Bank Clerk Questions

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The given number series follows a pattern that,

$$\begin{gathered} 19 + 12 = 20 (not 21) \\ 20 + 22 = 24 \\ 24 + 32 = 33 \\ 33 + 42 = 49 \\ 49 + 52 = 74 \end{gathered}$$

Hence, the odd number in the given number series is 21.

The given number series 22, 26, 42, 78, 142, 244 follows a pattern that,

$$\begin{gathered} 22 \\ 22 + 22 = 22 + 4 = 26 \\ 26 + 42 = 26 + 16 = 42 \\ 42 + 62 = 42 + 36 = 78 \\ 78 + 82 = 78 + 64 = 142 \\ 142 + 102 = 142 + 100 = 242 \end{gathered}$$ 

Hence, the odd one in the given number series is 244.

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 2, 19, 0, 21, –2, ? follows a pattern,

2

2 + 17 = 19

19 - 19 = 0

0 + 21 = 21

21 - 23 = -2

-2 + 25 = 23

Hence, the next number in the series is 23.

Milk = 3/5 x 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 - 6 = 6 liters

Remaining water = 8 - 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed,
then amount of milk removed = 4/5 x 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 - 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

Now 10 lts milk is added => total milk = 18 lts

The required ratio of milk and water in the final mixture obtained

= (8 + 10):2 = 18:2 = 9:1.

When you develop a product, you assess its performance for proper functioning.

Similarly, when you train a person to do a job, you analyze his/her performance to understand how good he/she is at it.

Hence, Develop : Assess :: Train : Analyze.

HIS RECENT SUCCESS HAS MADE HIM MORE ARROGANT THAN WHAT HE USED.