UGC NET Questions (Paper – 1)

In a positional system with radix r and k digits, the smallest value is all zeros and the largest is all digits equal to r−1. The numeric value of k digits all set to r−1 is (r−1)(1 + r + r² + … + r^(k−1)), which sums to (r−1)(r^k − 1)/(r−1) = r^k − 1. This is the highest value attainable without increasing the number of digits.

Option A:
Option A precisely captures the algebraic result of a geometric series of place values. It recognises that with k digits, the largest magnitude is one less than the next power of the base, r^k. Therefore the formula r^k − 1 is logically consistent with positional representation.

Option B:
Option B, r^(k−1) − 1, only considers up to the (k−1)-th power and would correspond to the maximum value of a (k−1)-digit number instead. It underestimates the representable range for k digits and thus is incorrect.

Option C:
Option C, r^k, corresponds to the value requiring k+1 digits in general, because r^k is written as 1 followed by k zeros in base r. It therefore exceeds the maximum value that fits in exactly k digits.

Option D:
Option D, r^k + 1, extends the overestimation even further and represents a number that is not just larger but two units beyond the next power of r. Such a value cannot be expressed using only k digits in that base.

To convert 555₈ to decimal, we expand using powers of 8. The digits are 5×8² + 5×8¹ + 5×8⁰ = 5×64 + 5×8 + 5×1 = 320 + 40 + 5 = 365. Thus, the decimal representation of 555₈ is 365.

Option A:
Option A correctly computes each positional contribution and sums them. It uses 64, 8, and 1 as place values and multiplies by 5 each time, yielding a total of 365.

Option B:
Option B, 349, results from an arithmetic miscalculation or incorrect weighting of one of the positions. It does not match the standard expansion of 555₈.

Option C:
Option C, 373, is close but implies that one of the place values or digit weights was slightly misapplied, possibly adding an extra 8 or subtracting a 0 somewhere.

Option D:
Option D, 381, overshoots the correct result and similarly indicates a mis-evaluation of one or more terms in the expansion.

7B₁₆ = 7×16 + 11 = 112 + 11 = 123.

Option A:
Option A correctly interprets B as 11 and sums to 123.

Option B:
Option B is too large and not equal to 112 + 11.

Option C:
Option C is too small and indicates arithmetic error.

Option D:
Option D is too large and incorrect.

19A₁₆ = 1×16² + 9×16¹ + 10×16⁰ = 256 + 144 + 10 = 410.

Option A:
Option A correctly converts each hex digit using powers of 16 and sums to 410.

Option B:
Option B (394) is smaller and implies miscalculation of place values.

Option C:
Option C (402) is also incorrect due to arithmetic error.

Option D:
Option D (418) overshoots the correct total.