GATE CSE 1996 Question 1.25

Question

I can’t understand what is ment by that:

                           

I can’t understand what is the bit required by the exponent and the mantissa.

Question link: https://gateoverflow.in/2729/gate-cse-1996-question-1-25

        

Answer 1

In IEEE 754 single precision representation Mantissa consists of 23 bits, exponent consists of 8 bits and 1 bit for sign. 

Here it is given that Mantissa is in sign magnitude form so bit 0 -22 is for magnitude of mantissa and 23^rd bit for sign , bit 24 – 31 is for exponent. 

Comments

Thanks a lottt!!