Calculating Image Storage Space

To determine the total hard disk space required to store the images, we need to calculate the size of each type of image individually and then sum them up. The storage space for an image depends on its resolution (measured in megapixels) and its color depth (measured in bits or bytes per pixel).

Storage for Black and White Images

We have eight 'black and white' images, each taken with a 1-megapixel camera. In computing, 'black and white' can sometimes refer to 1 bit per pixel (binary), but often in the context of storage, it implies grayscale which is 8 bits per pixel (1 byte per pixel) to represent 256 shades from black to white. Given the options and the nature of the second image, let's assume 'black and white' here means 8 bits per pixel (1 byte per pixel) for storage purposes.

• Number of black and white images: 8
• Resolution per image: 1 megapixel ($1 \times 10^6$ pixels)
• Color depth (assumed): 8 bits per pixel (1 byte per pixel)

Size per black and white image:

$\text{Size} = \text{Resolution} \times \text{Color Depth}$

$\text{Size} = 1 \times 10^6 \text{ pixels} \times 8 \frac{\text{bits}}{\text{pixel}} = 8 \times 10^6 \text{ bits}$

Convert bits to bytes (since 8 bits = 1 byte):

$\text{Size in Bytes} = \frac{8 \times 10^6 \text{ bits}}{8 \frac{\text{bits}}{\text{byte}}} = 1 \times 10^6 \text{ bytes}$

Total size for eight black and white images:

$\text{Total Size (BW)} = 8 \times (1 \times 10^6 \text{ bytes}) = 8 \times 10^6 \text{ bytes}$

Storage for the 8-Color Image

We have one '8-color' image taken by an 8-megapixel camera. An 8-color image means there are 8 possible colors. Since $2^3 = 8$, this theoretically requires only 3 bits per pixel to represent the color. However, images with limited colors are often stored in an indexed color format, typically using 8 bits per pixel (1 byte per pixel) to index a palette of up to 256 colors, even if fewer colors are used. Let's assume the image is stored in an 8-bit indexed format (1 byte per pixel).

• Number of 8-color images: 1
• Resolution: 8 megapixels ($8 \times 10^6$ pixels)
• Color depth (assumed for format): 8 bits per pixel (1 byte per pixel)

Size of the 8-color image:

$\text{Size} = \text{Resolution} \times \text{Color Depth}$

$\text{Size} = 8 \times 10^6 \text{ pixels} \times 8 \frac{\text{bits}}{\text{pixel}} = 64 \times 10^6 \text{ bits}$

Convert bits to bytes:

$\text{Size in Bytes} = \frac{64 \times 10^6 \text{ bits}}{8 \frac{\text{bits}}{\text{byte}}} = 8 \times 10^6 \text{ bytes}$

Calculating Total Hard Disk Space

The total hard disk space needed is the sum of the storage required for the black and white images and the 8-color image.

$\text{Total Size} = \text{Total Size (BW)} + \text{Size (8-Color)}$

$\text{Total Size} = (8 \times 10^6 \text{ bytes}) + (8 \times 10^6 \text{ bytes}) = 16 \times 10^6 \text{ bytes}$

Now, we convert the total size from bytes to megabytes (MB). In the context of data storage and MCQs, 1 megabyte is often considered as $10^6$ bytes, although in some contexts $1 \text{ MB} = 2^{20}$ bytes (1024 * 1024 bytes). Using the $10^6$ conversion gives a value that matches one of the options exactly.

Using the conversion $1 \text{ MB} = 10^6 \text{ bytes}$:

$\text{Total Size in MB} = \frac{16 \times 10^6 \text{ bytes}}{10^6 \frac{\text{bytes}}{\text{MB}}} = 16 \text{ MB}$

If we used the $2^{20}$ bytes/MB conversion:

$\text{Total Size in MB} = \frac{16 \times 10^6 \text{ bytes}}{1024 \times 1024 \frac{\text{bytes}}{\text{MB}}} \approx \frac{16,000,000}{1,048,576} \text{ MB} \approx 15.26 \text{ MB}$

Since 16 MB is an option, the assumption of 8 bits/pixel for both types and using the $10^6$ conversion for Mega is consistent with the intended solution.

Based on these calculations, the total hard disk space required is 16 MB.

Analysis of Options

The calculation assuming 8 bits per pixel for both black and white and 8-color images, and using $10^6$ bytes per MB, results in exactly 16 MB, which is one of the options.

Revision Table: Key Concepts

Additional Information: Image Storage Concepts

Image file size calculations often rely on resolution and color depth. However, real-world image files (like JPEGs or PNGs) are usually compressed, meaning their actual file size is smaller than the raw uncompressed data size calculated above. This problem likely asks for the uncompressed raw data size.

• Common Color Depths:
  • 1 bit: Black and white (binary)
  • 8 bits: Grayscale (256 shades) or Indexed color (up to 256 colors from a palette)
  • 16 bits: High color
  • 24 bits: True color (typically 8 bits each for Red, Green, Blue)
  • 32 bits: True color + Alpha channel (transparency)
  • 48 bits: Deep color (16 bits per channel)
• Megapixel Calculation: A 1-megapixel camera captures an image with approximately one million pixels. For example, a resolution of 1280x800 pixels is $1280 \times 800 = 1,024,000$ pixels, which is about 1 megapixel.
• Units of Storage:
  • Bit (b): The smallest unit, 0 or 1.
  • Byte (B): 8 bits.
  • Kilobyte (KB): $10^3$ bytes (1000) or $2^{10}$ bytes (1024).
  • Megabyte (MB): $10^6$ bytes ($1,000,000$) or $2^{20}$ bytes ($1,048,576$).
  • Gigabyte (GB): $10^9$ bytes or $2^{30}$ bytes.
  The problem likely uses the $10^6$ definition for MB to yield a clean integer result from the options.

Understanding these concepts helps in calculating storage requirements for various digital media, including images.