Analyst Questions

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 series follows a pattern that,

$$\begin{gathered} 4 + 32 = 4 + 9 = 13 \\ 13 + 52 = 13 + 25 = 38 \\ 38 + 72 = 38 + 49 = 87 \\ 87 + 92 = 87 + 81 = 198 \end{gathered}$$ 

Hence, the odd man in the given number series is 12.

Firstly Rain falls and then Sun comes and Rainbow forms. After that Child sees the rainbow and feels Happy.

Java is a general-purpose computer-programming language that is concurrent, class-based, object-oriented, and specifically designed to have as few implementation dependencies as possible.

Java is used to develop mobile apps, web apps, desktop apps, games and much more...

If you add all the terms of the LHS of equality, then the last letter of the sum written in words would give you the RHS.

1 + 1 + 1 + 1 = 4 = FOUR

2 + 2 + 2 + 2 = 8 = EIGHT

3 + 3 + 3 + 3 = 12 = TWELVE

4 + 4 + 4 + 4 = 16 = SIXTEEN

So, following the same pattern.

5 + 5 + 5 + 5 = 20 = TWENTY

That gives us our answer to 5 5 5 5 =  "Y".

Let breadth = x m

Then, length = (x+5)m

Area of a rectangle = Length x Breadth

x(x+5) = 750 

x² + 5x - 750= 0

(x+30)(x-25)= 0

x = 25 or x = -30

Hence, breadth x = 25m

=> Length = x + 5 = 25 + 5 = 30m. 

Given Transition is to Change, and we know that Transition and change are Synonyms. Similarly, Immobility means not moving and synonym of immobility are motionlessness, stillness,...

Hence, Transition is to change as immobility is to stillness.

Tipu Sultan was the first indian ruler to apply the western methods to his administration. Sultan Fateh Ali Sahab Tipu, also known as the Tipu Sahib, was a ruler of the Kingdom of Mysore. He was the eldest son of Sultan Hyder Ali of Mysore.

Tipu Sultan introduced a number of administrative innovations during his rule, including his coinage, a new Mauludi lunisolar calendar, and a new land revenue system which initiated the growth of the Mysore silk industry.

He was one of the few South Indian kings to provide stiff resistance to British imperialism, along with Hyder Ali. He is applauded as a ruler who fought against British colonialism. Similarly he has been a controversial figure and criticized for his atrocities against Hindus, Christians, and Mappla Muslims.