PUNE MCA Exam 2016 Question Paper with answers for online practice

Concept:Each letter is encoded by its alphabet position using a transformation based on modulo 3 division.Explanation:• Let the alphabet position of a letter be $x$ ($A=1$, $B=2$, ...).• The output letter's position is $((x-1) \bmod 3) + 1$, i.e., remainder when dividing $x-1$ by 3, plus 1. This gives values $1,2,3$ for letters $A,B,C$.• The output number is $\lfloor (x-1)/3 \rfloor + 1$, i.e., integer division of $x-1$ by 3, plus 1.Apply to COP:• $C=3$: $(3-1)\bmod3 =2$, output letter $C$; $\lfloor 2/3 \rfloor +1 = 0+1=1$ → $C1$.• $O=15$: $(15-1)\bmod3 =14\bmod3=2$, output letter $C$; $\lfloor 14/3 \rfloor +1 = 4+1=5$ → $C5$.• $P=16$: $(16-1)\bmod3 =15\bmod3=0$, output letter $A$; $\lfloor 15/3 \rfloor +1 = 5+1=6$ → $A6$.Thus the code is $C1C5A6$.Answer:The code for COP is $C1C5A6$, corresponding to option D.