Search Previous Year Questions

The correct answer is x' + y.

To find the equivalent of  x'y' + xy + x'y,

First take x' common from x'y' + x'y

= xy + x'(y + y')

Use A + A' = 1

= xy + x'. 1 

Use (A+BC)= (A+B)(A+C)

= (x + x’)(y + x') 

Use A + A' = 1

= 1.(x’ + y) 

• Hence (x’ + y) is the answer.

The correct answer is Primary Memory

• The memory unit which directly communicates with the CPU is known as Primary Memory.

Key Points

• Primary memory is computer memory that a processor or computer accesses first or directly.
• It allows a processor to access running execution applications and services that are temporarily stored in a specific memory location.
• Primary memory is also known as primary storage or main memory.
• Random Access Memory (RAM) is an example of primary memory.

Additional Information

• Auxiliary Memory:-
• Also known as secondary storage or external memory.
• It is a non-volatile memory (does not lose stored data when the device is powered down).
• It is not directly accessed by the CPU, because it is not accessed via input/output channels.
• Secondary memory:-
• It works slower as compared to the main memory.
• These are installed for holding data/information permanently.
• CPU does not access these memories directly, rather they are accessed through input-output routines.
• The data hold in secondary memories are transferred to the main memory, and then the CPU can access that data.
• For instance, disk, CD-ROM, DVD, etc.

The correct answer is More, Slower.

Key Points

• Static RAM is fast and expensive, and dynamic RAM is less expensive and slower. Therefore static RAM is used to create the CPU's speed-sensitive cache, while dynamic RAM forms the larger system RAM space.
• SRAM module consumes less power than a DRAM module. This is because SRAM only requires a small steady current while DRAM requires bursts of power every few milliseconds to refresh. This refresh current is several orders of magnitude greater than the low SRAM standby current. Thus, SRAM is used in most portable and battery-operated equipment.

Important Points

• Dynamic RAM is the most common type of memory in use today. Inside a dynamic RAM chip, each memory cell holds one bit of information and is made up of two parts: a transistor and a capacitor. These are, of course, extremely small transistors and capacitors so that millions of them can fit on a single memory chip. The capacitor holds the bit of information -- a 0 or a 1. The transistor acts as a switch that lets the control circuitry on the memory chip read the capacitor or change its state.
• Static RAM uses a completely different technology. In static RAM, a form of flip-flop holds each bit of memory. A flip-flop for a memory cell takes 4 or 6 transistors along with some wiring but never has to be refreshed. This makes static RAM significantly faster than dynamic RAM. However, because it has more parts, a static memory cell takes a lot more space on a chip than a dynamic memory cell. Therefore you get less memory per chip, and that makes static RAM a lot more expensive.

The correct answer is (11101010.001)2

CONCEPT:

• Conversion steps (for the non-Fraction part):

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the binary digit.
4. Repeat the steps until the quotient is equal to 0.

• Conversion steps (for the Fraction part):

1. Multiply the decimal number by 2.
2. Get the result for the next iteration.
3. Get the number received before decimal for the binary digit.
4. Repeat the steps until the result is equal to 1.

Explanation:

• First Convert 234₁₀ to binary:

• Here, 23410 = 11101010  (Writing remainders from bottom to top)
• Now Convert 0.125₁₀to Binary:

• Here, 0.12510 = .001 (Writing Carry from top to bottom)
• Now after combining the number and its fraction part, we get 11101010.001.
• Hence, (234.125)10 = (11101010.001)2

The correct answer is None of the above

CONCEPT:

• Following are the steps to convert decimal to octal:

1. If the provided decimal number is less than the number 8 then the octal number is the same.
2. If the provided decimal number is larger than 7 then divide it by number 8.
3. take a note of the remainder, received after division
4. Repeat step 2 and 3 with the quotient until it is less than 8
5. Now, note the remainders in the reverse order (bottom to top)
6. The resultant thus received is the equivalent octal number to the provided decimal number.

