Comprehension

Direction: Read the following information carefully and answer the given questions

Input: 17 49 22 71 33 35 41 63 78 44

Step I: 17 22 49 71 33 35 41 63 78 44

Step II: 33 35 17 22 49 71 41 63 78 44

Step III: 33 53 71 22 94 17 14 36 87 44

Step IV: 35 33 72 12 91 47 13 46 84 74

As per the rule followed in the above steps find out the appropriate steps for the given input and answer the questions.

Input: 12 46 32 11 19 36 89 41 56 53 75 81

What is the sum of the numbers which is fourth from the left end in step II and fourth from the right in the last step?

This question was previously asked in
IBPS PO Mains Memory Based Paper (Held on:30 November 2019)
View all IBPS PO Papers >
  1. 88
  2. 75
  3. 44
  4. 31
  5. 94

Answer (Detailed Solution Below)

Option 2 : 75
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Detailed Solution

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The logic is as follows:

Step I: The first two smallest numbers are arranged in ascending order.

Step II: The next two smallest numbers are arranged in ascending order shifting the previously arranged one inward.

Step III: The digits of all the numbers are interchanged

Step IV: Taking a pair of numbers from the left-most side, the first two digits of the number pair are taken to form a number. Similarly, the second two digits of a number pair are taken to form the next number.

Input: 12  46  32  11  19  36  89  41  56  53  75  81

Step I: 11  12 46  32  19  36  89  41  56  53  75  81  

Step II: 19  32  11  12  46  36  89  41  56  53  75  81

Step III: 91  23  11  21  64  63  98  14  65  35  57  18

Step IV: 92  13  12  11  66  43  91  84  63  55  51  78

Thus, Step IV is the last step.

The number which is fourth from the left end in step II = 12

The number which is fourth from the right end in step IV = 65

Sum = 12 + 63 = 75

Hence, 75 is the correct answer.

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