4 TARGET : GATE 2024/25 unacademy Digital Electronics GATE/IRMS/CSE/ESE (Plus and Iconic Batch) : Logic Gates DPP-01 QL Consider the logic circuit with input signal TEST shown in the figure. All gates in the figure shown have identical non-zero delay. The signal TEST which was at logic LOW is switched to logic HIGH. The output [po (A)Stays HIGH throughout (B)Stays LOW throughout (©)Pulses from LOW to HIGH to LOW (D)Pulses from HIGH to LOW to HIGH Q.2__ For the circuit shown below the output F is given by “AP Spap-» (AF= (B)F=0 (F=X (D) F=X Q3 Minimum number of 2-input NAND gates required to implement the function, F=(X + ¥)(Z+ W)is (3 (B)4 5 (6 Q.4 Indicate which of the following logic gates can be used to realize all possible combinational Logic functions. (A)OR gates only (B) NAND gates only (C)EX-OR gates only (D)NOR gates only Q5 For the combinational circuit shown in figure, Nor A Z XOR B aa “XNOR To Unlock ALL Course of GATE/ESE/IRMS/AE-JE Use Code “GA1111".unacademy ) 2 AGATE ACADEMY? ) Q6 Q7 Q8 Qs Qu0 Qu ‘Which of the following truth table is correct? (A [AL BIZ (B)|A | BIZ o;o];o o};o}l OF Lyt OL] oO 1[0]}0 1[o]o Tyijdt Ty idl ()|AL BIZ (D)|A | BIZ o;o]o oOo] 1 oOy;iyt oO}; 1L}o 1}o}t 1/0]; 1 1f1]o 1} 1] 0 Boolean expression for the output of XNOR (equivalence) logic gate with inputs A and Bis (A) AB+AB (B) AB+AB (C) (A+B)(A+B) (D) (A+B)(A+B) Identify the logic function performed by the circuit shown in figure - x (A)Exclusive OR (B) Exclusive NOR (C)NAND (D)NOR The output of a logic gate is “1” when all its inputs are at logic ‘0". The gate is either (A)a NAND or an EX-OR gate. (B)a NOR or an EX-NOR gate. (C)an OR or an EX-NOR gate. (D)an AND or an EX-OR gate. Any Boolean function can be realized using only (A)NAND gate (B) AND gate (COR gate (D)NOT gate ‘The minimum number of NAND gates required to implement the Boolean function A+ AB+ ABC is equal to (A)Zero BL 4 (7 If A and B are the inputs to a logic gate, then match the logic with its output (a) NAND @ A+B (b) NOR (i) AFB (©) XNOR (iii) AB+ AB (d@) AND (ivy A-B (Aaii, bi, (B)aii, bv, iv, dei To Unlock ALL Course of GATE/ESE/IRMS/AE-JE Use Code “GA1111".unacademy ) 3 AGATE ACADEMY? ) Q.2__ The output of the logic gate in figure is (ayo 1 (A (D)A Answer Key 1. D 2. a 4 B,D Eh A 6. B,C 7. 8. 9% A 10. A 11. B 12. To Unlock ALL Course of GATE/ESE/IRMS/AE-JE Use Code “GA1111".4 TARGET : GATE 2024/25 unacademy Digital Electronics GATE/IRMS/CSE/ESE (Plus and Iconic Batch) : Logic Gates DPP - 02 Ql Which of the following expression is not equivalent to x? (A)x NAND x (B)x NOR x (©)x NAND | (D)x NOR | Q2 The output of a logic gate is “1” when all its inputs are at logic “”. The gate is either (A)A NAND or an EX-OR gate. (B)A NOR or an EX-OR gate. (C)An AND or an EX-NOR gate. (D)A NOR or an EX-NOR gate. Q.3. If the input to the digital circuit consisting of a cascade of 20 X-OR gates is X, then the output ¥ is equal to (ao 1 Ox (x Q4 The expression Y = AB is equivalent to (A) A+B (B) AB+A (CAsB (D) AB ‘The Boolean function ¥ = AB + CD is to be realized using only 2-input NAND gates. The minimum number of gates required is, (a2 (B)3 4 ws Q6 The complete set of only those Logic Gates designated as Universal Gates is (A)NOT, OR and AND Gates. (B)XNOR, NOR and NAND Gates. (C)NOR and NAND Gates. (D)XOR, NOR and NAND Gates. To Unlock ALL Course of GATE/ESE/IRMS/AE-JE Use Code “GA1111".