Analyst Questions

deleteCharAt() method deletes the character at the specified index location and returns the resulting StringBuffer object.

So after deleting the character at 0 ie 'H', the string returns 'ello' as the output.

The given number series is 152, 124, 96, 69, 40

152

152 - 28 = 124

124 - 28 = 96

96 - 28 = 68 (not 69)

68 - 28 = 40

Hence, the odd man in the given number series is 69.

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Given number series is 61, 66, 83, 121, 185

Here the given series follows a pattern that,

$$\begin{gathered} 61 \\ 61 + 22 + 1 = 66 \\ 66 + 42 + 1 = 83 \\ 83 + 62 + 1 = 120 (not 121) \\ 120 + 82 + 1 = 185 \end{gathered}$$

Hence, the odd man in the given series is 121.

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.

The given number series is 8, 16, 20, 36, 68

8

8 x 2 - 4 = 12 (NOT 16)

12 x 2 - 4 = 20

20 x 2 - 4 = 36

36 x 2 - 4 = 68

Hence, the odd man in the given series is 16.

A vector quantity is a quantity which has both magnitude and direction. Here in the given options, speed is a scalar quantity but not the vector quantity.

Here the given series is 123 112 133 112 143

It follows a pattern that,

123

123 - 11 = 112

112 + 21 = 133

133 - 31 = 102 (not 122)

102 + 41 = 143

Hence, the odd man in the given number sequence is 122 instead of 102.