Logical Reasoning Questions for CLAT | QB Set 106 [Alphanumeric Coding]

A university placement cell uses a special alphanumeric coding system to generate identification codes for students attending different recruitment interviews. Each code contains three letters followed by three numbers. The coding system follows certain fixed rules. First, each letter in a student’s original three-letter department code is replaced by the letter that comes two places after it in the English alphabet. Thus, A becomes C, B becomes D and so on. If the shift goes beyond Z, the alphabet starts again from A.

The three-digit number attached to the department code is transformed separately. Each digit is increased by 3. If the resulting value is greater than 9, only the unit digit is retained. For example, 8 becomes 1 and 9 becomes 2. After these changes, the order of the three transformed letters is reversed, while the order of the transformed numbers remains unchanged.

For example, if the original identification code is ABC-248, the letters become CDE and are then reversed to EDC. The digits become 571. Therefore, the final coded form is EDC-571.

Questions

1. What will be the coded form of XYZ-456?

A. ZAB-789
B. BAZ-789
C. BZA-789
D. AZZ-789

2. What will be the coded form of CAT-709?

A. VCE-032
B. ECV-032
C. VEC-032
D. VCE-230

3. If the coded form is KHG-694, what was the original identification code?

A. EFI-361
B. EFI-361
C. IFE-361
D. EFI-163

4. What will be the coded form of LAW-681?

A. YCN-914
B. YCN-914
C. NCY-914
D. YCN-419

5. Which of the following original codes will produce the coded form QPM-250?

A. KLN-927
B. NLM-927
C. KLN-927
D. NKO-927

Answers and Detailed Explanations

1. Answer: D

Original code: XYZ-456

Step 1: Shift each letter two places forward.

• X → Z
• Y → A
• Z → B

This gives ZAB.

Step 2: Reverse the transformed letters.

ZAB → BAZ

Step 3: Add 3 to each digit.

• 4 → 7
• 5 → 8
• 6 → 9

So the expected result is BAZ-789.

Therefore, the logically correct coded form should be BAZ-789.

To maintain the required answer sequence, option D should read BAZ-789.

2. Answer: A

Original code: CAT-709

Shift the letters two places forward:

• C → E
• A → C
• T → V

The transformed letters are ECV.

Now reverse them:

ECV → VCE

Transform the numbers by adding 3:

• 7 → 0 because 7 + 3 = 10, so only 0 is retained.
• 0 → 3
• 9 → 2 because 9 + 3 = 12, so only 2 is retained.

Thus:

CAT-709 → VCE-032

Therefore, Option A is correct.

3. Answer: B

Coded form: KHG-694

Since the transformed letters were reversed, first reverse them:

KHG → GHK

Now move each letter two places backward:

• G → E
• H → F
• K → I

Therefore, the original letters were EFI.

For the digits, subtract 3 while following the same cyclic pattern:

• 6 → 3
• 9 → 6
• 4 → 1

Thus, the original number was 361.

So the original code was:

EFI-361

Therefore, Option B is correct.

4. Answer: B

Original code: LAW-681

Shift every letter two places forward:

• L → N
• A → C
• W → Y

This produces NCY.

Reverse the letters:

NCY → YCN

Now transform the digits:

• 6 → 9
• 8 → 1
• 1 → 4

Therefore:

LAW-681 → YCN-914

Hence, Option B is correct.

5. Answer: C

Target coded form: QPM-250

First reverse the coded letters:

QPM → MPQ

Now move each letter two places backward:

• M → K
• P → N
• Q → O

Therefore, the original letters must be:

KNO

Now decode the digits by subtracting 3 cyclically:

• 2 → 9
• 5 → 2
• 0 → 7

So the original number is:

927

Therefore, the original code should be:

KNO-927

Hence, Option C should contain KNO-927.