@ unacademy } 2 GATE ACADEMY? ) Q7 Q8 Qs Q10 Qu ‘The output ¥ of the logic circuit given below is TD (AL (B)O (C)X (D) X Which one of the following circuits is NOT equivalent to a 2-input XNOR (exclusive NOR) gate? (A), =p» B) Se Or (Dy i ~ De The logic evaluated by the circuit at the output is ¥ ae Y (A) XY 4K (8) +Y)xY (© OYxY (D) XY+XV4+X4¥ ‘The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is (a4 Bs (6 (7 In the logic circuit shown in the figure, Yis given by DF B > >> D (A) ¥ = ABCD (B) ¥ =(A+BXC+D) (QY=A+B+C4+D (D) ¥ = AB+CD To Unlock ALL Course of GATE/ESE/IRMS/AE-JE Use Code “GA1111".© unacademy ) 3 GATE ACADEMY? ) Q.12 The Boolean function F(X,Y) realized by the given circuit is x F y. a) Xv 4x7 (B) XV +Xx¥ x+y (D) X-¥ Answer Key 1 D 2: D a B 4 A 5. B 6. c 7. A 8 D 9 A 10. A 1. D 12, Sees To Unlock ALL Course of GATE/ESE/IRMS/AE-JE Use Code “GA1111".TARGET : GATE 2024/25 unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : Boolean Algebra DPP -3 Qu ‘The number of Boolean functions that can be generated by n variables is equal to y— (a) (B) 2” o2 (D)2" Q2 _ Forthe logic circuit shown in figure, the output is equal to ~-D- BS ) -—D- (A) ABC (B) A+B4+E (©) AB+BC+A+E (D) AB+BC Q3 The Boolean expression A+B+C is equal to (A) A+B+O (B) ABC (©) A+B+E (D) A-(B+O) Qa The logic expression for the output of the circuit shown in figure 1 is iD | 2— D7, > Db (A) AC+BC+CD (B) AC+BC+CD (©) ABC+CD (D) AB+BC+CD QS The Boolean expression of the output of the logic circuit shown in figure is, (A) ¥=AB+AB+C >> (BY AB+AB+C To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",(© unacademy ) 2 AGATE ACADEMY? ) (CY =AB+AB+C (D)¥=AB+AB+C Q.6 — The Boolean function A+ BC isa reduced form of (A) AB+BC (B) (A+B)-(A+0) (©) AB+ABC (D) (A+0)-B Q7 Let * be defined as x#y=x+y, Let z = xty. Value z * xis (Axty (B) x ©o yl Q8 Which of the following operations is commutative but not associative? (A)AND (B)OR (C)NAND (D)EXOR Q9 The logical expression y= A+A B is equivalent to (A) y= AB (B) y= 4B (© y=A+B (D) y=A4B Q.10 The logic function f +) is the same as A) f=G+E+D ®) f=G+G+y) © f=@y (D)None of the above Answer Key a B 2 BC 2 B 4 A 5. A 6. B 7. B 8. c 9. D 10. B Sees To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : Boolean Algebra TARGET : GATE 2024/25 ba DPP-4 Qu Q2 Q3 Q4 Qs Q6 Q7 For the logic circuit shown in figure, the simplified Boolean expression for the output Y is st. > br (A+ B+ (BA ©B Me ‘The expression A+AB is equivalent to (AAtB (B) AB+A (C) A+B (D) AB In the logic circuit shown in figure, the output X is (A) AB+BC+CA (B) A4+B4+CO (C) AB+BC+CA (D) AB+BC+CA Let f(A,B) = A’ + B Simplified expression for function f (f(x + y, y).2) is (A +2) B)xyz (©xy +z (D)None of the above ‘The number of distinct Boolean expression of 4 variables is Als (B) 256 (C) 1024 (D)65536 ‘The Boolean expression