64K memory contains how many words of 8 bits each?
| (a) | 65,536 |
| (b) | 64,536 |
| (c) | 65,436 |
| (d) | 65,546 |
| (e) | 65,556. |
The correct answer and explanation is:
The correct answer is:
(a) 65,536
Explanation:
A memory size of 64K means it contains 64 × 1024 bytes because 1K = 1024 bytes.
Step-by-Step Calculation:
- Understanding the ‘K’ in 64K:
- ‘K’ stands for kilo, which is 1024 in computing (not 1000 as in standard metric notation).
- So, 64K = 64 × 1024 bytes
- 64×1024=65,53664 \times 1024 = 65,536 bytes
- 1 Byte = 8 Bits:
- Since each memory location stores 1 byte, and 1 byte = 8 bits, each memory location is an 8-bit word.
- Finding the Number of 8-bit Words:
- Since each byte is an 8-bit word, the number of 8-bit words in 64K memory is simply the number of bytes, which is 65,536.
Conclusion:
A 64K memory contains 65,536 words of 8 bits each.
This calculation applies to standard memory architectures where each addressable memory unit (byte) is 8 bits. If a system had a different word size (e.g., 16-bit or 32-bit), then the calculation would differ, but in this case, for 8-bit words, the answer remains 65,536.
Alternative Answer Choices Analysis:
- (b) 64,536 – Incorrect (off by 1000).
- (c) 65,436 – Incorrect (off by 100).
- (d) 65,546 – Incorrect (off by 10).
- (e) 65,556 – Incorrect (off by 20).
Thus, the correct choice is (a) 65,536.
Now, I will generate an image representing this concept.
I’ve provided an image illustrating the concept of 64K memory, showing its structure and breakdown into 8-bit words. Let me know if you need any modifications or further explanations!
