Here it is a case of shifting, And is a case of two-step shifting, i.e. the logic consists of two steps.This implies that the change from Batch II to Batch III is same as Input to Batch I.Therefore, the changes from Batch I to Batch II, Batch III to Batch IV and Batch V to Batch VI will be the same.Look at the changes from Input to Batch I; and from Batch I to Batch II.P. If input is           1 2 3 4 5 6 7Batch I becomes   1 7 2 6 3 5 4Q. And if Batch is   1 2 3 4 5 6 7Batch II becomes as follows: 7 1 6 2 5 3 4Using the above two-step logic. Let us make a chart.                              1 2 3 4 5 6 7Input:                   A B C D E F GBatch I: (10 - 11)  A G B F C E DBatch II: (11-12)   D A E G C B FBatch III: (12 - 1)  D F A B E C GBatch V: (3 - 4)     G B D A C E FBatch VI:               F G E B C D AThe pass code for the batch at 3 p.m. means the pass code for the Batch V.Input: eight friends are sitting in the circle                    A B C D E F GBatch V:     G B D A C E Fcircle friends sitting eight are in the 

Here it is a case of shifting, And is a case of two-step shifting, i.e. the logic consists of two steps.This implies that the change from Batch II to Batch III is same as Input to Batch I.Therefore, the changes from Batch I to Batch II, Batch III to Batch IV and Batch V to Batch VI will be the same.Look at the changes from Input to Batch I; and from Batch I to Batch II.P. If input is           1 2 3 4 5 6 7Batch I becomes   1 7 2 6 3 5 4Q. And if Batch is   1 2 3 4 5 6 7Batch II becomes as follows: 7 1 6 2 5 3 4Using the above two-step logic. Let us make a chart.                              1 2 3 4 5 6 7Input:                   A B C D E F GBatch I: (10 - 11)  A G B F C E DBatch II: (11-12)   D A E G C B FBatch III: (12 - 1)  D F A B E C GBatch V: (3 - 4)     G B D A C E FBatch VI:               F G E B C D A

Here it is a case of shifting, And is a case of two-step shifting, i.e. the logic consists of two steps.This implies that the change from Batch II to Batch III is same as Input to Batch I.Therefore, the changes from Batch I to Batch II, Batch III to Batch IV and Batch V to Batch VI will be the same.Look at the changes from Input to Batch I; and from Batch I to Batch II.P. If input is           1 2 3 4 5 6 7Batch I becomes   1 7 2 6 3 5 4Q. And if Batch is   1 2 3 4 5 6 7Batch II becomes as follows: 7 1 6 2 5 3 4Using the above two-step logic. Let us make a chart.                              1 2 3 4 5 6 7Input:                   A B C D E F GBatch I: (10 - 11)  A G B F C E DBatch II: (11-12)   D A E G C B FBatch III: (12 - 1)  D F A B E C GBatch V: (3 - 4)     G B D A C E FBatch VI:               F G E B C D A

Here it is a case of shifting, And is a case of two-step shifting, i.e. the logic consists of two steps.This implies that the change from Batch II to Batch III is same as Input to Batch I.Therefore, the changes from Batch I to Batch II, Batch III to Batch IV and Batch V to Batch VI will be the same.Look at the changes from Input to Batch I; and from Batch I to Batch II.P. If input is           1 2 3 4 5 6 7Batch I becomes   1 7 2 6 3 5 4Q. And if Batch is   1 2 3 4 5 6 7Batch II becomes as follows: 7 1 6 2 5 3 4Using the above two-step logic. Let us make a chart.                              1 2 3 4 5 6 7Input:                   A B C D E F GBatch I: (10 - 11)  A G B F C E DBatch II: (11-12)   D A E G C B FBatch III: (12 - 1)  D F A B E C GBatch V: (3 - 4)     G B D A C E FBatch VI:               F G E B C D A

I   62 world 23 new 47 major 13 deskII  62 desk world 23 new 47 major 13III  62 desk 47 world 23 new major 13IV  62 desk 47 major world 23 new 13V   62 desk 47 major 23 world new 13

From the last step it can be concluded that words and numbers are arranged alternately. Words are arranged alphabetically where as numbers are arranged in descending order. When the arrangement of all elements gets completed in a particular step that step is called last step.Input: box at 20 53 62 gift now 32
Step I: at box 20 53 62 gift now 32Step II: at 62 box 20 53 gift now 32Step III: at 62 box 53 20 gift now 32Step IV: at 62 box 53 gift 20 now 32

From the last step it is clear that there are two alternating series of numbers. One in descending order and the other in ascending order. When we go through input to step I, we find that the largest number because the first and remaining numbers shift rightward. In the next step the smallest number becomes the second and the rest shift rightward. These two steps continue alternately untill the two alternate series are formed.Step II: 765 42 183 289 542 65 110 350Step III: 765 42 542 183 289 65 110 350Step IV: 765 42 542 65 183 289 110 350

From the last step it can be concluded that words and numbers are arranged alternately. Words are arranged alphabetically where as numbers are arranged in descending order. When the arrangement of all elements gets completed in a particular step that step is called last step.Input: on at 33 27 42 sky mat 51Step I: at on 33 27 42 sky mat 51Step II: at 51 on 33 27 42 sky matStep III: at 51 mat on 33 27 42 skyStep IV: at 51 mat 42 on 33 27 skyStep V: at 51 mat 42 on 33 sky 27

From the last step it can be concluded that words and numbers are arranged alternately. Words are arranged alphabetically where as numbers are arranged in descending order. When the arrangement of all elements gets completed in a particular step that step is called last step.Step III: bring 63 desk 11 29 together fight 30Step IV: bring 63 desk 30 11 29 together fightStep V: bring 63 desk 30 fight 11 29 togetherStep VI: bring 63 desk 30 fight 29 11 togetherStep VII: bring 63 desk 30 fight 29 together 11Step VII is the last step. Hence, step VI is the second last step(penultimate step).

From the last step it can be concluded that words and numbers are arranged alternately. Words are arranged alphabetically where as numbers are arranged in descending order. When the arrangement of all elements gets completed in a particular step that step is called last step.It cannot be determined.