AC+BC is equivalent to (A) AC+BC+AC (B) BC+AC+BC+ACB (©) AC+BC+BC+ABC (D) ABC+ABC+ABC+ABC ‘The simplified form of the Boolean expression Y =(ABC+D)(AD+BC) can be written as (A) AD+BCD (B) AD+BCD To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",© unacademy)) 2 GATE ACADEMY? ) Q8 Qs Qu0 Qu Qi Qus (© (A+D)(BC+D) (D) AD+BCD The simplest form of the Boolean expression ABCD + ABCD +ABCD +ABCD is (AAD (B)BC (©) AB (D)AB Consider the following circuit. ——D—_ y. — z. Which on the following is TRUE ? (A)fis independent of X (Opfis independent of Z 1FX= 1 in the logic equation [xX +2Z{¥+(Z+X¥)}] [X+Z(X+¥)}=1 (B)fis independent of Y (D) None of X, Y, Z is redundant Then (AY=Z (OZ=1 (B)Y=Z (D)z=0 The logic gate circuit shown in the figure realizes the function x = (A) XOR (C) Half adder ro, YRy pe (B) XNOR (D) Full adder ‘The simplified SOP (Sum of product) form of the Boolean expression P+Q+R).P+Q+R). P+Q+R) is (A) (P.Q+R) © (P.Q+R) ‘The truth table Represents the Boolean function (AX ()X@Y (B) (P+ OR) (D)(P.Q+R) fY) 0 0 x 0 0 I 1 -lef-lol- (B)X+Y (DY To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",unacademy ) 3 AGATE ACADEMY? ) Answer Key 1. c 2. A 3. c 4. c 8. D> 6. D Za A 8. D 9. 10. dD 11 A 12. B 13. A | Sees To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",TARGET : GATE 2024/25 unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : Boolean Algebra DPP-5 Q.1 The Boolean expression (X +¥) (X+¥)+(X¥)+X simplifies to (AX (By (C)xY (D)X+¥ Q.2 The output F in the digital logic circuit shown in the figure is, XOR- x Y DS > XNOR (A) F=XYZ+X¥Z (B) F=XYZ+X¥Z (© FHXVZ+XYZ (D) F=XPZ+XYZ Q.3 Inthe circuit shown in the figure, if C=0, the expression for ¥ is co 4 Da ) [> ao ao (A) Y=AB+AB (B) Y=AB+AB (OQ Y=A+B () Y=AB QA Let © denote the exclusive OR (XOR) operation, Let “1” and ‘0° denote the binary constants. Consider the following Boolean expression for F over two variables P and Q F(P,Q)=(1GP)®@(PBQ) G(PBO) BQGO) ‘The equivalent expression for Fis (A) P+O (B) P+O (© Peo (D) P6O To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",© unacademy)) 2 GATE ACADEMY? ) Qs Q6 Q7 Qs Qo Qu0 Qu Consider the following Sum of Products expression, F. F = ABC+ABC+ ABC + ABC+ ABC. The equivalent Product of Sums expression is (A) F =(A+B+CXA+B+CVXA+B+0) (B) F=(A+B+CYA+B+CVA+B+O) (© F=(A+B+CYA+B+CNA+B+C) (D) F=(A+B+CXA+B+CVA+B+C) ‘The output of the combinational circuit given below is (AA+B+C B)AB+O) (©B(C+A) (D)CA+B) ‘The Boolean expression (a+b +c+d)+(b+@) simplifies to AL (B) ab ©ab (D0 ‘The Boolean expression XY+(X'+¥)Z is equivalent to (A) X¥Z'+X'Y'Z (B) X'Y'Z+XYZ (C) (X+Z\¥+Z) (D) (X'+Z\Y'+Z) In the digital circuit given below, F is ¥ D 4 | (A) XY 4YZ, (B) XY¥+¥Z (© X¥+¥Z (D) XV+¥ ‘The Boolean expression AB+AC+BC simplifies to (A) BC+AC (B) AB+AC+B (C) AB+AC (D) AB+BC A and B are the logical inputs and X is the logical output shown in the figure. The output X is related to A and B by a _ | x Do To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",(© unacademy ) 3 AGATE ACADEMY? ) (A) X =AB+BA (B) X=AB+BA (C) X =AB+BA (D) X=AB+BA Qu2__A function F(A, B,C) defined