## Electronics Engineering Multiple Choice Questions and Answers (MCQs) Part 7:

277. What is the octal equivalent of FF16?

(a) 255
(b) 377
(c) 256
(d) 11111111

(a) 7E
(b) 7C
(c) 6B
(d) F1

279. If (212)x = (23)10 then what is the value x?

(a) 2
(b) 3
(c) 4
(d) 5

280. If √61 = 7, the base of the number system is

(a) 4
(b) 8
(c) 9
(d) 6

281. Decimal 43 in hexadecimal and BCD number system is respectively

(a) B2, 0100 0011
(b) 2B, 0100 0011
(c) 2B, 0011 0100
(d) B2, 0100 0100

Answer : (b) 2B, 0100 0011

282. Which is the in valid code in Excess-3 code?

(a) 0001 and 0000
(b) 0110 and 01000
(c) 1010 and 1100
(d) None of these

Answer : (a) 0001 and 0000

283. 48421 is

(a) Non-weight
(b) Weighted
(c) Complementary code
(d) All of these

284. Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter

(a) 1000
(b) 1001
(c) 0011
(d) 1100

285. The binary representation 100110 is numerically equivalent to the

1. Decimal representation 46.
2. Octal representation 46.
4. Excess-3 representation 13.

Select the correct answer using the codes given below.

Codes :

(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 2 and 4

Answer : (b) 2 and 3

286. If (2.3)base 4 + (1.2)base 4  = (y)base 4; what is the value of y?

(a) 10.1
(b) 10.01
(c) 10.2
(d) 1.02

287. The range of signed decimal numbers that can be represented by 6-bit 1’s complement number is

(a) -31 to +31
(b) -63 to +64
(c) -64 to +63
(d) -32 to +31

Answer : (a) -31 to +31

288. The number of distinct Boolean expressions of 4 variables is

(a) 16
(b) 256
(c) 1024
(d) 65536

289. Positive logic in a logic circuit is one in which

(a) logic 0 and 1 are represented by 0 and positive voltage respectively.
(b) logic 0 and 1 are represented by the negative and positive voltages respectively.
(c) logic 0 voltage level is higher than logic 1 voltage level.
(d) logic 0 voltage level is lower than logic 1 voltage level.

Answer : (d) logic 0 voltage level is lower than logic 1 voltage level.

290. Three Boolean operators are:

(a) NOT, OR, AND
(b) NOT, NAND, OR
(c) NOR, OR, NOT
(d) NOR, NAND, NOT

Answer : (a) NOT, OR, AND

291. A bulb in a stair case has two switches, one switch being at the ground floor and the other at the first floor. The bulb can be turned on and also can be turned off by any one of the switches. The logic of switching of the bulb resembles

(a) an AND gate.
(b) an OR gate.
(c) an XOR gate.
(d) a NAND gate.

Answer : (c) an XOR gate.

292. 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.

Answer : (c) NOR and NAND Gates.

293. If a 3-input NOR gate has 8-input possibilities, how many of these possibilities will result in a high input

(a) 1
(b) 2
(c) 7
(d) 8

294. How is inversion achieved using EX-OR gate?

(a) Giving input signal to the two input lines of the gate tied together.
(b) Giving input to one input line and logic zero to the other line.
(c) Giving input to one input line and logic one to the other line.
(d) Inversion cannot be achieved using EX-OR gate.

Answer : (c) Giving input to one input line and logic one to the other line.

295. The output of a logic gate is ‘1’ when all its inputs are at logic ‘0’. Then 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.

Answer : (b) a NOR or an EX-NOR gate.

296. AND operation of (79)10 and (-56)10 results in:

(a) 50 H
(b) 48 H
(c) 42 H
(d) 08 H

297. If A and B are Boolean variables, then what is (A + B) . (A + B̅ ) equal to ?

(a) B
(b) A
(c) A + B
(d) AB

298. If X = 1 in the logic equation

[X + Z {Y̅ + (Z̅ + XY̅)}] {X̅ + Z̅(X + Y)} = 1, then

(a) Y = Z
(b) Y = Z̅
(c) Z = 1
(d) Z = 0

Answer : (d) Z = 0

299. If the Boolean expression P̅Q + QR + PR is minimized, the expression becomes

(a) P̅Q + QR
(b) P̅Q + PR
(c) QR + PR
(d) P̅Q + QR + PR

Answer : (b) P̅Q + PR

300. Consider the Boolean expression

X = ABCD + AB̅CD + A̅BCD + A̅CB̅D

The simplified form of X is

(a) C̅ + D̅
(b) BC
(c) CD
(d) B̅C

301. The complement of the Boolean expression AB.(B̅C + AC) is

(a) (A̅ + B̅) + (B + C̅)(A̅ + C̅)
(b) (A̅ · B̅) + (BC̅ + A̅C̅)
(c) (A̅ + B̅) · (B + C̅)(A̅ + C̅)
(d) (A + B) · (B̅ + C) (A + C)

Answer : (a) (A̅ + B̅) + (B + C̅)(A̅ + C̅)

302. For the identity AB + A̅C + BC = AB + A̅C the dual form is

(a) (A + B)(A̅ + C)(B + C)=(A + B)(A̅ + C)
(b) (A̅ + B̅)(A + C̅)(B̅ + C̅) = (A̅ + B̅)(A + C̅)
(c) (A + B) (A̅ + C) (B + C) = (A̅ + B̅)(A + C̅)
(d) A̅B̅ + AC̅ + B̅C̅ = A̅B̅ + AC̅

Answer : (a) (A + B)(A̅ + C)(B + C)=(A + B)(A̅ + C)

303. Normally in HA circuit which gate is used for sum part

(a) XOR
(b) NAND
(c) OR
(d) AND

304. To add two m-bit numbers, the required number of half adders is

(a) 2m – 1
(b) 2m -1
(c) 2m + 1
(d) 2 m

Answer : (a) 2m – 1

305. The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B. The number of combinations for which the output is logic 1, is

(a) 4
(b) 6
(c) 8
(d) 10