CBSE Class 11 Computer Science Chapter 3: Data Representation NCERT Solutions

The Octal number system, also known as base 8, uses eight unique digits: 0, 1, 2, 3, 4, 5, 6, and 7. It was historically popular for representing digital circuit numbers because it is more compact than binary. Counting in octal proceeds as follows: 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15, 16, 17, 20, 21, and so on.

The Hexadecimal number system, also known as base 16, is widely used today for representing digital circuit numbers compactly. It uses sixteen distinct symbols: the digits 0 through 9 and the letters A through F, where A represents 10, B represents 11, C represents 12, D represents 13, E represents 14, and F represents 15. Hexadecimal numbers are often prefixed with '0x' (e.g., 0x1B84). Counting in hexadecimal proceeds as: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, and so on.

The Decimal number system, commonly known as base 10, is the standard system used by humans for counting and calculations. It utilizes ten unique digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Decimal numbers are sometimes abbreviated as 'dec'. Counting in decimal proceeds as: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, and so on.

The Binary number system, or base 2, is the fundamental language of digital computers. It uses only two digits: 0 and 1. Binary numbers are often prefixed with '0b' (e.g., 0b1011). Binary is sometimes abbreviated as 'bin'. Counting in binary proceeds as: 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, and so on.

To convert the decimal number 259 to binary, we use the method of successive division by 2 and record the remainders.

We perform the divisions as follows:

259 ÷ 2 = 129 remainder 1 (LSB)

129 ÷ 2 = 64 remainder 1

64 ÷ 2 = 32 remainder 0

32 ÷ 2 = 16 remainder 0

16 ÷ 2 = 8 remainder 0

8 ÷ 2 = 4 remainder 0

4 ÷ 2 = 2 remainder 0

2 ÷ 2 = 1 remainder 0

1 ÷ 2 = 0 remainder 1 (MSB)

Reading the remainders from bottom to top (MSB to LSB), we get the binary equivalent.

Therefore, (259)_{10} = (100000011)_2.

To convert the decimal number 17.35 to binary, we convert the integer part (17) and the fractional part (0.35) separately.

1. Converting the integer part (17):

We use successive division by 2:

17 ÷ 2 = 8 remainder 1 (LSB)

8 ÷ 2 = 4 remainder 0

4 ÷ 2 = 2 remainder 0

2 ÷ 2 = 1 remainder 0

1 ÷ 2 = 0 remainder 1 (MSB)

Reading the remainders from top to bottom (MSB to LSB), we get (17)_{10} = (10001)_2.

2. Converting the fractional part (0.35):

We use successive multiplication by 2 and take the integer part of the result:

0.35 \times 2 = 0.70 -> Integer part is 0

0.70 \times 2 = 1.40 -> Integer part is 1

0.40 \times 2 = 0.80 -> Integer part is 0

0.80 \times 2 = 1.60 -> Integer part is 1

0.60 \times 2 = 1.20 -> Integer part is 1

0.20 \times 2 = 0.40 -> Integer part is 0

The process can continue, but we observe that the fractional part 0.40 repeats, indicating a recurring binary fraction. Reading the integer parts from top to bottom, we get (0.35)_{10} \approx (0.010110\overline{0101})_2. The provided source shows (0.35)_{10} = (0.010110)_2 which implies a truncation or a specific stopping point.

Combining the integer and fractional parts:

(17.35)_{10} = (10001.010110\overline{0101})_2. The source indicates (17.35)_{10} = (10001.01\overline{0110})_2, which is a slightly different recurring pattern based on the stopping point of the calculation.

To convert the decimal number 39286 to octal (base 8), we use the method of successive division by 8 and record the remainders.

We perform the divisions as follows:

39286 ÷ 8 = 4910 remainder 6 (LSB)

4910 ÷ 8 = 613 remainder 6

613 ÷ 8 = 76 remainder 5

76 ÷ 8 = 9 remainder 4

9 ÷ 8 = 1 remainder 1

1 ÷ 8 = 0 remainder 1 (MSB)

Reading the remainders from bottom to top (MSB to LSB), we get the octal equivalent.

Therefore, (39286)_{10} = (114566)_8.