Number systems are where a digital electronics paper opens, and where careless aspirants leak easy marks: a base conversion done in the wrong direction, a 2's complement range off by one bit, or a signed-number question read as unsigned. The same handful of ideas, conversion between binary, octal and hexadecimal, solving for an unknown radix, and 1's versus 2's complement, come back in GATE year after year. Below are 12 solved MCQs pulled from KnowledgeGate's published question bank, every one a previous-year GATE problem with the year marked. Attempt each before reading the explanation, which stays short on purpose. Every question here links its exact solved page; if the underlying idea feels shaky, the full theory sits in the Digital Electronics learn module.
Binary, octal and hexadecimal conversions
Q1. The number (123456) in base 8 is equivalent to (GATE 2004)
(a) (A72E) base 16 and (22130232) base 4
(b) (A72E) base 16 and (22131122) base 4
(c) (A73E) base 16 and (22130232) base 4
(d) (A62E) base 16 and (22120232) base 4
Answer: (a). The fastest route between power-of-two bases is through binary. Write each octal digit as three bits to get 001 010 011 100 101 110, then regroup those bits into fours for hex (A72E) and into twos for base 4 (22130232). Only option (a) matches on both regroupings, which is the whole point of listing two bases at once (see the solved page).
Q2. The hexadecimal representation of (657) in base 8 is (GATE 2005)
(a) 1AF
(b) D78
(c) D71
(d) 32F
Answer: (a) 1AF. Convert each octal digit to three bits: 6 is 110, 5 is 101, 7 is 111, giving 110101111. Now group the bits into fours from the right, 1 1010 1111, and read each group as a hex digit to get 1AF. Never convert octal to hex digit-by-digit directly; always pass through binary (see the solved page).
Q3. (1217) in base 8 is equivalent to (GATE 2009)
(a) (1217) base 16
(b) (028F) base 16
(c) (2297) base 10
(d) (0B17) base 16
Answer: (b) (028F) base 16. Compute the decimal value first: 1×512 + 2×64 + 1×8 + 7 = 655. Then 655 in hex is 0x28F. The distractors reuse the same digit string in another base to catch anyone who compares symbols instead of values (see the solved page).
Q4. Let the representation of a number in base 3 be 210. What is the hexadecimal representation of the number? (GATE 2021)
(a) 15
(b) 21
(c) D2
(d) 528
Answer: (a) 15. Go through decimal for any conversion between non-related bases: 210 in base 3 is 2×9 + 1×3 + 0 = 21 in decimal, and 21 in decimal is 0x15. Option (b) is the trap for anyone who copies the decimal value straight across without converting to hex (see the solved page).
Q5. The value of a 16-bit unsigned integer X in hexadecimal is BCA9. The representation of X in octal is (GATE 2017)
(a) 571244
(b) 736251
(c) 571247
(d) 136251
Answer: (d) 136251. Expand the hex digits to binary: B C A 9 becomes 1011 1100 1010 1001. Octal needs groups of three, so pad with two leading zeros to 18 bits and regroup: 001 011 110 010 101 001, which reads as 136251. The padding step is where most slips happen (see the solved page).
Radix equations and base arithmetic
Q6. If (73) in base x equals (54) in base y, the possible values of x and y are (GATE 2004)
(a) 8, 16
(b) 10, 12
(c) 9, 13
(d) 8, 11
Answer: (d) 8, 11. Turn both numerals into values: 73 in base x is 7x + 3, and 54 in base y is 5y + 4. Setting them equal gives 7x - 5y = 1, and checking the options, x = 8 and y = 11 gives 56 - 55 = 1. This kind of question is solved fastest by substituting the choices rather than solving the Diophantine equation cold (see the solved page).
Q7. In the equation (7526) base 8 minus (Y) base 8 equals (4364) base 8, find Y. (GATE 2014)
(a) 1634
(b) 1737
(c) 3142
(d) 3162
Answer: (c) 3142. Subtract directly in octal, borrowing 8 rather than 10 at each column, or convert, subtract in decimal, and convert back. The cleanest check is addition: 4364 + 3142 in octal is 7526, so Y is 3142. Verifying by adding the answer back to the subtrahend catches borrow mistakes instantly (see the solved page).
Signed numbers, 1's and 2's complement
Q8. The number 43 in 8-bit 2's complement representation is (GATE 2000)
(a) 01010101
(b) 11010101
(c) 00101011
(d) 10101011
Answer: (c) 00101011. For a positive number the 2's complement form is just its ordinary binary value with a 0 sign bit. 43 is 32 + 8 + 2 + 1, which is 101011, and padding to eight bits gives 00101011. The complement steps only matter for negatives (see the solved page).
Q9. The 2's complement representation of the decimal value -15, using the minimum number of bits, is (GATE 2002)
(a) 1111
(b) 11111
(c) 111111
(d) 10001
Answer: (d) 10001. Four bits of 2's complement only span -8 to +7, so -15 does not fit and you need five bits. Write +15 as 01111, invert to 10000, and add 1 to get 10001. The "minimum number of bits" phrase is the real question here, not the arithmetic (see the solved page).
Q10. The range of integers that can be represented by an n-bit 2's complement number system is (GATE 2005)
(a) -2^(n-1) to 2^(n-1) - 1
(b) -(2^(n-1) - 1) to 2^(n-1) - 1
(c) -2^(n-1) to 2^(n-1)
(d) -(2^(n-1) + 1) to 2^(n-1) + 1
Answer: (a) -2^(n-1) to 2^(n-1) - 1. With n bits there are 2^n patterns; exactly half encode negatives, from -2^(n-1) up to -1, and the other half cover 0 through 2^(n-1) - 1. The asymmetry, one extra negative value with no positive twin, is the fact that Q9 quietly relied on (see the solved page).
Q11. Zero has two representations in which schemes? Consider (a) sign-magnitude, (b) 1's complement, (c) 2's complement. (GATE 1999)
(a) Only a
(b) a and b
(c) a and c
(d) a, b and c
Answer: (b) a and b. Sign-magnitude has +0 as all zeros and -0 as a leading 1 with zeros, and 1's complement has +0 as all zeros and -0 as all ones. Two's complement collapses these into a single all-zero zero, which is exactly why it wins one extra negative value on the range (see the solved page).
Q12. In 2's complement addition, overflow (GATE 2002)
(a) is flagged whenever there is a carry from the sign-bit addition
(b) cannot occur when a positive value is added to a negative value
(c) is flagged when the carries from the sign bit and the previous bit match
(d) none of the above
Answer: (b). Overflow in 2's complement can only happen when both operands share a sign and the result comes out with the opposite sign. Adding a positive and a negative always produces a result that fits, so no overflow is possible there. Equivalently, overflow is detected when the carry into the sign bit differs from the carry out of it, not when a carry is merely present (see the solved page).
How this set is examined
The 12 questions above trace how examiners actually test number systems: pure conversions between binary, octal and hex (Q1 to Q5), a radix equation and octal arithmetic (Q6 to Q7), and the signed-number family built on 1's and 2's complement (Q8 to Q12). Notice how Q9, Q10 and Q11 are really one idea seen from three angles, the asymmetric 2's complement range. Once the Boolean algebra and K-map MCQs feel easy too, you are through the first two units of the subject.
If conversions still feel slow, the full theory with worked base-change examples sits in our Number systems and base conversions deep dive, which rebuilds every method from scratch. GATE aspirants get the full digital electronics sequence inside GATE Guidance by Sanchit Sir, and you can place the subject in the wider syllabus on the GATE CS Exam page. Solve, review the ones you missed, and come back a week later; the second pass is where the marks get locked in.
