Testbook

Question

How to solve this type of question?

Answer 1

It is a hypothetical 16 bit floating point representation
sign bit =1 bit exponent=7 bit it is excess-64 means add 64 (1000000) in actual exponent 
mantissa=8 bits
no is (0.2 *2¹³)₁₀
= (.001100110011.... *2¹³)  is not normalize
=(.100110011..... *2¹⁰)  is normalized now 
exponent field = actual exponent + bias= (10+64)=(74)₁₀ =(1001010)₂ 
 now 8 bits mantissa = 10011001
sign bit field =0 (means positive no)
now put all these values in given format(sign bit-exponent-mantisssa) nd convert it into base-16
=(0100101010011001)₂
=(4A99)