A mono 44.1 kHz audio file consumes approximately 5 MB per minute. How much does a 88.2 kHz, 16-bit audio file consume?

A mono 44.1 kHz audio file consumes approximately 5 MB per minute. How much does a 88.2 kHz, 16-bit audio file consume?

• A. 15 MB
• B. 5 MB
• C. 10 MB
• D. 20 MB

Answer: (C)

To understand the storage requirements of audio files, it's important to know how sampling rate, bit depth, and track configuration influence file size. A standard mono audio file at 44.1 kHz and 16-bit resolution consumes approximately 5 MB per minute. The calculation for this size is based on the following formula: File size (MB per minute) = (Sampling Rate × Bit Depth × Tracks) / (8 bits/byte × 60 seconds) For a mono file at 44.1 kHz and 16 bits: - Sampling Rate: 44,100 samples/second - Bit Depth: 16 bits - Tracks: 1 (for mono) Applying the formula: File size = (44,100 samples/second × 16 bits/sample × 1) / (8 bits/byte × 60 seconds) This simplifies and leads to the approximate size of 5 MB per minute. Now, when considering a mono audio file at 88.2 kHz and 16 bits, we can see that the sampling rate has doubled while the bit depth remains the same. Thus, the calculation would be: File size = (88,200 samples/second × 16 bits/sample × 1) /

To understand the storage requirements of audio files, it's important to know how sampling rate, bit depth, and track configuration influence file size.

A standard mono audio file at 44.1 kHz and 16-bit resolution consumes approximately 5 MB per minute. The calculation for this size is based on the following formula:

File size (MB per minute) = (Sampling Rate × Bit Depth × Tracks) / (8 bits/byte × 60 seconds)

For a mono file at 44.1 kHz and 16 bits:

• Sampling Rate: 44,100 samples/second

• Bit Depth: 16 bits

• Tracks: 1 (for mono)

Applying the formula:

File size = (44,100 samples/second × 16 bits/sample × 1) / (8 bits/byte × 60 seconds)

This simplifies and leads to the approximate size of 5 MB per minute.

Now, when considering a mono audio file at 88.2 kHz and 16 bits, we can see that the sampling rate has doubled while the bit depth remains the same. Thus, the calculation would be:

File size = (88,200 samples/second × 16 bits/sample × 1) /