Question

A 3 x 8 decoder with two enable inputs is to be used to address 8 blocks of memory. What will be the size of each memory block when addressed from a sixteen bit bus with two msbs used to enable the decoder?

Answer 1

Answer - 2K
Explanation....
A 16-bit address space contains 64K addresses.

Since you're using the two msb's for enable inputs, you can select among 4 16K address regions.

Since each of those regions is split into 8 blocks, so each block contains 2K addresses.

Answer 2

Answer:

It is question of math 24÷8= 3