a.Understanding the customer profile : the tastes,preferences, habits of the customers may differ from person to person.

b.Determining the objective as to who would be the target market, whether the company wants to go for cost minimisation or not etc.

c.Type of channel members :Who all would be involved? Wholesalers, retailers?

d.The criteria on which the channel members would be evaluated and selected.

e.How many intermediaries will be involved in the overall distribution?

The following are the factors to be considered before deciding the target audience for any advertisements:

•       Age


•       Working class (professionals)


•       Non working class (Retired/Home makers)


•       Socio economic status


•       Region specific


•       Religion/community etc. 

1. ANSWER : C 

 Explanation -  Number of students who passed half-yearly exams in the school 

= (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)
= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true. 

 2. ANSWER : D 

 Explanation -  Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class: 

 = (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430. 

 3. ANSWER : D 

 Explanation- Pass percentages in at least one of the two examinations for different sections are:

  For Section A = 14+6+6428+14+6+64×100 = 84112×100% = 75%

 For Section B =12+17+5523+12+17+55×100 % = 78.5%

 For Section C = 8+9+4617+8+9+46×100%= 78.75%

 For Section D = 13+15+7627+13+15+76×100%= 79.39%

  Clearly ,the pass percentage is maximum for Section D 

 4. ANSWER : A

 Explanation - Total number of students passed in annual exams in a section 

= [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams) ] in that section

 Success rate in annual exams in Section A= 14+64112 × 100% = 69.64%

 Similarly, success rate in annual exams in:

 Section B = 12+55107×100% =  62.62% 

 Section C = 8+4680×100% = 67.5% 

 Section D = 89131×100% = 67.94% 

 Clearly, the success rate in annual examination is maximum for Section A. 

 5. ANSWER : D 

 Explanation - Total number of failures in half-yearly exams in a section

  = [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams) ] in that section

 Failure rate in half-yearly exams in Section A %= 37.5 %

 Similarly, failure rate in half-yearly exams in:

  Section B = 32.71%

  Section C = 31.25%

  Section D = 30.53%

  Clearly, the failure rate is minimum for Section D.

 = (Number of students passed in half-yearly but failed in annual exams) + (Number of students passed in both exams)

 = (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

 = 288.

 Also, Number of students who passed annual exams in the school

 = (Number of students failed in half-yearly but passed in annual exams) + (Number of students passed in both exams)

 = (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

 = 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

Answer : D