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📌 Section 1 is Logical reasoning specially designed for CAT
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SET 1 — 🔍 CAT Level
A cybersecurity firm, NovaSec, has a vault with 6 encrypted modules — M1, M2, M3, M4, M5, M6. Each module can be decrypted only by specific cipher keys from the set {K1, K2, K3, K4, K5, K6, K7}. The decryption table is as follows:
Modules | M1 | M2 | M3 | M4 | M5 | M6 |
|---|---|---|---|---|---|---|
Valid keys | K3, K7 | K1, K5 | K2, K4, K6 | K4, K7 | K1, K3 | K2, K5 |
An analyst attempts to decrypt all six modules over six successive attempts.
In each attempt, exactly one distinct key is assigned to each module (i.e., no key is repeated within an attempt). A key, once used for any module in an attempt (whether successful or not), cannot be reused for any other module in that attempt.
If a wrong key is assigned to a module, an alert (beep) is triggered.
The sequence of assignments within an attempt follows the order M1 to M6.
A maximum of 6 attempts is made.
The table below shows the partial information.
Modules / Attempts | M1 | M2 | M3 | M4 | M5 | M6 |
|---|---|---|---|---|---|---|
Attempt 1 | K1 | K6 | ||||
Attempt 2 | K7 | |||||
Attempt 3 | K2 | K6 | ||||
Attempt 4 | K2 | K4 | ||||
Attempt 5 | K3 | |||||
Attempt 6 | K7 | K1 |
The additional information is also known:
All six modules were successfully decrypted only in Attempt 6.
At least four modules were successfully decrypted in each attempt.
For Module 4 (M4), a different key was used in every attempt.
For every module, the same key cannot be used in three consecutive attempts.
Question 1
How many times is cipher key K7 used in all attempts?
Question 2
Which cipher key was used in Attempt 4 for Module 4 (M4)?
K1
K5
K7
Either K1 or K5
Question 3
What is the ratio of the number of times cipher key K3 is used in all attempts to the number of times cipher key K5 is used in all attempts?
1 : 1
6 : 5
5 : 6
2 : 3
Question 4
For how many modules are at least 4 distinct cipher keys used across all six attempts?
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✅ Answers & Explanations
Final Table
Attempt | M1 | M2 | M3 | M4 | M5 | M6 |
|---|---|---|---|---|---|---|
Attempt 1 | K1 | K5 | K4 | K6 | K3 | K2 |
Attempt 2 | K3 | K5 | K6 | K2 | K1 | K7 |
Attempt 3 | K3 | K1 | K4 | K7 | K2 | K6 |
Attempt 4 | K7 | K2 | K4 | K1 | K3 | K5 |
Attempt 5 | K4 | K5 | K6 | K3 | K1 | K2 |
Attempt 6 | K7 | K5 | K6 | K4 | K1 | K2 |
Set 1
Valid keys:
M1 → K3, K7
M2 → K1, K5
M3 → K2, K4, K6
M4 → K4, K7
M5 → K1, K3
M6 → K2, K5
Attempt 6 must have all correct keys.
Given Attempt 6 has:
M1 = K7
M5 = K1
So the remaining correct keys without repetition are:
M2 = K5
M3 = K6
M4 = K4
M6 = K2
Hence Attempt 6 is fixed.
Since M4 uses a different key in every attempt, and already has:
Attempt 1 → K6
Attempt 3 → K7
Attempt 5 → K3
Attempt 6 → K4
The remaining keys for M4 in Attempts 2 and 4 must be K1, K2, or K5.
But each attempt needs at least 4 correct modules, and after checking row-wise possibilities, the only complete arrangement is the table above.
Answers
K7 is used 4 times.
In Attempt 4, Module 4 used K1.
Answer: 1) K1K3 is used 5 times.
K5 is used 5 times.
Ratio = 5 : 5 = 1 : 1
Answer: 1) 1 : 1Modules with at least 4 distinct keys:
M1 → 4 distinct keys
M4 → 6 distinct keys
M6 → 4 distinct keys
Answer: 3
⚡ Brain teasers are not just about getting the right answer — they train you to think differently under pressure.
Stay consistent with your daily practice, challenge your assumptions, and keep sharpening your reasoning skills.
See you in tomorrow’s challenge!