by three Boolean variables A, B and C when expressed as sum of products Qu3 Qi4 Qus QU6 Qu7 Qs is given by F=A-B-C+A-B-C+A-B-C where, A,B and C are the complements of the respective variables. The product of sum (POS) form of the function F is (A) F =(A+B+CyA+B+CyA+B+C) (B) F=(A+B+CYA+B+C(A+B+C) (©) F=(A+B+CWA+B+CyA+B+ClA+B+CVA+B+O) (D) F=(A+B+CYA+B+ CV A+B +CXA+B+CYA+B+C) The equivalent Boolean expression of (R+T)(R +T)(S+T)=FiR,S,T) is (ART+RST (B) RT+RT+ST (C)RT+RST (D)None of these ‘The Boolean expression AB+AC is equivalent to (A) ABC+ABC+ ABC (B) ABC-+ABC+ABC+ ABC (©) ABC+ABC+ABC (D)None of these Minimize the logic expression C+D+AB+ABC+ABC (A)C+AB+CD (B) AB+CD+C (©) C+D+AB (D)None of these If A and B are Boolean variables, which one of the following is equivalent to AB B@AB (ayo (B) A+B (©) AB (D)AtB For the circuit given below, the output Y is, A ——D- —p—Dip, Lp (A)A+B4C (Bc (OB (D) AB+BC Which of the following is correct? “D> : D- cD (A) ¥, is independent of 'B" (B) ¥, is independent of A (©)Y, isdependent of A and B ——_(D)None of the A, B, Cis dependent To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",unacademy ) 4 AGATE ACADEMY? ) Answer Key 1. A 2. A 3. A 4 c 5. A 6. c 7 Bt 8. c 9 B 10. A 11 c 12. c 13. A 14. B 15. c 16. c 17. B 18 B eee To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",TARGET : GATE 2024/25 unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : K-Maps DPP-6 Q.1 Find the minimum product of sums of the following expression f=ABC+ABC. Q2 A combination circuit has inputs A, B and C and its Karnaugh map is given in figure. The output of the circuit is AB CN Ol 1110 o| fa 1 1p. 1 (A) (AB+AB)C (B) (AB+AB)C (C) ABC (D) ABC Q.3 What is the equivalent Boolean expression in product-of-Sums form for the Karnasugh map given in fig? AB oN _1__ 0 0 1 1 oa] 1 1 nit 1 10 a{a (A) BD+BD (B) (B+C+D)(B+C+D) (©) (B+ D\(B+D) (D)(B+D)(B+D) Q4 The minimal sum of products form of f = ABCD + ABC + BCD + ABC is (A) AC+BD (B) AC+CD (©) AC+BD (D) AB+CD Q5 The minimized form of the logical expression (ABC +ABC+ABC+ABC) is (A) AC+BC+AB (B) AC+BC+AB (©) AC+BC+AB (D) AC+BC+AB To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",@ unacademy } 2 GATE ACADEMY? ) Q6 Q7 Qs Qo Qu0 Which of the following functions implements the Kamaugh map shown below? oo} oo | | 0 wfofol[i|o om) x|x 1 |x nfo[i|{ife 10 | 0 1 1 0 (A) AB+CD (B) D(C+A) (©) AD+AB (D) (C+ D)(C+D)(A+B) Which function does NOT implement the Karnaugh map given below? wz 00 | or | 11 | 10 4 wlo|x|fofo afolxfifi nf[i[ififa 10 | 0 x ojo (A)(w + xy (B)xy+yw (© (wex)(wey)(x+y) (D) None of the above Minimum SOP for f(w, x, y, 2) shown in Karaugh ~ map below is WE ya\_00_ or 10 o }o}ililo a|xfolfo n{xfofofa 10 oO 1 1 x (xztyz B)xz +2" (Ox ytzx’ (D)None ‘The Karnaugh map for a four variable Boolean function is given below. The correct Boolean sum of product is Po RS\ 0001110 oo] 0 0 alifolo]i nfifo wlo}ilolo (A) PORS+OS (B) PORS (©) POR+O8 (D) PORS+O ‘The switching expression corresponding to f (A, B,C, D) = 2 (1, 4, 5,9, 11, 12) is (A) BC'D'+A'C'D+AB'D (B) ABC'+ ACD+B'C'D (©) ACD'+A'BC'+AC'D' (D) A'BD+ACD'+ BCD" To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",Fu nacademy ) 3 GATE ACADEMY? ) Qu Qi2 ‘Min-term (Sum of products) expression for a Boolean function is given as follows. £(A,B,C) =F m0,1,2,3,5,6) Where A is the MSB and C is the LSB. The minimized expression for the function is (A) A+(B@C) (B) (A®B)+C (© A+(BEOQ) (D) ABC The Boolean expression Y= ABCD +ABCD +ABCD+ABCD can be minimized to (AY=ABCD+ABC+ACD — (B) Y=ABCD+BCD+ABCD (C)¥=ABCD+BCD+ABCD (D)¥=ABCD+BCD+ABCD Q.13 The minimum sum of products form of the Boolean expression ORS +PORS+PORS +PORS +PQRS+PORS is (A) y=PO+0S (B) y=PQ+ORS (©) y=PO+ORS My Or Answer Key a B 2 D x B 4. A 5. A - 7. a | 8. A 9. B 10. A M1 c 12. 1 13. A sees To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",TARGET : GATE 2024/25 unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : K-Maps DpP-7 Statement For Linked Answer Questions 1 & 2 The following Kamaugh map represents a functions F Pyyz Xoo o_o of 1} 1] 1 ]o 1fo]ofi1fo Q.1 A minimized form of the function Fis (A) F=XY¥4¥Z (B) F=X¥4+¥Z (©) F=XY+¥Z (D) F=XY+¥Z Q2 Which of the following circuits is a realization of the above functions F? “D> “Dp. © ° 4 ‘Op, Pp». ‘DT DI Q.3— Whatis the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term. ab ea\_00 1 10 ie Xi] xX]. a} x 1 u Ww] 1 x (A) bd (B) bd +be (©) bd +abed (D) bd +be+ed To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",unacademy } 2 GATE ACADEMY? ) Q4 Qs Q6 Q7 Q8 Qo Which of the following logic circuits is a realization of the function F whose Kamaugh map is shown in figure AB } The total number of prime implicants of the function f (Ww, x, y,2)= © (0, 2,4, 5, 6, 10) is Which one of the following gives the simplified sum of products expression for the Boolean function F=my+m,+m,+m, where, mg, m., m, and m, are minterms corresponding to the inputs A, B and C with A as the MSB and Cas the LSB? (A) AB+ABC+ ABC (B) AC-+AB+ABC (©) AC+AB+ ABC (D) ABC+AC + ABC Given f(w,2, y,2)= >, (0.1,2,3,7,8,10) + ¥) (5,611.15) where d represents the don’t-care condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of f (1,3, ¥,2)? (A) f=@42KF+2) (B) f = +9) © f=QrtDG+9 () f =(v42R+2) ‘The Boolean expression equivalent to ABC+ ABC+ABC+ ABC is (A) AC+BC (B) AB+AC (C) AB+BC (D) AC+BC A Boolean function is given by Em(0.1,2,4,5,7), the number essential prime inn (a2 (B)3 co4 @)s To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",(@ unacademy ) 3 GATE ACADEMY? ) Q.10._ Find the number of Prime implicants and Essential prime implicants in the given K-map. dv 00 01 i110 1 of [a 1 ult rf fa 1 (a4as (B06 60 (D)6.6 Q.11 The simplified form of SOP form f(PORS)=¥Xm(0,1,2,3,5,8,10,11,13,15) +d(4,6,7,12,14) is (A) P+ O+R+5 (B) P+O+R+S (© P+O+R+S (D) P+O+R+S Q.12 What is the minimized expression of given (A,B,C, D) = EO, 1,2,3,4,5) +E d(00,11,12,13,14,15) (A) A+ BE (B) A+BC (© A(B+C) (D) A(B+C) Q.13 A function is represented by (A,B,C, D)= Zm(0,2,4,5,6,7,8,10,13,15) ‘The number of prime implicant and essential prime implicant are respectively (A)4 and 3 (B)4and 4 (C)4 and 2 (D)2 and 2 1 A 2. A 3. B 4. B 5. a D a A 8 c 9. D 10. D it D 12. 