Problem Statement:  

Convert a 2-digit BCD number stored at memory address 2200H into its binary equivalent number and store the result in a memory location 2300H.

Sample problem 1:

(2200H) = 67H

(2300H) = 6 x OAH + 7 = 3CH + 7 = 43H

Source program :

    * LDA 2200H : Get the BCD number
    * MOV B, A : Save it
    * ANI OFH : Mask most significant four bits
    * MOV C, A : Save unpacked BCDI in C register
    * MOV A, B : Get BCD again
    * ANI FOH : Mask least significant four bits
    * RRC : Convert most significant four bits into unpacked BCD2
    * RRC
    * RRC
    * RRC
    * MOV B, A : Save unpacked BCD2 in B register
    * XRA A : Clear accumulator (sum = 0)
    * MVI D, 0AH : Set D as a multiplier of 10
    * Sum:   ADD D : Add 10 until (B) = 0
    * DCR B : Decrement BCD2 by one
    * JNZ SUM : Is multiplication complete? i if not, go back and add again
    * ADD C : Add BCD1
    * STA 2300H : Store the result
    * HLT : Terminate program execution