UGC NET Questions (Paper – 1)

To convert binary to octal, we group the bits in triples from the right. The number 111011110 becomes 111 011 110. Each group is then converted: 111₂ = 7₈, 011₂ = 3₈, and 110₂ = 6₈. Combining these octal digits gives 736₈ as the equivalent of 111011110₂.

Option A:
Option A, 756, would correspond to binary groups 111 101 110, which is different from the original 111 011 110.

Option B:
Option B, 7360, implies an extra 3-bit group (or trailing zeros) that are not present in the given binary numeral.

Option C:
Option C, 734, would correspond to 111 011 100; the last group does not match 110.

Option D:
Option D, 736, comes directly from converting 111, 011, and 110 into 7, 3, and 6.

To convert 10011010010₂ to octal, we group the bits into triples from the right: 010 011 010 010, which corresponds to octal digits 2, 3, 2 and 2. This gives the octal number 2322₈. If we convert 2322₈ back to decimal, we obtain 1234, confirming the correctness of the conversion. Therefore, option A is the correct octal equivalent of the given binary number.

Option A:
Option A correctly groups the binary digits and interprets each triple as an octal digit, reproducing the original binary when reversed. Its decimal value also matches 1234, which confirms that the mapping is accurate.

Option B:
Option B differs in one of the octal digits, changing one three-bit group and leading to a different binary pattern. As a result, the corresponding decimal value is not 1234, so it cannot match the given binary representation.

Option C:
Option C modifies another octal digit relative to the correct answer, so its underlying binary pattern differs in at least one three-bit block. This leads to a decimal value that is higher than 1234 and makes it an incorrect choice.

Option D:
Option D rearranges the octal digits in a way that changes multiple three-bit groups, producing a binary string distinct from 10011010010₂. Therefore, it does not give the correct octal representation of the given binary number.

To convert binary to octal, we group the binary digits into triples from the right and convert each group to its octal digit. The binary 1110110011₂ can be written as 001 110 110 011, giving octal digits 1, 6, 6 and 3. This results in the octal number 1663₈. Since this corresponds to the same value as the original binary, option D is correct.

Option A:
Option A corresponds to a different octal pattern whose binary form is not 1110110011₂. When regrouped into binary, it yields a distinct set of bits and therefore a different decimal value. Thus, it does not match the given binary number.

Option B:
Option B produces a larger octal number which, when converted back to binary and then decimal, exceeds the value of 1110110011₂. It involves a higher octal digit in one position, making it inconsistent with the required binary pattern. Hence, it is not correct.

Option C:
Option C differs from the correct option in one of its octal digits, which changes the underlying three-bit group. This leads to a binary value that is not 1110110011₂ and therefore not equal to the given binary number. Thus, it is an incorrect octal equivalent.

Option D:
Option D correctly groups the binary digits and converts each group to its octal counterpart, yielding 1663₈. Its reverse conversion reproduces the original binary number exactly. Therefore, this option represents the accurate octal equivalent of 1110110011₂.