Match List I with List II. List I Question List II Value A If the broadband…
2023
Match List I with List II.
List I | Question | List II | Value |
|---|---|---|---|
A | If the broadband data download rate is 40 megabits per second, then how many seconds will it take to download a 60-megabyte file? | I | 64000 |
B | Base-10 (denary) value of the binary number 1000001 | II | 12 |
C | Number of 8-megabyte files that can be stored on a backup memory device having 500 gigabytes of storage | III | 17 |
D | What hexadecimal value is obtained when the two hexadecimal digits B and C are added together? | IV | 65 |
Choose the correct answer from the options given below:
Answer: B. A-II, B-IV, C-I, D-III — ConceptDigital data calculations require consistent units. One byte equals 8 bits, and in the binary storage convention used by the listed values, one…
- A.
A-III, B-IV, C-I, D-II
- B.
A-II, B-IV, C-I, D-III
- C.
A-IV, B-III, C-II, D-I
- D.
A-II, B-I, C-IV, D-III
Show answer & explanation
Correct answer: B
Concept
Digital data calculations require consistent units. One byte equals 8 bits, and in the binary storage convention used by the listed values, one gigabyte equals 1024 megabytes. Positional notation gives a number by multiplying each digit by its base raised to the corresponding place value.
Application
For A, 60 megabytes equals 480 megabits. At 40 megabits per second, time = 480 ÷ 40 = 12 seconds, so A maps to II.
For B, binary 1000001 equals 1 × 26 + 1 × 20 = 64 + 1 = 65, so B maps to IV.
For C, 500 gigabytes equals 500 × 1024 = 512000 megabytes. The number of 8-megabyte files is 512000 ÷ 8 = 64000, so C maps to I.
For D, hexadecimal B is 11 and C is 12. Their sum is 23 in decimal, which is 1 × 16 + 7 = 17 in hexadecimal, so D maps to III.
Item | Derived value | Match |
|---|---|---|
A | 12 seconds | II |
B | 65 | IV |
C | 64000 files | I |
D | 17 (hexadecimal) | III |
Cross-check
12 × 40 megabits = 480 megabits = 60 megabytes.
65 in decimal is 64 + 1, giving binary 1000001.
64000 × 8 megabytes = 512000 megabytes = 500 × 1024 megabytes.
Hexadecimal 17 equals decimal 23, the same as 11 + 12.
Therefore, the complete mapping is A-II, B-IV, C-I, D-III.