Past Year QuestionsJIPMATLRDICoding & Decoding

JIPMAT Coding & Decoding — PYPs

3 solved Coding & Decoding previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.

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Q1:jipmat 2025LRDICoding & DecodingMediumLR · MCQ
If '-' stands for 'division', '+' for 'multiplication', '/' for 'subtraction' and 'x' for 'addition', then which one of the following equations is correct?
  • A6+2012÷71=386+20-12\div7-1=38
  • B620÷12×7+1=576-20\div12\times7+1=57
  • C6+2012÷7×1=626+20-12\div7\times1=62
  • D6÷20×12+71=706\div20\times12+7-1=70
Pick an option to attempt
The Setup: We map the replacement operators as stated in the rules. (÷)(- \Rightarrow \div), (+×)(+ \Rightarrow \times), (÷)(\div \Rightarrow -), and (×+)(\times \Rightarrow +). Apply standard BODMAS rules. Step 1: Test the options Let's directly decode Option D: Original: 6÷20×12+71=706\div20\times12+7-1=70 Substituted: 620+12×7÷16 - 20 + 12 \times 7 \div 1 Step 2: Solve via BODMAS First, handle division and multiplication: 12×7÷1=8412 \times 7 \div 1 = 84 Equation becomes: 620+846 - 20 + 84 Add and subtract from left to right: 620=146 - 20 = -14 14+84=70-14 + 84 = 70 The logic holds up perfectly. Option D is the mathematically sound equation. Final Answer: 6÷20×12+71=706\div20\times12+7-1=70
Q2:jipmat 2025LRDICoding & DecodingEasyLR · MCQ
If ROAST is coded as PQYUR in a certain code language, then how will SLOPPY be coded in that language?
  • AMRNAQN
  • BNRMNQA
  • CQNMRNA
  • DQANNMR
Pick an option to attempt
Step 1: Dox the pattern in the first word Let's map the letter shifts from ROAST to PQYUR: * R \rightarrow P (2-2) * O \rightarrow Q (+2+2) * A \rightarrow Y (2-2, wrapping around the alphabet backward) * S \rightarrow U (+2+2) * T \rightarrow R (2-2) The hidden mechanic is an alternating shift sequence of 2-2 and +2+2. Step 2: Apply the mechanic to the target word Now we run the exact same 2,+2-2, +2 sequence on the word SLOPPY: * S (2-2) \rightarrow Q * L (+2+2) \rightarrow N * O (2-2) \rightarrow M * P (+2+2) \rightarrow R * P (2-2) \rightarrow N * Y (+2+2) \rightarrow A (wrapping past Z) The final generated string is QNMRNA. Final Answer: QNMRNA
Q3:jipmat 2025LRDICoding & DecodingEasyLR · MCQ
If DRIVER = 12, PEDESTRIAN = 20 and ACCIDENT = 16, then CAR is equal to:
  • A3
  • B6
  • C8
  • D10
Pick an option to attempt
The Setup: The numbers given are way too small to be based on the alphabetical positional values (A=1, B=2, etc.). The absolute meta strat here is to simply count the number of letters in each word. Step 1: Dox the hidden mechanic * DRIVER has 6 letters. 6×2=126 \times 2 = 12. * PEDESTRIAN has 10 letters. 10×2=2010 \times 2 = 20. * ACCIDENT has 8 letters. 8×2=168 \times 2 = 16. The rule is completely locked in: count the letters and multiply by 2. Step 2: Apply to the target * CAR has exactly 3 letters. * 3×2=63 \times 2 = 6. Final Answer: 6

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