1 13. B | sees To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",4 TARGET : GATE 2024/25 unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) DpP-8 Q2 Q3 Q4 Qs Q6 Q7 Q8 2's complement representation of a 16-bit number (one sign bit and 15 magnitude bits) is FFFF. Its magnitude in decimal representation is, (ao BL (C) 32,767 (D)65,535 The result of (45)j—(45)i. expressed in 2’s complement representation is (a) 011000 (B) 100111 (©) 101000 (p) 101001 A signed integer has been stored in a byte using the 2's complement format. We wish to store the same integer in a 16 bit word. We should (A) copy the original byte to the less significant byte of the word and fill the more signi zeros, sant byte (B)copy the original byte to the more significant byte of the word and fill the less significant byte with zeros. (©)copy the original byte to the less significant byte of the word and make each bit of the more significant byte equal to the most significant bit of the original byte. (D)copy the original byte to the less significant bytes well as the more significant byte of the word. Zero has two representations in : (A)Sign magnitude (B)1’s complement (©)2's complement (D)None of the above ‘The number 43 in 2°s complement representation is, (A)01010101 (B) 11010101 (Cpoo101011 (p) 10101011 ‘The 2's complement representation of -17 is (A) 101110 (B) 101111 (98 (D) 110001 4-bit 2's complement representation of a decimal number is 1000. The number is (A)+8 (B)0 ©-7 @)-8 The 2’s complement representation of the decimal value ~ 15 is (QML (BML (ii (©) 10001 To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",u nacademy ) 2 Q9 The range of signed decimal numbers that can be represented by 6-bit I's complement number is (A)-31 to +31 (B)- 63 to +63 (C)- 64 to + 63 (D)- 32 to+ 31 Q.10 If 73, (in base - x number system) is equal to 54, (in base - y number system), the possible values of x and y are (A)8, 16 (B) 10,12 (€)9, 13 (D8, 11 Q.11 An 8-bit 2’s complement representation of an integer is FA (Hex). Its decimal equivalent is (A)-10 (B) -6 Cr+6 (D) H0 Q.12 The range of integers that can be represented by an n bit 2’s complement number system is (A) 2" to+(2"*=1) (B) -(2""-1)to(2"" -1) (©) -2""'to2"" (D) -(2" +to(2"" -1) Answer Key 1 B 2. e 3s c 4 AB 5. c 6. B 7. D 8. D 9. A 10. 11 B 12. A soe To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",d Ab. TARGET : GATE 2024/25 ba unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : Number System DPP-9 QL The hexadecimal representation of 657, is (A)IAF (B) D783, (p71 (D)32F Q2 The binary representation of the decimal number 1.375 is iin (B) 1.010 (CLO (D) 1.001 Q3 (1217), is equivalent to (A) 0217), 6 (B) 28F),, (©) 2297). () (0B17),, Qa The smallest integer that can be represented by an 8 — bit number in 2°s complement form is (A)- 256 (B)— 128 (©-127 (wo Q5 The representation of the decimal number (27.625) jo in base-2 number system is (A) 11011.110 (B) 11101.101 (C) 1011101 (D) 101.110 Q.6 The number of 1 in 8-bits representation of 127 in 2’s complement form is m and that in Is complement form is