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·           Consider that a word of data is present in the AX register and 2^nd word of data is present in the BX register.

·           We have to multiply the word in AX with the word in BX. Using MUL instruction, multiply the contents of the 2 registers.

·           The multiplication of the two 16 bit numbers may result into a 32 bit number. So result is stored in the DX and AX register.

·            Consider that a word of data is present in the AX register and second word of data is present in the BX register.

·            We have to subtract the word in BX with the word AX.

·            Using subtract (SUB) instruction, subtract the contents of two registers.

·            Result will be stored in the AX register.

·            Display the result using display routine.

·            Consider that a word of data is present in the AX register and a 2^nd word of data is present in the BX register.

·            We have to add word in AX with the word in BX. Using ADD instruction, add the contents.

·            Result will be stored in the AX register. Display the result using display routine.

·            Consider that a word of data is present in the AX register and byte of data is present in the BL register, we have to divide word in AX with the byte in BL.

·            Using DIV instruction, divide the contents of two register. Result of division is stored in the AX register. AL contains the quotient and AH contains the remainder.

·            Consider that two words are available in registers AX and BX. We have to add these two words.

·            Using add instruction, add the contents, of the lower two bits i.e. add AL and BL.

·            The result of this addition is stored in the AL register.

·            DAA instruction is then used to convert the result to valid BCD. Now, add the contents of MSB alongwith carry if generated in LSB addition.

·            Consider that two unpacked numbers are present in the AL and BL registers. We have to multiply the two numbers. The result is stored in the AX register. The AAM (BCD adjust after Multiply) is used to adjust the product to two unpacked BCD digits in AX.

·           Consider that a byte of data is present in the AL register and a second byte of data is present in the BL register. We have to subtract byte in BL with byte in AL. Using SUB instruction, we will subtract the contents. The result is stored in the AL register. Using DAS instruction we will make sure that the result is valid BCD.

·            Consider that a byte of data is present in the AL register and a second byte of data is present in the BL register. We have to add byte in AL with the byte in BL. Using add instruction, add the contents of 2 registers.

·            Result will be stored in the AL register.

·            Use DAA instruction that will check if BCD is valid, if it is not valid then 6 is added to give proper BCD result. Display the result.

·            Consider that one byte is present in the AL register and another byte is present in the BL register.

·            We have to multiply the byte in AL with the byte in BL.

·            We will multiply the numbers using add and shift method. In this method, you add number with itself and rotate the other number each time and shift it by one bit to left alongwith carry. If carry is present add the two numbers.

·            Initialise the count to 4 as we are scanning for 4 digits. Decrement counter each time the bits are added. The result is stored in AX. Display the result.

·           Consider that a byte is present in the AL register and second byte is present in the BL register.

·           We have to multiply the byte in AL with the byte in BL.

·           We will multiply the numbers using successive addition method.

·           In successive addition method, one number is accepted and other number is taken as a counter. The first number is added with itself, till the counter decrements to zero.

·           Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.

·           We have to multiply the byte in AL with the byte in BL.

·           Using MUL instruction, multiply the contents of two registers.

·            Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.

·            We have to subtract the byte in BL from the byte in AL. Using sub instruction subtract the contents of two registers. Result will be stored in the AL register.

·            Consider that a byte of data is present in the AL register and second byte of data is present in the BL register.

·            We have to add the byte in AL with the byte in BL.

·            Using ADD instruction add the contents of 2 registers.

·            Result will be stored in the AL register.

·           We have a word that is stored in AX register.

·           Initialize the counter 1 = 16.

·           Initialize counter 2 = 0 to count the number of 1’s.

·           We will rotate the number in AX alongwith carry by 1 bit to the right. If there is a carry we will increment counter 2. Decrement counter 1. This process with continue till all the bits are checked. The counter 2 will indicate the number of 1’s present in the word. The result of counter 2 is stored in BL. Display the result.

·            A digit BCD number is available in register AL. We have to unpack this BCD number i.e. we have to separate the BCD digits. e.g : If the number = 92 H then in unpack form the two digits will 02 H and 09 H. i.e. we have to mask the lower nibble, first and rotate four times to the right to get the MSB digit. Then to get the LSB digit mask the upper nibble. Display the result. Masking lower nibble means ANDing the number with OF0 to get MSB.

·            We have two digits available in the AL and BL registers in unpacked BCD form. The digit in BL is the LSB, whereas digit in AL is MSB. To pack the two numbers, means to combine the two numbers.

·            To make the digit in AL as MSB we rotate it 4 times to the left. i.e. if digit = 09 H, after rotating it by a 4 times it becomes 90 H. Now add the two numbers. Result will be packed BCD. Display the result.