A weather display uses a small 8-pixel-wide monochrome icon. The icon is stored row by row as binary patterns. A bit value of 1 represents a lit pixel and a...

Assessment: Computer Science 4CP0 | Paper 1 Mock 01 | Written Paper 1 Subject: Computer Science - 4CP0

Question 1 Report

A weather display uses a small 8-pixel-wide monochrome icon. The icon is stored row by row as binary patterns. A bit value of 1 represents a lit pixel and a bit value of 0 represents an unlit pixel. The display hardware reads one binary pattern for each row. The diagram shows four stored rows and the corresponding pixels. A technician needs to check the binary data, calculate storage requirements and compare it with an icon that uses more colours. Assume that no file headers, metadata or compression are used. A byte contains 8 bits.

00111100010000101010010101000010© EAGLE BEACON GLOBAL

(a) State the number of bits used to store one row. [1]

(b) Complete the table for the first two rows. [2]

Row binary patternNumber of lit pixels
00111100
01000010

(c) Convert 10100101 to denary. [2]

(d) Give the number of bits needed for each pixel in an image with 16 possible colours. [2]

(e) Identify the total number of bits needed to store an uncompressed 8 by 4 monochrome icon. [2]

(f) Explain why increasing image resolution can increase file size, even when the colour depth remains unchanged. [3]

(g) Convert denary 60 to an 8-bit binary pattern and state how many lit pixels that pattern would display. [3]

Answer Details

(a) Each row is eight pixels wide and one bit represents each monochrome pixel, so it uses 8 bits [1].

(b)

Row patternLit pixels
001111004
010000102

Count the 1s because 1 means lit [2].

(c) \(10100101_2=128+32+4+1=\textbf{165}\) [2].

(d) \(2^4=16\), so 4 bits per pixel are needed [2].

(e) \[8\times4\times1=\textbf{32 bits}\] [2].

(f) Higher resolution means more pixels. With unchanged colour depth, each pixel still needs the same number of bits; more pixels therefore mean more stored bits and a larger file [3].

(g) \(60=32+16+8+4\), so the pattern is 00111100. It contains 4 lit pixels [3].

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