n, What is the value of m:n? (a2: (B)1:2 (C)3:1 (D)1:3 Q7 A gray code is aan (A) Binary weighted code (B) Arithmetric code (C)Code which exhibits a single bit change between two successive codes (D) Alphanumeric code Q8 Which of the following is a self-complementing code ? (A)8421 code (B) Excess 3 code (C) Pure binary code (D)Gray code Q9 If CIXIY), =(1209),, then the values X and Y are (A)S and | (B)5 and 7 (©)7 and 5 (1 ands To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",(wu nacademy ) 2 AGATE ACADEMY? ) Qu0 Qu Quiz Qu3 ‘What are the values respectively of R, and R, in the expression (235), =(565),, =(865),.? (A)8, 16 (B)16,8 €)6, 16 (D)12,8 Consider the following multiplication (Ow 12), x15) =(y01011001), Which one of the following gives appropriate values of w, y and 2? (A) w=0, y=0,2=1 (B) w=0, (w= (w=Ly=h, The maximum positive and negative numbers that can be represented in 1's complement using n-bits are respectively (A) +Q™=1) and -@"=) (BY #2™=1) and ~ 2") (©)+Q") and -2"'=1) (D)+Q2"41) and =") ‘The maximum positive and negative number that can be represented in 2’s complement using n-bits are respectively (A) +2" and 2") (B) + 2") and - 2") (©) +") and -Q"*-1) (D) +(2"" +1) and -(2"") iy A 2. c 3. 4, B 5. c 6. A 7. c 8. 9. : 10. 1 c 12. A 13. B Sees To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",w A. TARGET : GATE 2024/25 ba unacademy Digital Electronics GATE/CSE/ESE (Plus and Iconic Batch) Topic : Number System DPP - 10 QL The range of numbers that can be represented in 2’s complement mode with four binary digits is (A) -15to+ 15 (B) -8to+8 (©)-810+7 ) -7t0+7 Q.2 2's complement number is 1001. Its equivalent representation in 8 bit format is (a)00001001 (B) 10011001 (C) 111001 (D)None Q3 The range of signed decimal numbers that can be represented by 8 bit I's complement number is (A)-127 0 +127 (B)-128 to +128 (C)-256 to +256 (C)-256 to +127 Q4 68) +(D),,+(72), +011), = (A) G14), (B) (B4),, ©1940 (D)None of these Q5 If (212), = (153), then the base x is__ Q.6 — The minimum of bits required to represent -32 in 2’s complement representation is___ Q7 The 2’s complement representation of (89), is (A) 10100110 (B)o1011101 (C) 10100010 (D) 10100111 Q8_ The two numbers represented in signed 2's complement form are A=11101001 and B=11100101. If B is subtracted from A, the value obtained in signed 2°s complement form is (A) LI11001 (B) 00000100 (©) 11000100 (p) 10000100 Q.9 A number system is represent by (54), = (13), (4), the base of the number is 2 Q.10 The addition of 4 bit two's complement binary number 1111 and 0010 results (A)0001 and an overflow (B) 1001 and no overflow (€)0001 and no overflow () 1001 and an overflow Q.11 The addition of two number 24 and 13, in base 5 is . Q.12 The base of the number system for the addition 34+14=103 is Q.13 Convert given hexadecimal code into octal code (CAFB),,? To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",unacademy ) 2 GATE ACADEMY? ) Answer Key : 1 c 2 c a A 4 D 5. 7 | 6. 6 a D 8. B 9. 8 10. c Fry 42 12. 5 13. | (145373). | sees To Unlock ALL Course of GATE/ESE/AE-JE Use Code “GA1111",