सादृश्यता टेस्ट टाइप 2

Fifth and third letters of the first term are first and second letters of the second term and first two letters of the first term are third and fourth letters of the second term.

First two letters of the first term are in reverse order in the second term and so are the next two letters.

The letters are consecutive and written in reverse order.

C and F have 2 letters in between, so do I and L. In each set of letters, the 1st and 3rd letters are consecutive.

In each group the first three letters are consecutive and they follows the fourth letter.

The letters in first set are jumping two letters i.e. moving three steps forward and in the second they are jumping one letter i.e. moving two step backward.

Here B and D are skipped

CEG is formed by skipping the letters in between them, the second set EGC is formed by simply putting the first letter of CEG at last to form EGC, and so on.

If A corresponds to X (the third from the end) then B should correspond to the fourth letter W.

The 5th letter from A correspond to 5th from Z and therefore 9th letter ‘I’ from A would correspond 9th letter ‘R’ from Z.

Because KLM are assigned No. 11, 12 & 13 from
A onwards, this corresponds to PON, which are also numbered 11, 12 and 13 from Z to A in reverse order. Hence NOP will correspond to MLK.

The three letters moved 5, 4, and 3 and steps forward respectively.

In the both terms, two letters are skipped in between. Similar relationship exists in (c).

BDF forms a sequence of alternate letters, while in HIL, the first two letters are consecutive and there is a gap of two letters between second and third letters.

Choice (a) is the correct answer as letters of 1st term go one step forward to be the 2nd term. Similarly, the letters of 3rd term will go one step forward to be the 4th term (Letters of step go one step backward to be the 3rd term).