Explanation:

• Hence, (432267)10 = (1514213)₈

The correct answer is All of the above.

One Exabyte is equal to: 

= 1018 Bytes.

= 1000 x 1 PB

= 1 ZB / 1000

Additional Information

The correct answer is 4

EXPLANATION:

• As per the question, we need to create a NOR gate (Question Diagram) using NAND gates.
• The boolean expression for NAND is \(Y = \widehat {A.B}\)
• The boolean expression for NOR is \(Y = \widehat{A+B}\)
• So, using the above boolean expression, the following is the possible diagram:

Important Points

• If inputs are joined together, then it acts like a NOT gate.

The correct answer is CPU cycle.

• The time required for the fetching and execution of one simple machine instruction is CPU cycle time.
• The fundamental sequence of steps that a CPU performs is called CPU Cycle, it is also known as the "fetch-execute cycle,".
• In simpler CPUs, the instruction cycle is executed sequentially, each instruction being processed before the next one is started.
• In most modern CPUs, the instruction cycles are instead executed concurrently, and often in parallel, through an instruction pipeline
• A clock cycle, or simply a "cycle," is a single electronic pulse of a CPU.
• During each cycle, a CPU can perform a basic operation such as fetching an instruction, accessing memory, or writing data.

Additional Information

• ​Real-time is a level of computer responsiveness that a user senses as sufficiently immediate or that enables the computer to keep up with some external process.
• Seek time is the time which is required by the drive to pick the read/write head and take it to the proper track of the disk.
• The seek time is a part of the access time where the latter also includes latency time and overhead time.
• Delay time is the time required for a digital signal to travel from the input(s) of a logic gate to the output.

The correct answer is ​x + y.

To solve X+X'Y

Use X.1 = X, 

= X.1 + X'Y 

Use 1 + Y = 1

= X(1 + Y) + X'Y 

= X + XY + X'Y

= X + Y(X + X')

Use X + X' = 1 

= X + Y.1

Use Y.1 = Y

= X + Y

• Hence answer is X+Y.

The correct answer is FFFF

Explanation:

• Here we have to perform an XOR operation on two Hexadecimal numbers.

• The boolean expression for XOR is A \(\bigoplus\) B = \(A\bar B\) + \(\bar AB\)

• The truth table for XOR is:

• | Symbol | Truth Table | | | | --- | --- | --- | --- | | 2-input Ex-OR Gate | B | A | Q | | 0 | 0 | 0 | | | 0 | 1 | 1 | | | 1 | 0 | 1 | | | 1 | 1 | 0 | | | Boolean Expression Q = A ⊕ B | A OR B but NOT BOTH gives Q | | |

• Means A and B are the expressions given to us in Hex code. So, we need to find the binary equivalent of the Hexadecimal numbers. This is because these codes will be executed in 1 and 0 in the CPU.

• The following table is to be memorized in order to convert Hexadecimal to Binary:

• | Hex | Binary | | --- | --- | | 0 | 0000 | | 1 | 0001 | | 2 | 0010 | | 3 | 0011 | | 4 | 0100 | | 5 | 0101 | | 6 | 0110 | | 7 | 0111 | | 8 | 1000 | | 9 | 1001 | | A | 1010 | | B | 1011 | | C | 1100 | | D | 1101 | | E | 1110 | | F | 1111 |

• Therefore, (4AC0)₁₆ = 0100101011000000

• And (B53F)16 = 1011010100111111

• Now follow the Truth Table to perform the operation on every bit in XOR gate:

• (4AC0)16   \(\bigoplus\)  (B53F)16  = 0100101011000000   \(\bigoplus\)   1011010100111111

• 0100101011000000   \(\bigoplus\)   1011010100111111   =  1111111111111111

• So the result obtained is (1111111111111111)₂ which is the Binary Equivalent of (FFFF)₁₆ (using the table given above).

• Hence the correct answer is FFFF.