vehicles on all locations on Tuesday exceeding the speed limit = 2+4+7+10 = 23

3. Wednesday - Total vehicles on all locations on Wednesday exceeding the speed limit = 2+6+5+1 = 14

4. Thursday - Total vehicles on all locations on Thursday exceeding the speed limit = 7+4+2+8 = 21

5. Friday - Total vehicles on all locations on Friday exceeding the speed limit = 4+5+3+4 = 16

Therefore, only option A is true - Records for Friday $$\geq 15$$